text:
Nitrogen and phosphorus fertilization consistently favor pathogenic over mutualistic fungi in grassland soils
Ylva Lekberg, Carlos A. Arnillas, Elizabeth T. Borer, Lorinda S. Bullington, Noah Fierer, Peter G. Kennedy, Jonathan W. Leff, Angela D. Luis, Eric W. Seabloom, Jeremiah A. Henning (2021). 12, 3484 (2021). https://doi.org/10.1038/s41467-021-23605-y.
Ultrasound imaging identifies life history variation in resident Cutthroat Trout
Kellie J. Carim, Scott Relyea, Craig Barfoot, Lisa A. Eby, John A. Kronenberger, Andrew R. Whiteley, Beau Larkin (2021). 16(2): e0246365. https://doi.org/10.1371/journal.pone.0246365 .
Positive interactions between an exotic invader and moss biocrusts vary across life stage and correspond with the effect of water pulses on soil nitrogen
Mandy L. Slate, Morgan Luce McLeod, Ragan M. Callaway (2021). 00, 1– 11. https://doi.org/10.1111/1365-2435.13831.
Integrative Organismal Biology
Metabolic Rates Predict Baseline Corticosterone and Reproductive Output in a Free-Living Passerine
Blanca Jimeno, Mackenzie R Prichard, Devin Landry, Cole Wolf, Beau Larkin, Zachary Cheviron, Creagh Breuner (2020). Volume 2, Issue 1 https://doi.org/10.1093/iob/obaa030.
Avian Ecology
We seek to understand how birds use the habitats available and how that will change as we work to create more diverse plant communities. We also host researchers that document migrations of raptors and songbirds across MPG.
In this section of the research pages, you will find links to reports and updates from all the researchers involved with avian ecology, posted chronologically. The links will show you more in-depth reports on our findings. The three main projects covered here are:
Songbird Counts- A grid of sampling points covers MPG with 560 points. We visit each point 3 times a year, once in winter and twice during the songbird breeding season. We record, by ear or by sight, all the birds near that point for 10 minutes.
Songbird Banding- The University of Montana Bird Ecology Lab, UMBEL, runs several trapping stations at MPG as part of their regional songbird monitoring program. UMBEL sets up very fine nets that are nearly invisible to birds in brushy habitats. Songbirds fly into the nets and become entangled. The researchers take the birds from the nets and affix a numbered band to their leg before releasing them.
Raptor Research- The Raptor View Research Institute monitors raptor populations on MPG and counts raptors that migrate past MPG in the spring and fall. Raptor View researchers have placed transmitters on osprey and golden eagles that use the Bitterroot Valley.
The Journal of Wildlife Management
Widespread Lead Exposure in Golden Eagles Captured in Montana
Robert Domenech, Adam Shreading, Philip Ramsey, Michael McTee (2020). https://doi.org/10.1002/jwmg.21980.
Golden Eagle Project, Avian Ecology
Ecological Engineering
Producing moss-colonized burlap fabric in a fog chamber for restoration of biocrust
Kyle D. Doherty, Henry S. Grover, Matthew A. Bowker, Rebecca A. Durham, Anita J. Antoninka, Philip W. Ramsey, (2020). Volume 158, https://doi.org/10.1016/j.ecoleng.2020.106019.
Biocrusts, Lichen and Moss
Biocrusts Lichen and Moss
The term biological soil crust (or biocrust) encompasses the diverse community of moss, algae, lichens, and cyanobacteria living within the top inch of soil. Biocrusts form in all terrestrial ecotypes, but people often overlook them in favor of larger species. Biocrusts stabilize the soil by aggregation, reducing erosion. They increase soil fertility by enhancing nutrient cycling. A healthy biocrust increases infiltration and plays a role in succession towards a diverse and functioning ecosystem. They are especially important in arid and semi-arid ecosystems.
Towards the goal of restoring land to healthy ecosystem function, we consider all facets of the land, not just the vascular plants that may inhabit it. Unaided, biocrusts may take from decades to a century to reestablish after degradation. We would like to determine the most effective and practical methods to restore biological soil crusts.
Research and updates found here include projects on moss, lichens, and biocrust as we delve into this charismatic and important part of the ecosystem.
Codependency between plant and arbuscular mycorrhizal fungal communities: what is the evidence?
Vasilis Kokkoris, Ylva Lekberg, Pedro M. Antunes, Catherine Fahey, James A. Fordyce, Stephanie N. Kivlin, Miranda M. Hart (2020). 228: 828-838. https://doi.org/10.1111/nph.16676.
Journal of Mammalogy
Response of a remnant marmot population to habitat enhancement yields insights into marmot ecology
Melissa J Reynolds-Hogland, Alan B Ramsey, August T Seward, Kristine L Pilgrim, Cory Engkjer, Philip W Ramsey (2020). Volume 101, Issue 3, 22 May 2020, Pages 658–669, https://doi.org/10.1093/jmammal/gyaa021.
The Journal of Wlldlife Management
Elk forage and risk tradeoffs during the fall archery season
Jesse D. DeVoe Kelly M. Proffitt Michael S. Mitchell Craig S. Jourdonnais Kristin J. Barker (2019). 83: 801-816. https://doi.org/10.1002/jwmg.21638.
Elk Research, Mammals
Forestry: An International Journal of Forest Research
A century of changing fire management alters ungulate forage in a wildfire-dominated landscape.
Proffitt, K. M., DeVoe, J., Barker, K., Durham, R., Hayes, T., Hebblewhite, M., ... & Shamhart, J. (2019). 00, 1-15. https://doi.org/10.1093/forestry/cpz017.


text:
\section{Introduction}
The purpose of this paper is to study equalities involving the norm
of operators on Banach spaces, and to discuss the possibility of
defining isometric properties for Banach spaces by requiring that
all operators of a suitable class satisfy such a norm equality.

The interest in this topic goes back to 1963, when the Russian
mathematician I.~Daugavet \cite{Dau} showed that each compact
operator $T$ on $C[0,1]$ satisfies the norm equality
\begin{equation*}\label{DE}\tag{DE}
\|\mathrm{Id} + T\|= 1 + \|T\|.
\end{equation*}
The above equation is nowadays referred to as \emph{Daugavet
equation}. Few years later, this result was extended to various
classes of operators on some Banach spaces, including weakly compact
operators on $C(K)$ for perfect $K$ and on $L_1(\mu)$ for atomless
$\mu$ (see \cite{Wer0} for an elementary approach). These results
were left without much attention until the beginning of the
eighties, when a new wave of interest in this topic surfaced, and
the Daugavet equation was studied by many authors in various
contexts. We refer the reader to the books \cite{AbrAli-1,AbrAli-2}
for a brief study of this equation from different points of view.

In the late nineties, new ideas were infused into this field and,
instead of looking for new spaces and new classes of operators on
them for which \eqref{DE} is valid, the geometry of Banach spaces
having the so-called Daugavet property was studied. Following
\cite{KSSW0,KSSW}, we say that a Banach space $X$ has the
\emph{Daugavet property} if every rank-one operator $T\in L(X)$
satisfies \eqref{DE} (we write $L(X)$ for the Banach algebra of all
bounded linear operators on $X$). In such a case, every operator on
$X$ not fixing a copy of $\ell_1$ also satisfies \eqref{DE}
\cite{Shv}; in particular, this happens to every compact or weakly
compact operator on $X$ \cite{KSSW}. There are several
characterizations of the Daugavet property which does not involve
operators (see \cite{KSSW,WerSur}). For instance, a Banach space $X$
has the Daugavet property if and only if for every $x\in S_X$ and
every $\varepsilon>0$ the closed convex hull of the set
$$
B_X\setminus\big(x+(2-\varepsilon)\,B_X\big)
$$
coincides with the whole $B_X$. Here and subsequently, $B_X$ and
$S_X$ stand, respectively, for the closed unit ball and the unit
sphere of a Banach space $X$. Let us observe that the above
characterization shows that the Daugavet property is somehow
extremely opposite to the Radon-Nikod\'{y}m property.

Although the Daugavet property is clearly of isometric nature, it
induces various isomorphic restrictions. For instance, a Banach
space with the Daugavet property does not have the Radon-Nikod\'{y}m
property \cite{Woj} (actually, every slice of the unit ball has
diameter $2$ \cite{KSSW}), it contains $\ell_1$ \cite{KSSW}, it does
not have unconditional basis \cite{Kadets} and, moreover, it does
not isomorphically embed into an unconditional sum of Banach spaces
without a copy of $\ell_1$ \cite{Shv}. It is worthwhile to remark
that the latter result continues a line of generalization
(\cite{K97}, \cite{KS99}, \cite{KSSW}) of the known theorem of
A.~Pe\l czy\'nski \cite{Pelcz} from 1961 saying that neither
$C[0,1]$ nor $L_1[0,1]$ embeds into a space with unconditional basis

The state-of-the-art on the Daugavet property can be found in
\cite{WerSur}; for very recent results we refer the reader to
\cite{BM,BKSW,I-K-W} and references therein.

In view of the deep consequences that the Daugavet property has on
the geometry of a Banach space, one may wonder whether it is
possible to define other interesting properties by requiring all
rank-one operators on a Banach space to satisfy a suitable norm
equality. This is the aim of the present paper.

Let us give some remarks on the question which will also serve to
present the outline of our further discussion. First, the Daugavet
property clearly implies that the norm of $\mathrm{Id} + T$ only depends on
the norm of $T$. Then, a possible generalization of the Daugavet
property is to require that every rank-one operator $T$ on a Banach
space $X$ satisfies a norm equality of the form
$$
\|\mathrm{Id} + T\|=f(\|T\|)
$$
for a fixed function $f:\mathbb{R}^+_0\longrightarrow \mathbb{R}$. It is easy to
show (see Proposition~\ref{prop1}) that the only property which can
be defined in this way is the Daugavet property. Therefore, we
should look for equations in which $\mathrm{Id} + T$ is replaced for other
function of $T$, i.e.\ we fix functions $g$ and $f$ and we require
that every rank-one operator $T$ on a Banach space $X$ satisfies the
norm equality
$$
\|g(T)\|=f(\|T\|).
$$
We need $g$ to carry operators to operators and to apply to
arbitrary rank-one operators, so it is natural to impose $g$ to be a
power series with infinite radius of convergence, i.e.\ an analytic
function defined on the whole $\mathbb{K}$ ($=\mathbb{R}$ or $\mathbb{C}$). Again, the only
non trivial possibility is the Daugavet property, as we will show in
section~\ref{sec:g(T)=f(T)}. Section~\ref{sec:Id+g(T)=f(g(T))} is
devoted to the last kind of equations we would like to study.
Concretely, we consider an analytic function $g$, a continuous
function $f$, and a Banach space $X$, and we require each rank-one
operator $T\in L(X)$ to satisfy the norm equality
\begin{equation}\label{eq:introduction-id+g=fg}
\|\mathrm{Id} + g(T)\|=f(\|g(T)\|).
\end{equation}
If $X$ is a Banach space with the Daugavet property and $g$ is an
analytic function, then it is easy to see that the norm equality
$$
\|\mathrm{Id} + g(T)\|=|1+g(0)|-|g(0)| + \|g(T)\|
$$
holds for every rank-one $T\in L(X)$. Therefore, contrary to the
previous cases, our aim here is not to show that only few functions
$g$ are possible in \eqref{eq:introduction-id+g=fg}, but to prove
that many functions $g$ produce the same property. Unfortunately, we
have to separate the complex case and the real case, and only in the
first one we are able to give fully satisfactory results. More
concretely, we consider a complex Banach space $X$, an entire
function $g$ and a continuous function $f$, such that
\eqref{eq:introduction-id+g=fg} holds for every rank-one operator
$T\in L(X)$. If $\re g(0)\neq -1/2$, then $X$ has the Daugavet
property. Surprisingly, the result is not true when $\re g(0)=
-1/2$, where another family of properties strictly weaker than the
Daugavet property appears: there exists a modulus one complex number
$\omega$ such that the norm equality
\begin{equation}\label{eq:introduction-omega0}
\|\mathrm{Id} + \omega\,T\|=\|\mathrm{Id}+ T\|
\end{equation}
holds for every rank-one $T\in L(X)$. In the real case, the
discussion above depends upon the surjectivity of $g$, and there are
many open questions when $g$ is not onto. Finally, we give in
section~\ref{sec:newproperties} some remarks about the properties
defined by norm equalities of the form given in
\eqref{eq:introduction-omega0}.

We finish the introduction by commenting that, although we have been
unable to find any result as the above ones in the literature, there
are several papers in which the authors work on inequalities which
remind the Daugavet equation. For instance, it is proved in
\cite{Ben-Lin} (see also \cite[\S9]{Pli-Pop}) that for every
$1<p<\infty$, $p\neq 2$, there exists a function
$\varphi_p:(0,\infty)\longrightarrow (0,\infty)$ such that the inequality
\begin{equation}\label{eq:pseudoDE}
\|\mathrm{Id} + T \| \geq 1 + \varphi_p(\|T\|)
\end{equation}
holds for every nonzero compact operator $T$ on $L_p[0,1]$ (see
\cite{Boy-Kad} for the $\varphi_p$ estimates). This result has been
recently carried to non-commutative $\mathcal{L}_p$-spaces and to
some spaces of operators \cite{Oik-cstar,Oik04}. Finally,
inequalities as \eqref{eq:pseudoDE} in which $\varphi_p$ is linear
are studied in \cite{Mar-daugavetian}.

\section{Preliminaries}\label{sec:preliminares}

Let us start by fixing some notation. We use the symbols $\mathbb{T}$ and
$\mathbb{D}$ to denote, respectively, the unit sphere and the closed unit
ball of the base field $\mathbb{K}$, and we write $\re(\cdot)$ to denote the
real part function, which is nothing than the identity when $\mathbb{K}=\mathbb{R}$.
Given a Banach space $X$, the dual space of $X$ is denoted by
$X^{*}$ and, when $X$ is a complex space, $X_\mathbb{R}$ is the underlying
real space. Given $x\in X$ and $x^*\in X^*$, we write $x^*\otimes x$
to denote the bounded linear operator
$$
y\longmapsto x^*(y)\,x \qquad (y\in X),
$$
whose norm is equal to $\|x\|\,\|x^*\|$.

It is straightforward to check that if an operator $T$ on a Banach
space $X$ satisfies \eqref{DE}, then all the operators of the form
$\alpha\,T$ for $\alpha>0$ also satisfy \eqref{DE} (see
\cite[Lemma~2.1]{AAB} for instance). We will use this fact along the
paper without explicit mention.

Next, let us observe that the definition of the Daugavet property
for complex spaces may be a bit confusing, since the meaning of a
rank-one operator can be understood in two different forms. Indeed,
an operator $T$ on a Banach space $X$ is a rank-one operator if
$\dim T(X)\leq 1$. Then, when $X$ is a complex space, two different
classes of rank-one operators may be considered: the complex-linear
operators whose ranks have complex-dimension~$1$ (i.e.\ operators of
the form $x^*\otimes x$ for $x^*\in X^*$ and $x\in X$) and, on the
other hand, the real-linear operators whose ranks have
real-dimension~$1$ (i.e.\ operators of the form $\re x^*\otimes x$
for $\re x^*\in \left(X_\mathbb{R}\right)^*$ and $x\in X_\mathbb{R}$). As a matter
of facts, the two possible definitions of the Daugavet property that
one may state are equivalent.

\begin{remark}
Let $X$ be a complex Banach space. Then, the following are
equivalent:
\begin{equi}
\item Every (real) rank-one operator on $X_\mathbb{R}$ satisfies
\eqref{DE}.
\item Every (complex) rank-one operator on $X$ satisfies \eqref{DE}.
\end{equi}
\end{remark}

\begin{proof}
$(i)\Rightarrow (ii)$ follows immediately from
\cite[Theorem~2.3]{KSSW}, but a direct proof is very easy to state.
Given a complex rank-one operator $T=x_0^*\otimes x_0$ with
$\|x_0^*\|=1$ and $\|x_0\|=1$, we have $\|\mathrm{Id}+\re x_0^*\otimes
x_0\|=2$. Therefore, given $\varepsilon>0$, there exists $x\in S_X$ such
that $\|x+\re x_0^*(x)x_0\|\geq 2-\varepsilon$. It follows that
$$
|\re x_0^*(x)|\geq 1-\varepsilon, \qquad \text{and so} \qquad |\im
x_0^*(x)|\leq \varepsilon.
$$
Now, it is clear that
\begin{align*}
\|\mathrm{Id}+T\|&\geq \|x + T x\| =\|x+x_0^*(x)x_0\| \\ &\geq \|x+\re
x_0^*(x)x_0\| -\|\im x_0^*(x)x_0\|\geq 2-2\varepsilon.
\end{align*}
$(ii)\Rightarrow (i)$. Let $T=\re x_0^* \otimes x_0$ be a real
rank-one operator with
$$
\|\re x_0^*\|=\|x_0^*\|=1 \qquad \text{and} \qquad \|x_0\|=1.
$$
By $(ii)$, we have that $\|\mathrm{Id}+x_0^*\otimes x_0\|=2$ so, given
$\varepsilon>0$, there exists $x\in S_X$ satisfying
$$
\|x+x_0^*(x)x_0\|\geq2-\varepsilon.
$$
If we take $\omega\in\mathbb{T}$ such that $\omega x_0^*(x)=|x_0^*(x)|$,
then
$$
x_0^*(\omega x)=\omega x_0^*(x)=\re x_0^*(\omega x).
$$
Therefore,
\begin{align*}
\|\mathrm{Id}+T\| &\geq \|\omega x+\re x_0^*(\omega x)x_0\| \\ &= \|\omega
x+ \omega x_0^*(x)x_0\|= \|x+x_0^*(x)x_0\|\geq 2-\varepsilon.\qedhere
\end{align*}
\end{proof}

From now on, by a rank-one operator on a Banach space over $\mathbb{K}$, we
will mean a bounded $\mathbb{K}$-linear operator whose image has
$\mathbb{K}$-dimension less or equal than $1$.

As we commented in the introduction, the aim of this paper is to
discuss whether there are other isometric properties apart from the
Daugavet property which can be defined by requiring all rank-one
operators on a Banach space to satisfy a norm equality. A first
observation is that the Daugavet property implies that for every
rank-one operator $T$, the norm of $\mathrm{Id} + T$ only depends on the
norm of $T$. It is easy to check that the above fact only may happen
if the Banach space involved has the Daugavet property. We state and
prove a slightly more general version of this result which we will
use later on.

\begin{prop}\label{prop1}
Let $f:\mathbb{R}^+_0\longrightarrow\mathbb{R}^+_0$ be an arbitrary function.
Suppose that there exist $a, b\in \mathbb{K}$ and a non-null Banach space
$X$ over $\mathbb{K}$ such that the norm equality
$$
\|a\mathrm{Id}+b\,T\|=f(\|T\|)
$$
holds for every rank-one operator $T\in L(X)$. Then,
$f(t)=|a|+|b|\,t$ for every $t\in\mathbb{R}^+_0$. In particular, if $a\neq
0$ and $b\neq 0$, then $X$ has the Daugavet property.
\end{prop}

\begin{proof}
If $a\,b=0$ we are trivially done, so we may assume that $a\neq0$,
$b\neq0$ and we write
$\omega_0=\frac{\overline{b}}{|b|}\frac{a}{|a|}\in \mathbb{T}$. Now, we fix
$x_0\in S_X$, $x_0^*\in S_{X^*}$ such that $x_0^*(x_0)=\omega_0$
and, for each $t\in\mathbb{R}^+_0$, we consider the rank-one operator
$T_t=t\,x_0^* \otimes x_0\in L(X)$. Observe that $\|T_t\|=t$, so we
have
$$
f(t)=\|a \mathrm{Id}+ b\, T_t\| \qquad (t\in \mathbb{R}^+_0).
$$
Then, it follows that
\begin{align*}
|a|+|b|\,t \geq f(t) &= \|a \mathrm{Id} + b\, T_t\|\geq\bigl\|[a\mathrm{Id}+b\,
T_t](x_t)\bigr\| \\ & = \|a\,x_t+b\,\omega_0t\,x_t\|=|a +
b\,\omega_0 t|\,\|x_t\|\\ & =\left|a + b
\frac{\overline{b}}{|b|}\frac{a}{|a|}\, t \right| = |a|+|b|\,t.
\end{align*}
Finally, if the norm equality
$$
\|a \mathrm{Id} + b\, T\|= |a| + |b|\, \|T\|
$$
holds for every rank-one operator on $X$, then $X$ has the Daugavet
property. Indeed, we fix a rank-one operator $T\in L(X)$ and apply
the above equality to $S=\frac{a}{b}\,T$ to get
\begin{equation*}
|a|\bigl(1 + \|T\|\bigr)=|a|+ |b|\,\|S\| =\|a\mathrm{Id} + b\, S\|= |a|\,
\|\mathrm{Id} +T\|.\qedhere
\end{equation*}
\end{proof}

With the above property in mind, we have to look for Daugavet-type
norm equalities in which $\mathrm{Id} + T$ is replaced by other function of
$T$. If we want such a function to carry operators to operators and
to be applied to arbitrary rank-one operators on arbitrary Banach
spaces, it is natural to consider $g$ to be a power series with
infinite radius of convergence. In other words, we consider an
analytic function $g:\mathbb{K}\longrightarrow \mathbb{K}$ and, for each operator $T\in L(X)$,
we define
$$
g(T)=\sum_{k=0}^\infty a_k\, T^k,
$$
where $g(\zeta)=\sum_{k=0}^\infty a_k\, \zeta^k$ is the power series
expansion of $g$. The following easy result shows how to calculate
$g(T)$ when $T$ is a rank-one operator.

\begin{lemma}\label{lemma}
Let $g:\mathbb{K} \longrightarrow \mathbb{K}$ be an analytic function with power series
expansion
$$
g(\zeta)=\sum_{k=0}^\infty a_k\,\zeta^k \qquad (\zeta\in \mathbb{K}),
$$
and let $X$ be a Banach space over $\mathbb{K}$. For $x^*\in X^*$ and $x\in
X$, we write $T=x^*\otimes x$ and $\alpha=x^*(x)$. Then, for each
$\lambda\in\mathbb{K}$,
$$
g(\lambda T)=\begin{cases} a_0 \mathrm{Id} + a_1\lambda T & \text{if $\alpha=0$} \\
a_0 \mathrm{Id} + \frac{\widetilde{g}(\alpha\lambda)}{\alpha}T & \text{if
$\alpha\neq 0$,}
\end{cases}
$$
where
$$
\widetilde{g}(\zeta)=g(\zeta)-a_0 \qquad \bigl(\zeta\in \mathbb{K}\bigr).
$$
\end{lemma}

\begin{proof}
Given $\lambda\in\mathbb{K}$, it is immediate to check that
$$
(\lambda T)^k=\alpha^{k-1}\lambda^k\,T \qquad (k\in\mathbb{N}).
$$
Now, if $\alpha=0$, then $T^2=0$ and the result is clear. Otherwise,
we have
\begin{align*}
g(\lambda T) & =a_0 \mathrm{Id} + \sum_{k=1}^\infty
a_k\,\alpha^{k-1}\,\lambda^k\,T \\ & = a_0 \mathrm{Id} +
\left(\frac{1}{\alpha}\sum_{k=1}^\infty
a_k\,\alpha^k\lambda^k\,\right)T = a_0 \mathrm{Id} +
\frac{\widetilde{g}(\alpha\lambda)}{\alpha}T.\qedhere
\end{align*}
\end{proof}

\section{Norm equalities of the form
$\|g(T)\|=f(\|T\|)$} \label{sec:g(T)=f(T)}

We would like to study now norm equalities for operators of the form
\begin{equation}\label{eq:normg(T)=f(normT)}
\|g(T)\|=f(\|T\|),
\end{equation}
where $f:\mathbb{R}^+_0\longrightarrow \mathbb{R}^+_0$ is an arbitrary function and $g:\mathbb{K}\longrightarrow \mathbb{K}$
is an analytic function.

Our target is to show that the Daugavet property is the only
non-trivial property that it is possible to define by requiring all
rank-one operators on a Banach space of dimension greater than one
to satisfy a norm equality of the form \eqref{eq:normg(T)=f(normT)}.
We start by proving that $g$ has to be a polynomial of degree less
or equal than $1$, and then we will deduce the result from
Proposition~\ref{prop1}.

\begin{theorem}\label{theorem}
Let $g:\mathbb{K}\longrightarrow \mathbb{K}$ be an analytic function and $f: \mathbb{R}_0^+
\longrightarrow \mathbb{R}_0^+$ an arbitrary function. Suppose that there is
a Banach space $X$ over $\mathbb{K}$ with $\dim(X)\geq 2$ such that the norm
equality
\begin{equation*}
\|g(T)\|=f(\|T\|)
\end{equation*}
holds for every rank-one operator $T$ on $X$. Then, there are $a,\,
b\in \mathbb{K}$ such that
$$
g(\zeta)=a+b\zeta \qquad \bigl(\zeta \in \mathbb{K}\bigr).
$$
\end{theorem}

\begin{proof} Let $g(\zeta)= \sum_{k=0}^\infty a_k\, \zeta^k$ be the
power series expansion of $g$ and let \mbox{$\widetilde{g}=g-a_0$}.
Given $\alpha\in \mathbb{D}$, we take $x_\alpha^*\in S_{X^*}$ and
$x_\alpha\in S_X$ such that $x_\alpha^*(x_\alpha)=\alpha$ (we can do
it since $\dim(X)\geq2$), and we write $T_\alpha=x_\alpha^*\otimes
x_\alpha$, which satisfies $\|T_\alpha\|=1$. Using
Lemma~\ref{lemma}, for each $\lambda\in \mathbb{K}$ we obtain that
\begin{align*}
g(\lambda T_0) &=a_0\mathrm{Id}+a_1\lambda\, T_0 \intertext{and} g(\lambda
T_\alpha) &=
a_0\mathrm{Id}+\frac{1}{\alpha}\widetilde{g}(\lambda\alpha)\,T_\alpha \qquad
(\alpha\neq 0).
\end{align*}
Now, fixed $\lambda\in \mathbb{K}$, we have
\begin{align*}
f(|\lambda|)&=\|g(\lambda T_0)\|=\|a_0\mathrm{Id} +a_1\lambda T_0\|,
\intertext{and} f(|\lambda|)&=\|g(\lambda T_\alpha)\|=
\left\|a_0\mathrm{Id}+\frac{1}{\alpha}\widetilde{g}(\lambda\alpha)T_\alpha\right\|.
\end{align*}
Therefore, we have the equality
\begin{equation}\label{eq:opzero=opalpha}
\left\|a_0\mathrm{Id}+\frac{1}{\alpha}\widetilde{g}(\lambda\alpha)T_\alpha\right\|=\|a_0\mathrm{Id}
+a_1\lambda T_0\|\qquad (\lambda\in\mathbb{K},\ 0<|\alpha|\leq1).
\end{equation}
In the complex case is enough to consider the above equality for
$\alpha=1$ and to use the triangle inequality to get that
\begin{equation}\label{eq:th1-complex}
|\widetilde{g}(\lambda)|\leq 2|a_0| + |a_1|\,|\lambda| \qquad
(\lambda\in \mathbb{C}).
\end{equation}
From this, it follows by just using Cauchy's estimates, that
$\widetilde{g}$ is a polynomial of degree one (see \cite[Exercise 1,
p.~80]{Conway} or \cite[Theorem~3.4.4]{Gre-Kra}, for instance), and
we are done.

In the real case, it is not possible to deduce from inequality
\eqref{eq:th1-complex} that $\widetilde{g}$ is a polynomial, so we
have to return to \eqref{eq:opzero=opalpha}. From this equality, we
can deduce by just applying the triangle inequality that
\begin{equation*}
\left|\frac{\widetilde{g}(\lambda\alpha)}{\alpha}\right|-|a_0|\leq|a_0|+|a_1|\,|\lambda|
\qquad \text{and} \qquad
|a_1|\,|\lambda|-|a_0|\leq\left|\frac{\widetilde{g}(\lambda\alpha)}{\alpha}\right|+|a_0|
\end{equation*}
for every $\lambda\in\mathbb{R}$ and every $\alpha\in [-1,1]\setminus\{0\}$.
It follows that
\begin{equation}\label{th:eq-desigualdad-importante}
\left|\left|\frac{\widetilde{g}(\lambda\alpha)}{\alpha}\right|-|a_1|\,|\lambda|\right|\leq
2|a_0|\qquad \bigl(\lambda\in\mathbb{R},\ \alpha\in
[-1,1]\setminus\{0\}\bigr).
\end{equation}
Next, for $t\in ]1,+\infty[$ and $k\in\mathbb{N}$, we use
\eqref{th:eq-desigualdad-importante} with $\lambda=t^k$ and
$\alpha=\frac{1}{t^{k-1}}$ to obtain that
$$
\big|\,|\widetilde{g}(t)|-|a_1|\,t\big|\leq\frac{2|a_0|}{t^{k-1}}
$$
so, letting $k\longrightarrow\infty$, we get that
$$
|\widetilde{g}(t)|=|a_1|\,t \qquad \bigl(t\in ]1,+\infty[\bigr).
$$
Finally, an obvious continuity argument allows us to deduce from the
above equality that $\widetilde{g}$ coincides with a degree one
polynomial in the interval $]1,+\infty[$, thus the same is true in
the whole $\mathbb{R}$ by analyticity.
\end{proof}

We summarize the information given in Proposition~\ref{prop1} and
Theorem~\ref{theorem}.

\begin{corollary}
Let $f: \mathbb{R}_0^+ \longrightarrow \mathbb{R}_0^+$ be an arbitrary function and
$g:\mathbb{K}\longrightarrow \mathbb{K}$ an analytic function. Suppose that there is
a Banach space $X$ over $\mathbb{K}$ with $\dim(X)\geq 2$ such that the norm
equality
\begin{equation*}
\|g(T)\|=f(\|T\|)
\end{equation*}
holds for every rank-one operator $T$ on $X$. Then, only three
possibilities may happen:
\begin{enumerate}
\item[(a)] $g$ is a constant function (trivial case).
\item[(b)] There is a non-null $b\in \mathbb{K}$ such that
$g(\zeta)=b\,\zeta$ for every $\zeta\in \mathbb{K}$ (trivial case).
\item[(c)] There are non-null $a,b\in \mathbb{K}$ such that $g(\zeta)=a +
b\, \zeta$ for every $\zeta\in \mathbb{K}$, and $X$ has the Daugavet
property.
\end{enumerate}
\end{corollary}

\begin{remark}
{\slshape The above result is not true when the dimension of the
space is one.\ } Indeed, for every $a\in L(\mathbb{K})\equiv \mathbb{K}$, one has
$|a^2|=|a|^2$.
\end{remark}

\section{Norm equalities of the form $\|\mathrm{Id} + g(T)\|=f(\|g(T)\|)$}
\label{sec:Id+g(T)=f(g(T))}

Let $X$ be a Banach space over $\mathbb{K}$. Our next aim is to study norm
equalities of the form
\begin{equation}\label{eq:DEwith-g-and-f(g)}
\|\mathrm{Id} + g(T)\| = f(\|g(T)\|)
\end{equation}
where $g:\mathbb{K} \longrightarrow \mathbb{K}$ is analytic and $f:\mathbb{R}^+_0 \longrightarrow \mathbb{R}^+_0$ is
continuous.

When $X$ has the Daugavet property, it is clear that
Eq.~\eqref{eq:DEwith-g-and-f(g)} holds for every rank-one operator
if we take $g(\zeta)=\zeta$ and $f(t)=1 + t$. But, actually, every
analytic function $g$ works with a suitable $f$.

\begin{remark}\label{remark:4.1}
{\slshape If $X$ is a real or complex Banach space with the Daugavet
property and $g:\mathbb{K} \longrightarrow \mathbb{K}$ is an analytic function, the norm
equality
$$
\|\mathrm{Id} + g(T)\|=|1+g(0)|-|g(0)| + \|g(T)\|
$$
holds for every weakly compact operator $T\in L(X)$.\ } Indeed,
write $\widetilde{g}=g-g(0)$ and observe that $\widetilde{g}(T)$ is
weakly compact whenever $T$ is. Since $X$ satisfies the Daugavet
property, we have
\begin{align*}
\|\mathrm{Id} + g(T)\| &=\|\mathrm{Id} + g(0)\,\mathrm{Id} + \widetilde{g}(T)\|=|1+g(0)| +
\|\widetilde{g}(T)\| \\ & = \bigl(|1+g(0)| -|g(0)|\bigr) +
\bigl(|g(0)| + \|\widetilde{g}(T)\|\bigr) \\ & =|1+g(0)|-|g(0)| +
\|g(0)\,\mathrm{Id} + \widetilde{g}(T)\| \\ &= |1+g(0)|-|g(0)| + \|g(T)\|.
\end{align*}
\end{remark}

With the above result in mind, it is clear that the aim of this
section cannot be to show that only few $g$'s are possible in
\eqref{eq:DEwith-g-and-f(g)}, but it is to show that many $g$'s
produce only few properties. At first, before formulating our
results, let us discuss the case when the Banach space we consider
is one-dimensional.

\begin{remark}$ $\label{remark-dim-one}
\begin{enumerate}
\item[(a)] \textsc{Complex case:\ } {\slshape It is not possible
to find a non-constant entire function $g$ and an arbitrary function
$f:\mathbb{R}^+_0\longrightarrow \mathbb{R}$ such that the equality
$$
|1+g(\zeta)|=f(|g(\zeta)|)
$$
holds for every $\zeta\in \mathbb{C}\equiv L(\mathbb{C})$.\ } Indeed, we suppose
otherwise that such functions $g$ and $f$ exist, and we use Picard
Theorem to assure the existence of $\lambda>0$ such that
$-\lambda,\, \lambda \in g(\mathbb{C})$. We get
$$
f(\lambda)=|1+\lambda|\neq |1-\lambda|=f(|-\lambda|)=f(\lambda),
$$
a contradiction.
\item[(b)] \textsc{Real case:\ } {\slshape The equality
$$
|1+t^2|=1 + |t^2|
$$
holds for every $t\in \mathbb{R}\equiv L(\mathbb{R})$.}
\end{enumerate}
\end{remark}

It follows that real and complex spaces do not behave in the same
way with respect to equalities of the form given in
\eqref{eq:DEwith-g-and-f(g)}. Therefore, from now on we study
separately the complex and the real cases. Let us also remark that
when a Banach space $X$ has dimension greater than one, it is clear
that
$$
\|g(T)\|\geqslant |g(0)|
$$
for every analytic function $g:\mathbb{K}\longrightarrow \mathbb{K}$ and every rank-one operator
$T\in L(X)$. Therefore, the function $f$ in
\eqref{eq:DEwith-g-and-f(g)} has to be defined only in the interval
$[|g(0)|,+\infty[$.

\vspace{0.5cm}

\noindent \textbf{\large $\bullet$} \textsc{\large Complex case:}

Our key lemma here states that the function $g$ in
\eqref{eq:DEwith-g-and-f(g)} can be replaced by a degree one
polynomial.

\begin{lemma}\label{lemma:4.3}
Let $g:\mathbb{C}\longrightarrow\mathbb{C}$ be a non-constant entire function, let\linebreak
$f:[|g(0)|,+\infty[\longrightarrow \mathbb{R}$ be a continuous function and let $X$ be a
Banach space with dimension greater than $1$. Suppose that the norm
equality
$$
\|\mathrm{Id} + g(T)\| = f(\|g(T)\|)
$$
holds for every rank-one operator $T\in L(X)$. Then,
$$
\bigl\|\big(1+g(0)\big)\,\mathrm{Id} + T\bigr\|=|1+g(0)|-|g(0)| +
\|g(0)\,\mathrm{Id} + T\|
$$
for every rank-one operator $T\in L(X)$.
\end{lemma}

\begin{proof}
We claim that the norm equality
\begin{equation}\label{eq:gtopoldegree1}
\bigl\|\big(1+g(0)\big)\,\mathrm{Id} + T\bigr\|=f\big(\|g(0)\,\mathrm{Id} + T\|\big)
\end{equation}
holds for every rank-one operator $T\in L(X)$. Indeed, we write
$\widetilde{g}=g-g(0)$ and we use Picard Theorem to assure that
$\widetilde{g}(\mathbb{C})$ is equal to $\mathbb{C}$ except for, eventually, one
point $\alpha_0\in \mathbb{C}$. Next, we fix a rank-one operator
$T=x^*\otimes x$ with $x^*\in X^*$ and $x\in X$. If $x^*(x)\neq 0$
and $x^*(x)\neq \alpha_0$, we may find $\zeta\in \mathbb{C}$ such that
$\widetilde{g}(\zeta)=x^*(x)$, and we use Lemma~\ref{lemma} to get
that
$$
g\left(\frac{\zeta}{x^*(x)}\,T\right)= g(0)\,\mathrm{Id} +
\frac{\widetilde{g}(\zeta)}{x^*(x)}\,T=g(0)\,\mathrm{Id} +T.
$$
We deduce that
\begin{align*}
\left\|\bigl(1 + g(0)\bigr)\,\mathrm{Id} + T\right\|& =\left\|\mathrm{Id} +
g\left(\frac{\zeta}{x^*(x)}\,T\right)\right\| \\ & =
f\left(\left\|g\left(\frac{\zeta}{x^*(x)}\,T\right)\right\|\right) =
f\bigl(\|g(0)\,\mathrm{Id} + T\|\bigr).
\end{align*}
The remaining cases in which $x^*(x)=0$ or $x^*(x)=\alpha_0$ follow
from the above equality thanks to the continuity of $f$.

To finish the proof, we have to show that
\begin{equation*}
f(t)=|1+g(0)|-|g(0)| + t \qquad \bigl(t\geq |g(0)|\bigr).
\end{equation*}
Suppose first that $g(0)=-1$. We take $x\in S_X$ and $x^*\in
S_{X^*}$ such that \mbox{$x^*(x)=1$} and, for every $t\geq 1$, we
define the rank-one operator
$$
T_t=(1-t)\,x^*\otimes x.
$$
It is immediate to show that
$$
\|-\mathrm{Id} + T_t\|=t \qquad \text{and} \qquad \|T_t\|=t-1.
$$
Then, it follows from \eqref{eq:gtopoldegree1} that $f(t)=t-1$, and
we are done.

Suppose otherwise that $g(0)\neq -1$. We take $x\in S_X$ and $x^*\in
S_{X^*}$ such that $x^*(x)=1$ and, for every $t\geq |g(0)|$, we
define the rank-one operator
$$
T_t=\frac{1+g(0)}{|1+g(0)|}\,\big(t-|g(0)|\big)\,x^*\otimes x.
$$
It is routine to show, by just evaluating at the point $x$, that
\begin{equation*}
\left\|\bigl(1 + g(0)\bigr)\,\mathrm{Id} + T_t\right\|=|1+g(0)|+t-|g(0)|.
\end{equation*}
Therefore, if follows from \eqref{eq:gtopoldegree1} that
\begin{equation}\label{eq:gtopoldegree1(2)}
f\bigl(\|g(0)\,\mathrm{Id} + T_t\|\bigr)=|1+g(0)|+t-|g(0)|,
\end{equation}
and the proof finishes by just proving that
$$
\|g(0)\,\mathrm{Id}+T_t\|=t.
$$
On the one hand, it is clear that
$$
\|g(0)\,\mathrm{Id}+T_t\|\leq |g(0)| + \|T_t\|=|g(0)| + t - |g(0)|=t.
$$
On the other hand, the converse inequality trivially holds when
$g(0)=0$, so we may suppose \mbox{$g(0)\neq 0$} and we define the
rank-one operator
$$
S_t=\frac{g(0)}{|g(0)|}\,\bigl(\|g(0)\,\mathrm{Id}+T_t\|-|g(0)|\bigr)\,x^*\otimes
x.
$$
It is routine to show, by using that $\|g(0)\,\mathrm{Id}+S_t\|\geq |g(0)|$
and evaluating at the point $x$, that
$$
\|g(0)\,\mathrm{Id}+S_t\|=\|g(0)\,\mathrm{Id}+T_t\|.
$$
From the above equality, \eqref{eq:gtopoldegree1}, and
\eqref{eq:gtopoldegree1(2)}, we deduce that
\begin{multline*}
|1+g(0)|+t-|g(0)| = f\bigl(\|g(0)\,\mathrm{Id} +
S_t\|\bigr)=\left\|\bigl(1+g(0)\bigr)\,\mathrm{Id}+S_t\right\| \\
\leq |1+g(0)| + \|S_t\| =|1+g(0)|+\|g(0)\,\mathrm{Id} + T_t\|-|g(0)|,
\end{multline*}
so $t\leq \|g(0)\,\mathrm{Id}+T_t\|$ and we are done.
\end{proof}

In view of the norm equality appearing in the above lemma, two
different cases arise either $|1+g(0)|\neq |g(0)|$ or $|1+g(0)|=
|g(0)|$; equivalently, $\re g(0)\neq -1/2$ or $\re g(0)=-1/2$. In
the first case, we get the Daugavet property.

\begin{theorem}\label{th:4.4}
Let $X$ be a complex Banach space with $\dim(X)\geq2$. Suppose that
there exist a non-constant entire function $g:\mathbb{C}\longrightarrow\mathbb{C}$
with $\re g(0)\neq-\frac{1}{2}$ and a continuous function
$f:[|g(0)|,+\infty[\longrightarrow\mathbb{R}^+_0$, such that the norm
equality
$$
\|\mathrm{Id}+g(T)\|=f(\|g(T)\|)
$$
holds for every rank-one operator $T\in L(X)$. Then, $X$ has the
Daugavet property.
\end{theorem}

\begin{proof}
Let us first suppose that $\re g(0)>-1/2$ so that
$$
M=|1+g(0)|-|g(0)|>0.
$$
Then, dividing by $M$ the equation given by Lemma~\ref{lemma:4.3},
we get that the norm equality
\begin{equation}\label{eq:reg(0)>-1/2---1}
\left\|\textstyle\frac{1+g(0)}{M}\,\mathrm{Id}+\frac{1}{M}\,T\right\|=1+\left\|\textstyle
\frac{g(0)}{M}\,\mathrm{Id}+ \frac{1}{M}\,T\right\|
\end{equation}
holds for every rank-one operator $T\in L(X)$. Now, we take
$\omega,\,\xi\in\mathbb{T}$ such that
$$
\omega\,(1+g(0))=|1+g(0)|\qquad \text{and} \qquad \xi\, g(0)=|g(0)|,
$$
we fix a rank-one operator $T\in L(X)$, and we observe that
$$
\left\|\textstyle{\frac{1+g(0)}{M}}\,\mathrm{Id}+\frac{1}{M}\,T\right\|=
\left\|\textstyle{\frac{|1+g(0)|}{M}}\,\mathrm{Id}+\frac{\omega}{M}\,T\right\|=
\left\|\mathrm{Id}+\textstyle{\frac{|g(0)|}{M}}\,\mathrm{Id}+\frac{\omega}{M}\,T\right\|
$$
and
$$
\left\|\textstyle\frac{g(0)}{M}\,\mathrm{Id}+ \frac{1}{M}\,T\right\|=
\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+\frac{\xi}{M}\,T\right\|.
$$
Therefore, from \eqref{eq:reg(0)>-1/2---1} we get
$$
\left\|\textstyle\mathrm{Id}+\frac{|g(0)|}{M}\,\mathrm{Id}+\frac{\omega}{M}\,T\right\|=
1+\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+\frac{\xi}{M}\,T\right\|.
$$
It follows straightforwardly from the arbitrariness of $T$ that the
norm equality
\begin{equation}\label{eq:teor-complejo-igualdad-xi-omega}
\left\|\textstyle\mathrm{Id}+\frac{|g(0)|}{M}\,\mathrm{Id}+T\right\|=
1+\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+\frac{\xi}{\omega}\,T\right\|
\end{equation}
holds for every rank-one operator $T\in L(X)$. Now, we claim that
\begin{equation}\label{eq:teor-complejo-----2}
\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+T\right\|=
\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+\frac{\xi}{\omega}\,T\right\|
\end{equation}
for every rank-one operator $T\in L(X)$. Indeed, by the triangle
inequality, we deduce from
\eqref{eq:teor-complejo-igualdad-xi-omega} that
$$
\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+T\right\|\geq
\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+\frac{\xi}{\omega}\,T\right\|
$$
for every rank-one operator $T\in L(X)$ and, for every $n\in \mathbb{N}$,
applying the above inequality $n$ times, we get that
$$
\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+T\right\|\geq
\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+\frac{\xi}{\omega}\,T\right\|\geq
\cdots \geq
\left\|\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+\left(\frac{\xi}{\omega}\right)^n\,T\right\|;
$$
the claim follows from the easy fact that the sequence
$\left\{\left(\frac{\xi}{\omega}\right)^n\right\}_{n\in\mathbb{N}}$ has a
subsequence which converges to $1$.

Now, given a rank-one operator $T\in L(X)$, it follows from
\eqref{eq:teor-complejo-igualdad-xi-omega} and
\eqref{eq:teor-complejo-----2} that
$$
\left\|\textstyle\mathrm{Id}+\frac{|g(0)|}{M}\,\mathrm{Id}+T\right\|= 1+
\left\|\textstyle\ \frac{|g(0)|}{M}\,\mathrm{Id}+T\right\|
$$
holds and, therefore, for every $n\in\mathbb{N}$ we get
$$
\left\|\textstyle\mathrm{Id}+\frac{1}{n}\left(\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+T\right)\right\|=
1 + \left\|\textstyle\frac{1}{n}
\left(\textstyle\frac{|g(0)|}{M}\,\mathrm{Id}+T\right)\right\|.
$$
To finish the proof, we fix a rank-one operator $S\in L(X)$ and
$n\in\mathbb{N}$, and we apply the above equality to $T=n\,S$ to get that
$$
\left\|\textstyle\mathrm{Id}+S+\textstyle\frac{|g(0)|}{n\,M}\,\mathrm{Id}\right\|=
\left\|\textstyle\frac{|g(0)|}{n\,M}\,\mathrm{Id}+S\right\|.
$$
We let $n\to \infty$ to deduce that
\begin{equation*}
\|\mathrm{Id}+S\|=1+\|S\|,
\end{equation*}
so $X$ has the Daugavet property.

In case that $\re g(0)<-\frac{1}{2}$, we write
$$
M=-|1+g(0)|+|g(0)|>0
$$
and we deduce from Lemma~\ref{lemma:4.3} that the norm equality
\begin{equation*}
\left\|\textstyle \frac{g(0)}{M}\,\mathrm{Id}+
\frac{1}{M}\,T\right\|=1+\left\|\textstyle\frac{1+g(0)}{M}\,\mathrm{Id}+\frac{1}{M}\,T\right\|
\end{equation*}
holds for every rank-one operator $T\in L(X)$. The rest of the proof
is completely analogous, using the above equality instead of
\eqref{eq:reg(0)>-1/2---1}.
\end{proof}

When $\re g(0)=-\frac{1}{2}$, the above proof does not work.
Actually, contrary to all the previous cases, another family of
properties apart from the Daugavet property appears. More
concretely, let $X$ be a complex Banach space with dimension greater
than two, let $g:\mathbb{C}\longrightarrow\mathbb{C}$ be a non-constant entire function with
$\re g(0)=-1/2$, and let $f:[|g(0)|,+\infty[\longrightarrow \mathbb{R}$ be a continuous
function, such that the norm equality
$$
\|\mathrm{Id} + g(T)\|=f(\|g(T)\|)
$$
holds for every rank-one operator $T\in L(X)$. Then, we get from
Lemma~\ref{lemma:4.3} that
$$
\bigl\|\big(1+g(0)\big)\,\mathrm{Id} + T\bigr\|=\|g(0)\,\mathrm{Id} + T\|
$$
for every rank-one operator $T\in L(X)$. Therefore, being
$|1+g(0)|=|g(0)|$, we deduce that there are $\omega_1,\,\omega_2\in
\mathbb{C}$ with $\omega_1\neq\omega_2$ and $|\omega_1|=|\omega_2|$, such
that
$$
\|\mathrm{Id}+\omega_1\,T\|=\|\mathrm{Id}+\omega_2\,T\|
$$
for every rank-one operator $T\in L(X)$ or, equivalently, that there
is $\omega\in\mathbb{T}\setminus\{1\}$ such that
\begin{equation*}
\|\mathrm{Id} + \omega\,T\|=\|\mathrm{Id}+T\|
\end{equation*}
for every rank-one operator $T\in L(X)$. It is routine to check
that, fixed a Banach space $X$, the set of those $\omega\in \mathbb{T}$
which make true the above equality for all rank-one operators on $X$
is a multiplicative closed subgroup of $\mathbb{T}$. Recall that such a
subgroup of $\mathbb{T}$ is either the whole $\mathbb{T}$ or the set of those
$n^{\rm th}$-roots of unity for a positive entire $n\geq 2$. Let us
state the result the we have just proved.

\begin{theorem}\label{th:4.5}
Let $X$ be a complex Banach space with $\dim(X)\geq2$. Suppose that
there exist a non-constant entire function $g:\mathbb{C}\longrightarrow\mathbb{C}$
with $\re g(0)=-\frac{1}{2}$ and a continuous function
$f:[|g(0),+\infty[\longrightarrow\mathbb{R}^+_0$, such that the norm
equality
$$
\|\mathrm{Id}+g(T)\|=f(\|g(T)\|)
$$
holds for every rank-one operator $T\in L(X)$. Then, there is
$\omega\in \mathbb{T}\setminus\{1\}$ such that
$$
\|\mathrm{Id} + \omega T\|=\|\mathrm{Id} + T\|
$$
for every rank-one operator $T\in L(X)$. Moreover, two possibilities
may happen:
\begin{enumerate}
\item[(a)] If $\omega^n\neq 1$ for every $n\in \mathbb{N}$, then
$$
\|\mathrm{Id} + \xi\,T\|=\|\mathrm{Id} + T\|
$$
for every rank-one operator $T\in L(X)$ and every $\xi\in \mathbb{T}$.
\item[(b)] Otherwise, if we take the minimum $n\in\mathbb{N}$ such that
$\omega^n=1$, then
$$
\|\mathrm{Id} + \xi\,T\|=\|\mathrm{Id} + T\|
$$
for every rank-one operator $T\in L(X)$ and every $n^{\rm th}$-root
$\xi$ of unity.
\end{enumerate}
\end{theorem}

The next example shows that all the properties appearing above are
strictly weaker than the Daugavet property.

\begin{example}\label{example-new-property}
{\slshape The real or complex Banach space $X=C[0,1]\oplus_2C[0,1]$
does not have the Daugavet property. However, the norm equality
\begin{equation*}
\|\mathrm{Id}+\omega T\|=\|\mathrm{Id}+T\|
\end{equation*}
holds for every rank-one operator $T\in L(X)$ and every
$\omega\in\mathbb{T}$.}
\end{example}

\begin{proof}
$X$ does not have the Daugavet property by
\cite[Corollary~5.4]{BKSW}. For the second assertion, we fix
$\omega\in\mathbb{T}$ and a rank-one operator $T=x^*\otimes x$ on $X$ (we
take $x\in S_X$ and $x^*\in X^*$), and it is enough to check that
$$
\|\mathrm{Id} + \omega\,T\|^2\geq \|\mathrm{Id} + T\|^2.
$$
To do so, we write $x^*=(x_1^*,x_2^*)$ and $x=(x_1,x_2)$, with
$x_1^*$, $x_2^*\in C[0,1]^*$ and $x_1$, $x_2\in C[0,1]$, and we may
and do assume that
$$
x_1^*=\mu_1+\sum_{j=1}^{n_1}\alpha_j\delta_{r_j}\qquad
x_1^*=\mu_2+\sum_{j=1}^{n_2}\beta_j\delta_{s_j},
$$
where $\alpha_1$,$\dots$, $\alpha_{n_1}$, $\beta_1$,$\dots$,
$\beta_{n_2}\in \mathbb{C}$, $r_1$,$\dots$, $r_{n_1}$, $s_1$,$\dots$,
$s_{n_2}\in [0,1]$, and $\mu_1$, $\mu_2$ are non-atomic measures on
$[0,1]$ (indeed, each rank-one operator can be approximated by
operators satisfying the preceding condition).

Now, we fix $0<\varepsilon<1$ and we consider $y=(y_1,y_2)\in S_X$ such
that
\begin{equation}\label{example:norm(Id+T)}
\|y+Ty\|^2=\|y_1+x^*(y)x_1\|^2+\|y_2+x^*(y)x_2\|^2\geq\|\mathrm{Id}+T\|^2-\varepsilon.
\end{equation}
Since $x_1$, $x_2$, $y_1$, and $y_2$ are continuous functions and
$[0,1]$ is perfect, we can find open intervals $\Delta_1$,
$\Delta_2\subset [0,1]$ so that
\begin{equation}\label{open-interval-property}
|y_i(t)+x^*(y)x_i(t)|\geq\|y_i+x^*(y)x_i\|-\varepsilon \qquad \big(t\in
\Delta_i,\ i=1,2\big)
\end{equation}
and
\begin{equation}\label{open-interval-property2}
\Delta_1\cap\{r_j \ : \ j=1,\dots,n_1\}=\emptyset, \qquad
\Delta_2\cap\{s_j \ : \ j=1,\dots, n_2\}=\emptyset
\end{equation}
Furthermore, using that $\mu_1$, $\mu_2$ are non-atomic and reducing
$\Delta_1$, $\Delta_2$ if necessary, we can assume that they also
satisfy
\begin{equation}\label{open-interval-property2-bis}
|\mu_1(f)|<\varepsilon\,\|f\|, \qquad |\mu_2(g)|<\varepsilon\,\|f\|
\end{equation}
for every $f,g\in C[0,1]$ with $\text{supp}(f)\subset \Delta_1$,
$\text{supp}(g)\subset \Delta_2$.

Now, we fix $t_1\in\Delta_1$ and $t_2\in\Delta_2$. For $i=1,2$, we
take piece-wise linear continuous functions
$\phi_i:[0,1]\longrightarrow [0,1]$ such that
$$
\phi_i(t_i)=1 \qquad \text{and} \qquad
\phi_i\big([0,1]\setminus\Delta_i\big)=\{0\},
$$
and we define
$$
\widetilde{y_i}=y_i(1-\phi_i+\omega\phi_i).
$$
It is easy to check that
$$
|1-\phi_i(t)+\omega\phi_i(t)|\leq 1 \qquad \big(t\in[0,1]\big),
$$
so, $\|\widetilde{y_i}\|\leq\|y_i\|$. This implies that
$\|\widetilde{y}\|=\|(\widetilde{y_1},\widetilde{y_2})\|\leq1$ and,
therefore,
\begin{equation}\label{example:norm(Id+omegaT)}
\|\mathrm{Id}+\omega T\|^2\geq\|\widetilde{y}+\omega
T(\widetilde{y})\|^2=\|\widetilde{y_1}+\omega
x^*(\widetilde{y})x_1\|^2+\|\widetilde{y_2}+\omega
x^*(\widetilde{y})x_2\|^2.
\end{equation}
Since, clearly,
$$
y_i-\widetilde{y_i}=(1-\omega)y_i\phi_i,\qquad (i=1,2)
$$
we deduce from \eqref{open-interval-property2},
\eqref{open-interval-property2-bis}, and the fact that
$\text{supp}(\phi_i)\subset \Delta_i$, that
$$
|x^*(y-\widetilde{y})|\leq
|x_1^*(y_1-\widetilde{y_1})|+|x_2^*(y_2-\widetilde{y_2})|\leq 4\,
\varepsilon M,
$$
where $M=\max\{\|x_1^*\|,\|x_2^*\|\}$. Using this,
\eqref{open-interval-property}, and the fact that $\|x\|=1$, we
obtain the following for $i=1,2$:
\begin{align*}
\|\widetilde{y_i}+\omega\,
x^*(\widetilde{y})x_i\|&\geq\|\widetilde{y_i}+\omega\,
x^*(y)\,x_i\|-|x^*(y-\widetilde{y})|\\
 &\geq |\widetilde{y_i}(t_i)+\omega\,
x^*(y)\,x_i(t_i)|-4\,\varepsilon M\\ & = |\omega\,y_i(t_i) +
\omega\,x^*(y)\,x_i(t_i)|-4\,\varepsilon M\\
&= |y_i(t_i)+ x^*(y)\,x_i(t_i)|-4\,\varepsilon M\\ &\geq
\|y_i+x^*(y)\,x_i\|-\varepsilon-4\,\varepsilon M.
\end{align*}
Finally, using this together with \eqref{example:norm(Id+T)} and
\eqref{example:norm(Id+omegaT)}, it is not hard to find a suitable
constant $K>0$ such that
$$
\|\mathrm{Id}+\omega T\|^2\geq\|\mathrm{Id}+T\|^2-K\,\varepsilon,
$$
which finishes the proof.
\end{proof}

\vspace{0.5cm}

\noindent \textbf{\large $\bullet$} \textsc{\large Real case:}

The situation in the real case is far away from being so clear. On
the one hand, the proof of Lemma~\ref{lemma:4.3} remains valid if
the function $g$ is surjective (this substitutes Picard Theorem) and
then, the proofs of Theorems \ref{th:4.4} and \ref{th:4.5} are
valid. In addition, Example~\ref{example-new-property} was also
stated for the real case. The following result summarizes all these
facts.

\begin{theorem}\label{theorem-real-case}
Let $X$ be a real Banach space with dimension greater or equal than
two. Suppose that there exists a \textbf{surjective} analytic
function $g:\mathbb{R}\longrightarrow\mathbb{R}$ and a continuous function
$f:[|g(0),+\infty[\longrightarrow\mathbb{R}^+_0$, such that the norm
equality
$$
\|\mathrm{Id}+g(T)\|=f(\|g(T)\|)
$$
holds for every rank-one operator $T\in L(X)$.
\begin{enumerate}
\item[(a)] If $g(0)\neq -1/2$, then $X$ has the Daugavet property.
\item[(b)] If $g(0)=-1/2$, then the norm equality
$$
\|\mathrm{Id} - T\|=\|\mathrm{Id} + T\|
$$
holds for every rank-one operator $T\in L(X)$.
\item[(c)] The real space $X=C[0,1]\oplus_2 C[0,1]$ does not have the
Daugavet property but the norm equality
$$
\|\mathrm{Id} - T\|=\|\mathrm{Id} +T\|
$$
holds for every rank-one operator $T\in L(X)$.
\end{enumerate}
\end{theorem}

On the other hand, we do not know if a result similar to the above
theorem is true when the function $g$ is not onto. Let us give some
remarks about two easy cases:
$$
g(t)=t^2\quad (t\in\mathbb{R}) \qquad \text{and} \qquad g(t)=-t^2\quad
(t\in\mathbb{R}).
$$
In the first case, it is easy to see that if the norm equality
$$
\|\mathrm{Id} + T^2\|=f(\|T^2\|)
$$
holds for every rank-one operator, then $f(t)=1 + t$ and, therefore,
the interesting norm equality in this case is
\begin{equation}\label{eq:T2}
\|\mathrm{Id} +T^2\|=1+\|T^2\|.
\end{equation}
This equation is satisfied by every rank-one operator $T$ on a
Banach space $X$ with the Daugavet property. Let us also recall that
the equality
$$
|1+t^2|=1+|t^2|
$$
holds for every $t\in L(\mathbb{R})\equiv\mathbb{R}$ (Remark~\ref{remark-dim-one}).
For the norm equality
$$
\|\mathrm{Id} -T^2\|=f(\|T^2\|),
$$
we are not able to get any information about the shape of the
function $f$. Going to the $1$-dimensional case, we get that
$$
|1-t^2|=\max\{1-|t^2|,|t^2|-1\}
$$
for every $t\in L(\mathbb{R})\equiv\mathbb{R}$, but it is not possible that the
corresponding norm equality holds for all rank-one operators on a
Banach space with dimension greater than one (in this case,
$\|\mathrm{Id}-T^2\|\geq 1$). On the other hand, if a Banach space $X$ has
the Daugavet property, then
\begin{equation}\label{eq:-T2}
\|\mathrm{Id}-T^2\|=1+\|T^2\|
\end{equation}
for every rank-one operator $T\in L(X)$. Therefore, an interesting
norm equality of this form could be the above one.

Let us characterize the properties which flow out from the norm
equalities \eqref{eq:T2} and \eqref{eq:-T2}.

\begin{prop}
Let $X$ be a real Banach space.
\begin{enumerate}
\item[(a)] The following are equivalent:
\begin{equi}
\item $\|\mathrm{Id}+T^2\|=1+\|T^2\|$ for every rank-one operator $T$.
\item $\|\mathrm{Id}+x^*\otimes x\|=1+\|x^*\otimes x\|$ for $x^*\in
X^*$, $x\in X$ with $x^*(x)\geq0$.
\item For every $x\in S_X$, $x^*\in S_{X^*}$ with $x^*(x)\geq 0$,
and every $\varepsilon>0$, there exists $y\in S_X$ such that
$$
\|x + y\| > 2-\varepsilon \qquad \text{and} \qquad x^*(y)> 1-\varepsilon.
$$
\end{equi}
\item[(b)] The following are equivalent:
\begin{equi}
\item $\|\mathrm{Id} - T^2\|=1+\|T^2\|$ for every rank-one operator $T$.
\item $\|\mathrm{Id}+x^*\otimes x\|=1+\|x^*\otimes x\|$ for $x^*\in
X^*$, $x\in X$ with $x^*(x)\leq 0$.
\item For every $x\in S_X$, $x^*\in S_{X^*}$ with $x^*(x)\leq 0$,
and every $\varepsilon>0$, there exists $y\in S_X$ such that
$$
\|x + y\| > 2-\varepsilon \qquad \text{and} \qquad x^*(y)> 1-\varepsilon.
$$
\end{equi}
\end{enumerate}
\end{prop}

\begin{proof}
We give our arguments just for item (a), following (b) in the same
way.

$(i)\Rightarrow (ii)$. We consider $x\in X$ and $x^*\in X^*$ with
$x^*(x)>0$, and we write $T=x^*\otimes x$. Let us observe that the
rank-one operator $S=\big(x^*(x)\big)^{-1/2}\,T$ satisfies
$S^{\,2}=T$, so we get
$$
1+\|T\|=1+\|S^{\,2}\|=\|\mathrm{Id}+S^{\,2}\|=\|\mathrm{Id}+T\|.
$$
To finish the argument, it suffices to observe that any $x^*\otimes
x$ with $x^*(x)=0$ can be approximated in norm by operators of the
form $y^*\otimes y$ with $y^*(y)>0$, and that the set of rank-one
operators satisfying \eqref{DE} is closed.

$(ii)\Rightarrow(i)$. Just observe that for every rank-one operator
$T=x^*\otimes x$, it is clear that $T^2=x^*(x)\,T$ and, therefore,
$T^2=y^*\otimes x$ where $y^*=x^*(x)\,x^*$ with
$y^*(x)=\big(x^*(x)\big)^2\geq 0$.

Finally, for the equivalence between $(ii)$ and $(iii)$ just follow
the proof of \cite[Lemma~2.1]{KSSW} or \cite[Lemma~2.2]{WerSur}.
\end{proof}

\section{The new properties}\label{sec:newproperties}

Our aim in this section is to give some remarks concerning the
properties appearing at Theorems \ref{th:4.5} and
\ref{theorem-real-case}.b, i.e.\ we consider a non-trivial
multiplicative subgroup $A$ of $\mathbb{T}$ and study those Banach spaces
for which all rank-one operators satisfy the norm equality
$$
\|\mathrm{Id} + \omega\,T\|=\|\mathrm{Id}+T\|.
$$
In the real case, only one property arises; in the complex case,
there are infinitely many properties and we do not know if all of
them are equivalent.

Our first (easy) observation is that all these properties pass from
the dual of a Banach space to the space. We will see later that the
converse result is not valid.

\begin{remark}\label{remark:dualtospace}
{\slshape Let $X$ be a Banach space and let $\omega\in \mathbb{T}$. Suppose
that the norm equality
$$
\|\mathrm{Id} + \omega\,T\|=\|\mathrm{Id}+T\|
$$
holds for every rank-one operator $T\in L(X^*)$. Then, the same is
true for every rank-one operator on $X$.\ } Indeed, the result
follows routinely by just considering the adjoint operators of the
rank-one operators on $X$.
\end{remark}

Our next results deal with the shape of the unit ball of the Banach
spaces having any of the properties

By a \emph{slice} of a subset $A$ of a normed space $X$ we mean a
set of the form
$$
S(A,x^*,\alpha)=\bigl\{x\in A\ : \ \re x^*(x)>\sup \re x^*(A)-\alpha
\bigr\}
$$
where $x^*\in X^*$ and $\alpha\in \mathbb{R}^+$. If $X$ is a dual space, by
a \emph{weak$^*$-slice} of a subset $A$ of $X$ we mean a slice of
$A$ defined by a weak$^*$-continuous functional or, equivalently, a
weak$^*$-open slice $A$.

\begin{prop}
Let $X$ be a real or complex Banach space and let $A$ be a
non-trivial closed subgroup of $\mathbb{T}$. Suppose that the norm equality
$$
\|\mathrm{Id} + \omega\,T\|=\|\mathrm{Id}+T\|
$$
holds for every rank-one operator $T\in L(X)$ and every $\omega\in
A$. Then, the slices of $B_X$ and the weak$^*$-slices of $B_{X^*}$
have diameter greater or equal than
$$
2-\inf \bigl\{|1+\omega|\ : \ \omega\in A\bigr\}.
$$
\end{prop}

\begin{proof} We give the arguments only for slices of $B_X$, being
the proof for weak$^*$-slices of $B_{X^*}$ completely analogous. We
fix $x^*\in S_{X^*}$ and $0<\alpha<2$. Given $0<\varepsilon<\alpha$, we
take $x\in S_{X}$ such that $\re x^*(x)>1-\varepsilon$ and so, in
particular, $x\in S(B_X,x^*,\alpha)$. We define the rank-one
operator $T=x^*\otimes x$ and observe that
$$
\|\mathrm{Id}+T\|\geq\|x + x^*(x)\,x\|=\|x\|\,|1+x^*(x)|\geq|1+\re
x^*(x)|>2-\varepsilon.
$$
By hypothesis, for every $\omega\in A$ we may find $y\in S_X$ such
that
$$
\|y+\omega\,x^*(y)\,x\|>2-\varepsilon
$$
so, in particular,
$$
|x^*(y)|>1-\varepsilon.
$$
We take $\xi\in \mathbb{T}$ such that $\xi\,x^*(y)=|x^*(y)|$ and we deduce
that $$\xi\,y\in S(B_X,x^*,\alpha).$$ From the inequalities
\begin{align*}
\|\xi y-x\| &=\big\|\xi y +\omega x^*(\xi y) x-\omega x^*(\xi y) x -x\big\|\\
&\geq \big\|\xi y +\omega x^*(\xi y) x\big\| -\big|1+\omega x^*(\xi y)\big| \\
&> 2-\varepsilon -\big|1+\omega |x^*(y)|\big|
\end{align*}
and
\begin{align*}
\big|1+\omega |x^*(y)|\big|&\leq\big|1+\omega
+\omega(|x^*(y)|-1)\big|\\ &\leq |1+\omega|+
\big|1-|x^*(y)|\big|<|1+\omega|+\varepsilon,
\end{align*}
we get
\begin{equation*}
\|\xi y-x\|\geq 2-2\,\varepsilon - |1+\omega|.\qedhere
\end{equation*}
\end{proof}

It is well-known that the unit ball of a Banach space $X$ with the
Radon-Nikod\'{y}m property has many denting points and the unit ball of
the dual of an Asplund space has many weak$^*$-denting points.
Recall that $x_0\in B_X$ is said to be a \emph{denting point} of
$B_X$ if it belongs to slices of $B_X$ with arbitrarily small
diameter. If $X$ is a dual space and the slices can be taken to be
weak$^*$-open, then we say that $x_0$ is a \emph{weak$^*$-denting
point}. We refer to \cite{Bou,D-U} for more information on these
concepts.

\begin{corollary}\label{cor-newpropertynodenting}
Let $X$ be a real or complex Banach space and let $\omega\in
\mathbb{T}\setminus\{1\}$. If the norm equality
$$
\|\mathrm{Id} + \omega\,T\|=\|\mathrm{Id}+T\|
$$
holds for every rank-one operator $T\in L(X)$, then $B_X$ does not
have any denting point and $B_{X^*}$ does not have any $w^*$-denting
point. In particular, $X$ is not an Asplund space and it does not
have the Radon-Nikod\'{y}m property.
\end{corollary}

In view of this result, it is easy to show that the converse of
Remark~\ref{remark:dualtospace} is not true.

\begin{example}
{\slshape The real or complex space $X=C[0,1]$ satisfies that the
norm equality
$$
\|\mathrm{Id}+\omega\,T\|=\|\mathrm{Id}+T\|
$$
holds for every rank-one operator $T\in L(X)$ and every $\omega\in
\mathbb{T}$, in spite of the fact that for every $\omega\in \mathbb{T}\setminus
\{1\}$, there is a rank-one operator $S\in L(X^*)$ such that}
$$
\|\mathrm{Id} + \omega\, S\|<\|\mathrm{Id} + S\|.
$$
Indeed, the first assertion follows from the fact that $X$ has the
Daugavet property; the second one follows from
Corollary~\ref{cor-newpropertynodenting} since the unit ball of
$X^*$ is plenty of denting points.
\end{example}

One of the properties we are dealing with is related to a property
for one-codimensional projections.

\begin{remark}
Consider a Banach space $X$ such that the equality
$$\|\mathrm{Id} + T\|=
\|\mathrm{Id} - T\|
$$
holds true for every rank-one operator $T\in L(X)$. If we apply this
equality to an operator $P$ which is a rank-one projection, we get
$$
\|\mathrm{Id} - P\| \geqslant 2,
$$
i.e. every one-codimensional projection in $L(X)$ is at least of
norm $2$. Such spaces were introduced recently \cite{IvKa} and are
called ``spaces with bad projections''.
\end{remark}

redpajama_set_name: RedPajamaArXiv


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Sometimes it's difficult for us to determine the scope of the consequences that result from our actions–or someone else's. The current presidential administration has made no secret of its distaste for the Islamic faith in addition to immigration in general. With a base made up of far-right wing nut Republicans, what effect is the Trump extremism having on Muslim Americans who have lived here all their lives? It's a combination of good and bad.
Trump (in)famously declared that he thinks "Islam hates us" during his 2016 presidential campaign when asked if Islam is engaged in war with the western world. He claimed that Muslims are rapists. What it really boils down to is his and his supporters' belief that America is for white people only. When confronted with white supremacists in Charlottesville, Trump implied that they were "very fine people." He proceeded to tweet and retweet videos made by nationalistic political organizations with a knack for spewing hate at Muslims.
Not surprisingly, members of these same crazed right wing groups carried out a whopping 71 percent of extremist-related murders here at home. Right wing groups. Not Muslims. White supremacy was directly tied to over half of all extremist-related murders last year. It might as well be Trump policy, because he and his supporters are certainly responsible for it. They should be held accountable as well.
His words have had predictable consequences for the Muslims here at home, most of whom have faced an increased rate of harassment and racially motivated violence ever since. This type of violence skyrocketed about 45 percent during 2017. Naturally, these people are incapable of discriminating only against their targeted group. They've mistakenly identified South Asians, Sikhs, Hindus, and other Arabs as belonging to the Islamic faith during their deranged attacks. We haven't seen levels of violence this high since the year after 9/11.
This wave of extremism has also led to more Muslim-American candidates, who are fighting tooth and nail alongside many other like-minded individuals to take back this country's values. There are more educators and women and varied ethnicities and religious backgrounds in the mix, too. Not only that, but the number of registered voters in these demographics is climbing substantially, and so is the financial situation thanks to donors. The tide is shifting to swallow the right wing. It's about time.
This entry was posted in General on January 30, 2019 by Vicki Black.


If you are upcoming or established IT Company, who is majorly into providing IT services to companies within the IT sector or to other ancillary industries then, you shall always be required to provide customized software development services to your varied list of clientele. Sometimes it may so happen that a few particular clients may have some specific requirements, which may be outside the core competence area of your company. This is when you have to consider outsourcing, in order to fulfill your contractual obligations and the specific need of your client. We are sure you would want to be considered a one-stop solution of IT services by all your clients and not be known as a specific IT service provider and allow this to be a hurdle in the multifaceted growth of your company.
One of the other important reason as to why you should consider outsourcing is that; when it comes to meeting the specific or unique needs of in order to fulfill the IT requirements of your client, it makes bad business sense to invest your precious man-power and divert your financial resources towards unfamiliar technologies and the paraphernalia that come with these technologies. This would also cut down your profits and have an adverse effect on the quality of your team's performance, we should not deny the R&D that goes around when we venture into unknown waters and get into experimental mode. On the other hand backed with a rich experience and deep-rooted technological knowhow Prospect Softech Pvt. Ltd. is ever ready to fulfill your requirements irrespective of the size and budget of the project. Prospect Softech Pvt. Ltd. shall always be at your disposal and ready to deliver at the standards laid down by you. We look forward to having a ever enriching partnership alliances.
Following shall be considered as the strategic advantages of out sourcing.
Outsourcing adds to the list of services that you are able to provide, increases your horizon in terms of client serviceability. Give your company the added edge in terms of competition, increases the efficiency and most importantly gets you benefits in terms of time management and profitability.
Allows you focus in the area of your core competence and keep the cost of operation low by avoiding your hassle to recruit additional manpower for tasks which may not required to utilize in the future.
Our policies of the partner project give you the assurance that the implementation of the project is fully transparent, and proceeds as per your expectations and gives you clear idea about the management steps required to take things in the right direction. All this is made possible because of our completely documented and controlled software development process.


If you scroll to the bottom of the record you'll find plenty of recommended similar articles from other repositories such as the Norwegian Open Research Archives, PLoS and LSE's Open Access repository. There are also Suggested Articles from LSHTM Research Online as a separate tab.
CORE blimey is all we can say!
You can read more about how the CORE Recommender works from their blog.


Just getting started? Review the PharmGrad Quick Start Guide.
Access the PharmCAS.org site, hosted by the American Association of Colleges of Pharmacy (AACP), for additional school-specific admissions requirements and pharmacy resources for applicants.


Read the following introductions for attractions in Wolong. You will know the top attractions there and you can also get to know the city attractions by category.
Wolong Giant Panda Museum exhibits the history, present and future of the giant panda. It gives all detailed information about the living environment, historical evolution, living habits and characteristics, the breeding, reproduction, protection research and natural reserves of the giant panda.
Nestled inside Wolong Nature Reserve with favorable climate and conditions, Wolong Giant Panda Center is a perfect site for preservation and reproduction of Chinese pandas. Before 2008, the center housed 100 wild giant pandas and over 40 captive ones, and had artificially produced 33 pandas in success.


text:
Electronic Arts Signs On For 'E For All'
Author: Chris KohlerChris Kohler
The E For All gaming fan expo, set to rock the face off the Los Angeles Convention Center later this year, has officially announced its second exhibitor: Electronic Arts.
Nobody's saying exactly which of EA's games will be shown off at the convention, which will take place from October 18-21. But since the general public is invited in, expect the lineup to be focused mostly on EA's near-term holiday releases.
In March, E For All announced that Nintendo would be one of the show's "anchor" exhibitors.

redpajama_set_name: RedPajamaCommonCrawl


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Director/Choreographer
Acoustic Voice
Graham Clarke – Artistic Director
My initial training at undergraduate level was as a concert pianist. I was taught by Roy Shepherd, Reader, Faculty of Music, University of Melbourne. My final performance examinations were in singing. I have worked professionally as a singer, accompanist, assistant conductor, a musical director, and as an arranger. I was the first voice teacher engaged by Mr John Sumner, AO, CBE, the founding director of Melbourne Theatre Company. I continued teaching there for four and a half years and also worked as Musical Director for several shows.
This led to being asked by Ms Alisa Piper (now of the Victorian Arts Centre Trust) to give voice workshops for professional users and aspiring actors at Playbox Theatre Company. As a result of accompanying singers at auditions in Her Majesty's Theatre for JC Williamsons, Ms Sue Nattras, AO, engaged me to accompany auditionees for "Evita" with Mr Hal Prince as director. This began a long association with theatre companies.
The Adelaide Festival Centre, for example, engaged me as a company vocal teacher on several occasions, to provide general and rehabilitation vocal techniques. The Edgley group have engaged me as an accompanist for auditions and for concert appearances. Highlights of this time also include working as audition pianist and vocal advisor on the casting of the national touring company for "Joseph", "Oklahoma!" with Jamie Hammerstein as director, "Barnum" with Baayork Lee as director, "The Sound of Music", Best Little Whorehouse", and "Westside Story" conducted by Dobbs Frank.
I have been a musical consultant and performing artist (including pianist) for companies such as Crawford Productions (television) and the musical director for many music theatre events including concerts, cabarets, theatre restaurants, recording studio sessions, and theatrical productions such as the national touring companies of "They're Playing Our Song" and "Peach Melba". My students have taught singing and voice production to fellow professionals throughout the world, including England, Japan (alongside Jo Estill), and in America at Shakespeare and Company (the Linklater group).
This demonstrates that they are accepted as world class educators who were trained in my studio. As further evidence of my students' international recognition, my private practice student roll includes names (and skill ranges) as diverse as:
Kylie Minogue (pop)
Helen Noonan (opera)
Daryl Braithwaite (pop)
Stephen Housden ('Little River Band'),
David Hobson (Opera Australia, 'It Takes Two', and 'Dancing with the Stars')
Evelyn Krape (comedian)
Beverley Dunn (actor)
Daniel MacPherson (UK 'Godspell' 'City Homicide')
Jonathan Hardy (actor)
Alan Dale ('The OC' 'Ugly Betty')
Jesse Spencer ('House')
Former students of mine are now on staff at the Melbourne Theatre Company, Playbox, the Victorian Arts Centre, and teach in major schools at both secondary and tertiary levels. Many students perform internationally in centres such as London's West End, D'Oly Carte Opera, the professional jazz scene in Paris, the German pop scene and operatic houses, as well as theatre and cabaret in Asia. My students have appeared in West End productions such as "Kiss Me Kate" (Brenda Clarke, understudy for Kate); "Sugar Babies" (Joy Bishop, ensemble), and Janice Torrens understudied Elaine Page in "Anything Goes", to name just several.
Helen Noonan's one-woman show, "Recital", has played some of the major houses including Lincoln Center, Covent Garden, and toured to Tokyo and Spain with critical success. In addition to achieving international success directly from my studio as performers, some of my students have taught at tertiary institutions in England (for example, Junewyn Jones), USA (for example, Beth Child at 'Shakepspeare and Company' and The University of South Carolina), and Japan (for example, Denise Wharmby) as well as presenting masterclasses for professional voice users throughout Australia (for example, Anna McCrossin-Owen). They have taught at institutions such as Rusden College, Victorian College of Arts, Monash University, Ballarat Academy of Performing Arts, and are employed as consultants for the Melbourne Theatre Company (for example, as accent coach, Michael Daley).
Commercially released recordings of my students include pop, folk, musical comedy, contemporary art song, and the spoken word. I served the Australian National Association of Teachers of Singing (ANATS) as their national Secretary and Editor for one year. My four-year membership on the Victorian Chapter Committee of ANATS allowed me to make many contributions by organising events such as weekend workshops, student seminars, teacher seminars, and professional development workshops.
In 1994, I was the Principal examiner, singing, for the Melba Memorial Conservatorium of Music, and I was appointed by the University of Melbourne as an examiner for a Master of Music thesis in 1999. I coached and then played piano accompaniment in 2013 for a former student's CD recording of unfamiliar music theatre songs by Berlin, Gershwin, and Porter. I founded Acoustic Voice as a performance opportunity in 2012.
I devised and accompanied three concerts with former students in the historic venue of Melbourne City Conference Centre and the recently restored facility of the Ballarat Mechanics Institute to great acclaim. Then, we gave the concerts, 'Morning Melodies with Acoustic Voice' in September, 2013, and 'Singing the Sacred' in December, 2013 in the Northcote Town Hall. In 2012,
I worked as a mentor and consultant in 2012 for the original music theatre book and several songs (as examples of the whole), Melbourne Cup—The Musical. Also, I worked as a mentor and musician for the CD recording of 17 original songs that make up the national touring cabaret show (commencing October, 2011), What's Love Got To Do With It? I served on the Australian Ballet Society Committee, May 2011 until 2013. I was one of the team of pianists for the Summer School, 2014 and 2015 of The Australian Ballet School.
Food science, in particular food combining for well-being and health.
Recipe collection and development for athletes including performers.
Researching non-drug clinical studies that test nutritional science.
Digital sound technologies. Ahlbourn-Galanti used my designs in their Series B instruments.
I counter the constant sitting involved in my work by a program of training at a gym.
Twice weekly I attend adult ballet classes for personal growth and discipline.
Diploma in Music, University of Melbourne, granted in 1976.
Master of Music (Preliminary), University of Melbourne, granted in 1990.
Doctor of Philosophy, Monash University, granted in 2000.
For more information on The Merman Theater Songbook please contact Graham at:
graham@acousticvoice.net


Continued digitization of societal and industrial processes produces vast volumes of data. Meanwhile, the transportation sector is facing profound disruption due to developments such as electric cars, autonomous cars, and sharing economy. My project will study how to best utilize massive trajectory data that describes the driving of cars to support the increasing needs for greener, cheaper, more efficient, and more predictable transportation. More specifically, my project will establish a dynamic and uncertain network model by analyzing massive trajectory data, which enables highly accurate, efficient, and scalable travel cost (e.g., travel time and greenhouse gas emission) modeling and path finding.
My grandfather bought a computer for me and my cousin in the early 90s. I played many computer games and also made some simple programs such as drawing triangles on the computer, which I really enjoyed. During my university education on computer science, I started to realize that computers can help people solve very complex and important problems, which motivated me to pursue a Ph.D. in computer science. Now, in the era of big data, my research focuses on computer algorithms that are able to learn knowledge from large amounts of data to help people solve problems that can benefit society.
Many transportation applications rely on path finding algorithms on weighted graphs, where weights are in the form of static values, which represent, e.g., distances or average travel times of roads. However, traffic is often dynamic and uncertain, rendering the existing weighted graph model unrealistic. The challenges are (1) how to build a new model that is able to capture dynamic and uncertain traffic conditions from massive trajectory data; (2) how to utilize the new model to enhance the efficiency and quality of path finding algorithms. From an application perspective, my project has potential to profoundly change the foundation for vehicular path finding. From a scientific perspective, my project will extend graph theory by opening a new direction on stochastic path finding.
Both the EU and Denmark are very ambitious on making transportation green. The application aspect of my project is well aligned with that ambition. My project will also bring solutions relating to dynamic, uncertain networks into the open domain and thus will facilitate open innovation in enabling greener transportation. Due to the data-driven nature of my project, it will also support emerging data science education activities. Further, the project will improve the internationalization of Danish research by collaborating with researchers from U.S., U.K., Australia, and Singapore.
The Sapere Aude programme will greatly boost my research career as an independent researcher. The grant enables me to establish and consolidate a research group on data analytics and machine learning, which offers me a very good foundation to achieve ground-breaking research outcomes with substantial societal benefits in the years to come. In addition, it offers good opportunities to enhance my research network through collaborations with leading international research groups.
Currently, I live in Aalborg.
I like soccer, basketball, and rock music. Now both my wife and I spend most of our spare time with our daughter and son.
High school affiliated to Shaanxi Normal University (also known as Xi'an NO. 84 high school), Xi'an, Shaanxi, China.


text:
Applegreen ramps up operations in New Jersey
/ 25th April 2022 /
George Morahan
Irish petrol station operator Applegreen will invest $126m in its operations in New Jersey, opening a new headquarters and adding hundreds of jobs in the US state over the next several years.
The company, which is the largest forecourt retailer and service provider in the US, operating 250 sites across 18 states, plans to redevelop all its 21 service stations in New Jersey in a move that will create 800 jobs.
A further 100 jobs in finance, administration, development and logistics will be added with the opening of the offices in Glen Rock, and New Jersey Governor Phil Murphy travelled to Dublin to take part in the announcement.
"As a proud Irish-American, I am thrilled to be in Dublin and joined by Applegreen, Ireland's foremost service plaza operator, to announce the relocation of its US Travel Plaza Headquarters to New Jersey," Murphy said.
"Applegreen is a world-class travel infrastructure partner who will generate significant economic growth for the state and bring innovation and expertise to the redevelopment of New Jersey's 21 on-highway services plazas.
"We welcome the hundreds of new jobs that Applegreen will bring to New Jersey and look forward to building a strong partnership that drives our state's innovation economy forward."
Applegreen, which was taken into private ownership last March by its founders Bob Etchingham and Joe Barrett, and Blackstone Infrastructure, will also install infrastructure for electric vehicles across its branches in the north-east of the US.
New Jersey Governor Phil Murphy - Applegreen will invest $126m in its operations in New Jersey. Pic Julien Behal
"Our investment in a new Travel Plaza Headquarters in the town of Glen Rock and rebuilding of 21 service plazas underscores our commitment to the state of New Jersey," said Etchingham, CEO of Applegreen.
"We look forward to continuing to be a good partner to New Jersey and to helping the Garden State maintain its national transport connected competitiveness for many years to come."
"We're delighted to play a lead role in updating New Jersey's infrastructure to better serve the needs of modern [travellers] and state residents," said Sebastien Sherman, senior managing director of Blackstone Infrastructure.
"Governor Murphy's leadership is a national model for effective public-private partnerships. Blackstone Infrastructure looks forward to deepening our ties to the Garden State, working with government agencies and local stakeholders to deliver high-quality infrastructure benefiting local communities and the environment for decades to come."
Elizabeth Pierce, president of Applegreen's US operations, said New Jersey was "an ideal location" for the company's new headquarters and that the creation of hundreds of jobs would deliver "a significant boost to the New Jersey economy," calling the announcement "a genuine Irish-American win-win."
Applegreen acquired the 21 New Jersey services plazas last summer, and the company now manages a workforce of 5,000 in the US. The company operates 630 sites across Ireland, the UK and the US and employs 14,000 people globally.
The Irish firm said that it currently operates 91 "travel plazas," its term for large forecourt retail stores and concessions located along motorways, across New Jersey, New York, Connecticut, Pennsylvania, Delaware, Maine, Massachusetts, Ohio, Indiana and West Virginia.
According to its most recently available accounts, Applegreen saw revenues fall 24% to €2.3bn and adjusted earnings drop 30.8% to €97.2m in 2020, with the US accounting for revenues of €426.4m and Ireland making up €709.4m.
Healy Communications appoints former Sky News journalist Enda Brady
Impact of sport on women in business


text:
Tanabe Helps ENA Adopt Sickle Cell Pain Guidelines
Paula Tanabe was standing in a nondescript convention center ballroom in Texas on an otherwise unremarkable Sunday morning in late September 2019. Looking back now, she describes that moment as "the highlight of my career." What happened in that room, at the General Assembly of the Emergency Nurses Association (ENA), will change the lives of thousands of patients with sickle cell disease for years to come.
That morning was the culmination of more than a decade of research, planning and advocacy. Tanabe, the Laurel Chadwick Distinguished Professor of Nursing at Duke University School of Nursing, has been working with colleagues across the state and the nation to solve one of the most frustrating problems facing sickle cell patients: helping them receive treatment for severe pain in the emergency department. At that September meeting, the ENA took an enormous step to address that problem when it voted to adopt a resolution, drafted by Tanabe and two other emergency department nurses from North Carolina, to disseminate evidence-based guidelines and education on sickle cell treatment to all EDs in the United States.
Sickle cell disease is still as misunderstood as it is debilitating. This genetic disease affects the production of hemoglobin, resulting in misshapen red blood cells. Instead of carrying much-needed oxygen throughout the body, these "sickle cells" can get stuck in blood vessels. Organs and tissues are choked off in an enormously painful episode called a "vaso-occlusive crisis." People with sickle cell disease can suffer through several of these crises a year, and their only recourse for pain management often comes from receiving opioids in the emergency department.
Tanabe saw the problem first-hand when she was working on pain management as an emergency department nurse and encountered patients with sickle cell disease. "As an ED nurse," she noted, "all you see is that they come in with this horrible pain, and they need opioids – often a lot of them – to manage that pain. Patients can't really predict when those crises are going to come, and when they do, the pain is horrible, and they need relief. So they usually have to go to an emergency department."
It seems simple enough: a patient comes into the ED with crippling pain, so you give them the pain medicines they require. But Tanabe was seeing a complicating factor with deep social roots: approximately 100,000 Americans have sickle cell disease, and the majority of them are of African descent. "If you're a 20-year-old black male coming into the ED for opioids, the stigma and bias against the disease in the population is huge," said Tanabe. "Patients are being told that they're addicted to opioids and that they're just drug-seeking. ED providers are trying to do the right thing, but that bias is still there."
Tanabe and her team knew that the key to dispelling this bias was making sure that all ED providers had reliable, evidence-based pain management guidelines for sickle cell patients. They surveyed the latest research, conducted new research of their own, and spoke with experts and patients from all over. "We learned more about the disease, and we met more patients to hear their stories and really understand what their life was like," she said. Two sickle cell patients in particular, Darryl Smith and Gail Aiken, made immeasurable contributions to Tanabe's research over these years.
After more than ten years of nationwide research, advocacy and education, it was time to create the ENA resolution. Tanabe worked with Emilia Frederick of UNC-Chapel Hill and Angie Alexander of Atrium Health in Charlotte to craft a comprehensive resolution that would transform the treatment of patients with sickle cell disease in emergency departments across the United States. The resolution lays out background information on sickle cell, and charges the Emergency Nurses Association to widely disseminate the evidence-based recommendations and other important education.
Finally, Tanabe and her colleagues were ready to bring the resolution to the ENA General Assembly in Austin. She expected the worst. "For the last ten years, all I've heard is negative, negative, negative," she said. But to her shock, nearly all of the comments were in support of the resolution. She had been careful to lay the groundwork by reaching out to hundreds of colleagues at other state ENAs beforehand, but what happened next far surpassed her expectations. After a negative comment or two, one member got up and galvanized the crowd: "We should be embarrassed. Why are we even having this discussion? These individuals have severe pain, and there are evidence-based guidelines to treat it. What are we talking about?" According to Tanabe, that was the final straw. "We didn't have to say anything else, and then the vote was 87.6% for the resolution."
Even as Tanabe and her team celebrated this enormous achievement, their joy was tempered because two of their own weren't there to celebrate with them. Darryl Smith and Gail Aiken had both passed away in July 2018 of complications from sickle cell disease. In an almost mystical coincidence, Aiken's family and friends were holding a dedication service at her church the same morning the ENA resolution came up for a vote. Tanabe had become close with Aiken through their research and advocacy work, and regretted that she couldn't be present at the service to speak about her friend. She sent DUSON Assistant Professor Mariam Kayle in her stead, and texted her with the good news only moments before. "When she got up to speak at the service, she was able to say that the resolution just passed," said Tanabe. "The whole church clapped – it was all really emotional."
With this transformative resolution in hand, Paula Tanabe now wants to use it to ensure that thousands of sickle cell patients get the pain treatment they need no matter which ED they walk into. "I think this will be huge," she declares, "because when your professional organization says, 'this is the evidence-based practice, this is what you should do,' then you'll make a change. The goal is to have the guidelines be as strong as possible, with great evidence to back them up. And then providers will use them and believe patients, and there should be better patient outcomes."
Paula Tanabe