text:
\section{Introduction.}

The notion of Birkhoff-James orthogonality (B-J orthogonality) plays a very important role in the geometry of Banach spaces. In \cite{J},  James illustrated the role of  B-J orthogonality in characterizing geometric properties like smoothness, strict convexity etc. of the space. It is quite straightforward to observe that the notion of B-J orthogonality extends to the space of all bounded linear operators on a Banach space.  The role of B-J orthogonality in the study of geometry of Banach spaces has been explored by several researchers, from various points of view. We refer the readers to \cite{C,CSW,K,T}, and the references therein, for a detailed study in this regard. Recently, in \cite{S}, Sain characterized B-J orthogonality of linear operators on finite-dimensional real Banach spaces. Although B-J orthogonality can be defined for either real or complex Banach spaces, till now most of the operator theoretic study of B-J orthogonality \cite{S,SP} has been conducted exclusively in the context of real Banach spaces. In this paper, our aim is to initiate an analogous study of B-J orthogonality of linear operators in the complex case and to obtain a characterization of the same. It is interesting to observe that the results already known in the context of real Banach spaces follow quite easily from these new results. It is in this sense, that our present study can be considered as an extension of our earlier studies \cite{S,SGP}. Without further ado, let us establish the relevant notations and terminologies to be used throughout the paper.

\medskip

Let $\mathbb{X},~ \mathbb{Y}$ be complex Banach spaces. Let $B_{\mathbb{X}} = \{x \in \mathbb{X} \colon \|x\| \leq 1\}$ and
$S_{\mathbb{X}} = \{x \in \mathbb{X} \colon \|x\|=1\}$ be the unit ball and the unit sphere of $\mathbb{X}$, respectively. Let $\mathbb{L}(\mathbb{X},\mathbb{Y})(\mathbb{K}(\mathbb{X},\mathbb{Y}))$ denote the Banach space of all bounded (compact) linear operators from $\mathbb{X}$ to $\mathbb{Y}$, endowed with the usual operator norm. We write $\mathbb{L}(\mathbb{\mathbb{X}, \mathbb{Y}})= \mathbb{L}(\mathbb{X})$ and $\mathbb{K}(\mathbb{X},\mathbb{Y})=\mathbb{K}(\mathbb{X})$ if $\mathbb{X}= \mathbb{Y}$.

For any two elements  $x,y \in {\mathbb{X}}$, $x$ is said to be B-J orthogonal to $y$, written as $x \perp_B y$, if $ \|x+\lambda y\|\geq\|x\|$
$\forall \lambda \in \mathbb{C}$.

Similarly, for any two elements  $T,A \in \mathbb{L}(\mathbb{X})$,
$T$ is said to be B-J orthogonal to $A$, written as $T \perp_B A$,
if $ \|T + \lambda A\| \geq \|T\|$ $\forall \lambda \in  \mathbb{C}$.

For a linear operator $T$ defined on a Banach space $\mathbb{X}$, let $M_T$ denote the collection of all unit vectors in $\mathbb{X}$ at which $T$ attains norm, i.e,
\[ M_T= \{  x \in S_ \mathbb{X} \colon\|Tx\| = \|T\| \}. \]

In order to characterize B-J orthogonality of bounded linear operators on finite-dimensional real Banach spaces, Sain \cite{S} introduced the notions of $x^+ $ and $x^-$ in the following way: \\
For any two elements $ x, y $ in a real Banach space $ \mathbb{X}, $ let us say that $ y \in x^{+} $ if $ \| x + \lambda y \| \geq \| x \| $ for all $ \lambda \geq 0. $ Following similar motivations, we say that $ y \in x^{-} $ if $ \| x + \lambda y \| \geq \| x \| $ for all $ \lambda \leq 0. $ Using these notions, Sain \cite{S} characterized B-J orthogonality of linear operators defined on finite-dimensional real Banach spaces.

\begin{theorem}[Theorem 2.2 of \cite{S}]
	Let $ \mathbb{X} $ be a finite-dimensional real Banach space. Let $T, A \in \mathbb{L}(\mathbb{X}).$ Then  $T \perp_B A $  if and only if there exists $x, y \in M_T $ such that $ Ax \in Tx^{+} $ and $ Ay \in Ty^{-} $.
\end{theorem}

The characterization of B-J orthogonality of bounded linear operators on finite-dimensional real Banach spaces was later extended to real Banach spaces of arbitrary dimension in \cite{SPM}. The following two notions were introduced in \cite{SPM}, in order to accomplish the goal.\\
For any two elements $x,~y$ in a real Banach space $\mathbb{X}$ and $\epsilon \in [0,1)$, we say that $y \in x^{+ (\epsilon)}$ if $\|x+ \lambda y\| \geq \sqrt{1-{\epsilon}^2}\|x\|$ for all $\lambda \geq 0$. Similarly, we say that  $y \in x^{- (\epsilon)}$ if $\|x+ \lambda y\| \geq \sqrt{1-{\epsilon}^2}\|x\|$ for all $\lambda \leq 0$.\\
 In this paper, in order to obtain an analogous result for complex Banach spaces, let us introduce the following notions:  \\

Let  $ x \in \mathbb{X}$ and $ U = \{ \alpha \in \mathbb{C} : | \alpha | = 1, ~\arg \alpha \in [0,\pi) \}.$ For $ \alpha \in U $  and $\epsilon \in [0,1)$ define

\[ x_\alpha^{+}=\{y \in\mathbb{X}: \|x+\lambda y\|\geq\|x\|~ \forall~ \lambda = t\alpha,  t\geq 0 \}. \]
\[ x_\alpha^{-}=\{y \in\mathbb{X}:\|x+\lambda y\|\geq\|x\|~ \forall~ \lambda = t\alpha,  t\leq 0 \}.\]
\[ x_\alpha^{\perp}  =  \{y \in\mathbb{X}:\|x+\lambda y\|\geq\|x\|~ \forall~ \lambda = t\alpha,   t\in\mathbb{R}\}.\]
\[ x_\alpha^{+(\epsilon)}=\{y \in\mathbb{X}: \|x+\lambda y\|\geq \sqrt{1-{\epsilon}^2}\|x\|~ \forall~ \lambda = t\alpha,  t\geq 0 \} .\]
\[ x_\alpha^{-(\epsilon)}=\{y \in\mathbb{X}:\|x+\lambda y\|\geq \sqrt{1-{\epsilon}^2}\|x\|~ \forall~ \lambda = t\alpha,  t\leq 0 \}.\]
\[ x_\alpha^{\perp (\epsilon)}  =  \{y \in\mathbb{X}:\|x+\lambda y\|\geq \sqrt{1-{\epsilon}^2}\|x\|~ \forall~ \lambda = t\alpha,   t\in\mathbb{R}\}.\]

If $ \beta = e^{i\pi} \alpha $ then we define $ x_\beta^{+} = x_\alpha^{-}, ~ x_\beta^{-} = x_\alpha^{+} $, $x_\beta^{\perp} = x_\alpha^{\perp}$, $ x_\beta^{+(\epsilon)} = x_\alpha^{-(\epsilon)},~ x_\beta^{-(\epsilon)} = x_\alpha^{+(\epsilon)} $ and $x_\beta^{\perp(\epsilon)} = x_\alpha^{\perp(\epsilon)}$. \\
If $ y \in x_\alpha^{\perp} $ then we write $ x \bot_{\alpha} y.$  Let us define the notions of $x^+$, $x^-$ and $x^{\perp}$ in a complex Banach space in the following way:
 \[x^{+}   =  \bigcap \{ x_\alpha^{+} : \alpha \in U \} .\]
 \[x^{-}  =  \bigcap \{ x_\alpha^{-} : \alpha\in U \} .\]
 \[x^{\perp}  =  \bigcap \{ x_\alpha^{\bot} : \alpha\in U \} .\]
 If the space $\mathbb{X}$ is a real Banach space, then we must have, $\alpha \in U$ implies that $\alpha=1$. Therefore, $ x_\alpha^{+} = x^+, $  $ x_\alpha^{-} = x^- $  and $ x_\alpha^{\perp} = x^{\bot}.$

\smallskip
In this paper, we completely characterize B-J orthogonality of bounded linear operators between complex Banach spaces. The characterization assumes a particularly nice form if the domain space is reflexive and the operators are compact. In order to illustrate the importance of our study, we show that earlier characterizations of operator B-J orthogonality \cite{S,SPM} in the real case follow as simple corollaries to our present study. Next we consider the left symmetry of B-J orthogonality of linear operators defined on a finite-dimensional complex Banach space $ \mathbb{X}. $ For an element $ x \in  \mathbb{X}, ~ x $ is said to be left symmetric (with respect to B-J orthogonality) if $ x \perp_{B} y $ implies $ y \perp_{B} x $ for any $ y \in \mathbb{X}.$ A study of left symmetric linear operators on a finite-dimensional real Banach space was carried out in \cite{SGP}. In this paper, we study the left symmetric linear operators on complex two-dimensional $l_p$ spaces. We prove that $ T $  is a left symmetric linear operator on $ \ell_p^2{(\mathbb{C})}$ if and only if $ T $ is the zero operator.

\section{Main Results}
Let us begin with two easy propositions, that would be useful in obtaining the desired characterization of B-J orthogonality of bounded linear operators between complex Banach spaces.

\begin{prop}\label{prop:prop1 }
Let $\mathbb{X}$ be a complex Banach space, $x, y \in \mathbb{X} $ and $ \alpha \in U.$  Then the following are true:

	(i) Either $y \in x_{\alpha}^{+}$ or $y \in x_{\alpha}^{-}$.

	(ii)  $x\perp_{\alpha} y$ if and only if  $y \in x_\alpha^{+}$ and $y \in x_\alpha^{-}$.

	(iii)  $y \in x_\alpha^{+}$ implies that $\eta y \in (\mu x)_\alpha^{+}$ for all $\eta, \mu>0$.

	(iv)  $y \in x_\alpha^{+}$ implies that $ - y \in  x_\alpha^{-}$ and $y\in(-x)_\alpha^{-}$.

	(v)  $y \in x_\alpha^{-}$ implies that $\eta y \in (\mu x)_\alpha^{-}$ for all $\eta, \mu>0$.

	(vi)   $y \in x_\alpha^{-}$ implies that $ - y \in  x_\alpha^{+}$ and $y\in(-x)_\alpha^{+}$.

	(vii) $y \in x_\alpha^{+}$ implies that $\beta y \in (\beta x)_\alpha^{+}$ for all $\beta \in \mathbb{C}$.

	(viii) $y \in x_\alpha^{-}$ implies that $\beta y \in (\beta x)_\alpha^{-}$ for all $\beta \in \mathbb{C}$.
\end{prop}
\begin{proof}
$(i)$ If $ y \notin x_\alpha^+ $ then we show that $ y \in x_{\alpha}^-.$ Since $ y \notin x_\alpha^+ $ we have $\|x+\lambda_{0} y\|<\|x\|$ for some $\lambda_{0}=t_{0}\alpha$ with $t_{0}>0$. Let  $\lambda =t \alpha$ with $t< 0.$ Then there exists $ s \in [0,1] $ such that
	$x= s(x+\lambda_{0}y) + (1-s)(x+\lambda y) \Rightarrow \|x\|\leq s\|x+\lambda_{0}y\| +(1-s)\|x+\lambda y\| \Rightarrow  \|x\|< s\|x\| +(1-s)\|x+\lambda y\| \Rightarrow \|x\|< \|x+\lambda y\|$. 	 Therefore,  $ \|x\|\leq \|x+\lambda y\|  ~\forall \lambda=t\alpha$ with $t\leq 0 \Rightarrow y\in x_{\alpha}^-$. \\

The proofs of $(ii)-(viii)$ can be easily completed using similar approach.
\end{proof}

\begin{prop}\label{prop:prop2 }
Let $\mathbb{X}$ be a complex Banach space and $x, y \in \mathbb{X}$. Then the following are true:

	(i)  	$x\perp_{B} y$ if and only if  $y \in x^{+}$ and $y \in x^{-}$.

	(ii)  $y \in x^{+}$ implies that $\eta y \in (\mu x)^{+}$ for all $\eta, \mu>0$.

	(iii)  $y \in x^{+}$ implies that $ - y \in  x^{-}$ and $y\in(-x)^{-}$.

	(iv)  $y \in x^{-}$ implies that $\eta y \in (\mu x)^{-}$ for all $\eta, \mu>0$.

	(v)   $y \in x^{-}$ implies that $ - y \in  x^{+}$ and $y\in(-x)^{+}$.
\end{prop}
\begin{proof}

	 	$ (i)$ The proof follows from the definitions of $x^+ $ and $x^-$.

	 $(ii)$ Let $y \in x^+$. Then $y \in x_{\alpha}^+$ for each $\alpha $ with $ \arg \alpha \in [0, \pi).$ We show that $ \eta y \in (\mu x)_{\alpha}^+ $ for each $\alpha $ with $ \arg \alpha \in [0, \pi).$  Now, $\|\mu x + (t \alpha) \eta y\|= |\mu|\|x+    (\frac{t \eta}{\mu}) \alpha y\| \geq|\mu| \|x\|= \|\mu x\|~ \forall ~t,\mu, \eta> 0$  and so $ \eta y\in (\mu x)_\alpha^{+} ~\forall ~ \mu , \eta> 0$. 	Thus, $ \eta y \in (\mu x)^+   ~\forall ~ \mu , \eta> 0$.

	  $(iii)$ Suppose $y \in x^{+}$. Then for each $\alpha $ with $ \arg \alpha \in [0,\pi), ~\|x+ t \alpha y\|\geq\|x\| ~\forall~ t\geq 0.$ So $ \|x+ (-t) \alpha (-y)\| \geq \|x\| ~\forall~ t \geq 0. $ This shows that $-y \in x_{\alpha}^{-}$ for each $\alpha $ with $ \arg \alpha \in [0,\pi)$ and so $ -y \in x^{-}$.

	  Again, for each $\alpha $ with $ \arg \alpha \in [0,\pi), ~\|x+ t \alpha y\|=\|(-x) + (-t) \alpha (y)\|\geq\|x\|=\|-x\| ~\forall~ t\geq 0$.   This shows that $y \in (-x)_{\alpha}^{-}$ for each $\alpha $ with $ \arg \alpha \in [0,\pi)$ and therefore, $ y \in (-x)^{-}$.

	  $(iv)$ Follows similarly as $(ii).$

	 $ (v)$ Follows similarly as $(iii).$
\end{proof}

Let us now obtain the promised characterization theorem.

\begin{theorem}\label{theorem:bounded}
	Let $\mathbb{X}$ and $\mathbb{Y}$ be two complex Banach spaces. Let $T \in \mathbb{L} (\mathbb{X}, \mathbb{Y})$ be non-zero. Then for  any $A \in \mathbb{L} (\mathbb{X}, \mathbb{Y}) ,~ T \bot_B A$ if and only if either $(a)$ or  $(b)$ holds: \\
	$(a)$ There exists a sequence $\{x_n\}$ in $S_\mathbb{X}$ such that $\| Tx_n\| \rightarrow \|T\|$ and $\|Ax_n\| \rightarrow 0$ as $n\rightarrow \infty$. \\
	$(b)$ For each $\alpha \in U$, there exist two sequences $\{x_n= x_n(\alpha)\},~\{y_n=y_n(\alpha)\}$ in $S_\mathbb{X}$ and two sequences of positive real numbers $\{\epsilon_n=\epsilon_n(\alpha)\}$, $\{\delta_n=\delta_n(\alpha)\}$ such that \\
	(i) $ \epsilon_n \rightarrow 0$, $\delta_n\rightarrow 0$ as $n\rightarrow \infty$,\\
	(ii) $\| Tx_n\| \rightarrow \|T\|$ and $\|Ty_n\|\rightarrow\|T\|$ as $n\rightarrow \infty$,\\
	(iii) $Ax_n\in(Tx_n)^{+(\epsilon_n)}_{\alpha}$ and $Ay_n\in(Ty_n)^{-(\delta_n)}_{\alpha}$ for all  $n\in \mathbb{N}$.
\end{theorem}
\begin{proof}
	We first prove the easier sufficient part.\\
	Suppose $(a)$ holds. Now, for any  $\lambda \in \mathbb{C}$, $\|T + \lambda A\| \geq \|Tx_n + \lambda Ax_n\|\geq \|Tx_n\|- |\lambda | \|Ax_n\| \rightarrow \|T\|$ as $n \rightarrow \infty$. Therefore, $T \bot_{B} A$. \\
	Now,  suppose $(b)$ holds. Let $\lambda \in \mathbb{C}$. Then $\lambda= t \alpha$ for some $\alpha \in U$ and $t \in \mathbb{R}$. If $t \geq 0$ then  $ Ax_n\in(Tx_n)^{+(\epsilon_n)}_{\alpha} $ for all $n\in \mathbb{N},$ gives that  $\|Tx_n+\lambda Ax_n\|=\|Tx_n+ t \alpha Ax_n\| \geq \sqrt[]{1-\epsilon_n^2}~\|Tx_n\|$.\\
	This implies, $ \|T+\lambda A\|\geq\|(T+\lambda A)x_n\|=\|Tx_n+ t \alpha Ax_n\|  \geq\sqrt[]{1-\epsilon_n^2} \|Tx_n\|$. Since $ \epsilon_n \rightarrow 0 $ and $ \|Tx_n\| \rightarrow \| T \| $ as $ n \rightarrow \infty, $ we obtain,
	\[ \|T+\lambda A \|\geq \|T\|. \]
	Similarly, if $t \leq 0$ then $ Ay_n\in(Ty_n)^{-(\delta_n)}_{\alpha} $ for all $n\in \mathbb{N}$ implies that
	\[ \|T+\lambda A \|\geq \|T\|. \]
	This completes the proof of the sufficient part. \\
	Let us now prove the comparatively trickier necessary part. \\
	Suppose $(a)$ does not hold. \\
	Without loss of generality let us assume that $\|A\| \leq 1$. \\
	Since $T \bot_{B} A$, for any nonzero scalar $\lambda$, $\|T + \lambda A\| \geq \|T\|$. In particular, for $\alpha \in U$ and for each $n \in \mathbb{N}$,
	\[\|T + \frac{\alpha}{n} A\| > \|T\| - \frac{1}{n^3}. \]
	Therefore, for each $n \in \mathbb{N}$, there exists a sequence $\{x_n\} $ in $ S_{\mathbb{X}}$ such that $\|(T+ \frac{\alpha}{n} A) x_n \| > \|T\|- \frac{1}{n^3} \geq \|Tx_n\|- \frac{1}{n^3}$. \\
	We claim that $\|Tx_n\|\rightarrow \|T\|$. Indeed, $\|Tx_n\|=\|(T+\frac{\alpha}{n}A)x_n - \frac{\alpha}{n}Ax_n\|$ $\geq \|(T+\frac{\alpha}{n}A)x_n\| - |\frac{\alpha}{n}| \|Ax_n\|$ $>\|T\|- \frac{1}{n^3} - \frac{1}{n} \|A\|$ $\rightarrow \|T\|$ as $n \rightarrow \infty. $ Since $ x_n \in S_{\mathbb{X}},$ $ \| Tx_n \| \leq \| T \|  $. This proves our claim. \\
	Since $(a)$ does not hold, we assume that, $\inf \limits_{n \in \mathbb{N}}\|Ax_n\|= c > 0$. \\
	Choose $n_1 \in \mathbb{N}$ such that $n_1 > \frac{2 \|T\|}{c}$. Since $\|Tx_n\| \rightarrow \|T\|>0,$ there exists $n_2\in \mathbb{N}$ such that $\|Tx_n\| > \frac{\|T\|}{2} > 0$ for all $n \geq n_2$. Choose $n_3 \in \mathbb{N}$ such that $n_3 > \frac{2}{\|T\|}$. Let $n_0 =$ max$\{ n_1,n_2, n_3\}$. Then for all $n\geq n_0$, $0< \frac{1}{n \|Tx_n\|} < \frac{2}{n \|T\|}< 1$, which implies that for all $n\geq n_0$, $0< 1 - \frac{1}{n \|Tx_n\|}< 1$. \\
	Choose $\epsilon_n=\sqrt[]{1- (1- \frac{1}{n \|Tx_n\|})^2}$. Then clearly $\epsilon_n \rightarrow 0$ as $n \rightarrow \infty$.\\
	We claim that $Ax_n\in(Tx_n)^{+(\epsilon_n)}_{\alpha}$ for all $n\geq n_0$.\\
	Let $n \geq n_0$. Then for $0 \leq t < \frac{1}{n}$, \\
	$\|Tx_n + t \alpha Ax_n\| \geq \|Tx_n\|- t \|Ax_n\| \geq \|Tx_n\|-\frac{1}{n}$. \\
	For $\frac{1}{n} \leq t \leq n$, we claim that $\|Tx_n +t \alpha Ax_n\| \geq \|Tx_n\|- \frac{1}{n}$. \\
	Suppose on the contrary, we have, $\|Tx_n +t \alpha Ax_n\| < \|Tx_n\|- \frac{1}{n}$ for some $\frac{1}{n} \leq t \leq n$. Now, $Tx_n + \frac{\alpha}{n} Ax_n = (1- \frac{1}{n t})Tx_n + \frac{1}{n t}(Tx_n +t \alpha Ax_n)$. This implies that, $\|Tx_n\|- \frac{1}{n^3} < \|Tx_n + \frac{\alpha}{n} Ax_n\| \leq  (1- \frac{1}{n t})\|Tx_n\| + \frac{1}{n t}\|(Tx_n +t \alpha Ax_n)\| < (1- \frac{1}{n t})\|Tx_n\| + \frac{1}{n t}(\|Tx_n\| - \frac{1}{n})= \|Tx_n\|- \frac{1}{n^2 t}$. This implies that $t > n$, a contradiction. \\
	Thus, for $0 \leq t \leq n$, $\|Tx_n +t \alpha Ax_n\| \geq \|Tx_n\|- \frac{1}{n}$. Therefore, $0 \leq t \leq \frac{2 \|T\|}{c}$ gives that $\|Tx_n +t \alpha Ax_n\| \geq \|Tx_n\|- \frac{1}{n}$. \\
	Now, for $t > \frac{2 \|T\|}{c}$, $\|Tx_n +t \alpha Ax_n\| \geq t \|Ax_n\|- \|Tx_n\| \geq t c- \|Tx_n\| > 2\|T\|-\|Tx_n\| \geq \|Tx_n\|- \frac{1}{n}$. \\
	Therefore, for all $t \geq 0$, $\|Tx_n +t \alpha Ax_n\| \geq \|Tx_n\|- \frac{1}{n}= \sqrt[]{1-\epsilon_n^2} ~\|Tx_n\|$. This completes the proof of our claim. \\
	Similarly, considering $\|T-\frac{\alpha}{n} A\|> \|T\|- \frac{1}{n^3}$ for each $n \in \mathbb{N}$, we can find the desired sequences $\{y_n\} $ in $ S_{\mathbb{X}}$ and $ \{\delta _n\} $ in $ \mathbb{R}^+$ such that all the conditions of $(b)$ are satisfied. This completes the proof of the necessary part of the theorem.
\end{proof}

The corresponding characterization of Birkhoff-James orthogonality of bounded linear operators between real Banach spaces, now follow as a simple corollary to the previous theorem.
\begin{cor}(Theorem 2.4 of \cite{SPM})
	Let $\mathbb{X}$ and $\mathbb{Y}$ be two real Banach spaces. Let $T \in \mathbb{B} (\mathbb{X}, \mathbb{Y})$ be non-zero. Then for  any $A \in \mathbb{B} (\mathbb{X}, \mathbb{Y}),~ T \bot_B A$ if and only if either $(a)$ or $(b)$ holds: \\
	$(a)$ There exists a sequence $\{x_n\}$ in $S_\mathbb{X}$ such that $\| Tx_n\| \rightarrow \|T\|$ and $\|Ax_n\| \rightarrow 0$ as $n\rightarrow \infty$. \\
	$(b)$ There exist two sequences $\{x_n\},~\{y_n\}$ in $S_\mathbb{X}$ and two sequences of positive real numbers $\{\epsilon_n\}$, $\{\delta_n\}$ such that \\
	(i)  $ \epsilon_n \rightarrow 0$, $\delta_n\rightarrow 0$ as $n\rightarrow \infty$,\\
	(ii)  $\| Tx_n\| \rightarrow \|T\|$ and $\|Ty_n\|\rightarrow\|T\|$ as $n\rightarrow \infty$,\\
	(iii) $Ax_n\in(Tx_n)^{+(\epsilon_n)}$ and $Ay_n\in(Ty_n)^{-(\delta_n)}$ for all  $n\in \mathbb{N}$.
\end{cor}
\begin{proof}
	we note that in real Banach space, we must have, $\alpha \in U$ implies that $\alpha=1$. Therefore, the result follows easily from Theorem \ref{theorem:bounded}.
\end{proof}

In particular, in the statement of Theorem \ref{theorem:bounded}, if $\mathbb{X}$ is reflexive and $T,~ A$ from $\mathbb{X}$ to $\mathbb{Y}$ are compact, then we have the following theorem:

\begin{theorem}\label{theorem:reflexive}
	Let $\mathbb{X}$ be a reflexive complex Banach space, $\mathbb{Y}$ be any complex Banach space. Let $ T, A \in \mathbb{K}(\mathbb{X}, \mathbb{Y}).$  Then $ T\perp_{B}A$ if and only if for each $\alpha \in U$ there exist $x = x(\alpha), ~y= y(\alpha) \in M_{T} $ such that $Ax \in (Tx)_\alpha^{+}$ and $Ay \in (Ty)_\alpha^{-}.$
\end{theorem}

\begin{proof}
	Suppose $T\bot_B A$. Then either $(a)$ or $(b)$ of Theorem \ref{theorem:bounded} holds. If $(a)$ holds then there exists a sequence $\{x_n\}$ such that $\|Tx_n\|\longrightarrow \|T\|$ and $\|Ax_n\|$ converges to $0$. Then since $\mathbb{X}$ is reflexive, $\{x_n\}$ has a weakly convergent subsequence, say $\{x_{n_k}\}$ converging weakly to $x$ (say). Since $T,A$ are compact operators, $\{Tx_{n_k}\},~ \{Ax_{n_k}\}$ converges to $Tx,~ Ax$ respectively. Hence $x\in M_T$ and $Ax=0$. Thus, $Ax \in (Tx)_\alpha^{+}$ and $Ax \in (Tx)_\alpha^{-}$ for each $\alpha \in U$. If $(b)$ holds, then by using similar arguments as in case $(a)$, we can  find $x=x(\alpha), ~y=(\alpha) \in M_T$ such that $Ax \in (Tx)_\alpha^{+}$ and $Ay \in (Ty)_\alpha^{-}.$
\end{proof}
In particular, if $ \mathbb{X}, \mathbb{Y} $ are finite-dimensional complex Banach spaces, then we have the following corollary:

\begin{cor}\label{corollary:ortho complex}
	Let $\mathbb{X}, \mathbb{Y}$ be finite-dimensional complex Banach spaces. Let $ T, A \in \mathbb{L}(\mathbb{X}, \mathbb{Y}).$  Then $ T\perp_{B}A$ if and only if for each $\alpha \in U$ there exist $x = x(\alpha), ~y= y(\alpha) \in M_{T} $ such that $Ax \in (Tx)_\alpha^{+}$ and $Ay \in (Ty)_\alpha^{-}.$
\end{cor}

\begin{proof}
	Since every finite-dimensional complex Banach space is reflexive and every  linear operator on a finite-dimensional complex Banach space is compact, the proof of the corollary follows from Theorem \ref{theorem:reflexive}.
\end{proof}

We would further like to comment that the proofs of the corresponding characterization theorems in the real case are now obvious:
\begin{cor}\label{cor:reflexive}(Theorem 2.1 of \cite{SPM})
		Let $\mathbb{X}$ be a reflexive real Banach space, $\mathbb{Y}$ be any real Banach space. Let $ T, A \in \mathbb{K}(\mathbb{X}, \mathbb{Y}).$  Then $ T\perp_{B}A$ if and only if  there exist $x , y \in M_{T} $ such that $Ax \in (Tx)^{+}$ and $Ay \in (Ty)^{-}.$
\end{cor}
\begin{proof}
	Let $T \bot_B A$. Since in real Banach space, $\alpha \in U$ implies that $\alpha=1$, by Theorem \ref{theorem:reflexive}, there exist $x,y \in M_T$ such that  $Ax \in (Tx)^{+}$ and $Ay \in (Ty)^{-}.$
\end{proof}

\begin{cor}\label{corollary:ortho real}(Theorem 2.2 of \cite{S})
	Let $\mathbb{X}, \mathbb{Y}$ be finite-dimensional real Banach spaces. Let $ T, A \in \mathbb{L}(\mathbb{X}, \mathbb{Y}).$  Then $ T\perp_{B}A$ if and only if  there exist $x, y \in M_{T} $ such that $Ax \in (Tx)^{+}$ and $Ay \in (Ty)^{-}.$
\end{cor}
\begin{proof}
		Since every finite-dimensional complex Banach space is reflexive and every  linear operator on a finite-dimensional complex Banach space is compact, the proof of the corollary follows from Corollary \ref{cor:reflexive}.
\end{proof}

\begin{remark}
In view of the method employed in proving Theorem $ 2.3, $ it should be mentioned that our proof essentially follows the same line of arguments, as outlined in the proof of Theorem $ 2.4 $ of \cite{SPM}. Indeed, the novelty lies in introducing the corresponding notations in the complex case, analogous to the real case. Moreover, as we will see in the next theorem, in spite of being a complete characterization of B-J orthogonality of compact linear operators on a reflexive complex Banach space, Theorem \ref{theorem:reflexive} does not capture the  full strength of the complex number system. Indeed, in our opinion, Theorem \ref{theorem:reflexive} should be regarded as a stepping stone towards our next theorem, that also distinguishes the complex case from the real case. First we need the following geometric lemma:
\end{remark}

\begin{lemma}\label{lemma:complex}
	Let $\mathbb{X}$ be a complex Banach space. Let $x,~y \in \mathbb{X}$ and $\alpha=e^{i \theta}$, where $\theta \in [0,\pi]$. If $y \in x^{+}_{\alpha}$ then either $y \in x^{+}_{\beta}$ for all $\beta$ with $\arg \beta \in [0, \theta]$ or $y \in x^{+}_{\beta}$ for all $\beta$ with  $\arg \beta \in [\theta, \pi]$.
\end{lemma}
\begin{proof}
Suppose  $y \notin x^{+}_{\alpha_1}$ for some $\alpha_1$ with  $\arg \alpha_1 \in [0, \theta]$. Then there exists $t_1 > 0$ such that $\|x+ t_1 \alpha_1 y\|< \|x\|$. We claim that $y \in x^{+}_{\beta}$ for all $\beta$ with $\arg \beta \in [\theta, \pi]$. If possible suppose that  $y \notin x^{+}_{\alpha_2}$ for some $\alpha_2$ with  $\arg \alpha_2 \in [ \theta, \pi]$. Then there exists $t_2 > 0$ such that $\|x+ t_2 \alpha_2 y\|< \|x\|$. Then it is easy to verify that there exist $0< s < 1$ and $t > 0$ such that $(1-s) t_1 \alpha_1 + s t_2 \alpha_2= t \alpha$. Therefore, $(1-s)[x+t_1 \alpha_1 y]+ s [x+ t_2 \alpha_2 y]= x+ t \alpha y$. This implies that $\|x+ t \alpha y\|\leq (1-s)\|x+ t_1 \alpha_1 y \|+ s \|x+ t_2 \alpha_2 y\|< (1-s)\|x\| + s \|x\|= \|x\|$, a contradiction. This proves our claim.
\end{proof}

Let us now prove the following characterization theorem, that improves the necessary part of Theorem \ref{theorem:reflexive}.

\begin{theorem}
	Let $\mathbb{X}$ be a reflexive complex Banach space and $\mathbb{Y}$ be any complex Banach space. Let $ T,A\in K(\mathbb{X},\mathbb{Y})$. Then $T\bot_{B}A$ if and only if there exist $x,~ y,~ z,~ w \in M_{T}$ and $\phi_{1}, \phi_{2} \in [0,\pi]$ such that\\
	 (i) $ Ax \in (Tx)^{+}_{\alpha} ~~\forall~~ \alpha $ with $\arg \alpha \in [0,\phi_{1}]$,\\
	(ii) $ Ay \in (Ty)^{+}_{\alpha} ~~\forall~~ \alpha$ with $\arg \alpha \in [\phi_{1}, \pi],$ \\
	(iii) $  Az \in (Tz)^{-}_{\alpha} ~~\forall~~ \alpha $ with $\arg \alpha \in [0,\phi_{2}]$, \\
	(iv) $Aw \in (Tw)^{-}_{\alpha} ~~\forall~~ \alpha $ with $\arg \alpha \in [\phi_{2}, \pi]$.

\end{theorem}
\begin{proof}
	We first prove the easier sufficient part. Suppose there exist $x,~ y,~ z,~ w \in M_{T}$ and $\phi_{1}, \phi_{2} \in [0,\pi]$ such that all the conditions in $(i), (ii), (iii)$ and $(iv)$ are satisfied. Let $\lambda \in \mathbb{C}$. Then either there exist $t_1 \geq 0$ and $ \alpha_1 $ with $\arg \alpha_1 \in [0, \phi_1]$ such that $\lambda = t_1 \alpha_1$ or there exist $t_2 \geq 0$ and $\alpha_2 $ with $\arg \alpha_2  \in [ \phi_1, \pi]$ such that $\lambda = t_2 \alpha_2$ or there exist $t_3 \leq 0$ and $ \alpha_3 $ with $\arg \alpha_3 \in [0, \phi_2]$ such that $\lambda = t_3 \alpha_3$ or there exist $t_4 \leq 0$ and $ \alpha_4 $ with $\arg \alpha_4 \in [ \phi_2, \pi]$ such that $\lambda = t_4 \alpha_4$. Now $\lambda= t_1 \alpha_1$ implies that $\|T + \lambda A\| = \|T + t_1 \alpha_1 A\| \geq \|Tx + t_1 \alpha_1 Ax\| \geq \|Tx\|= \|T\|$. Similarly, in the other cases, it can be shown that $\|T+ \lambda A\|\geq \|T\|$.  Hence $T\bot_B A$.\\
    For the necessary part, suppose that $T\perp_{B}A$. Then from Theorem \ref{theorem:reflexive}, we have, for each $\alpha $ with $ \arg \alpha   \in [0,\pi]$, there exists $x_{\alpha}\in M_{T}$ such that $ Ax_{\alpha}\in (Tx_{\alpha})^{+}_{\alpha}$.
	Now, consider
	\[V_1=\{\theta \in [0, \pi]: \exists ~ x\in M_T : Ax \in (Tx)^{+}_{\alpha}~\forall ~ \alpha ~ with ~ \arg \alpha  \in [0,\theta]  \},\]
	\[V_2= \{\theta \in [0, \pi]: \exists ~ x\in M_T : Ax \in (Tx)^{+}_{\alpha}~\forall ~ \alpha ~ with ~ \arg \alpha  \in [\theta, \pi]  \}.\]
	Clearly, $ 0 \in V_1$ and $\pi \in V_2$ and therefore, $V_1,~ V_2$ are non-empty. Moreover, $V_1,~V_2 $ are bounded. Suppose $\xi=\sup V_1$ and $\eta= \inf V_2$. Now, we claim that $\xi \geq \eta$.  If possible suppose that $\xi < \eta$. Then consider $\zeta= \frac{\xi + \eta}{2} \in [0, \pi]$. Now, from Theorem \ref{theorem:reflexive}, we have, for $\alpha = e^{i \zeta}$ there exists $x_{\alpha} \in M_T$ such that $Ax_{\alpha}\in (Tx_{\alpha})^{+}_{\alpha}$. Using Lemma \ref{lemma:complex}, we have, either $Ax_{\alpha}\in (Tx_{\alpha})^{+}_{\beta}$ for all $\beta$ with $\arg \beta \in [0,\zeta]$ or $Ax_{\alpha}\in (Tx_{\alpha})^{+}_{\beta}$ for all $\beta$ with $\arg \beta \in [\zeta, \pi]$. But  $Ax_{\alpha}\in (Tx_{\alpha})^{+}_{\beta}$ for all $\beta$ with $ \arg \beta \in [0,\zeta]$ implies that $\zeta \in V_1$. This contradicts that $\xi= \sup V_1 $. Again, $Ax_{\alpha}\in (Tx_{\alpha})^{+}_{\beta}$ for all $\beta$ with $\arg \beta \in [\zeta, \pi]$ implies that $\zeta \in V_2$. This contradicts that $\eta= \inf V_2$. Hence $\xi \geq  \eta$.  Now, there exist sequences $\{\xi_n\} \subseteq V_1,~ \{\eta_{n} \}\subseteq V_2$ such that $\{\xi_n\}$ converges to $\xi$ and $\{\eta_{n}\}$ converges to $\eta$. Since $\xi_n \in V_1,~ \eta_{n} \in V_2$, there exist $x_{n},~ y_{{n}} \in M_T$ such that $Ax_{n} \in (Tx_{n})^{+}_{\alpha}~\forall ~ \alpha $ with $\arg \alpha \in [0,\xi_n] $ and $Ay_n  \in (Ty_n)^{+}_{\alpha}~\forall ~ \alpha $ with $\arg \alpha \in [\eta_{n}, \pi] $. Since $\mathbb{X}$ is reflexive, $\{x_{n}\},~ \{y_n \}$ have weakly convergent subsequences. Without loss of generality assume that $\{x_{n}\}$ weakly converges to $x$ and $\{y_n\}$  weakly converges to $y$. Since $T,A$ are compact operators, $Tx_n \longrightarrow Tx,~ Ty_n \longrightarrow Ty,~Ax_n \longrightarrow Ax,~ Ay_n \longrightarrow Ay.$ Clearly $x, y \in M_T$. Now, $\|Tx_{n}+ t \alpha Ax_{n}\|\geq \|T\|$ for all $t \geq 0$ and for all $\alpha$ with $ \arg \alpha \in [0, \xi_n] \Rightarrow \|Tx+ t \alpha Ax\|\geq \|T\|$ for all $t \geq 0$ and for all $\alpha$ with $\arg \alpha \in [0, \xi]$. Similarly, $\|Ty_n+ t \alpha Ay_n\|\geq \|T\|$ for all $t \geq 0$ and for all $\alpha$ with $\arg \alpha \in [ \eta_n, \pi] \Rightarrow \|Ty+ t \alpha Ay\|\geq \|T\|$ for all $t \geq 0$ and for all $\alpha$ with $\arg \alpha \in [ \eta, \pi]$. Since $\xi \geq \eta$, $\|Ty+ t \alpha Ay\|\geq \|T\|$ for all $t \geq 0$ and for all $\alpha$ with  $\arg \alpha \in [ \xi, \pi]$.  Let $\xi=\phi_{1}$. Then $Ax \in (Tx)_{\alpha}^+$ for all $\alpha$ with $\arg \alpha \in [0, \phi_1]$ and $Ay \in (Ty)_{\alpha}^+$ for all $\alpha $ with $\arg \alpha \in [\phi_1, \pi]$. \\
	Similarly, for each $\alpha$ with $\arg \alpha \in [0,\pi]$ there exists $z_{\alpha}\in M_{T}$ such that $ Az_{\alpha}\in (Tz_{\alpha})^{-}_{\alpha}$ gives that there exist $\phi_2 \in [0, \pi]$ and $z, w \in M_T$ such that 	$  Az \in (Tz)^{-}_{\alpha} ~~\forall~~ \alpha $ with $\arg \alpha \in [0,\phi_{2}]$ and
	$Aw \in (Tw)^{-}_{\alpha} ~~\forall~~ \alpha $ with $\arg \alpha \in [\phi_{2}, \pi]$.
\end{proof}

 Sain and Paul proved in \cite{SP} that if $T$ is a linear operator on  a finite-dimensional real Banach  space  $\mathbb{X}$, with $M_T = \pm D$ ($D$ being a closed connected subset of $S_{\mathbb{X}}$), then $ T \bot_{B} A $ if and only if there exists $ x \in D $ such that $Tx \bot_{B} Ax.$ In the following theorem we prove an analogous result for complex Banach spaces. Before proving the theorem, let us observe that if $\mathbb{X}$ is a complex Banach space, $ T \in \mathbb{L}(\mathbb{X})$ and $D$ is a closed connected subset of $S_{\mathbb{X}}$ such that $ D \subset M_T, $  then we must have, $ \bigcup_{\theta \in [0, 2\pi)} e^{i\theta}D \subset M_T $ and $ \bigcup_{\theta \in [0, 2\pi)} e^{i\theta}D $ is also a connected subset of $S_{\mathbb{X}}$. Note that, it is not true in general, if $\mathbb{X}$ is a real Banach space. This explains the change in the statement of Theorem \ref{theorem:connected}, compared to the corresponding real case.

\begin{theorem}\label{theorem:connected}
Let $\mathbb{X}$ be a finite-dimensional complex Banach space. Let $ T \in \mathbb{L}(\mathbb{X})$ be such that $M_T$ is a closed connected subset of $S_{\mathbb{X}}.$ Then for $A \in \mathbb{L}(\mathbb{X})$, $ T \bot_B A$ if and only if for each $ \alpha \in U $, there exists $x = x(\alpha) \in M_T $ such that $ Tx \bot_{\alpha} Ax.$
\end{theorem}
\begin{proof} The sufficient part of the theorem follows trivially.
Since $ T \bot_B A $, applying Corollary \ref{corollary:ortho complex}, it follows that for each $\alpha \in U$, there exist $x = x(\alpha), ~y= y(\alpha) \in M_{T} $ such that $Ax \in (Tx)_\alpha^{+}$ and $Ay \in (Ty)_\alpha^{-}.$
 Then considering the line passing through $\alpha $ and $ - \alpha$ in the complex plane and  following the same arguments as in Theorem 2.1 of \cite{SP}, it can be concluded that there exists $u \in M_T$ such that $Au \in (Tu)_\alpha^+ $ and $ Au \in (Tu)_\alpha^-$, by using the connectedness of $M_T.$ This implies that $Tu \bot_{\alpha} Au.$ This establishes the theorem.
\end{proof}

Once again, in contrast to the real case, we would like to sharpen the necessary part of Theorem \ref{theorem:connected} in the complex case. First we need the following lemma.

\begin{lemma}\label{lemma:connected}
	Let $\mathbb{X}$ be a complex Banach space. Let $x, y \in \mathbb{X} $ and $\alpha \in U$ with $\arg \alpha = \theta$ be such that  $x\bot_{\alpha} y$. Then either $y \in (x)^{+}_{\beta}$ for all $\beta$ with $\arg \beta \in [\theta- \pi, \theta]$ or $y \in (x)^{+}_{\beta}$ for all $\beta$ with  $\arg \beta \in [\theta, \theta + \pi]$.
\end{lemma}

\begin{proof}
	Let  $x\bot_{\alpha} y$. Suppose that $y \notin (x)^{+}_{\beta_1}$ for some $\beta_1$ with $\arg \beta_1 \in [\theta- \pi, \theta]$. Then there exists $t_1 > 0$ such that $\|x+ t_1 \beta_1 y\|< \|x\|$. If possible suppose that $y \notin (x)^{+}_{\beta_2}$ for some $\beta_2$ with $\arg \beta_2 \in [\theta, \theta + \pi]$.  Then there exists $t_2 > 0$ such that $\|x+ t_2 \beta_2 y\|< \|x\|$. Then it is easy to verify that there exist $0<s<1$ and $t\in \mathbb{R}$ such that
	\begin{eqnarray*}
	        t \alpha &=& (1-s)t_1 \beta_1+ s t_2 \beta_2 \\
	        \Rightarrow x+ t \alpha y &=&(1-s)(x+t_1 \beta_1 y)+ s (x+ t_2 \beta_2 y)  \\
	        \Rightarrow \|x+ t \alpha y\| &\leq &(1-s)\|(x+t_1 \beta_1 y)\|+ s \|(x+ t_2 \beta_2 y)\|  \\
	        \Rightarrow \|x+ t \alpha y\| &<&(1-s)\|x\|+ s \|x\| \\
	        &=&\|x\|,
	\end{eqnarray*}
	this leads to a contradiction and so $y \in (x)^{+}_{\beta}$ for all $\beta$ with $\arg \beta \in [\theta, \theta + \pi]$. This proves the lemma.
\end{proof}

Now, the promised theorem:
\begin{theorem}\label{theorem:conn}
   Let $\mathbb{X}$ be a finite-dimensional complex Banach space. Let $ T \in \mathbb{L}(\mathbb{X})$ be such that $M_T$ is a closed connected subset of $S_{\mathbb{X}}.$ Then for $A \in \mathbb{L}(\mathbb{X})$, $ T \bot_B A$ if and only if there exist some $ \theta \in [0,\pi] $ and $x ,y \in M_T $ such that $ Ax \in (Tx)^{+}_{\alpha}$ for all $\alpha$ with $\arg \alpha \in [\theta- \pi, \theta]$ and  $ Ay \in (Ty)^{+}_{\alpha}$ for all $\alpha$ with $\arg \alpha \in [\theta, \theta+ \pi]$.
\end{theorem}

\begin{proof}
	Let us first prove the sufficient part of the theorem. Suppose there exists some $ \theta \in [0,\pi] $ and $x ,y \in M_T$ such that $ Ax \in (Tx)^{+}_{\alpha}$ for all $\alpha$ with $\arg \alpha \in [\theta- \pi, \theta]$ and  $ Ay \in (Ty)^{+}_{\alpha}$ for all $\alpha$ with $\arg \alpha \in [\theta, \theta+ \pi]$. Let $\lambda \in \mathbb{C}$. Then either there exist $t_1 \geq  0$ and $ \alpha_1$ with $\arg \alpha_1 \in [\theta- \pi, \theta]$ such that $\lambda= t_1 \alpha_1$ or there exist $t_2 \geq  0$ and $\alpha_2$ with $\arg \alpha_2 \in [\theta, \theta + \pi]$ such that $\lambda= t_2 \alpha_2$. Now, $\lambda= t_1 \alpha_1 $ implies that $\|T+ \lambda A\|=\|T + t_1 \alpha_1 A\|\geq \|(T + t_1 \alpha_1 A)x\|\geq \|Tx\|=\|T\|$ and $\lambda= t_2 \alpha_2 $ implies that $\|T+ \lambda A\|=\|T + t_2 \alpha_2 A\|\geq \|(T + t_2 \alpha_2 A)y\|\geq \|Ty\|=\|T\|$. Therefore, $T\bot_B A$. This completes the proof of the sufficient part of the theorem.\\
	For the necessary part, suppose that $T\bot_B A$. Let us consider the following two sets
	\[V_1=\{\theta \in [0, \pi]: \exists~ x \in M_T ~:~ Ax ~ \in (Tx)^{+}_{\alpha}~ \forall ~ \alpha ~ with ~ \arg  \alpha  \in [\theta - \pi, \theta] \},\]
	 \[V_2=\{\theta \in [0, \pi]: \exists~ x \in M_T ~:~  Ax ~ \in (Tx)^{+}_{\alpha}~ \forall ~ \alpha ~ with ~ \arg \alpha \in [\theta, \theta+\pi]  \}.\]
	 We first show that $[0,\pi]= V_1 \cup V_2$. Let $\theta \in [0,\pi]$ and $\alpha= e^{i \theta}$. Since $T\bot_B A$, by Theorem \ref{theorem:connected}, we have, there exists $x=x(\alpha) \in M_T$ such that $Tx \bot_{\alpha} Ax$. Therefore, applying Lemma \ref{lemma:connected}, we have, either $Ax \in (Tx)^{+}_{\beta}$ for all $\beta$ with $\arg \beta \in [\theta- \pi, \theta]$, i.e, $\theta \in V_1$ or $Ax \in (Tx)^{+}_{\beta}$ for all $\beta$ with $\arg \beta \in [\theta, \theta + \pi]$, i.e, $\theta \in V_2$. Hence $[0,\pi]= V_1 \cup V_2$.\\
	  We claim that $V_1 \neq \emptyset$. Let $0 \notin V_1 $. Then $0 \in V_2$. Hence there exists $z \in M_T$ such that $Az ~ \in (Tz)^{+}_{\beta}~ \forall ~ \beta $ with $\arg \beta \in [0,\pi] $. This implies that $\pi \in V_1$. Hence $V_1 \neq \emptyset$. Similarly, it can be shown that $V_2 \neq \emptyset$.\\
	   We next show that $V_1$ is closed. Let $\{\theta_n\}$ be a sequence in $V_1$ such that $\{\theta_n\}$ converges to $\theta$. Let $\beta=e^{i \theta}$. Then there exists $x_n \in M_T$ such that $ Ax_n ~ \in (Tx_n)^{+}_{\alpha}~ \forall ~ \alpha$ with $\arg \alpha \in [\theta_n - \pi, \theta_n]$. Since $\mathbb{X}$ is finite-dimensional, $\{x_n\}$ has a convergent subsequence. Without loss of generality assume that $\{x_n\}$ converges to $x$ (say). Clearly, $x \in M_T$. Now, $Ax_n \in (Tx_n)^{+}_{\alpha}$ for all $\alpha$ with $\arg \alpha \in [\theta_n -\pi, \theta_n]$ gives that $\|Tx_n + t e^{i \theta_n} Ax_n\| \geq \|T\|$ for all $t \geq 0$. Letting $n \longrightarrow \infty$, we have $\|Tx+ t e^{i \theta}Ax\|=\|Tx + t\beta Ax\|\geq \|T\| \Rightarrow Ax \in (Tx)^{+}_{\beta}$. Similarly, $Ax \in (Tx)^{+}_{\gamma}$, where $\arg \gamma =\theta- \pi$. Now, let $\theta- \pi < \phi < \theta$. If possible suppose that there does not exist any $n_0 \in \mathbb{N}$ such that $\phi \in [\theta_n - \pi, \theta_n]$ for all $n \geq n_0$. Then without loss of generality we may assume that $\phi > \theta_n$ for all $n \in \mathbb{N}$. Letting $n \longrightarrow \infty$, we have $\phi \geq \theta$, a contradiction. Hence  there exists $n_0 \in \mathbb{N}$ such that $\phi \in [\theta_n - \pi, \theta_n]$ for all $n \geq n_0$. This implies that $\|Tx_n + t e^{i \phi}Ax_n\| \geq \|T\| $ for all $t \geq 0$ and for all $n \geq n_0$. Therefore, as $n \longrightarrow \infty$, we have $\|Tx+ te^{i \phi}Ax \| \geq \|T\|$ for all $t \geq 0$. This implies that $Ax \in (Tx)^{+}_{\delta}$, where $\delta= e^{i \phi}$.  Thus, $Ax\in (Tx)^{+}_{\alpha}$ for all $\alpha$ with $\arg \alpha \in [\theta -\pi, \theta]$. Hence $\theta \in V_1$. Thus, $V_1$ is closed. Similarly, it can be shown that $V_2$ is closed. \\
	   Now, since $[0,\pi]$ is connected, $V_1 \cap V_2 \neq \emptyset$. Let $\theta \in V_1 \cap V_2$. Then there exist $x, y \in M_T$ such that $ Ax ~ \in (Tx)^{+}_{\alpha}~ \forall ~ \alpha $ with $\arg \alpha \in [\theta - \pi, \theta]$ and $ Ay ~ \in (Ty)^{+}_{\alpha}~ \forall ~ \alpha$ with $\arg \alpha \in [\theta, \theta+ \pi]$. This establishes the theorem.
\end{proof}

Next, in the context of complex Banach spaces we explore the structure of $M_T$ in connection with B-J orthogonality. We would like to invite the reader to have a look at Theorem 2.2 and Corollary 2.2.1 of \cite{Sa}, for an analogous result in the real case.
\begin{theorem}
Let $\mathbb{X} $ be a complex Banach space,  $0 \neq T\in \mathbb{L}(\mathbb{X})$  and  $x\in M_{T}.$ \\
(i)   If $y\in \mathbb{X}$ is such that $ Tx\perp_{B}Ty $ then $ x\perp_{B}y$.\\
(ii) $T(x_\alpha^{+}\setminus x_\alpha^{\perp})\subset (Tx)_\alpha^{+}\setminus (Tx)_\alpha^{\perp}$, for $\alpha \in U.$\\
(iii)  $T(x_\alpha^{-}\setminus x_\alpha^{\perp})\subset (Tx)_\alpha^{-}\setminus (Tx)_\alpha^{\perp}$, for $\alpha \in U.$\\
(iv) $\ker T \subset \bigcap_{x\in M_{T}}x^{\perp}$.
\end{theorem}
\begin{proof}
$(i)$ 	Suppose $ Tx\perp_{B}Ty $. Then $\|T\|\|x\|= \|Tx\| \leq \|Tx+\lambda Ty\| \leq \|T\| \|x+\lambda y\| ~\forall \lambda \in \mathbb{C}$. This implies that $ \| x + \lambda y \| \geq \|x\| ~\forall \lambda \in \mathbb{C}$. Therefore, $ x \bot_B y.$ \\

$(ii)$  Let $y \in x_\alpha^{+}\setminus x_\alpha^{\perp}.$ Then there exist some $t<0$ such that $ \|x+ t \alpha y\|< \|x\|.$ Now, $\|Tx + t \alpha Ty\| \leq \|T\| \|x+ t \alpha y\|< \|T\|\|x\|=\|Tx\|$. This implies that $  Ty\notin (Tx)_{\alpha}^{-}$. It now follows from Proposition 2.1 that $Ty\in (Tx)_\alpha^{+}\setminus (Tx)_\alpha^{\perp}$. \\

$(iii)$ Follows similarly as $(ii)$. \\

$(iv)$ If   $M_{T}=\phi$, then the theorem follows trivially. Let us assume that  $M_{T}\neq\phi$. Let $y\in \ker T.$ Then for any $x\in M_{T}$, we have $Ty\in(Tx)^{\perp}$, since $Ty=0$. From (i) it follows that $y\in x^{\perp}$. This implies that
$\ker T \subset \bigcap_{x\in M_{T}}x^{\perp}$.
\end{proof}

\begin{remark}
In addition, if $x,~Tx$ are smooth points in $\mathbb{X}$ then for any $y \in \mathbb{X}$, we have, $x\bot_B y \Rightarrow Tx \bot_B Ty$. This can be proved following the same line of arguments, as in Lemma 2.1 of \cite{SGP}.

\end{remark}

We next study left symmetric linear operator(s) defined on a finite-dimensional complex Banach space $\mathbb{X}.$ An element $ T \in \mathbb{L}(\mathbb{X}) $ is said to be left symmetric if $ T \bot_B A $ implies $ A \bot_B T $ for any $ A \in \mathbb{L}(\mathbb{X}).$ In \cite{S}, Sain proved a nice connection between the left symmetry of bounded linear operators and the left symmetric points in the corresponding norm attainment set, for real Banach spaces. Following the same line of arguments, these results can be proved for complex Banach spaces. We simply state the following two theorems for complex Banach spaces.
\begin{theorem}\label{theorem:left1}
Let $\mathbb{X} $ be a finite-dimensional strictly convex complex Banach space. If $ T \in \mathbb{L}(\mathbb{X})$ is left symmetric then for each $x \in M_T$, $Tx$ is a left symmetric point.
\end{theorem}
\begin{theorem}\label{theorem:left2}
	Let $\mathbb{X}$ be a finite-dimensional strictly convex and smooth complex Banach space. Let $ T \in \mathbb{L}(\mathbb{X})$
be such that there exists $x, y \in S_\mathbb{X}$ satisfying\\
 $(i)$ $x\in M_{T}$, $(ii)$ $y\perp_{B} x$, $(iii)$ $Ty\neq 0$. Then $T$ can not be left symmetric.
\end{theorem}
In \cite{GSP}, it was proved that if $T$ is a compact linear operator on a real Hilbert space $\mathbb{H}$ then $T$ is left symmetric if and only if $T$ is the zero operator. Sain proved in \cite{S} that if $T \in \ell_p^2(\mathbb{R}) , 1 < p < \infty,$ then $T$ is left symmetric if and only if $T$ is the zero operator. The situation is a little bit complicated when we consider $ \ell_p^2(\mathbb{C})$ instead of $\ell_p^2(\mathbb{R})$. We first find the symmetric points of $ \ell_p^2(\mathbb{C})$ in the following propositions. We would like to mention that the proofs of the propositions follow from elementary calculations in each case.

\begin{prop}
Let $\mathbb{X}=\ell_{p}^{2}(\mathbb{C}),1 < p <  \infty$. If $(z_{1},z_{2}) \in S_{\mathbb{X}}$ then \\
(i) $(z_{1},z_{2})\perp_{B} (1,-\frac{|z_{1}|^{p-2}\bar{z_{1}}}{|z_{2}|^{p-2}\bar{z_{2}}})$, where  $ z_{2} \neq 0.$ \\
(ii)  $(z_{1},z_{2})\perp_{B} (-\frac{|z_{2}|^{p-2}\bar{z_{2}}}{|z_{1}|^{p-2}\bar{z_{1}}}, 1)$, where  $ z_{1} \neq 0.$
\end{prop}

\begin{proof}
	$(i)$ Let $z_2 \neq 0$. Then it is easy to see from elementary calculations that the Gateaux derivative $G((z_{1},z_{2}),(1,-\dfrac{|z_{1}|^{p-2}\bar{z_{1}}}{|z_{2}|^{p-2}\bar{z_{2}}}))=0$. Therefore, $(z_{1},z_{2})\perp_{B} (1,-\frac{|z_{1}|^{p-2}\bar{z_{1}}}{|z_{2}|^{p-2}\bar{z_{2}}})$, where  $ z_{2} \neq 0.$ \\
	$(ii)$ Follows similarly as $(i)$.
\end{proof}

\begin{prop}
Let $\mathbb{X}=\ell_{p}^{2}(\mathbb{C}), 1 < p <  \infty$. If $ x, y\in S_{\mathbb{X}}$ are such that $x\perp_{B}y $ and $y\perp_{B}x, $ then either of the following is true:

  	(i) $x=(e^{i\theta_{1}},0)$ and $y= (0,e^{i\theta_{2}})$, where $\theta_i \in [0, 2\pi), i=1,2$.

  	(ii) $x= (0,e^{i\theta_{1}})$ and $y=(e^{i\theta_{2}},0)$, where $\theta_i \in [0, 2\pi), i=1,2$.

  	(iii) $x=(\dfrac{1}{2^{1/p}} e^{i\theta_{1}},\dfrac{1}{2^{1/p}} e^{i\theta_{2}})$ and $y= (\dfrac{1}{2^{1/p}} e^{i\theta_{1}},-\dfrac{1}{2^{1/p}} e^{i\theta_{2}})$, where $\theta_i \in [0, 2\pi), i=1,2$.

  	(iv) $x=(\dfrac{1}{2^{1/p}} e^{i\theta_{1}},-\dfrac{1}{2^{1/p}} e^{i\theta_{2}})$ and $y= (\dfrac{1}{2^{1/p}} e^{i\theta_{1}},\dfrac{1}{2^{1/p}} e^{i\theta_{2}})$, where $\theta_i \in [0, 2\pi), i=1,2$.
\end{prop}

\begin{prop}\label{prop:left}
	Let $\mathbb{X}$ be  the two-dimensional complex Banach space $l_{p}^{2}(\mathbb{C}), 1< p < \infty$. Then $x\in S_{\mathbb{X}}$ is a left symmetric point in $\mathbb{X}$ if and only if

	$x\in \{(e^{i\theta_{1}},0),(0,e^{i\theta_{2}}),(\dfrac{1}{2^{1/p}} e^{i\theta_{3}},\dfrac{1}{2^{1/p}} e^{i\theta_{4}}),(\dfrac{1}{2^{1/p}} e^{i\theta_{3}},-\dfrac{1}{2^{1/p}} e^{i\theta_{4}}) \}$, where $\theta_i \in [0, 2\pi), i=1,2,3,4$.
\end{prop}

\begin{theorem}
	Let $\mathbb{X}=\ell_{p}^{2}(\mathbb{C}),1 < p <  \infty$. Then $T\in \mathbb{L}({\mathbb{X}})$ is left symmetric if and only if $T$ is the zero operator.
\end{theorem}
\begin{proof}
	If possible suppose that $T\in \mathbb{L}({\mathbb{X}})$ is a non-zero left symmetric point. Since B-J orthogonality is homogeneous and $T$ is non-zero, we may assume without loss of generality, that $\|T\|=1$. Suppose $T$ attains norm at $x\in S_{\mathbb{X}}$. From Theorem 2.3 of James \cite{J}, it follows that there exists $y\in S_{\mathbb{X}}$  such that $y\perp_{B}x$. Since $\mathbb{X}$ is strictly convex and smooth, applying Theorem \ref{theorem:left2}, we see that $Ty=0$. Theorem \ref{theorem:left1}, ensures that $Tx$ must be a left symmetric point in $\mathbb{X}$. Thus, applying Proposition \ref{prop:left}, we have that

	$Tx\in \{(e^{i\theta_{1}},0),(0,e^{i\theta_{2}}),(\dfrac{1}{2^{1/p}} e^{i\theta_{3}},\dfrac{1}{2^{1/p}} e^{i\theta_{4}}),(\dfrac{1}{2^{1/p}} e^{i\theta_{3}},-\dfrac{1}{2^{1/p}} e^{i\theta_{4}}) \}$, where $\theta_i \in [0, 2\pi),~ i=1,2,3,4$.

	We claim that $x\perp_{B}y$.

	From Theorem 2.3 of James \cite{J}, it follows that there exists $\alpha\in\mathbb{C} $ such that $\alpha y+x\perp_{B}y$. Since $y\perp_{B}x$ and $x, y\neq0$, $\{x,y\}$ is linearly independent and hence $\alpha y+x\neq0$. Consider $ z=\dfrac{\alpha y+x}{\|\alpha y+x\|}$. Now, if $Tz=0$ then $T$ is zero operator. Let $Tz\neq0$. Clearly, $\{y,z\}$ is a basis of $\mathbb{X}$.

	Let $\|c_{1}z+c_{2}y\|=1$ for some $c_{1}, c_{2}\in\mathbb{C}$. Then $1=\|c_{1}z+c_{2}y\|=|c_{1}|\|z+\dfrac{c_{2}}{c_{1}}y\|\geq|c_{1}|$. Since $\mathbb{X}$ is strictly convex, $|c_{1}|<1$, if $c_{2}\neq0$. We also have, $\|T(c_{1}z+c_{2}y)\|=\|c_{1}Tz\|\leq\|Tz\|$ and $\|T(c_{1}z+c_{2}y)\|=\|Tz\|$  if and only if $|c_{1}|=1$ and $c_{2}=0$. This shows that $M_{T}=\{e^{i \theta} z: \theta \in [0,2 \pi)\}$.Thus, we must have $x= e^{i \theta}z$ for some $\theta \in [0,2 \pi)$. Hence $x\perp_{B}y$.

	Thus, $x, y\in\mathbb{X}$ are such that $x\perp_{B}y$ and $y\perp_{B}x$. Therefore, by Proposition \ref{prop:left}, we see that  we have the following informations about $T$:

	$(i)$ $T$ attains norm at $x,~ x\perp_{B}y,~ y\perp_{B}x,~ Ty =0$,~ $Tx$ is left symmetric.

	$(ii)$ $ x,~ y,~ Tx\in\{(e^{i\theta_{1}},0),(0,e^{i\theta_{2}}),(\dfrac{1}{2^{1/p}} e^{i\theta_{3}},\dfrac{1}{2^{1/p}} e^{i\theta_{4}}),(\dfrac{1}{2^{1/p}} e^{i\theta_{3}},-\dfrac{1}{2^{1/p}} e^{i\theta_{4}})\}$, where $\theta_i \in [0, 2\pi), i=1,2,3,4$.

	In order to prove that $T\in\mathbb{L}({\mathbb{X}})$ is left symmetric if and only if $T$ is the zero operator, we only need to consider 16 different types of operators that satisfy $(i)$ and $(ii)$ and show that none of them is left symmetric.

	Let us first consider one such typical linear operator and prove that it is not left symmetric.

	Let $T\in\mathbb{L}({\mathbb{X}})$ be defined by $T(e^{i\theta},0)=(e^{i\theta},0),~ T(0,e^{i\theta})=(0,0)$. Define $A\in\mathbb{L}(\mathbb{X})$ by $A(e^{i\theta},0)=(0,e^{i\theta}),~ A(0,e^{i\theta})=(e^{i\theta},e^{i\theta})$. Then $M_{T}= (e^{i\theta_{1}},0)$. Since $T(e^{i\theta_{1}},0)=(e^{i\theta_{1}},0)\perp_{B} (0,e^{i\theta_{1}})=A(e^{i\theta_{1}},0)$, it follows that $T\perp_{B}A$. We claim that $A\not\perp_{B}T$.

	Now, $\|A(\dfrac{1}{2^{1/p}} e^{i\theta},\dfrac{1}{2^{1/p}} e^{i\theta})\|^p=\|(\dfrac{1}{2^{1/p}}e^{i\theta},\dfrac{1}{2^{1/p}}e^{i\theta}+\dfrac{1}{2^{1/p}}e^{i\theta})\|^p=\dfrac{1}{2}+\dfrac{1}{2}|1+1|^p=\dfrac{1}{2} +2^{p-1}>2=\|A(0,e^{i\theta_{2}})\|^p$. This proves that $ (e^{i\theta_{1}},0), (0,e^{i\theta_{2}})\notin M_{A}$.

	Let $(\beta_{1} ,\beta_{2})\in M_{A}$, where $\beta_{1}=a+ib, \beta_{2}=c+id$. Since $ \|A(\beta_{1} ,\beta_{2})\|^p=|\beta_{2}|^p + |\beta_{1} + \beta_{2}|^p =|c+id|^p + |(a+c)+i(b+d)|^p $, one of the following holds: $(i)$ $a,b,c,d >0$,  $(ii) a,b,c,d<0$, $(iii) a,c>0, b,d<0$, $(iv) a,c<0, b,d>0$. Suppose $\lambda=\alpha t$, where $\alpha=\alpha_{1} + i\alpha_{2}$, with $\alpha \in U$ and $t \in \mathbb{R}$. Now, $ \|A(\beta_{1} ,\beta_{2})+\lambda T(\beta_{1} ,\beta_{2})\|^p=\|(\beta_{2}, \beta_{1}+\beta_{2}), \lambda(\beta_{1},0)\|^p=|\beta_{2}+\lambda\beta_{1}|^P+|\beta_{1}+\beta_{2}|^p =[(c+\alpha_{1}ta-\alpha_{2}tb)^2 +(d+\alpha_{1}tb+\alpha_{2}ta)^2]^\frac{p}{2}+ [(a+c)^2+(b+d)^2]^\frac{p}{2}$. If we take $t < - \frac {2[\alpha_{1}(ac+bd)+\alpha_{2}(ad-bc)]}{|\beta_{1}|^2}$, we have, $ (c+\alpha_{1}ta-\alpha_{2}tb)^2 +(d+\alpha_{1}tb+\alpha_{2}ta)^2 < c^2 + d^2$. Therefore, for $t < - \frac {2[\alpha_{1}(ac+bd)+\alpha_{2}(ad-bc)]}{|\beta_{1}|^2}$, we have, $ \|A(\beta_{1} ,\beta_{2})+\lambda T(\beta_{1} ,\beta_{2})\|^p=\|(\beta_{2}, \beta_{1}+\beta_{2}), \lambda(\beta_{1},0)\|^p=|\beta_{2}+\lambda\beta_{1}|^P+|\beta_{1}+\beta_{2}|^p<|\beta_{2}|^p+|\beta_{1}+\beta_{2}|^p= \|A(\beta_{1} ,\beta_{2})\|^p$.

	 So we can find $\lambda= \alpha t, ~\alpha \in U$, with $t < 0$ such that $\|A(\beta_{1} ,\beta_{2})+\lambda T(\beta_{1} ,\beta_{2})\|^p=\|(\beta_{2}, \beta_{1}+\beta_{2}), \lambda(\beta_{1},0)\|^p=|\beta_{2}+\lambda\beta_{1}|^P+|\beta_{1}+\beta_{2}|^p<|\beta_{2}|^p+|\beta_{1}+\beta_{2}|^p$. This proves that for any $w\in M_{A},~ Tw\notin (Aw)_\alpha^{-}$ for each $\alpha \in U$. Applying Corollary \ref{corollary:ortho complex}, it now follows that $A\not\perp_B T$. However, this proves that $T$ is not left symmetric in $\mathbb{L}(\mathbb{X})$, contradicting our initial assumption.

	Next, we describe a general method to prove that none among these 16 types of linear operators are left symmetric.

	Let $T$ attains norm at $x, x\perp_{B}y, y\perp_{B}x, Ty =0$ and

	 $ x,~y,~ Tx\in\{(e^{i\theta_{1}},0),(0,e^{i\theta_{2}}),(\dfrac{1}{2^{1/p}} e^{i\theta_{3}},\dfrac{1}{2^{1/p}} e^{i\theta_{4}}),(\dfrac{1}{2^{1/p}} e^{i\theta_{3}},-\dfrac{1}{2^{1/p}} e^{i\theta_{4}})\}$, where $\theta_i \in [0, 2\pi), i=1,2,3,4$.

	 Define a linear operator $A\in \mathbb{L}(\mathbb{X})$ by $Ax=y,~ Ay=(e^{i\theta},0)$ or $(e^{i\theta},e^{i\theta})$ such that the following two conditions are satisfied:

	 $(i) A$ does not attain its norm at $ x,~  y$.
	 $(ii)Tw\notin (Aw)_\alpha^{-}$ for each $\alpha \in U$ and for any $w\in M_{A}$.

	 Then as before it is easy to see that $T\perp_{B}A$ but $A\not\perp_{B}T$. Thus, $T$ is not left symmetric. This completes the proof of the theorem.

\end{proof}

\begin{remark}
	The general method described in the previous theorem to prove that $T \in \mathbb{L}(\ell_p^2({\mathbb{C}}))(1<p < \infty)$ is left symmetric if and only if $T$ is the zero operator has been verified separately in each possible case mentioned in the proof of the theorem. In this paper, only one particular case has been dealt with explicitly. The details have been omitted in other cases since the method remains same in each case.
\end{remark}
\begin{acknowledgement}
Dr. Debmalya Sain would like to lovingly acknowledge the monumental role played by Shri Sakti Prasad Mishra, a National Award winning teacher from his school Ramakrishna Mission Vidyapith, Purulia, in rightly shaping the philosophies of so many young learners including himself.

\end{acknowledgement}

redpajama_set_name: RedPajamaArXiv


Complete p. VA17 and 18 on INp. 32 and 33. This assignment was started in class by most students. Study for SOL tests.
Study for SOL tests. INp. 30 and 31 Complete p. VA 15 and 16. Show ALL work!!!
Study for SOL test. Test schedule: 1st block Friday, June 3rd 2nd block Thursday, June 2nd 3rd block Thursday, June 2nd Study old notebooks and blue review booklets from each of the 4 nine weeks.
INp. 24 p. VA8 INp. 25 p. VA9 Make sure you show work for every problem even if it is just an explanation.
Midpoint test was given today. Only a handful of students finished, so we will have to finish this test tomorrow. Please continue to study.
Midpoint test will be on Wednesday, May 11th. Study SOLs 6.18, 6.9, 6.10, 6.6a, 6.7, 6.8, 6.17 and 6.19.


Comments: 3-cam: two on the balcony stage right, one just right of center about stage height. Partially shot on screen. The second balcony right angle, more to the right, is clearly visible during Shock me where both cams are shown simultaneously (see bottom right screenshot). Picture quality is mostly crap with the center cam being by far the worst. Audio is noisy and lacking in bass. There's some tape warble during Rock bottom.


In order for DB Energy Solutions to confirm whether you are entitled to ECO Help to Heat funded measures, we will share your personal data with our trusted third party service provider, the Energy Saving Trust, and with the Department for Work and Pensions.
• Get in touch with the Department for Work and Pensions via the Energy Saving Trust to check if you may be eligible for this assistance.
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• Share your details with the scheme administrator, OFGEM, so they can review the scheme.


Our gîte is located in the Morvan Nature Park, 220km from Paris, 120km from Dijon, and only 15km from Avallon.
Our gite is an old barn renovated and furnished with all the comfort necessary for your stay. You will discover, on the first level, an equipped kitchen (fridge / freezer, dishwasher, large electric oven, hotplates, filter coffee maker and espresso, toaster, microwave, kettle …) and a cozy living room with fireplace (wood available). On the second floor you will find two bedrooms with shower with three and four beds, respectively, as well as two other showers / washbasins, four washbasins and two toilets. At the last level of the gîte, there are two four-bedded rooms. All beds are pine beds of one person, 2m x 1m (comfortable dimensions). The total capacity of the lodging is fifteen people.
The Gîte du Vézelien also hosts the horses and donkeys off the Grande Randonnée footpath 13 (GR13) and the Riding Trail of the Morvan, as well as pilgrims on the way to Santiago de Compostela. The horses and donkeys can be kept in the stalls located in the courtyard of the gite or in a field or paddock within 200m. You will find horse feed (grass, hay, straw and grains) and water at will as well as a saddle room and a shower.


Each January we see a round of diets and exercise programs aimed at those of us who have indulged over the festive season. The New Year is a great time to think about changes we can make, but rather than opting for crazy diets, try to introduce simple changes that will help you to lead a healthy lifestyle. To help you get started, we've put together our top five tips for 'detoxing' your life!
Take everything off the shelves and count how many tins of tomatoes or jars of olives you have 'just in case' and decide if you'll actually use them. If not, consider donating any extras to a food aid organisation, like Foodbank, so they can be put to good use. This is also the perfect time to check use-by dates and any open packets for weevils or pantry moths. Give everything a wipe down and a re-shuffle.
A clean, organised pantry helps you instantly see what you have and makes meal planning and shopping much easier.
Consider setting up a weekly walking date with a friend, joining a social sporting club or trying out local yoga or pilates classes. Exercise is one of the first things we drop when life gets busy. But if you find an enjoyable way to move your body, it becomes easier to squeeze it in.
Moving more is also great for your mental health, particularly if 'me' time is in short supply.
Facebook, Instagram, Snapchat...they're all fun and it's hard to imagine life without them. But they can also be a time vacuum. You don't need to cancel your accounts but look at who you're following. Do you feel good or bad after reading their posts? Unfollow or remove yourself from groups that don't bring you joy.
Follow accounts on social media that leave you feeling inspired and informed.
Look out for local food suppliers - nothing beats the big supermarkets for convenience but if you have some extra time, visit your local butcher or fruit and veg retailer. Markets are also a great place to source locally produced food as well as connect with growers and retailers. Check with your council for what's on in the area.
Eating local and seasonal produce (where possible) can be cheaper, and helps build relationships within your community.
Instead of changing WHAT you eat, try changing HOW you eat. Mindfulness is a hot topic right now and it's easy to make it part of how you eat. Turn eating a meal into an event instead of something you fit in between tasks. Switch the TV off or move from your desk. Put your phone away or eat with a friend.
When you focus on just the eating, you enjoy and appreciate your food and are less likely to ignore your body's signs of fullness.
The start of a new year doesn't have to be about guilt and best intentions. Try making just one of these changes and give yourself the best possible start to 2018! For more tips on leading a healthy lifestyle visit LiveLighter.
Would you like to be part of a WA based research study helping people eat well and move more?
Our LiveLighter team is working with Curtin University to help people eat healthier, be more active and lose weight using the latest food and fitness tracking apps. Find out how you can get involved here.


Another little video featuring my wife and a warm day in the Adirondack Mt's of NY. All shot with the SONY A7III, a SONY 10-16mm, Sony 35mm F1.8 and a Sony 50mm F1.8 lens. Used a gimbal for only a few shots the rest were handheld.
I frequently head out to shoot stock footage to sell online. It's one of my favorite things to do and occasionally I can bring my wife out with me! She is a real trooper about carrying tripods and lenses but also for being in my shots. I've found nature shots with people in them tend to sell a little better than just pure nature shots so it's fantastic that she is so cooperative.
This is real short, the song I picked for it was only 1;15 long but I really liked it and felt it matched the mood of the footage and the day.
Short film to test out my new SONY A7III. Sort of a love story to my wife! The early shots represent life without her, sort of stark, lonely, struggling to "get off the ground" (leaf), barely hanging on (blue tarp) but then when she arrived in my life all was "wiped clean" (car wipers) and we've never looked back! Shot in 1080p 120fps. I only had crop lenses to use at this point, mostly used a SONY 50mm F1.8 and and SONY 35mm F1.8. Shot with EOSHD ProColor 3 settings and then added a slight film grade with filmconvert. All shot completely handheld, which when slowed down to 120 frames a second looks really smooth.
This is a promotional video I recently made for a beautiful Inn dating back to the early 1800's. Decided to try and do it more "Documentary style". I actually shot the couples first visit and then at the end of the stay did an interview to capture their feelings about it. Was meant to be mostly from the female perspective. It really is selling unplugging, relaxing and reconnecting with a spouse as much as this particular place. That's a little different than most promo videos but I think it worked overall. It tells a story and that is more interesting I think than just having a narrator tell stats and scripted material about the property. I also wanted it to be told mainly from the woman's perspective, finding the song was a key part to giving me the idea on how to put this all together.
The owners of the Inn are friends of mine that I got to know by filming their wedding in lake Placid in 2010! The couple visiting are also friends of mine that didn't know the owners or had ever stayed at the Inn before.
Tech details: Shot with the EOSHD Pro Color 3.0 Picture profile for the SONY A6500 and a SONY A6300 on a Zhiyun Crane gimbal. Also used a edlekrone motorized slider for several shots.
It was back on the road this summer to shoot a couple more episodes of our plant show! This time we went to the central part of Pennsylvania. These are always so much fun to make. It's a chance for myself, fellow producer Tim Kramer and host/producer Chris Martine to get back together. We all used to live in the same area when we started this series, but now Chris and Tim have moved away. Getting to see them and make these programs is like getting the "band" back together again!
We met some fantastic people making this one including a kindred spirit in Lance Schwartz who makes his own videos for ATV Rider magazine! He is in the video and was loads of help, and fast became a good friend!
Animals may get all the love on TV but……Plants are cool, too!
Had some time this summer to shoot some timelapse cloud formations in the Adirondack Mountains. All the original footage is shot in 4k, but I edited it in an HD timeline so i could add some pans and zooms without losing any quality. I didnt use an intervalometer to do this in still photo mode, (triggering a picture every few seconds) instead I just recorded for 10 minutes or so and then sped it up in post. I like this way as I can control the speed of the clouds better and it looks smoother to me when it's all edited.


text:
Allie Gemmill
'Doctor Strange 2' Cast, Release Date, Plot, Spoilers for the Trippy Sequel
Marvel Studios.
Out of all the (rumored) upcoming Marvel Cinematic Universe's Phase 4 movies, the one we're most excited for may be Doctor Strange 2, which is sure to be the trippiest entry in Marvel's Phase 4. The return of Benedict Cumberbatch as our favorite smarmy sorcerer also means some favorite characters will be returning and new characters, including a new villain (or two) and potential new allies for Doctor Strange, will be coming along for the ride.
Doctor Strange 2 is expected to take place after the life-altering events of Avengers: Endgame and fit nicely into the newer, larger story arc of the Marvel Cinematic Universe's Phase 4. Doctor Strange will no doubt be a changed man after Endgame but he (and his close friend, Wong) will be keen to get back to work as Master of the Mystic Arts, taking down magical foes as they pop up.
Doctor Strange might know what the future holds, but we're not on his level yet. So let's take a look at everything we know so far about Doctor Strange 2, from possible release dates to the latest casting rumors.
Doctor Strange experiments with the Time Stone in 'Doctor Strange' Marvel Studios.
When Is the Doctor Strange 2 Release Date?
With a production start date still unknown, we're not sure when Doctor Strange 2 will be released. With the exception of Spider-Man: Far From Home, no MCU Phase 4 films have a confirmed released date. However, we might be able to make an educated guess.
There are nine confirmed weekends that have been reserved by Marvel through 2022:
July 21, 2020 (previously August 7)
Doctor Strange was released November 4, 2016. The film made back half of its reported $165 million budget, with a little over $85 million earned domestically on opening weekend and a cumulative domestic gross of more than $232 million. In movie speak, the first Doctor Strange did well enough to warrant a sequel in addition to Cumberbatch's contractual obligations to appear in other MCU films, including Thor: Ragnarok, Avengers: Infinity War, and Avengers: Endgame.
With all of this in mind, we're willing to bet Marvel will want to recreate that opening weekend magic by scheduling Doctor Strange 2 for November 6, 2020. Early November will see the early stirrings of holiday moviegoers and the timing would match up if production begins on the film soon.
Is There a Trailer For Doctor Strange 2?
There are no trailers, teasers, or even early promotional images for Doctor Strange 2 at this time. But you can bet your Cloak of Levitation that when we get our hands on them, we'll share them with you.
Benedict Cumberbatch on the set of 'Doctor Strange' Marvel Studios.
Who Is In Doctor Strange 2's Cast?
There's been no official confirmations, but logic tells us that the following cast members will return for Doctor Strange 2:
Benedict Cumberbatch as "Doctor Strange"
Benedict Wong as "Wong"
Chiwetel Ejiofor as "Mordo"
In late June, it was reported by Discussing Film that casting for a Chinese-American male named "Will" for an untitled Marvel film believed to be Doctor Strange 2 had begun. According to the report, the character breakdown allegedly reads: "Mid-20s, Chinese-American (must be able to do a standard American accent)." Just two days after that report, Jeremy Conrad of MCU Cosmic posted a tweet seemingly refuting the Discussing Film's report, writing, "The Doctor Strange 2 casting news people were talking about this week apparently isn't for that movie. It seems to be for Shang-Chi instead."
It's worth noting that while it's still entirely possible for Asian and/or Asian-American actors and characters to be included in Doctor Strange 2, it's tough to tell whether this casting news is actually for the sequel since neither Derrickson nor Marvel Studios have confirmed it.
As for possible returns based on what happened in Doctor Strange and beyond, there's a chance Tilda Swinton could reprise her role as The Ancient One. Rachel McAdams could also return to the role of Dr. Christine Palmer, although Strange and Christine fail to resolve their rocky relationship before he has to go off and defeat Kaecilius in Doctor Strange.
Finally, Twitter leaker @RogerWardell, who revealed several key plot points for Endgame that were proven to be true upon the film's release, has also teased some casting news that has yet to be confirmed.
"Doctor Strange 2: 1980's Boogaloo," Wardell tweeted on May 7, 2019. "Tilda Swinton is in talks to return as the Ancient One. Jericho Drumm and Clea will make their MCU debuts, Clea will likely be portrayed by an Asian actress. Most importantly the gloves will be back!"
Although no names have been dropped for Jericho Drumm, a.k.a. Brother Voodoo, or Clea, the Sorceress Supreme of the Dark Dimension, just the idea of two powerful sorcerers being added into the mix is an exciting prospect.
Who Is the Villain in Doctor Strange 2?
Doctor Strange seemed to leave Mordo (Ejiofor) primed and ready to take over as the big bad for the inevitable Doctor Strange sequel.
Not so fast. A 2016 IGN interview with director Scott Derrickson revealed that Derrickson was toying with the idea of bringing one of the first Doctor Strange comic book villains to life for the sequel: Nightmare.
"I really like the character of Nightmare and the concept that the Nightmare Realm is a dimension. That's early — that's like the first Strange tale. I think that's in the introductory episode of Doctor Strange, and I always loved that."
In the comics, Nightmare is a demon from the dimension Everinnye and the ruler of the Dream Dimension and an anatgonist to both Strange and Ghost Rider. The MCU's version of Freddy Krueger, Nightmare prefers to torture people by turning their dreams into nightmares. Nightmare feeds off of the dreams of humans and needs those dreams to exist.
It hasn't been confirmed if Nightmare will be the actual Doctor Strange 2 villain, but you can't say he wouldn't add some intrigue to the sequel.
What Is the Plot of Doctor Strange 2?
No official plot has been confirmed. However, with Doctor Strange as an origin story, Doctor Strange 2 will be able to move deeper into the Doctor Strange world. Expect to see more of an exploration of the Mystic Arts, more exhibitions of Strange's powers, and more world-building
Presumably, Doctor Strange 2 will also take place 5 years after The Snap. As such, Strange and Wong will have to deal with the effect of The Snap as they return to their old lives and pick up the pieces.
If @RogerWardell's tweet about Brother Voodoo and Clea prove to be true, than expect even more insight into the world of the Mystic Arts and magic as a whole. In the comics, Brother Voodoo is the brother of Daniel Drumm, who was actually killed in Doctor Strange when Kaecilius invaded the New York Sanctum. It stands to reason Brother Voodoo would enter the scene to find out the truth about what happened to his brother.
In the comics, Clea is revealed to be the niece of Dormammu and estranged wife of Strange. Since Strange isn't known to be a married or divorced in his first story, Clea's relationship to him could be adjusted so that she is a new love interest as well as a possible threat thanks to her Dormammu connection.
Does Avengers: Infinity War & Avengers: Endgame Set Up Doctor Strange 2?
Doctor Strange plays a fairly important role in Infinity War and Endgame. At the beginning of Infinity War, Strange and Wong are in the Sanctum, guarding over the valuables as they track Thanos's movements across the universe. Strange is the protector of the Time Stone during this time, which makes him a person of interest when two members of the Children of Thanos, Ebony Maw and Black Dwarf, land in New York City. Tony Stark and Peter Parker end up protecting Strange from being tortured and killed by Ebony Maw for the Time Stone, with all three ending up on Titan and plotting a way to defeat Thanos. During their time on Titan, Strange reveals there more than 14 million ways things end between Thanos and the Avengers, with one one possible victory for the Avengers.
Because Strange gets dusted in Infinity War, he is actually not seen until the final Endgame battle when all of the fallen Avengers are brought back to life. We do, however, return to the world of Doctor Strange when Hulk, Iron Man, and Captain America travel through time to New York City in 2012 to retrieve the Space Stone, the Mind Stone, and the Time Stone. Hulk is tasked with going to the New York Sanctum to retrieve the Time Stone from Strange, but instead meets the Ancient One. As the two converse, the Ancient One reveals that Strange was not yet the hero he was meant to become, but he was destined to be the greatest of all practitioners of the Mystic Arts.
With this in mind, it's likely that The Ancient One's words on Strange and his destiny for greatness will manifest in some way in Doctor Strange 2. As previously mentioned, Strange and Wong will probably also have to deal with the effects of being dusted during Infinity War as well as any/all issues that arose from that time.
We'll keep you posted on Doctor Strange 2 details as they emerge.
'Doctor Strange' director Scott Derrickson leaves. 'WandaVision' to blame?
By Eric Francisco
'Westworld' Season 3 release date, trailer, cast, spoilers for HBO's show
By Jake Kleinman and Allie Gemmill
Study of 1,225 festival-goers pinpoints positive effects of trippy drugs
By Ali Pattillo

redpajama_set_name: RedPajamaCommonCrawl


Mel7em is a long-time Lebanese pianist and producer who kickstarted his new musical project in 2018. He released singles and EPs on various underground labels such as Dear Deer and Uxoa Dutxa Elite. We welcome him on Art Vibes with his 'Broken Windows' EP presenting three of his latest productions.
'Broken Windows', 'Uncompromising', and 'The Forgotten Tribe' are proper progressive melodic tracks blending oriental sounds with the power of all-electronic rhythms and grooves. Echoes of airy sounds mingling with stunning eastern melodies, yet remaining familiar to the ears of a contemporary electronic music enthusiast.
Turkish rising stars Lunar Plane gifted this EP with a club-oriented version of 'Broken Windows'. No need to say this one is going to shake the dance floor this winter. Nacim Gastli's talent has always fascinated our audience and we are excited to have him remix 'Uncompromising'. This time, he builds up this rework on a powerful trancy vibe. To complete this already eclectic tracklist, John Lead is back on Art Vibes to gift us with his own ethereal interpretation of 'The Forgotten Tribe'. As always, the Portuguese producer designs the sound on point and the immersion into his alternate version is fascinating.


Getting Started with Power BI is 2-day online training course designed for students who are just getting started with Power BI and who need to get up to speed on building and deploying custom solutions using the Power BI Service and Power BI Desktop. Students will work through hands-on lab exercises to gain and reinforce skills for using a project-based approach for building custom BI solutions using Power BI Desktop.
This workshop-style course will teach you how to build powerful SharePoint business solutions without having to engage in any custom development. You will be amazed at how far you can take things with Office 365 and SharePoint Online just using the out-of-the-box features and functionality. This course is designed for the experienced business user who wants to re-engineer business processes using SharePoint Online in the Office 365 environment.
Learn how to confidently deploy and manage the many hybrid features of Office 365 for an enterprise deployment of SharePoint. Workshop style course designed to teach seasoned SharePoint and Office 365 administrators how to prepare and deploy the various hybrid features of Office 365 to an on-premises SharePoint 2013 or 2016 server farm.


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The Convention sets forth a number of fundamental rights and freedoms. The Protocols expand on the set of fundamental rights. The European Court of Human Rights in Strasbourg has been set up to ensure the observance of the obligations.
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This is a very fun film, mixing thrilling aerobatics with good acting, a well-paced plot, and a murder mystery, all in one. The stunt flying in this film, is awesome and so cutting edge it was re-used in such future Warners programmers as Fly-Away Baby and Fugitive in the Sky. Again, my problem is in identifying the specific stunt pilots for this film. I can't find the information anywhere. It seems that the pilots names have been lost over time.
I think it's important to acknowledge that when a stunt pilot works, he doesn't do so by himself. The other one up there to capture the magic in the skies and also risk his life, I must add, is the aerial photographer. In this case it was Veteran cinematographer Elmer Dyer. He was viewed as one of the very best. He was nominated for an Academy Award for his photography in Air Force (1943). During World War II he was assigned to the Army's Motion Picture Unit and shot much aerial footage that was used in training films. He began assembling a library of stock footage during the war, and eventually left active photography to spend most of his time in that business.
Murder in the Clouds is directed by D. Ross Lederman for First National Pictures.
Lyle Talbot and Ann Dvorak star in "Murder in the Clouds" in 1934.
First National Pictures – First National was an association of independent theater owners in the United States that expanded from exhibiting movies to distributing them, and eventually to producing them as a movie studio, called First National Pictures, Inc. (1917−1936). In 1928 it merged with Warner Bros.


In the past year or so, I've pretty much completely switched to cream blushes. I still have a few special powders sticking around, but, eighty percent of the time, I reach for stick or a jar. I like the different consistencies and formulas, from a sheer stains to solid creams. I find that it not only looks more natural, but it's a lot more instinctual to apply blush with my hands, being able to really feel my cheekbones. Plus, there is no hassle of having to clean the brush all the time.
I'm slowly running out of my YSL's Creme de Blush in 'Babydoll' and it's incredibly difficult to find, so, if I'm in a mood for a baby pink, I alternate between it and Lise Watier's Blush Fondant Supreme in 'Rose'. They are both little wonders-in-a-jar, with a similar mousse-y formula that glides divinely on the skin and dries matte.
Estée Lauder's Bronze Goddess summer collection is a must every year. (That scent!) This upcoming summer, the brand is adding a Limited-Edition All-Over Illuminator stick to its lineup. I adore the shade, as it can act as both a blush on pale complexions and an illuminator for all. It gives one that sun-kissed look without having to reach for a bronzer.


We make sure our clients have perfect form for each exercise in order to prevent injuries, increase strength, and improve flexibility.
"I've been working out for most of my adult life, and I can say without hesitation that I've gotten the best results in terms of toning and aerobic fitness at Total Form Fitness. I am always excited to see the trainers and look forward to every session I have."
"It is easiest for me to focus in a place like Total Form Fitness. I receive consistent attention from the trainers, and there are few distractions. All of the machines and equipment I need are here and easy to access."
"For the past several years, I have exercised with a personal trainer for one hour sessions, twice a week. I cannot say enough good things about Total Form Fitness. Everybody is nice, respectful, and genuinely wants to help me. My whole family works out at TFF!"
"Total Form Fitness has been a great place for me and my family. We have exercised here for the past few years, and it has been a great experience. The trainers are friendly, intelligent, and incredibly accommodating. My family and I have appreciated getting to know the guys on a professional and personal level."
"I joined Total Form Fitness to lose weight, improve my conditioning, and increase my strength. The trainers have been great in helping me achieve these goals. I have noticed substantial improvements and am completely happy with my experience at TFF."
"My entire family has exercised at TFF for over 4 years, and each one of us pursues a different goal. The trainers help us achieve our individual goals, and they know when to turn it up a notch and get the most out of our sessions. While I have worked out at other places, I believe there is something special about the trainers at TFF. They have become like family to us."
"I've been a client of Total Form Fitness for over 2 years. Since I started working out at TFF, I feel and look healthier than I did 25 years ago, and I have lost 38 pounds. My waist size went from a size 38 to 32, and I have lowered my cholesterol. I truly cannot ask for anything more in a personal training studio."
"I began working out at Total Form Fitness in high school. TFF's four phase approach has helped me increase my strength, flexibility, and speed. I feel as if I am continually pushed to exercise harder, and it is with the gym's help that I played four years of college baseball."


text:
Heodes pallida är en fjärilsart som beskrevs av Fritsch 1912. Heodes pallida ingår i släktet Heodes och familjen juvelvingar. Inga underarter finns listade i Catalogue of Life.

Källor

Juvelvingar
pallida

redpajama_set_name: RedPajamaWikipedia


text:
Cagamas issues RM200m conventional commercial papers
Shawn Ng June 25, 2018
MoT eyes affordable housing projects on RAC's 9,192 acres of land
KAJANG (June 25): The Ministry of Transport (MoT) is planning to develop affordable housing projects on Railway Assets Corp's (RAC) 9,192 acres of land to generate better returns for the company.
"I have requested for RAC to come up with a list of their lands that are available for development by this Friday and I will submit the list to the Cabinet next Wednesday to seek approval for developing the lands. Once we have gotten the approval, we will then call for open tenders for the lands," said Transport Minister Anthony Loke at a press conference after his working visit to RAC here today.
"However, we don't encourage developers to develop office blocks on the lands as we don't want to see oversupply of office spaces in Kuala Lumpur. So we hope for them to do more housing developments, especially affordable housing projects to allow more people to live nearby the railways and use public transport to commute to work," he added.
As a corporation under the ministry, RAC owns assets including the land, rolling stocks and railways of Keretapi Tanah Melayu Bhd (KTMB), which are worth over RM35 billion, said Loke.
According to him, the company owns a total 31,712 acres of land along the KTM railways as well as pockets of land in prime areas, of which 9,192 acres are available for development, while the remaining 22,520 acres are reserve land.
"We see a lot of potential in RAC as it owns a lot of land. Many of the lands are situated in urban areas and are worth a lot.
"If RAC is able to generate more income through unlocking the value of its landbank, it will have more flexibility to develop our railway service and provide better service to the public," he pointed out.
He cited the Electric Train Services (ETS) as an example, whereby many facilities of the rail line are supposed to be upgraded and repaired, but the government do not have the capital to do so at the moment.
He reiterated that the development of RAC's landbank must go through open tender, which is expected to value the land at a higher price and thus bringing a higher return to the government.
"The previous government had given out many land parcels to private developers through direct negotiation instead of open tender. But now the new government wants to ensure all of them to go through the process of open tender," he said.
According to Loke, private developers have obtained approvals-in-principal from the previous administration to develop three plots of land in Pudu, Batu Tiga and Bangsar, which are worth "hundreds of millions" of ringgit without going through open tenders.
"The approvals were given out through direct negotiations. But there is no commitment given and we will review all the approvals and ensure all the development on the lands will go through the process of open tender," he said.
Going forward, Loke said that MoT hopes to transform RAC into an "assetco" for all the railway assets in the country while KTMB will focus on the operations of railways and service provision to commuters.
Anthony Loke Ministry of Transport Keretapi Tanah Melayu Bhd Railway Assets Corporation Open tender Affordable housing
VIVO Residential Suites
Jalan Klang Lama (Old Klang Road), Kuala Lumpur
Studio | 1 bath | 527 sqft
Plans to revive Seremban-Port Dickson rail line being studied — Loke
180,000 affordable housing units to be built in Penang — Jagdeep
Loke: I disinvited Mavcom to post-cab meeting because they refused to execute Cabinet decisions


Online PR News – 22-August-2017 – This year marks the 30th anniversary of Bitner/Hennessy Public Relations, which opened in downtown Orlando in 1987 as Gary Bitner Public Relations.
The company is a full-service public relations and content development firm offering marketing and public relations campaigns, media relations, website content and social media management.
For decades, Bitner/Hennessy professionals have successfully coordinated major public relations and marketing campaigns in Florida and beyond on behalf of a diverse client base.
Our team here in Orlando is proud to have reached this milestone, said partner Kimbra Hennessy. We owe our success to the strong relationships we have formed with our clients and the media, as well as the support from our home office in Fort Lauderdale.
"The Orlando office gives us a statewide orientation and provides local and regional insight and support for many of our clients," said Gary Bitner, Bitner Group founder and president.
Bitner Group is a highly-regarded public relations and integrated communications firm with offices in Fort Lauderdale and Orlando (Bitner/Hennessy). Bitner Group professionals have successfully coordinated major public relations and marketing campaigns in Florida and beyond on behalf of the firms diverse client base, including the Seminole Tribe of Florida, Seminole Hard Rock Hotels & Casinos in Tampa and Hollywood, Fla., Sims Crane & Equipment, Renovate America, and Show Technology, presenters of large home and garden shows around the country. The firm takes great pride in developing compelling content and strategically working with clients to achieve their public relations goals.


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New Portrait of County Reveals Five 'LAs' from 'Glittering' to 'Precarious'
Press Release | Wednesday, Nov 29, 2017
A new report shows that some groups of Los Angeles County residents have the highest levels of well-being in the United States while others have some of the lowest.
The findings are based on a cutting-edge formula that combines data on income, health, and education and offers an unprecedented portrait of LA County.
The American HD Index measures how people are doing, offering a more accurate perspective on opportunity and well-being.
"A Portrait of Los Angeles County," commissioned by the LA County Office of Child Protection and funded by a consortium of 14 grantmakers and the Los Angeles County Quality and Productivity Commission's Productivity Investment Fund, is expected to be used by county leaders in making future decisions about where to target resources.
"The portrait provides us with critical information about the well-being of communities that is greatly needed to effectively target prevention supports to the families who need them most," said Office of Child Protection executive director Judge Michael Nash.
LA County Supervisor and outgoing Board Chair Mark Ridley-Thomas said, "The county has been prioritizing data collection and analysis to enable us to make smarter decisions to support the well-being of our 10 million residents. This report provides us with a portrait that can begin to immediately inform County decision-making."
On the 10-point HD Index scale, the highest- and lowest-scoring communities are separated by an astounding seven points.
In LA County, San Marino, Manhattan Beach, and Palos Verdes Estates top the list, all scoring more than 9.3 on the HD Index. Lennox, East Rancho Dominquez, and Florence-Graham each score lower than 2.7.
Within the city of Los Angeles, the gap is even larger. Bel Air-Beverly Crest and Brentwood-Pacific Palisades are on top with scores of 9.51 and 9.24, respectively. Southeast LA is at the bottom with 2.26.
LA County Supervisor and incoming Board Chair Sheila Kuehl said, "This newer and smarter way of measuring well-being will allow us to understand how income, health and education all intersect to contribute to whether you and your community experience well-being or live at a great disadvantage. We knew there were grave disparities across the county, but now we have a more reliable way of geographically pinpointing the areas of greatest need."
"A Portrait of Los Angeles County" is a collaborative effort between Measure of America and partners that include Southern California Grantmakers and the Los Angeles County Office of Child Protection, Department of Children and Family Services, and Department of Public Health.
It was made possible with funding from the Annenberg Foundation, BCM Foundation, Blue Shield of California Foundation, California Community Foundation, The California Endowment, The California Wellness Foundation, Citi Community Development, Conrad N. Hilton Foundation, The James Irvine Foundation, Leonetti/O'Connell Family Foundation, Los Angeles County Quality and Productivity Commission—Productivity Investment Fund, The Ralph M. Parsons Foundation, Sony Pictures, United Way of Greater Los Angeles, and the Weingart Foundation.
The full report is available [here].


Exercise and Pregnancy? Yes! See the benefits of why they go together so well and exercise programming ideas for each trimester.
Remaining physically active during pregnancy can help improve posture, decrease back pain, swelling, constipation, and incontinence, and significantly reduce the risk of many other pregnancy-related ailments. Women who exercise throughout their pregnancy tend to have shorter, less difficult labors with fewer complications. They also have a decreased need for medical intervention with quicker, easier recoveries once the baby is born. Research has repeatedly shown that active women even have healthier, happier, smarter babies. Unfortunately, the majority of pregnant women do not exercise consistently, if at all, and those who do tend to reduce activity in the later months (Garland, 2017). Exercising during pregnancy is often mistakenly viewed as an unnecessary risk to a woman's unborn child. Expectant mothers are encouraged to "play it safe" and "take it easy" by misinformed loved ones, well-meaning strangers, and inaccurate websites. Outdated medical advice is still circulating, telling pregnant women to avoid elevating their heart rate, abstain from many types of exercise and spend their pregnancy resting instead of staying physically active. But, this could not be further from the truth!
The American Congress of Obstetricians and Gynecologists (ACOG) is the organization that releases the official guidelines regarding exercise during pregnancy. Their early recommendations were much more conservative and alarmist than necessary because they were based on assumptions and limited animal studies. However, the last few decades of extensive research have shown numerous incredible benefits to exercise, with no risk to mother or child. As a result, ACOG now encourages pregnant women to get at least 20-30 minutes of exercise each day in the absence of medical or obstetric complications (ACOG, 2015). Prenatal exercise is no longer just acceptable; it is a necessary component to achieving optimal health of both mother and child (Harris, Baer & Stanford, 2018). Staying active and avoiding excess weight gain during pregnancy offers lifelong health benefits for women and their children.
In a singleton pregnancy, without complications, women can typically maintain their pre-pregnancy routine, regardless of intensity, as long as it still feels comfortable. The number one rule for prenatal exercise is that women need to listen to their body and adjust workouts accordingly. When creating a program for a pregnant client, it is important to consider that the frequency, intensity, time, and type (FITT) of exercises will vary with each workout depending on prenatal symptoms. Working around fatigue, nausea, "growing pains," etc. means that trainers may need to adjust the routine at any given moment. Using the rate of perceived exertion (RPE) scale or "talk test" to measure intensity will help facilitate ongoing communication and keep workouts within a manageable range. Heart rate is typically not a reliable indicator during pregnancy. The NASM Optimum Performance Training™ (OPT™) with Phases 1 and 2 provide the structure for creating a safe and effective program throughout pregnancy and postpartum. While a client's fitness before pregnancy is typically a good starting point, the focus of prenatal workouts is to help manage changes in the body, prepare for labor and delivery, and build a solid foundation of balance, endurance, and strength to meet the demands of motherhood.
A personal trainer's responsibility is to guide expectant mothers through a safe fitness program by encouraging them to exercise while consistently checking in to make sure everything feels okay. Women should not push through any discomfort as they may have in the past. Pregnancy is the perfect time to strengthen a mind-body connection and allow women to take the lead in their workouts, with the trainer in a much-needed supportive role.
It is important for women to drink plenty of water and breathe deeply during exercise to avoid feeling dizzy or faint since these are common side effects of the added blood volume and hormone shifts. If any discomfort or dizziness occurs, make adjustments or take a break before trying again. Remember that the number one concern is always listening to the client's body cues. While there is no "one-size-fits-all" approach, the following circuits can help the typical pregnant client to improve balance, strength, and endurance. Most women should aim for about 2-3 sets of 15 repetitions, or 60 seconds each, and then do a bit more or less depending on how she feels during that particular workout. Do not be afraid to use some creativity and a unique training style when creating a prenatal program following the OPT™ model and FITT principle to vary workouts as needed.
The first trimester is notorious for intense nausea and exhaustion as well as strong emotions including fear of doing anything that may harm the baby. There is a common misconception that physical activity is unsafe in the early months of pregnancy, but rest assured that research has proven otherwise. It is essential that trainers are confident that prenatal exercise is safe and empower women to dictate the pace and intensity of their workouts. Always default to the recommendations of a woman's doctor and guidance of ACOG when faced with any doubt.
Works core, glutes and quads. Gentle on joints while challenging stabilization and endurance. Be sure to perform near something sturdy for safety.
Works core, glutes and hamstrings. Improves stabilization and requires mental focus which improves mind-body communication. Perform with one hand gently placed on a wall or bench to prevent falling. Instruct clients to fix their gaze several feet in front if balance is a challenge.
Strengthens abdominals and improves balance. Ensure ball is properly inflated for lower back support. Bracing abdominals, slowly exhale during contraction with a small range of motion. Crunches, performed correctly, are safe and effective for maintaining core muscles during pregnancy and postpartum.
Cardio – Stair climber at slow to moderate pace or high knees, quickly march in place elevating knees to waist at a pace that feels comfortable. Just 1-3 minutes as part of the circuit or at the end of a workout is often enough.
In the second trimester, the extreme fatigue, dizziness, and nausea of the preceding weeks will hopefully subside, and women will begin feeling more like themselves. This is a good opportunity to get the most out of workouts in preparation for the next trimester when exercise starts slowing down prior to labor.
Draw shoulders down and back to counterbalance added weight from the growing chest and belly. A strong back will also help maintain good posture and prevent pain. Be sure that clients are not shrugging as they pull down toward the chest. Consistently remind women to keep abs tight and exhale as they pull down.
Strengthens the inner-thighs, quads, core, glutes, and pelvic floor along with balance and endurance. Squat to bench for added safety and feedback that hips are back and low. Cue women to "zip up" through inner thighs, pelvic floor, and lower abs.
A full-body exercise that challenges a woman's balance, core strength, and coordination. Complete one set with an opposite arm and leg before switching. Bring the knee and elbow in, around the belly, during the exhale and round the back for added core work and back stretch. Return to a neutral spine, avoiding hyperextension of the lower back.
Cardio – Modified Burpee (perform slowly): Squat down to the floor. Place hands flat on the mat and step the feet out behind one at a time into the push-up position. Return to the squatted position and press body up while reaching hands overhead.
Exercises may feel more challenging during the third trimester. Pregnant women are at their heaviest which makes just getting around an act of endurance and power. The majority of a woman's weight is in front which will throw off her center of gravity, pull shoulders forward, increase the risk of falling, and strain the core, which makes even simple tasks more difficult. Women will also become out of breath and fatigued more easily. Do not rush through the workout, and modify wherever necessary.
A pelvic tilt helps women stretch a tight lower back while engaging the overstretched lower abdominals. Adding hip circles can relieve tension while also gently contracting core muscles.
Like the lat pulldown, a row strengthens the back and improves posture, which is very important for pregnant women. Be sure clients do not raise their shoulders up as they pull, but draw them back and down toward the spine with chest raised and core engaged.
Strengthens lower body while supporting lower back and improving balance and coordination. With the ball placed between the wall and the client's lower back, her body should move straight up and down with the feet far enough in front that the legs make a 90-degree angle at the lowest point of the squat.
Cardio – Walk or swim.
Regardless of a woman's BMI or previous activity level, physical activity during pregnancy offers many lifelong benefits for mother and child. Doing something, no matter how small, is always better than doing nothing. With so many obstacles and misguided messages, supporting expectant and new moms is critical. Working with pregnant clients is a uniquely challenging and rewarding experience that quite literally helps to reshape a family's genetics and have a positive impact on multiple generations.


We are aware that some users are having trouble in receiving the email to authentice their account.
Our IT team has identified that there is problem with email addresses that are using BT as they are blocking the authentication email from reaching the inbox.
IT is working on a fix between BT and ourselves to figure out why this is happening. Any users who are experincing these problems please contact our Inbound Team on 020 7267 2900 who are now better equipped to help you out.
Apologies for the frustrating experiences caused while we work on reoslving the issue.
We at Ringley have made a few changes to how you login into the Gateway to better protect your data.
Every user will be required to activate their account before they can continue to use the Gateway, this can be done by clicking on the button "Activate Account" and filling out the form found on the page.


There are few places on this planet more exclusive than a top-level golf club. Your goal as the owner or manager of such an environment is to attract the most desirable clientele. This is why you manage your club in such a professional style. However, it can be quite easy for you to get distracted. With so many tasks on your agenda, it can be very easy to lose focus. You are simply being spread too thin.
You Can't Do it All By Yourself, So Why Try?
It isn't a matter of your basic competence. You have proven yourself to be a resourceful and visionary owner or manager in a thousand different scenarios. It's more a matter of simply being placed in an impossible situation. You can't handle each and every detail of a busy and ever-evolving top rated golf course all on your own. This is why you need to turn to a team of reputable and professional management experts who can take much of the hard work off your hands.
This is why the time is now for you to bring in a team of top level golf club management experts. These are the pros who can take charge of your course and get it back on track. If you are concerned about the level of play being compromised by less than ideal turf, you can rely on your new managers to hire the best quality course management experts. There are a thousand benefits that will quickly come with this decision.
There are a thousand ways to improve the quality and reputation of your golf course. Even as imaginative and effective as you are, you can't possibly think of them all. This is why it's such a good idea to turn most of the hard work over to a team of qualified management experts. The time to make this move is now.