text:
\section{Introduction}

The purpose of this paper is to prove the weighted Hardy inequality for the quadratic form of the Landau Hamiltonian and for the magnetic Baouendi-Grushin operator with Aharonov-Bohm type magnetic field. In Part II of this paper we investigate and present the corresponding Caffarelli-Kohn-Nirenberg inequalities for the Landau Hamiltonian and for the Baouendi-Grushin operator, with and without magnetic fields.

The classical Hardy inequality for functions $f\in C_{0}^{\infty}(\mathbb R^{n}\backslash0)$ is
\begin{equation}\label{Intro_1}
\int_{\mathbb R^{n}}|\nabla f(w)|^{2}dw\geq \left(\frac{n-2}{2}\right)^{2}\int_{\mathbb R^{n}}\frac{|f(w)|^{2}}{|w|^{2}}dw, \quad n\geq3,
\end{equation}
where the constant $\left(\frac{n-2}{2}\right)^{2}$ is sharp but not attained. There exists a large literature concerning different versions of Hardy's inequalities and their applications. However, since we are interested in the inequalities associated with the Landau Hamiltonian and with the Baouendi-Grushin operator, let us only recall known results in these directions.

\subsection{Baouendi-Grushin operator}
\label{SEC:iGR}

The Hardy inequality \eqref{Intro_1} has been generalised for Baouendi-Grushin vector fields by Garofalo \cite{G},
\begin{equation}\label{Intro_2}
\int_{\mathbb R^{n}}(|\nabla_{x}f|^{2}+|x|^{2\gamma}|\nabla_{y}f|^{2})dxdy\geq \left(\frac{Q-2}{2}\right)^{2}
\int_{\mathbb R^{n}}\left(\frac{|x|^{2\gamma}}{|x|^{2+2\gamma}+(1+\gamma)^{2}|y|^{2}}\right)|f|^{2}dxdy,
\end{equation}
where $x\in\mathbb{R}^{m}$, $y\in\mathbb{R}^{k}$ with $n=m+k$, $m,k\geq1$, $\gamma\geq0$, $Q=m+(1+\gamma)k$ and
$f\in C_{0}^{\infty}(\mathbb{R}^{m}\times\mathbb{R}^{k}\backslash\{(0,0)\})$. Here, $\nabla_{x}f$ and $\nabla_{y}f$ are the gradients of $f$ in the variables $x$ and $y$, respectively. The inequality \eqref{Intro_2} recovers \eqref{Intro_1} when $\gamma=0$.

Let us put this result in perspective.
Let $z=(x_{1},...,x_{m}, y_{1},...,y_{k})=(x,y)\in \mathbb{R}^{m}\times\mathbb{R}^{k}$ with $k,m\geq1$, $k+m=n$ and $\gamma\geq0$. Let us consider the vector fields
$$X_{i}=\frac{\partial}{\partial x_{i}}, \;i=1,...,m, \;\;\; Y_{j}=|x|^{\gamma}\frac{\partial}{\partial y_{j}}, \;j=1,...,k.$$
The corresponding sub-elliptic gradient, which is the $n$ dimensional vector field, is then defined as
\begin{equation}\label{subgrad}
\nabla_{\gamma}:=(X_{1},...,X_{m}, Y_{1},...,Y_{k})=(\nabla_{x}, |x|^{\gamma}\nabla_{y}).
\end{equation}
The Baouendi-Grushin operator on $\mathbb{R}^{m+k}$ is defined by
\begin{equation}\label{Grush_op}
\triangle_{\gamma}=\sum_{i=1}^{m}X_{i}^{2}+\sum_{j=1}^{k}Y_{j}^{2}=\triangle_{x}+|x|^{2\gamma}\triangle_{y}=\nabla_{\gamma}\cdot \nabla_{\gamma},
\end{equation}
where $\triangle_{x}$ and $\triangle_{y}$ are the Laplace operators in the variables $x\in \mathbb{R}^{m}$ and $y\in \mathbb{R}^{k}$, respectively.  The Baouendi-Grushin operator for an even positive integer $\gamma$ is a sum of squares of $C^{\infty}$ vector fields satisfying
H\"{o}rmander condition
$${\rm rank} \;{\rm Lie} [X_{1},...,X_{m}, Y_{1},...,Y_{k}]=n.$$
We can define on $\mathbb{R}^{m+k}$ the anisotropic dilation attached to $\triangle_{\gamma}$ as
$$\delta_{\lambda}(x,y)=(\lambda x, \lambda^{1+\gamma} y)$$
for $\lambda>0$, and the homogeneous dimension with respect to this dilation is
\begin{equation}\label{hom_dim}
Q=m+(1+\gamma)k.
\end{equation}
A change of variables formula for the Lebesgue measure implies that
$$d\circ\delta_{\lambda}(x,y)=\lambda^{Q}dxdy.$$
It is easy to check that
$$X_{i}(\delta_{\lambda})=\lambda\delta_{\lambda}(X_{i}),\quad Y_{i}(\delta_{\lambda})=\lambda\delta_{\lambda}(Y_{i}),$$
and hence
$$\nabla_{\gamma}\circ \delta_{\lambda}=\lambda\delta_{\lambda} \nabla_{\gamma}.$$
Let $\rho(z)$ be the corresponding distance function from the origin for $z=(x,y)\in \mathbb{R}^{m}\times \mathbb{R}^{k}$:
\begin{equation}\label{dist}
\rho=\rho(z):=(|x|^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2})^{\frac{1}{2(1+\gamma)}}.
\end{equation}
By a direct calculation one obtains
\begin{equation}\label{formula}|\nabla_{\gamma}\rho|=\frac{|x|^{\gamma}}{\rho^{\gamma}}.
\end{equation}
The described setup may be thought of as a special case of the setting of homogeneous groups, see e.g. \cite{FR}.

The weighted $L^{p}$-versions of \eqref{Intro_2} have been obtained  by D'Ambrosio \cite{A2}:
Let $\Omega\subset\mathbb R^{n}$ be an open set. Let $p>1$, $k,m\geq1$, $\alpha, \beta\in\mathbb{R}$ be such that $m+(1+\gamma)k>\alpha-\beta$ and $m>\gamma p-\beta$. Then for every $f\in D_{\gamma}^{1,p}(\Omega, |x|^{\beta-\gamma p}
\rho^{(1+\gamma)p-\alpha})$ we have
\begin{equation}\label{Amb1}
\int_{\Omega}|\nabla_{\gamma}f|^{p}|x|^{\beta-\gamma p}\rho^{(1+\gamma)p-\alpha}dxdy\geq\left(\frac{Q+\beta-\alpha}{p}\right)^{p}\int_{\Omega}|f|^{p}\frac{|x|^{\beta}}{\rho^{\alpha}}dxdy,
\end{equation}
where $D_{\gamma}^{1,p}(\Omega,\omega)$ denotes the closure of $C_{0}^{\infty}(\Omega)$ in the norm $\left(\int_{\Omega}|\nabla_{\gamma}f|^{p}\omega dzdy\right)^{1/p}$ for $\omega\in L^{1}_{loc}(\Omega)$ with $\omega>0$ a.e. on $\Omega$.

If $0\in\Omega$, then the constant $\left(\frac{Q+\beta-\alpha}{p}\right)^{p}$ in \eqref{Amb1} is sharp.
The inequality \eqref{Amb1} has also been established in \cite{Kombe15}, and in \cite{ShY12} for $\Omega=\mathbb R^{n}$ with sharp constant. Moreover, in \cite{ShY12}, a Hardy-Rellich type inequality for the Baouendi-Grushin operator is obtained in $L^{2}$ with sharp constant:
$$\left(\frac{Q-\alpha-2}{2}\right)^{2}\int_{\mathbb R^{n}}|\nabla_{\gamma}f|^{2}
\rho^{\alpha}\leq\int_{\mathbb R^{n}}|\triangle_{\gamma}f|^{2}\rho^{\alpha+2}|\nabla_{\gamma}\rho|^{-2},$$
where $p>1$, $\frac{2-Q}{3}\leq \alpha \leq Q-2$, $f\in C^{\infty}_{0}(\mathbb R^{n}\backslash\{0\})$.

The inequalities of this type have been also studied for some sub-elliptic operators of different types (see e.g. \cite{G}, \cite{GL}, \cite{A1}, \cite{A2}, \cite{NCH}, \cite{K} and \cite{DGN}).
For Hardy and Caffarelli-Kohn-Nirenberg inequalities on more general homogeneous Carnot groups and the literature review including the Heisenberg group we refer to
\cite{Ruzhansky-Suragan:JDE,Ruzhansky-Suragan:Layers}, for the anisotropic versions of the usual $L^2$ and $L^p$ Cafarelli-Kohn-Nirenberg inequalities we refer to \cite{Ruzhansky-Suragan:L2-CKN} and \cite{ORS16}, respectively, and for Hardy inequalities for more general sums of squares of vector fields we refer to \cite{Ruzhansky-Suragan:squares}.

Here we obtain the following refinement of Hardy inequalities for the Baouendi-Grushin operator:

\begin{itemize}
\item ({\bf Weighted refined Hardy inequalities for Grushin operators}) Let $(x,y)=(x_{1},...,x_{m}, y_{1},...,y_{k})\in \mathbb{R}^{m}\times\mathbb{R}^{k}$ with $k,m\geq1$, $k+m=n$. Let $Q+\alpha_{1}-2>0$ and $m+\gamma \alpha_{2}>0$. Then we have the following Hardy type inequality for all complex-valued functions $f\in C_{0}^{\infty}(\mathbb R^{n}\backslash\{0\})$
\begin{multline}\label{EQ:rad}
\qquad \quad \int_{\mathbb R^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\left(
\left|\frac{d}{d|x|}f\right|^{2}+|x|^{2\gamma}|\nabla_{y}f|^{2}\right)dxdy \\
\geq \left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}
\int_{\mathbb R^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy,
\end{multline}
where the constant $\left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}$ is sharp.
\end{itemize}
Already in the absence of weights, i.e. for $\alpha_1=\alpha_2=0$, the estimate \eqref{EQ:rad} is new. Moreover, it is also new as a family of inequalities in the case $\gamma=0$: for $k=0$ it reduces to the classical Hardy inequality \eqref{Intro_1}, while for $m=0$ it reduces to the radial version established in  \cite{IIO:Lp-Hardy}, see also \cite{MOW:Hardy-Hayashi} (always for $\gamma=0$). We note that since we can estimate
$\left|\frac{d}{d|x|}f\right|\leq |\nabla_x f|$, inequality \eqref{EQ:rad} also gives a refinement to the inequality \eqref{Amb1} for $p=2$.

The estimate \eqref{EQ:rad}, in addition to its own interest, will play an important role in the derivation of estimates for magnetic operators.

\subsection{Magnetic Baouendi-Grushin operator}
\label{SEC:iGRm}

In \cite{LW} and \cite{AL11}, Hardy inequalities for some magnetic forms were obtained. For example, in \cite{AL11} for the quadratic form of the following magnetic Grushin operator
$$G_{\mathcal{A}}=-(\nabla_{G}+i\beta\mathcal{A}_0)^{2},$$
the following Hardy inequality for $-\frac{1}{2}\leq \beta \leq \frac{1}{2}$ was proved:
\begin{equation}\label{EQ:AL}
 \int_{\mathbb{R}^{3}}|(\nabla_{G}+i\beta\mathcal{A}_0)f|^{2}dzdt\geq (1+\beta^{2})\int_{\mathbb{R}^{3}}\frac{|z|^{2}}{d^{4}}|f|^{2}dzdt,
\end{equation}
where
$$\mathcal{A}_0=(\mathcal{A}_{1},\mathcal{A}_{2},\mathcal{A}_{3},\mathcal{A}_{4})=\left(-\frac{\partial_{y}d}{d},
\frac{\partial_{x}d}{d}, -2y\frac{\partial_{t}d}{d}, 2x \frac{\partial_{t}d}{d}\right),$$
$\nabla_{G}=(\partial_{x}, \partial_{y}, 2x\partial_{t}, 2y\partial_{t})$ with $z=(x,y)$, $|z|=\sqrt{x^{2}+y^{2}}$, $\beta\in\mathbb{R}$ is a \enquote{flux} and
$d(z,t)=(|z|^{4}+t^{2})^{\frac{1}{4}}$ is the Kaplan distance.

The following results extend the estimate \eqref{EQ:AL}.
While the most physical setting is $y\in \mathbb R^1$, we can obtain the results for any $y\in\mathbb R^k$.

\begin{itemize}
\item ({\bf Hardy inequality for the magnetic Baouendi-Grushin operator})
Let $(x,y)=(x_{1}, x_{2}, y)\in \mathbb{R}^{2}\times\mathbb{R}^{k}$. Let $\alpha_{1},\alpha_{2}, \beta\in\mathbb{R}$ be such that $\alpha_{1}+k(\gamma+1)>0$ and $\alpha_{2}\gamma+2>0$.
Let us define the
Aharonov-Bohm type magnetic field
\begin{equation}\label{another_mag2}
\mathcal{\widetilde{A}}:=
\left(-\frac{\partial_{x_{2}} \rho}{\rho}, \frac{\partial_{x_{1}} \rho}{\rho},-\frac{|x|^{\gamma}}{\sqrt{2}}
\frac{\nabla_{y} \rho}{\rho}, \frac{|x|^{\gamma}}{\sqrt{2}}
\frac{\nabla_{y} \rho}{\rho}\right),
\end{equation}
and the corresponding gradient
\begin{equation}\label{another_mag3}
\widetilde{\nabla_{\gamma}}=\left(\frac{\partial}{\partial x_{1}}, \frac{\partial}{\partial x_{2}}, \frac{|x|^{\gamma}}{\sqrt{2}}\nabla_{y}, \frac{|x|^{\gamma}}{\sqrt{2}}\nabla_{y}\right).
\end{equation}
Then for any complex-valued function $f\in C^{\infty}_{0}(\mathbb{R}^{2+k}\backslash\{0\})$ we have the following weighted Hardy inequality for the magnetic Baouendi-Grushin operator
$$\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}|(\widetilde{\nabla_{\gamma}}+i\beta\mathcal{\widetilde{A}})f|
^{2}dxdy
\geq\left(\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}\right)$$
$$
\times\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|
^{\alpha_{2}}\frac{|x|^{2\gamma}}{\rho^{2\gamma+2}}|f|^{2}dxdy,
$$
where the constant $\left(\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}\right)$ is sharp (so that the constant in \eqref{EQ:AL} is actually also sharp).
Moreover, we have the estimate for the remainder term in this inequality:
$$\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}|(\widetilde{\nabla_{\gamma}}+i\beta\mathcal{\widetilde{A}})f|
^{2}dxdy
\geq\left(\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}\right)$$
$$
\times\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|
^{\alpha_{2}}\frac{|x|^{2\gamma}}{\rho^{2\gamma+2}}|f|^{2}dxdy
$$
$$+\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|
^{\alpha_{2}}\frac{|f|^{2}-|f_{0}(|x|,y)|^{2}}{|x|^{2}}dxdy,
$$
where $f_{0}(|x|,y)=\frac{1}{2\pi}\int_{0}^{2\pi}f(|x|,\phi,y)d\phi$ with $(|x|,\phi)$ being the polar decomposition of $x$, so that the last term in the above inequality is nonnegative.
\item
({\bf Uncertainty type principle}) Let $(x,y)=(x_{1}, x_{2}, y)\in \mathbb{R}^{2}\times\mathbb{R}^{k}$. Let $\alpha_{1}, \alpha_{2}, \beta\in\mathbb{R}$ be such that $\alpha_{1}+k(\gamma+1)>0$ and $\alpha_{2}\gamma+2>0$. Then for any complex-valued function $f\in C^{\infty}_{0}(\mathbb{R}^{2+k}\backslash\{0\})$ we have
$$\|\rho^{\frac{\alpha_{1}}{2}}|\widetilde{\nabla_{\gamma}}\rho|^{\frac{\alpha_{2}}{2}}(\widetilde{\nabla_{\gamma}}+
i\beta\mathcal{\widetilde{A}})f\|_{L^{2}(\mathbb{R}^{2+k})}
\|f\|_{L^{2}(\mathbb{R}^{2+k})}$$
$$\geq\left(\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}\right)^{\frac{1}{2}}
\int_{\mathbb{R}^{2+k}}\rho^{\frac{\alpha_{1}}{2}}|\widetilde{\nabla_{\gamma}}\rho|^{\frac{\alpha_{2}}{2}}
\frac{|x|^{\gamma}}{\rho^{\gamma+1}}|f|^{2}dxdy.
$$
\item({\bf Magnetic Baouendi-Grushin operator with constant magnetic field})
In Remark \ref{Grushin_constant_mag}, we also give inequalities for the magnetic Baouendi-Grushin operator on $\mathbb{C}^{n}$ with the constant magnetic field
$$\mathcal{L}_{G}=\sum_{j=1}^{n}((i\partial_{x_{j}}+\psi_{1,j}(y_{j}))^{2}+(i|x|^{\gamma}\partial_{y_{j}}+\psi_{2,j}(x_{j}))^{2}).$$
\end{itemize}

\subsection{Landau Hamiltonian}
\label{SEC:introLH}

Let us recall that the Landau Hamiltonian (or the twisted Laplacian) on $\mathbb{C}^{n}$ is defined as
\begin{equation}\label{twist1}
\mathcal{L}=\sum_{j=1}^{n}\left[\left(i\partial_{x_{j}}+\frac{y_{j}}{2}\right)^{2}+
\left(i\partial_{y_{j}}-\frac{x_{j}}{2}\right)^{2}\right].
\end{equation}
Setting
$$\widetilde{X}_{j}=\partial_{x_{j}}-\frac{1}{2}iy_{j}\;\textrm{ and }\; \widetilde{Y}_{j}=\partial_{y_{j}}+\frac{1}{2}ix_{j},$$
we have
\begin{equation}\label{twist7}
\mathcal{L}=-\sum_{j=1}^{n}(\widetilde{X}_{j}^{2}+\widetilde{Y}_{j}^{2}).
\end{equation}
The twisted Laplacian can be also written as $\mathcal{L}=-\triangle+\frac{1}{4}(|x|^{2}+|y|^{2})+iN$, where
\begin{equation}\label{N}
N=\sum_{j=1}^{n}(y_{j}\partial_{x_{j}}-x_{j}\partial{y_{j}})
\end{equation}
is the rotation field.
Let $\nabla_{\mathcal{L}}$ be the gradient operator associated with $\mathcal{L}$:
\begin{equation}\label{twist2}
\nabla_{\mathcal{L}}f=(\widetilde{X}_{1}f,...,\widetilde{X}_{n}f, \widetilde{Y}_{1}f,...,\widetilde{Y}_{n}f).
\end{equation}
Let $W_{\mathcal{L}}^{1,2}(\mathbb{C}^{n})$ be the Sobolev space defined by
\begin{equation}\label{twist_def}W_{\mathcal{L}}^{1,2}(\mathbb{C}^{n})=\{f\in L^{2}(\mathbb{C}^{n}):\widetilde{X}_{j}f, \widetilde{Y}_{j}f\in L^{2}(\mathbb{C}^{n}), 1\leq j\leq n\}.
\end{equation}

In the recent paper \cite{ARS17}, a version of the Hardy inequality for the twisted Laplacian with Landau Hamiltonian magnetic field was established for {\em real-valued} functions $f\in W_{\mathcal{L}}^{1,2}(\mathbb{C}^{n})$, namely, the inequality
$$
\frac{1}{4}\int_{\mathbb{C}^{n}}|f|^{2}\omega(z)dz\leq\int_{\mathbb{C}^{n}}|\nabla_{\mathcal{L}}f|^{2}dz,
$$
with the weight
$$
\omega(z)=\left(\frac{|\nabla_{\mathcal{L}}\mathbb{E}|^{2}}{\mathbb{E}^{2}}+\frac{|z|^{2}}{4}\right),
$$
where $\mathbb{E}$ is a fundamental solution to the twisted Laplacian on $\mathbb{C}^{n}$.

In this paper we obtain the versions of Hardy inequalities for the Landau Hamiltonian for both {\em complex-valued} and {\em real-valued functions}, as well as estimates for the remainders.

In fact, we obtain also results for more general operators $\widetilde{\mathcal{L}}_\psi$
of the form
$$\widetilde{\mathcal{L}}_\psi=\sum_{j=1}^{n}\left[\left(i\partial_{x_{j}}+\psi(|z|)y_{j}\right)^{2}+
\left(i\partial_{y_{j}}-\psi(|z|)x_{j}\right)^{2}\right],$$
where $\psi(|z|)$ is a radial real-valued differentiable function, $z=(x,y)$.
Setting
$$\check{{X}_{j}}=\partial_{x_{j}}-i\psi(|z|)y_{j}\;\textrm{ and }\; \check{Y}_{j}=\partial_{y_{j}}+i\psi(|z|)x_{j},$$
we write
\begin{equation}\label{twist3_1_2}
\widetilde{\nabla_{\mathcal{L}_\psi}}f=(\check{X}_{1}f, \ldots, \check{X}_{n}f, \check{Y}_{1}f, \ldots, \check{Y}_{n}f).
\end{equation}
We note that for $\psi(|z|)=\frac12$ we recover the classical Landau Hamiltonian, i.e.
$$ \widetilde{\mathcal{L}}_{1/2}=-\mathcal{L} \textrm{ and }
\widetilde{\nabla_{\mathcal{L}_{1/2}}}=\nabla_{\mathcal{L}}.$$
As usual, we will identify $\mathbb{C}\cong\mathbb{R}^{2}$.
\begin{itemize}
\item ({\bf Hardy inequalities for the Landau-Hamiltonian $\widetilde{\mathcal{L}}_\psi$})
Let $\psi=\psi(|z|)$ be a radial real-valued function such that $\psi\in L^{2}_{loc}(\mathbb{C}\backslash\{0\})$.
For a function $f$ we will  use the notation $f_{0}(|z|):=\frac{1}{2\pi}\int_{0}^{2\pi}f(|z|,\phi)d\phi$.
Then we have the following inequalities:

\medskip
(i) {\bf (Hardy-Sobolev inequality)}  For any $\theta_{1}\in\mathbb{R}\backslash\{0\}$ we have
$$\int_{\mathbb{C}}\frac{|\widetilde{\nabla_{\mathcal{L}}}f|^{2}}{|z|^{2\theta_{1}}}dz-
\theta_{1}^{2}\int_{\mathbb{C}}\frac{|f|^{2}}{|z|^{2\theta_{1}+2}}dz\geq
\int_{\mathbb{C}}\frac{(\psi(|z|))^{2}}{|z|^{2\theta_{1}-2}}|f|^{2}dz$$
\begin{equation}\label{EQ:lh1}
+\int_{\mathbb{C}}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2\theta_{1}+2}}dz,
\end{equation}
for all complex-valued functions $f\in C_{0}^{\infty}(\mathbb{R}^{2}\backslash\{0\})$.

\medskip
(ii) {\bf (Logarithmic Hardy inequality)} We have
$$\int_{\mathbb{C}}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}|\log|z||^{2}dz-\frac{1}{4}\int_{\mathbb{C}}|f|^{2}dz\geq
\int_{\mathbb{C}}(\psi(|z|))^{2}|z|^{2}|\log|z||^{2}|f|^{2}dz$$
\begin{equation}\label{EQ:lh2}
+\int_{\mathbb{C}}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2}}|\log|z||^{2}dz,
\end{equation}
for all complex-valued functions $f\in C_{0}^{\infty}(\mathbb{R}^{2}\backslash\{0\})$.

\medskip
(iii) {\bf (Poincar\'e inequality)}
Let $\Omega$ be a bounded domain in $\mathbb{C}$ and $R=\underset{z\in\Omega}{\rm sup}\{|z|\}$. Then we have
$$\int_{\Omega}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}dz-\frac{1}{R^{2}}\int_{\Omega}|f|^{2}dz\geq
\int_{\Omega}(\psi(|z|))^{2}|z|^{2}|f|^{2}dz$$
\begin{equation}\label{EQ:lh3}
+\int_{\Omega}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2}}dz,
\end{equation}
for all complex-valued functions $f\in \widehat{\mathfrak{L}}_{0}^{1,2}(\Omega)$ satisfying $\frac{d}{d|z|}f\in L^{2}(\Omega)$, where the space $\widehat{\mathfrak{L}}_{0}^{1,2}(\Omega)$ is defined in \eqref{space}.

\medskip
(iv) {\bf (Hardy-Sobolev inequality with superweights)}
Let $\theta_{2}, \theta_{3}, \theta_{4}$, $a$, $b\in\mathbb{R}$ with $a,b>0$, $\theta_{2} \theta_{3}<0$ and $2\theta_{4}\leq \theta_{2}\theta_{3}$.
Then we have
$$\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}}}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}dz\geq
\frac{\theta_{2}\theta_{3}-2\theta_{4}}{2}\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}+2}}|f|^{2}dz$$
\begin{equation}\label{EQ:lh4}
\qquad +\int_{\mathbb{C}}\frac{(\psi(|z|))^{2}(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}-2}}|f|^{2}dz
+\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}(|f|^{2}-|f_{0}(|z|)|^{2})}{|z|^{2\theta_{4}+2}}dz,
\end{equation}
for all complex-valued functions $f\in C_{0}^{\infty}(\mathbb{R}^{2}\backslash\{0\})$.
Weights of this type has appeared in \cite{GM11} as well as in \cite{RSY17}, and are called the superweights due to the freedom in the choice of indices.

\medskip
{\em All terms on the right hand sides of inequalities \eqref{EQ:lh1}, \eqref{EQ:lh2}, \eqref{EQ:lh3} and \eqref{EQ:lh4} are non-negative}, therefore, they give the remainder estimates as well as Hardy's inequalities:
\begin{equation}\label{EQ:lh11}
\int_{\mathbb{C}}\frac{|\widetilde{\nabla_{\mathcal{L}_\psi}}f|^{2}}{|z|^{2\theta_{1}}}dz\geq
\theta_{1}^{2}\int_{\mathbb{C}}\frac{|f|^{2}}{|z|^{2\theta_{1}+2}}dz+
\int_{\mathbb{C}}\frac{(\psi(|z|))^2}{|z|^{2\theta_{1}-2}}|f|^{2}dz,
\end{equation}
\begin{equation}\label{EQ:lh22}
\int_{\mathbb{C}}|\widetilde{\nabla_{\mathcal{L}_\psi}}f|^{2}|\log|z||^{2}dz\geq \frac{1}{4}\int_{\mathbb{C}}|f|^{2}dz+
\int_{\mathbb{C}}(\psi(|z|))^2|z|^{2}|\log|z||^{2}|f|^{2}dz,
\end{equation}
\begin{equation}\label{EQ:lh33}
\int_{\Omega}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}dz\geq\frac{1}{R^{2}}\int_{\Omega}|f|^{2}dz+
\int_{\Omega}(\psi(|z|))^{2}|z|^{2}|f|^{2}dz
\end{equation}
and
$$\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}}}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}dz\geq
\frac{\theta_{2}\theta_{3}-2\theta_{4}}{2}\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}+2}}|f|^{2}dz$$
\begin{equation}\label{EQ:lh44}
+\int_{\mathbb{C}}\frac{(\psi(|z|))^{2}(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}-2}}|f|^{2}dz.
\end{equation}
For $\psi(|z|)=\frac12$ this yields Hardy inequalities and remainder estimates for the classical Landau Hamiltonian operator.
\end{itemize}

The Hardy inequalities for a magnetic Baouendi-Grushin operator with Aharonov-Bohm type magnetic field are proved in Section \ref{Sec3}. In Section \ref{Sec4} we prove the Hardy inequalities for the twisted Laplacian with the Landau-Hamiltonian type magnetic field.

\section{Weighted Hardy inequalities for magnetic Baouendi-Grushin operator with Aharonov-Bohm type magnetic field} \label{Sec3}

In this section we establish weighted Hardy inequalities for the quadratic form of the magnetic Baouendi-Grushin operator with Aharonov-Bohm type magnetic field.
We adapt all the notation introduced in Sections \ref{SEC:iGR} and \ref{SEC:iGRm}, namely,
$\nabla_{\gamma}$, $\rho$ and $Q$ defined in \eqref{subgrad}, \eqref{dist} and \eqref{hom_dim}, respectively. Recalling these for conveniece of the reader, we have
\begin{equation}\label{subgrad2}
\nabla_{\gamma}=(\nabla_{x}, |x|^{\gamma}\nabla_{y}),\quad
(x,y)\in \mathbb{R}^{m}\times\mathbb{R}^{k},\quad
Q=m+(1+\gamma)k,\; \gamma\geq 0,
\end{equation}
and the magnetic filed
 $\mathcal{A}$ is defined here as
\begin{equation}\label{mag_fieldi2}
\mathcal{A}=\frac{\nabla_{\gamma}\rho}{\rho}=
\left(\frac{\nabla_{x}\rho}{\rho},|x|^{\gamma}\frac{\nabla_{y}\rho}{\rho}\right)\in\mathbb{R}^{m}\times\mathbb{R}^{k}.
\end{equation}

We start with a simple inequality showing the best constants one can expect.

\begin{lem}\label{lem1}
Let $\Omega\subset \mathbb R^{n}$ be an open set. Let $(x,y)=(x_{1},...,x_{m}, y_{1},...,y_{k})\in \mathbb{R}^{m}\times\mathbb{R}^{k}$ with $k,m\geq1$, $k+m=n$. Let $\alpha_{1}, \alpha_{2}, \beta\in\mathbb{R}$ be such that
$$Q+\alpha_{1}-2>0 \textrm{ and } m+\alpha_{2}\gamma>0.$$
Then for any real-valued function $f\in C^{\infty}_{0}(\Omega)$ we have the following weighted Hardy inequality for the magnetic Baouendi-Grushin operator
$$\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}|(\nabla_{\gamma}+i\beta\mathcal{A})f|^{2}dxdy
\geq\left(\left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}+\beta^{2}\right)$$
\begin{equation}\label{Hardy1}
\times\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\frac{|x|^{2\gamma}}{\rho^{2\gamma+2}}|f|^{2}dxdy,
\end{equation}
Moreover, if $0\in\Omega$, then the constant in \eqref{Hardy1} is sharp.
\end{lem}
\begin{proof}[Proof of Lemma \ref{lem1}] By opening brackets we have
$$\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}|(\nabla_{\gamma}+i\beta\mathcal{A})f|^{2}dxdy=
\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}|\nabla_{\gamma}f|^{2}dxdy$$
\begin{equation}\label{Hardy2}
+\beta^{2}\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}|\mathcal{A}f|^{2}dxdy.
\end{equation}
Taking into account \eqref{formula} and putting $p=2$, $\alpha=\gamma(\alpha_{2}+2)+2-\alpha_{1}$, $\beta=\gamma(\alpha_{2}+2)$ in \eqref{Amb1}, we have for any $Q+\alpha_{1}-2>0$ and $m+\alpha_{2}\gamma>0$ that
\begin{equation}\label{Hardy3}\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}|\nabla_{\gamma}f|^{2}dxdy\geq \left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}
\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy.
\end{equation}
Taking into account the form of the magnetic field $\mathcal{A}=(\mathcal{A}_{1}, \mathcal{A}_{2})=\left(\frac{\nabla_{x}\rho}{\rho},|x|^{\gamma}\frac{\nabla_{y}\rho}{\rho}\right)$ in \eqref{mag_fieldi2}, and by a direct calculation one finds
$$\beta^{2}\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}|\mathcal{A}f|^{2}dxdy=
\beta^{2}\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\frac{|\nabla_{x}\rho|^{2}
+|x|^{2\gamma}|\nabla_{y}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy$$
\begin{equation}\label{Hardy4}
=\beta^{2}\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy.
\end{equation}
Then by \eqref{Hardy2}, \eqref{Hardy3} and \eqref{Hardy4} we obtain
$$\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}|(\nabla_{\gamma}+i\beta\mathcal{A})f|^{2}dz\geq
\left(\left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}+\beta^{2}\right)$$
\begin{equation}\label{Hardy5}
\times\int_{\Omega}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy.
\end{equation}
Then using \eqref{formula}, we observe that \eqref{Hardy5} yields \eqref{Hardy1}. Since the constant in \eqref{Hardy3} is sharp when $0\in\Omega$ by \eqref{Amb1}, then the constant in the obtained inequality is sharp when $0\in\Omega$.
\end{proof}

We obtain the following corollary in $\mathbb{R}^{2+k}$ for the Aharonov-Bohm potential of the type considered in \cite{AL11}.

\begin{cor}\label{another_mag} Let $\Omega\subset\mathbb{R}^{2+k}$ be an open set. Let $(x,y)=(x_{1},x_{2},y)\in \mathbb{R}^{2}\times\mathbb{R}^{k}$. Let $\alpha_{1}, \alpha_{2}, \beta\in\mathbb{R}$ be such that $\alpha_{1}+k(\gamma+1)>0$ and $\alpha_{2}\gamma+2>0$. Then for any real-valued function $f\in C^{\infty}_{0}(\Omega)$ and for the following Aharonov-Bohm type magnetic field
\begin{equation}\label{another_mag2}
\mathcal{\widetilde{A}}=
\left(-\frac{\partial_{x_{2}} \rho}{\rho}, \frac{\partial_{x_{1}} \rho}{\rho},-\frac{|x|^{\gamma}}{\sqrt{2}}
\frac{\nabla_{y} \rho}{\rho}, \frac{|x|^{\gamma}}{\sqrt{2}}
\frac{\nabla_{y} \rho}{\rho}\right),
\end{equation}
we have the following weighted Hardy inequality for the magnetic Baouendi-Grushin operator
$$\int_{\Omega}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}|(\widetilde{\nabla_{\gamma}}+i\beta\mathcal{\widetilde{A}})f|^{2}dxdy
\geq\left(\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}\right)$$
\begin{equation}\label{another_mag1}
\times\int_{\Omega}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}\frac{|x|^{2\gamma}}{\rho^{2\gamma+2}}|f|^{2}dxdy,
\end{equation}
where \begin{equation}\label{another_mag3}
\widetilde{\nabla_{\gamma}}=\left(\frac{\partial}{\partial x_{1}}, \frac{\partial}{\partial x_{2}}, \frac{|x|^{\gamma}}{\sqrt{2}}\nabla_{y}, \frac{|x|^{\gamma}}{\sqrt{2}}\nabla_{y}\right).
\end{equation} Moreover, if $0\in\Omega$, then the constant $\left(\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}\right)$ in \eqref{another_mag1} is sharp.
\end{cor}
\begin{proof}[Proof of Corollary \ref{another_mag}]
In this case $m=2$, then $Q=2+k(1+\gamma)$. Since we have $|\mathcal{\widetilde{A}}f|^{2}=|\mathcal{A}f|^{2}$, $|\widetilde{\nabla_{\gamma}}\rho|=|\nabla_{\gamma}\rho|$ and $|\widetilde{\nabla_{\gamma}}f|^{2}=|\nabla_{\gamma}f|^{2}$, then in the exact same way as in the proof of the Lemma \ref{lem1} one obtains \eqref{another_mag1}.
\end{proof}

As another corollary of Lemma \ref{lem1} we obtain the following uncertainty principle.

\begin{cor}[{\rm Uncertainty type principle}]\label{uncer}
Let $(x,y)=(x_{1},...,x_{m}, y_{1},...,y_{k})\in \mathbb{R}^{m}\times\mathbb{R}^{k}$ with $k,m\geq1$, $k+m=n$. Let $\Omega\subset\mathbb R^{n}$ be an open set. Let $\alpha_{1}, \alpha_{2}, \beta\in\mathbb{R}$ be such that $Q+\alpha_{1}-2>0$ and $m+\alpha_{2}\gamma>0$. Then for any real-valued function $f\in C^{\infty}_{0}(\Omega)$ we have
$$\|\rho^{\frac{\alpha_{1}}{2}}|\nabla_{\gamma}\rho|^{\frac{\alpha_{2}}{2}}(\nabla_{\gamma}+i\beta\mathcal{A})f\|_{L^{2}(\Omega)}
\|f\|_{L^{2}(\Omega)}\geq\left(\left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}+\beta^{2}\right)^{\frac{1}{2}}$$
\begin{equation}\label{uncer1}
\times\int_{\Omega}\rho^{\frac{\alpha_{1}}{2}}|\nabla_{\gamma}\rho|^{\frac{\alpha_{2}}{2}}
\frac{|x|^{\gamma}}{\rho^{\gamma+1}}|f|^{2}dxdy.
\end{equation}
\end{cor}
\begin{proof}[Proof of Corollary \ref{uncer}] By Lemma \ref{lem1} we get
$$\|\rho^{\frac{\alpha_{1}}{2}}|\nabla_{\gamma}\rho|^{\frac{\alpha_{2}}{2}}(\nabla_{\gamma}+i\beta\mathcal{A})f\|_{L^{2}(\Omega)}
\|f\|_{L^{2}(\Omega)}\geq\left(\left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}+\beta^{2}\right)^{\frac{1}{2}}
$$$$\times\left\|\rho^{\frac{\alpha_{1}}{2}}|\nabla_{\gamma}\rho|^{\frac{\alpha_{2}}{2}}
\frac{|x|^{\gamma}}{\rho^{\gamma+1}}f\right\|_{L^{2}(\Omega)}\|f\|_{L^{2}(\Omega)} $$
$$\geq \left(\left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}+\beta^{2}\right)^{\frac{1}{2}}
\int_{\Omega}\rho^{\frac{\alpha_{1}}{2}}|\nabla_{\gamma}\rho|^{\frac{\alpha_{2}}{2}}
\frac{|x|^{\gamma}}{\rho^{\gamma+1}}|f|^{2}dxdy.$$
The proof is complete.
\end{proof}

We now give the main theorem of this section for complex-valued functions $f$.

\begin{thm}\label{thm2}
Let $(x,y)=(x_{1}, x_{2}, y)\in \mathbb{R}^{2}\times\mathbb{R}^{k}$. Let $\alpha_{1},\alpha_{2}, \beta\in\mathbb{R}$ be such that $\alpha_{1}+k(\gamma+1)>0$ and $\alpha_{2}+2\gamma>0$. Recall $\widetilde{\nabla_{\gamma}}$, $\mathcal{\widetilde{A}}$ and $\rho$ defined in \eqref{another_mag3}, \eqref{another_mag2}, \eqref{dist}, respectively.

 Then for any complex-valued function $f\in C^{\infty}_{0}(\mathbb{R}^{2+k}\backslash\{0\})$ we have the following weighted Hardy inequality for the magnetic Baouendi-Grushin operator
\begin{multline}\label{thm2_0}
\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}|(\widetilde{\nabla_{\gamma}}+i\beta\mathcal{\widetilde{A}})f|
^{2}dxdy
\\ \geq\left(\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}\right)
\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|
^{\alpha_{2}}\frac{|x|^{2\gamma}}{\rho^{2\gamma+2}}|f|^{2}dxdy
\\ +\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|
^{\alpha_{2}}\frac{|f|^{2}-|f_{0}(|x|,y)|^{2}}{|x|^{2}}dxdy,
\end{multline}
with the remainder term, where $f_{0}(|x|,y)=\frac{1}{2\pi}\int_{0}^{2\pi}f(|x|,\phi, y)d\phi$,
$(|x|,\phi)$ is the polar decomposition of $x$. The latter remainder term is non-negative, therefore, we also have the following inequality for any complex-valued function $f\in C^{\infty}_{0}(\mathbb{R}^{2+k}\backslash\{0\})$
\begin{multline}\label{thm2_1}
\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}|(\widetilde{\nabla_{\gamma}}+i\beta\mathcal{\widetilde{A}})f|
^{2}dxdy
\\
\geq\left(\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}\right)
\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|
^{\alpha_{2}}\frac{|x|^{2\gamma}}{\rho^{2\gamma+2}}|f|^{2}dxdy,
\end{multline}
and the constant $\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}$ is sharp.
\end{thm}
The proof of Theorem \ref{thm2} will be based on the following theorem:
\begin{thm}\label{radial_ineq} Let $(x,y)=(x_{1},...,x_{m}, y_{1},...,y_{k})\in \mathbb{R}^{m}\times\mathbb{R}^{k}$ with $k,m\geq1$, $k+m=n$. Let $\alpha_1,\alpha_2\in\mathbb R$ be such that
$$Q+\alpha_{1}-2>0 \textrm{ and } m+\gamma \alpha_{2}>0.$$
Then we have the following Hardy type inequality for all complex-valued functions $f\in C_{0}^{\infty}(\mathbb R^{n}\backslash\{0\})$,
\begin{multline}\label{radial_ineq1}
\int_{\mathbb R^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\left(
\left|\frac{d}{d|x|}f\right|^{2}+|x|^{2\gamma}|\nabla_{y}f|^{2}\right)dxdy
\\ \geq \left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}
\int_{\mathbb R^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy,
\end{multline}
with sharp constant $\left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}$.
\end{thm}

\begin{rem}\label{rem_radial_der} Theorem \ref{radial_ineq} implies the following inequality
\begin{multline}\label{rem_radial_der1}
\int_{\mathbb R^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\left(
\left|\nabla_{x}f\right|^{2}+|x|^{2\gamma}|\nabla_{y}f|^{2}\right)dxdy
\\ \geq \left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}
\int_{\mathbb R^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy,
\end{multline}
with sharp constant, which gives the result of D'Ambrosio \eqref{Amb1} when $p=2$ and $\Omega=\mathbb R^{n}$. We also mention that inequality \eqref{rem_radial_der1} has been established in \cite{Kombe15} and \cite{ShY12}.
\end{rem}
\begin{proof}[Proof of Theorem \ref{radial_ineq}] We denote $r=|x|$ and $B(r,y)=\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}$. Then, using \eqref{dist} and \eqref{formula}, one has
\begin{equation}\label{B(r,y)}
B(r,y)=\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}
=r^{\alpha_{2}\gamma}\rho^{\alpha_{1}-\alpha_{2}\gamma}=r^{\alpha_{2}\gamma}(r^{2(1+\gamma)}+(1+\gamma)^{2}
|y|^{2})^{\frac{\alpha_{1}-\alpha_{2}\gamma}{2(1+\gamma)}}.
\end{equation}
Let us first calculate the following
$$\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\left(\left|\left(\partial_{r}+\alpha\frac{\partial_{r}\rho}{\rho}\right)f\right|^{2}+
r^{2\gamma}\left|\left(\nabla_{y}+\alpha\frac{\nabla_{y}\rho}{\rho}\right)f\right|^{2}\right)r^{m-1}B(r,y)drdy$$
$$=\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\left(|\partial_{r}f|^{2}+r^{2\gamma}|\nabla_{y}f|^{2}\right)r^{m-1}B(r,y)drdy$$
$$+\alpha^{2}\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\left(\left|\frac{\partial_{r}\rho}{\rho}\right|^{2}+
r^{2\gamma}\left|\frac{\nabla_{y}\rho}{\rho}\right|^{2}\right)|f|^{2}r^{m-1}B(r,y)drdy$$
$$+2\alpha{\rm Re}\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\frac{\partial_{r}\rho}{\rho}r^{m-1}B(r,y)\overline{\partial_{r}f}\cdot fdrdy$$
$$+2\alpha{\rm Re}\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\frac{\nabla_{y}\rho}{\rho}\cdot\overline{\nabla_{y}f}r^{2\gamma+m-1}B(r,y)fdrdy$$
\begin{equation}\label{radial_ineq2}=:I_{1}+I_{2}+I_{3}+I_{4}.
\end{equation}
Using
$$\frac{\partial_{r}\rho}{\rho}=\frac{r^{2\gamma+1}}{\rho^{2\gamma+2}} \;\;{\rm and} \;\;\frac{\nabla_{y}\rho}{\rho}=\frac{(\gamma+1)y}{\rho^{2\gamma+2}},$$
we calculate
\begin{equation}\label{calcul}
\left|\frac{\partial_{r}\rho}{\rho}\right|^{2}+
r^{2\gamma}\left|\frac{\nabla_{y}\rho}{\rho}\right|^{2}=\frac{r^{4\gamma+2}+r^{2\gamma}(\gamma+1)^{2}|y|^{2}}{\rho^{4\gamma+4}}
=\frac{r^{2\gamma}}{\rho^{2\gamma+2}}=\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}.
\end{equation}
Thus, we obtain
\begin{equation}\label{radial_ineq3}
I_{2}=\alpha^{2}\int_{-\infty}^{\infty}\int_{0}^{\infty}\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}r^{m-1}B(r,y)drdy.
\end{equation}
Now we proceed using integration by parts for $I_{3}$
$$I_{3}=-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}(2\gamma+m+\gamma \alpha_{2})\rho^{\alpha_{1}-\alpha_{2}\gamma-2\gamma-2}
r^{2\gamma+m-1+\gamma\alpha_{2}}|f|^{2}drdy$$
$$-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}
(\alpha_{1}-\alpha_{2}\gamma-2\gamma-2)\rho^{\alpha_{1}-\alpha_{2}\gamma-4\gamma-4}r^{4\gamma+m+\gamma \alpha_{2}+1}|f|^{2}drdy.$$
Since $B(r,y)=r^{\alpha_{2}\gamma}\rho^{\alpha_{1}-\alpha_{2}\gamma}$ by \eqref{B(r,y)}, then we have
$$I_{3}=-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\left((2\gamma+m+\gamma \alpha_{2})\frac{r^{2\gamma}}{\rho^{2\gamma+2}}+(\alpha_{1}-\alpha_{2}\gamma-2\gamma-2)\frac{r^{4\gamma+2}}{\rho^{4\gamma+4}}
\right)r^{m-1}B(r,y)|f|^{2}drdy$$
$$=-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\left(2\gamma+m+\gamma \alpha_{2}+(\alpha_{1}-\alpha_{2}\gamma-2\gamma-2)\frac{r^{2\gamma+2}}{\rho^{2\gamma+2}}
\right)\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}r^{m-1}B(r,y)drdy.$$
Similarly, we have for $I_{4}$
$$I_{4}=-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}{\rm div}_{y}\left(B(r,y)\frac{\nabla_{y}\rho}
{\rho}\right)r^{2\gamma+m-1}|f|^{2}drdy$$
$$=-\alpha(\gamma+1)\int_{\mathbb{R}^{k}}\int_{0}^{\infty}
{\rm div}_{y}\left(\rho^{\alpha_{1}-\alpha_{2}\gamma-2\gamma-2}y\right)r^{\alpha_{2}\gamma+2\gamma+m-1}|f|^{2}drdy$$
$$=-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\left((\alpha_{1}-\alpha_{2}\gamma-2\gamma-2)
\rho^{\alpha_{1}-\alpha_{2}\gamma-2\gamma-3}\frac{(\gamma+1)^{2}|y|^{2}}{\rho^{2\gamma+1}}\right)
r^{\alpha_{2}\gamma+2\gamma+m-1}|f|^{2}drdy$$
$$-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}k(\gamma+1)\rho^{\alpha_{1}-\alpha_{2}\gamma-2\gamma-2}
r^{\alpha_{2}\gamma+2\gamma+m-1}|f|^{2}drdy$$
Since $\frac{r^{2\gamma}}{\rho^{2\gamma+2}}=\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}$ and $B(r,y)=r^{\alpha_{2}\gamma}
\rho^{\alpha_{1}-\alpha_{2}\gamma}$ by \eqref{calcul} and \eqref{B(r,y)}, respectively, then we obtain
$$I_{4}=-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\left((\alpha_{1}-\alpha_{2}\gamma-2\gamma-2)
\frac{(\gamma+1)^{2}|y|^{2}}{\rho^{2\gamma+2}}+k(\gamma+1)\right)\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}r^{m-1}B(r,y)drdy.$$
Then, taking into account the definition \eqref{dist}, we get
$$I_{3}+I_{4}=-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}(\alpha_{1}-\alpha_{2}\gamma-2\gamma-2+2\gamma+m+\gamma \alpha_{2}+k(\gamma+1))\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}r^{m-1}B(r,y)drdy,$$
using that $Q=m+(1+\gamma)k$ in \eqref{hom_dim}, one has
\begin{equation}\label{radial_ineq4}
I_{3}+I_{4}=-\alpha\int_{\mathbb{R}^{k}}\int_{0}^{\infty}(Q+\alpha_{1}-2
)\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}r^{m-1}B(r,y)drdy.
\end{equation}
Putting \eqref{radial_ineq3} and \eqref{radial_ineq4} in \eqref{radial_ineq2}, we have
$$\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\left(\left|\left(\partial_{r}+\alpha\frac{\partial_{r}\rho}{\rho}\right)f\right|^{2}+
r^{2\gamma}\left|\left(\nabla_{y}+\alpha\frac{\nabla_{y}\rho}{\rho}\right)f\right|^{2}\right)r^{m-1}B(r,y)drdy$$
$$=\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\left(|\partial_{r}f|^{2}+r^{2\gamma}|\nabla_{y}f|^{2}\right)r^{m-1}B(r,y)drdy$$
$$-\left((Q+\alpha_{1}-2
)\alpha-\alpha^{2}\right)\int_{\mathbb{R}^{k}}\int_{0}^{\infty}\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}r^{m-1}B(r,y)drdy.$$
By substituting $\alpha=\frac{Q+\alpha_{1}-2}{2}$ and taking into account \eqref{B(r,y)}, we obtain \eqref{radial_ineq1}.

Let us now show the sharpness of the constant in \eqref{radial_ineq1}. Taking into account \eqref{Hardy3} and \eqref{radial_ineq1}, we have
$$\int_{\mathbb R^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\left(
\left|\nabla_{x}f\right|^{2}+|x|^{2\gamma}|\nabla_{y}f|^{2}\right)dxdy$$$$\geq\int_{\mathbb R^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}\left(
\left|\frac{d}{d|x|}f\right|^{2}+|x|^{2\gamma}|\nabla_{y}f|^{2}\right)dxdy$$
$$\geq\left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}\int_{\mathbb R^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}
\frac{|\nabla_{\gamma}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy,$$
which proves the constant $\left(\frac{Q+\alpha_{1}-2}{2}\right)^{2}$ in \eqref{radial_ineq1} is sharp.
\end{proof}

We are now ready to prove Theorem \ref{thm2}.
\begin{proof}[Proof of Theorem \ref{thm2}] By polar coordinates for $x$-plane $x_{1}=r\cos{\phi}$, $x_{2}=r\sin{\phi}$, we have $r=|x|$ and
$$ \frac{\partial_{x_{1}} \rho}{\rho}=\frac{r^{2\gamma+1}\cos{\phi}}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}},
\;\;\frac{\partial_{x_{2}} \rho}{\rho}=\frac{r^{2\gamma+1}\sin{\phi}}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}},$$
$$ \frac{\nabla_{y} \rho}{\rho}=\frac{(1+\gamma)y}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}}.$$
Thus, we can write
\begin{equation}\label{mag_Ahar1}
\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}
|(\widetilde{\nabla_{\gamma}}+i\beta\mathcal{\widetilde{A}})f|^{2}dx_{1}dx_{2}dy=I_{1}+I_{2},
\end{equation}
where
$$I_{1}=\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\left|\left(\cos{\phi}\partial_{r}
-\frac{\sin{\phi}}{r}\partial_{\phi}-i\beta\frac{r^{2\gamma+1}\sin{\phi}}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}}\right)
f\right|^{2}rB(r,y)dr d\phi dy$$
$$+\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\left|\left(\sin{\phi}\partial_{r}
+\frac{\cos{\phi}}{r}\partial_{\phi}+i\beta\frac{r^{2\gamma+1}\cos{\phi}}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}}\right)
f\right|^{2}rB(r,y)dr d\phi dy$$
and
$$I_{2}=\frac{1}{2}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\left|\left(\nabla_{y}-i\beta
\frac{(1+\gamma)y}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}}\right)f\right|^{2}r^{2\gamma+1}B(r,y)dr d\phi dy$$
$$+\frac{1}{2}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\left|\left(\nabla_{y}+i\beta
\frac{(1+\gamma)y}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}}\right)f\right|^{2}r^{2\gamma+1}B(r,y)dr d\phi dy.$$
By opening brackets we obtain
$$I_{1}=\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\left(|\partial_{r}f|^{2}+\frac{|\partial_{\phi}f|^{2}}{r^{2}}
+\beta^{2}\left|\frac{r^{2\gamma+1}}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}}f\right|^{2}\right)rB(r,y)dr d\phi dy$$
$$-2{\rm Re}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{r^{2\gamma}}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}}
\partial_{\phi}f\cdot\overline{f}rB(r,y)dr d\phi dy$$
and
$$I_{2}=\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}|\nabla_{y}f|^{2}r^{2\gamma+1} B(r,y)dr d\phi dy
$$$$+
\beta^{2}(1+\gamma)^{2}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{|y|^{2}|f|^{2}}
{(r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2})^{2}} r^{2\gamma+1}B(r,y)dr d\phi dy.$$
Integrating the second integral in $I_{1}$ by parts we see that
$$2{\rm Re}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{r^{2\gamma}}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}}
\partial_{\phi}f\cdot\overline{f}rB(r,y)dr d\phi dy$$
$$={\rm Re}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{r^{2\gamma}}{r^{2(1+\gamma)}+(1+\gamma)^{2}|y|^{2}}
\partial_{\phi}(|f|^{2})rB(r,y)dr d\phi dy=0.$$
Using the Fourier series for $f$ we can expand
$$f(r,\phi,y)=\sum_{k=-\infty}^{\infty}f_{k}(r,y)e^{ik\phi},$$
and by a direct calculation, we get
\begin{equation*}\begin{aligned}
\frac{1}{r^{2}}\int_{0}^{2\pi}|\partial_{\phi}f|^{2}d\phi & =\frac{2\pi}{r^{2}}\sum_{k}k^{2}|f_{k}(r,y)|^{2}
\\ & =
\frac{2\pi}{r^{2}}\sum_{k\neq0}k^{2}|f_{k}(r,y)|^{2} \\
& \geq\frac{2\pi}{r^{2}}\min_{k\neq0}(k^{2})\sum_{k\neq0}|f_{k}(r,y)|^{2} \\
& =\frac{1}{r^{2}}\int_{0}^{2\pi}|f|^{2}d\phi
-\frac{2\pi}{r^{2}}|f_{0}(r,y)|^{2} \\
& =\frac{1}{r^{2}}\int_{0}^{2\pi}(|f|^{2}-|f_{0}(r,y)|^{2})d\phi,
\end{aligned}
\end{equation*}
where $f_{0}(|z|,y)=\frac{1}{2\pi}\int_{0}^{2\pi}f(|z|,\phi,y)d\phi$.
Now putting the obtained estimates for $I_{1}$ and $I_{2}$ in \eqref{mag_Ahar1} we have
$$\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}
|(\widetilde{\nabla_{\gamma}}+i\beta\mathcal{\widetilde{A}})f|^{2}dx_{1}dx_{2}dy$$
$$\geq
\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}(|\partial_{r}f|^{2}+r^{2\gamma}|\nabla_{y}f|^{2})rB(r,y)dr d\phi dy$$
$$+\beta^{2}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\left(\frac{r^{4\gamma+2}
+(1+\gamma)^{2}|y|^{2}r^{2\gamma}}{(r^{2\gamma+2}+(1+\gamma)^{2}|y|^{2})^{2}}\right)|f|^{2}rB(r,y)dr d\phi dy$$
$$+\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{|f|^{2}-|f_{0}(r,y)|^{2}}{r^{2}}rB(r,y)dr d\phi dy$$
$$=\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}(|\partial_{r}f|^{2}+r^{2\gamma}|\nabla_{y}f|^{2})rB(r,y)dr d\phi dy$$
$$+\beta^{2}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{r^{2\gamma}}
{r^{2\gamma+2}+(1+\gamma)^{2}|y|^{2}}|f|^{2}rB(r,y)dr d\phi dy$$
\begin{equation}\label{mag_Ahar2}
+\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{|f|^{2}-|f_{0}(r,y)|^{2}}{r^{2}}rB(r,y)dr d\phi dy.
\end{equation}
Since $|\widetilde{\nabla_{\gamma}}\rho|=|\nabla_{\gamma}\rho|$, then putting $m=2$ in \eqref{Hardy3} and taking into account \eqref{B(r,y)} we obtain the following estimate for the first integral in \eqref{mag_Ahar2}
$$\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}(|\partial_{r}f|^{2}+r^{2\gamma}|\nabla_{y}f|^{2})rB(r,y)dr d\phi dy$$
\begin{equation}\label{mag_Ahar3}\geq \left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}
\frac{|\widetilde{\nabla_{\gamma}}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy.
\end{equation}
Let us calculate the second integral in \eqref{mag_Ahar2}
$$\beta^{2}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{r^{2\gamma}}
{r^{2\gamma+2}+(1+\gamma)^{2}|y|^{2}}|f|^{2}rB(r,y)dr d\phi dy$$
$$=\beta^{2}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{r^{4\gamma+2}\cos^{2}{\phi}}
{(r^{2\gamma+2}+(1+\gamma)^{2}|y|^{2})^{2}}|f|^{2}rB(r,y)dr d\phi dy$$
$$+\beta^{2}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{r^{4\gamma+2}\sin^{2}{\phi}}
{(r^{2\gamma+2}+(1+\gamma)^{2}|y|^{2})^{2}}|f|^{2}rB(r,y)dr d\phi dy$$
$$+\beta^{2}\int_{\mathbb{R}^{k}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{(1+\gamma)^{2}r^{2\gamma}|y|^{2}}
{(r^{2\gamma+2}+(1+\gamma)^{2}|y|^{2})^{2}}|f|^{2}rB(r,y)dr d\phi dy$$
$$=\beta^{2}\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}
\left|\left(\frac{\partial_{x_{1}} \rho}{\rho}, \frac{\partial_{x_{2}} \rho}{\rho},\frac{|x|^{\gamma}}{\sqrt{2}}
\frac{\nabla_{y} \rho}{\rho}, \frac{|x|^{\gamma}}{\sqrt{2}}
\frac{\nabla_{y} \rho}{\rho}\right)\right|^{2}|f|^{2}dxdy$$
\begin{equation}\label{mag_Ahar4}=
\beta^{2}\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|^{\alpha_{2}}
\frac{|\widetilde{\nabla_{\gamma}}\rho|^{2}}{\rho^{2}}|f|^{2}dxdy.
\end{equation}
Putting the estimates \eqref{mag_Ahar3} and \eqref{mag_Ahar4} in \eqref{mag_Ahar2}, we obtain \eqref{thm2_0}. Since we have
$$\int_{\mathbb{R}^{2+k}}\rho^{\alpha_{1}}|\widetilde{\nabla_{\gamma}}\rho|
^{\alpha_{2}}\frac{|f|^{2}-|f_{0}(|x|,y)|^{2}}{|x|^{2}}dxdy\geq0$$
and \eqref{thm2_1} with sharp constant for all real-valued functions by Corollary \ref{another_mag}, then this constant is sharp also in the class of complex-valued functions in \eqref{thm2_1}.
\end{proof}

We record the corresponding uncertainty principle.

\begin{cor}[{\rm Uncertainty type principle}]\label{uncer2} Let $(x,y)=(x_{1}, x_{2}, y)\in \mathbb{R}^{2}\times\mathbb{R}^{k}$.
Let $\alpha_{1},\alpha_{2}, \beta\in\mathbb{R}$ be such that $\alpha_{1}+k(\gamma+1)>0$ and $\alpha_{2}\gamma+2>0$. Then for any complex-valued function $f\in C^{\infty}_{0}(\mathbb{R}^{2+k}\backslash\{0\})$ we have
\begin{multline}\label{uncer21}
\|\rho^{\frac{\alpha_{1}}{2}}|\widetilde{\nabla_{\gamma}}\rho|^{\frac{\alpha_{2}}{2}}(\widetilde{\nabla_{\gamma}}+
i\beta\mathcal{\widetilde{A}})f\|_{L^{2}(\mathbb{R}^{2+k})}
\|f\|_{L^{2}(\mathbb{R}^{2+k})}
\\ \geq\left(\left(\frac{\alpha_{1}+k(\gamma+1)}{2}\right)^{2}+\beta^{2}\right)^{\frac{1}{2}}
\int_{\mathbb{R}^{2+k}}\rho^{\frac{\alpha_{1}}{2}}|\widetilde{\nabla_{\gamma}}\rho|^{\frac{\alpha_{2}}{2}}
\frac{|x|^{\gamma}}{\rho^{\gamma+1}}|f|^{2}dxdy.
\end{multline}
\end{cor}
\begin{proof}[Proof of Corollary \ref{uncer2}] Using \eqref{thm2_1} and in a similar way as in the proof of the Corollary \ref{uncer}, one obtains \eqref{uncer21}.
\end{proof}

\section{Hardy inequalities for
Landau-Hamiltonian} \label{Sec4}

In this section we show the Hardy inequalities for the twisted Laplacian with Landau-Hamiltonian type magnetic field.

We adapt the notation of Section \ref{SEC:introLH}, namely, we have
$\nabla_{\mathcal{L}}$ the gradient operator associated with $\mathcal{L}$:
\begin{equation}\label{twist2-2}
\nabla_{\mathcal{L}}f=(\widetilde{X}_{1}f,...,\widetilde{X}_{n}f, \widetilde{Y}_{1}f,...,\widetilde{Y}_{n}f),
\end{equation}
for the vector fields  $\widetilde{X}_{j}=\partial_{x_{j}}-\frac{1}{2}iy_{j}$ and $\widetilde{Y}_{j}=\partial_{y_{j}}+\frac{1}{2}ix_{j}$.

Let us introduce now the generalised form of the twisted Laplacian
$$\widetilde{\mathcal{L}}=\sum_{j=1}^{n}\left[\left(i\partial_{x_{j}}+\psi(|z|)y_{j}\right)^{2}+
\left(i\partial_{y_{j}}-\psi(|z|)x_{j}\right)^{2}\right],$$
where $\psi(|z|)$ is a radial real-valued differentiable function.
Setting
$$\check{{X}_{j}}=\partial_{x_{j}}-i\psi(|z|)y_{j} \textrm{ and }
\check{Y}_{j}=\partial_{y_{j}}+i\psi(|z|)x_{j},$$
we write
\begin{equation}\label{twist3_1_2}
\widetilde{\nabla_{\mathcal{L}}}f=(\check{X}_{1}f, \ldots, \check{X}_{n}f, \check{Y}_{1}f, \ldots, \check{Y}_{n}f).
\end{equation}
The classical Landau Hamiltonian is a special case of this with the choice of the constant function $\psi(|z|)=\frac12$.

\begin{thm}\label{twist_thm_1} Let $\theta_{1}, \theta_{2}, \theta_{3}, \theta_{4}, a, b \in\mathbb{R}$ with $a,b>0$, $\theta_{1}\neq0$, $\theta_{2} \theta_{3}<0$ and $2\theta_{4}\leq \theta_{2}\theta_{3}$. Let $\psi=\psi(|z|)$ be a radial real-valued function such that $\psi\in L^{2}_{loc}(\mathbb{C}\backslash\{0\})$. Let $\Omega$ be a bounded domain in $\mathbb{C}$ and $R=\underset{z\in\Omega}{\rm sup}\{|z|\}$.
For a function $f$, we will be denoting $f_{0}(|z|):=\frac{1}{2\pi}\int_{0}^{2\pi}f(|z|,\phi)d\phi$.

Then we have the following inequalities with remainders:

(i) the weighted Hardy-Sobolev inequality
\begin{equation}\label{twist3_1}
\int_{\mathbb{C}}\frac{|\widetilde{\nabla_{\mathcal{L}}}f|^{2}}{|z|^{2\theta_{1}}}dz-
\theta_{1}^{2}\int_{\mathbb{C}}\frac{|f|^{2}}{|z|^{2\theta_{1}+2}}dz\geq
\int_{\mathbb{C}}\frac{(\psi(|z|))^{2}}{|z|^{2\theta_{1}-2}}|f|^{2}dz+\int_{\mathbb{C}}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2\theta_{1}+2}}dz,
\end{equation}
for all complex-valued functions $f\in C_{0}^{\infty}(\mathbb{R}^{2}\backslash\{0\})$;

(ii) the logarithmic Hardy inequality
$$\int_{\mathbb{C}}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}|\log|z||^{2}dz-\frac{1}{4}\int_{\mathbb{C}}|f|^{2}dz\geq
\int_{\mathbb{C}}(\psi(|z|))^{2}|z|^{2}|\log|z||^{2}|f|^{2}dz$$
\begin{equation}\label{twist3_1_1}
+\int_{\mathbb{C}}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2}}|\log|z||^{2}dz,
\end{equation}
for all complex-valued functions $f\in C_{0}^{\infty}(\mathbb{R}^{2}\backslash\{0\})$;

(iii) the Poincar\'e inequality
$$\int_{\Omega}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}dz-\frac{1}{R^{2}}\int_{\Omega}|f|^{2}dz\geq
\int_{\Omega}(\psi(|z|))^{2}|z|^{2}|f|^{2}dz$$
\begin{equation}\label{twist3_1_1_Po}
+\int_{\Omega}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2}}dz,
\end{equation}
for all complex-valued functions $f\in \widehat{\mathfrak{L}}_{0}^{1,2}(\Omega)$ satisfying $\frac{d}{d|z|}f\in L^{2}(\Omega)$, where the space $\widehat{\mathfrak{L}}_{0}^{1,2}(\Omega)$ is defined in \eqref{space};

(iv) the Hardy-Sobolev inequality with more general weights
$$\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}}}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}dz\geq
\frac{\theta_{2}\theta_{3}-2\theta_{4}}{2}\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}+2}}|f|^{2}dz$$
\begin{equation}\label{twist3_1_1_sup}
+\int_{\mathbb{C}}\frac{(\psi(|z|))^{2}(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}-2}}|f|^{2}dz
+\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}(|f|^{2}-|f_{0}(|z|)|^{2})}{|z|^{2\theta_{4}+2}}dz,
\end{equation}
for all complex-valued functions $f\in C_{0}^{\infty}(\mathbb{R}^{2}\backslash\{0\})$.
\end{thm}
The proof of Theorem \ref{twist_thm_1} will be based on the following family of weighted Hardy inequalities and Poincar\'{e} type inequality that were obtained in \cite[Theorem 3.4 and Theorem 5.1]{RSY16} and \cite[Theorem 8.1]{RSY17}, where $\mathbb{E}=|x|\mathcal{R}$ is the Euler operator and $\mathcal{R}:=\frac{d}{d|x|}$ is the radial derivative.

\begin{thm}\cite[Theorem 3.4]{RSY16}\label{L_p_weighted_th}
Let $\mathbb{G}$ be a homogeneous group
of homogeneous dimension $Q$ and let $\theta\in \mathbb{R}$.
Then for any complex-valued function $f\in C^{\infty}_{0}(\mathbb{G}\backslash\{0\}),$ $1<p<\infty,$ and any homogeneous quasi-norm $|\cdot|$ on $\mathbb{G}$ for $\theta p \neq Q$ we have
\begin{equation}\label{L_p_weighted}
\left\|\frac{f}{|x|^{\theta}}\right\|_{L^{p}(\mathbb{G})}\leq
\left|\frac{p}{Q-\theta p}\right|\left\|\frac{1}{|x|^{\theta}}\mathbb{E} f\right\|_{L^{p}(\mathbb{G})}.
\end{equation}
If $\theta p\neq Q$ then the constant $\left|\frac{p}{Q-\theta p}\right|$ is sharp.
For $\theta p=Q$ we have
\begin{equation}\label{L_p_weighted_log}
\left\|\frac{f}{|x|^{\frac{Q}{p}}}\right\|_{L^{p}(\mathbb{G})}\leq
p\left\|\frac{\log|x|}{|x|^{\frac{Q}{p}}}\mathbb{E} f\right\|_{L^{p}(\mathbb{G})}
\end{equation}
with sharp constant.
\end{thm}
Let $\Omega \subset \mathbb{G}$ be an open set and let $\widehat{\mathfrak{L}}_{0}^{1,p}(\Omega)$ be the completion of $C^{\infty}_{0}(\Omega\backslash\{0\})$ with respect to
$$
\|f\|_{\widehat{\mathfrak{L}}^{1,p}(\Omega)}=\|f\|_{L^{p}(\Omega)}+\|\mathbb{E}f\|_{L^{p}(\Omega)}, \;\;1<p<\infty.
$$
In the abelian case, when $\mathbb{G}=(\mathbb{R}^{2},+)$ and $p=2$, let us give this definition: Let $\Omega \subset \mathbb{R}^{2}$ be an open set and let $\widehat{\mathfrak{L}}_{0}^{1,2}(\Omega)$ be the completion of $C^{\infty}_{0}(\Omega\backslash\{0\})$ with respect to
\begin{equation}\label{space}
\|f\|_{\widehat{\mathfrak{L}}^{1,2}(\Omega)}=\|f\|_{L^{2}(\Omega)}+\|\mathbb{E} f\|_{L^{2}(\Omega)}.
\end{equation}

\begin{thm}\label{Po}\cite[Theorem 5.1]{RSY16}
Let $\Omega$ be a bounded open subset of $\mathbb{G}$. If $1<p<\infty, \;f\in \widehat{\mathfrak{L}}_{0}^{1,p}(\Omega)$ and $\mathcal{R}f\in L^{p}(\Omega)$, then we have
\begin{equation}\label{Po01}
\|f\|_{L^{p}(\Omega)}\leq  \frac{R p}{Q}\|\mathcal{R}f\|_{L^{p}(\Omega)}=\frac{R p}{Q}\left\|\frac{1}{|x|}\mathbb{E}f\right\|_{L^{p}(\Omega)},
\end{equation}
where $R=\underset{x\in \Omega}{\rm sup}|x|$.
\end{thm}
\begin{thm}\label{1}\cite[Theorem 8.1]{RSY17}
Let $\mathbb{G}$ be a homogeneous group
of homogeneous dimension $Q$. Let $a,b>0$, $\theta_{2} \theta_{3}<0$ and $p\theta_{4}-\theta_{2}\theta_{3}\leq Q-p$. Then for any complex-valued function $f\in C^{\infty}_{0}(\mathbb{G}\backslash\{0\}),$ $1<p<\infty,$ and any homogeneous quasi-norm $|\cdot|$ on $\mathbb{G}$, we have
\begin{equation}\label{Lpweighted2}
\frac{Q-p\theta_{4}+\theta_{2}\theta_{3}-p}{p}
\left\|\frac{(a+b|x|^{\theta_{2}})^{\frac{\theta_{3}}{p}}}{|x|^{\theta_{4}+1}}f\right\|_{L^{p}(\mathbb{G})}
\leq\left\|\frac{(a+b|x|^{\theta_{2}})^{\frac{\theta_{3}}{p}}}{|x|^{\theta_{4}}}\mathcal{R}f\right\|_{L^{p}(\mathbb{G})}.
\end{equation}
If $Q\neq p\theta_{4}+p-\theta_{2}\theta_{3}$, then the constant $\frac{Q-p\theta_{4}+\theta_{2}\theta_{3}-p}{p}$ is sharp.
\end{thm}

We briefly recall their proof for the convenience of the reader but also since these will be useful in our argument.

\begin{proof}[Proof of Theorem \ref{L_p_weighted_th}]
Integrating by parts gives for $\theta p \neq Q$
\begin{multline*}
\int_{\mathbb{G}}\frac{|f(x)|^{p}}{|x|^{\theta p}}dx=\int_{0}^{\infty}\int_{\wp}|f(ry)|^{p}r^{Q-1-\theta p}d\sigma(y)dr\\
=-\frac{p}{Q-\theta p}\int_{0}^{\infty} r^{Q-\theta p} {\rm Re} \int_{\wp}|f(ry)|^{p-2} f(ry) \overline{\frac{df(ry)}{dr}}d\sigma(y)dr\\
\leq \left|\frac{p}{Q-\theta p}\right|\int_{\mathbb{G}}\frac{|\mathbb{E}f(x)||f(x)|^{p-1}}{|x|^{\theta p}}dx=
\left|\frac{p}{Q-\theta p}\right|\int_{\mathbb{G}}\frac{|\mathbb{E}f(x)||f(x)|^{p-1}}{|x|^{\theta+\theta (p-1)}}dx.
\end{multline*}
Using H\"{o}lder's inequality, we obtain
$$
\int_{\mathbb{G}}\frac{|f(x)|^{p}}{|x|^{\theta p}}dx\leq \left|\frac{p}{Q-\theta p}\right|\left(\int_{\mathbb{G}}\frac{|\mathbb{E}f(x)|^{p}}{|x|^{\theta p}}dx\right)
^{\frac{1}{p}}\left(\int_{\mathbb{G}}\frac{|f(x)|^{p}}{|x|^{\theta p}}dx\right)^{\frac{p-1}{p}},
$$
which implies \eqref{L_p_weighted}.

In order to show the sharpness of the constant, let us check the equality condition in above H\"older's inequality.
We consider the function
\begin{equation}\label{2}
g_{1}(x)=\frac{1}{|x|^{C}},
\end{equation}
where $C\in\mathbb{R}, C\neq 0$ and $\theta p\neq Q$. Then, a direct calculation implies
\begin{equation}\label{Holder_eq1}
\left|\frac{1}{C}\right|^{p}\left(\frac{|\mathbb{E}g_{1}(x)|}{|x|^{\theta }}\right)^{p}=\left(\frac{|g_{1}(x)|^{p-1}}
{|x|^{\theta (p-1)}}\right)^{\frac{p}{p-1}},
\end{equation}
which satisfies the equality condition in H\"older's inequality.
Thus, the constant $\left|\frac{p}{Q-\theta p}\right|$ is sharp in \eqref{L_p_weighted}.

Now we show \eqref{L_p_weighted_log}. Taking into account $\wp:=\{x\in \mathbb{G}:\,|x|=1\}$ and applying the polar decomposition on homogeneous groups $\mathbb{G}$, then using integration by parts, one calculates
\begin{multline*}
\int_{\mathbb{G}}\frac{|f(x)|^{p}}{|x|^{Q}}dx=\int_{0}^{\infty}\int_{\wp}|f(ry)|^{p}r^{Q-1-Q}d\sigma(y)dr\\
=-p\int_{0}^{\infty} \log r {\rm Re} \int_{\wp}|f(ry)|^{p-2} f(ry) \overline{\frac{df(ry)}{dr}}d\sigma(y)dr\\
\leq p \int_{\mathbb{G}}\frac{|\mathbb{E}f(x)||f(x)|^{p-1}}{|x|^{Q}}|\log|x||dx=
p\int_{\mathbb{G}}\frac{|\mathbb{E}f(x)|\log|x|||}{|x|^{\frac{Q}{p}}}\frac{|f(x)|^{p-1}}{|x|^{\frac{Q(p-1)}{p}}}dx.
\end{multline*}
Using again the H\"{o}lder's inequality, we get
$$
\int_{\mathbb{G}}\frac{|f(x)|^{p}}{|x|^{Q}}dx\leq p\left(\int_{\mathbb{G}}\frac{|\mathbb{E}f(x)|^{p}|\log|x||^{p}}{|x|^{Q}}dx\right)
^{\frac{1}{p}}\left(\int_{\mathbb{G}}\frac{|f(x)|^{p}}{|x|^{Q}}dx\right)^{\frac{p-1}{p}},
$$
which gives \eqref{L_p_weighted_log}.

As in the case of \eqref{L_p_weighted}, we consider the following function to show the sharpness of the constant in \eqref{L_p_weighted_log}
$$g_{2}(x)=(\log|x|)^{C},$$
where $C\in\mathbb{R}$ and $C\neq 0$.
Then, one has
\begin{equation}\label{Holder_eq2}
\left|\frac{1}{C}\right|^{p}\left(\frac{|\mathbb{E}g_{2}(x)||\log|x||}{|x|^{\frac{Q}{p}}}\right)^{p}=\left(\frac{|g_{2}(x)|^{p-1}}
{|x|^{\frac{Q (p-1)}{p}}}\right)^{\frac{p}{p-1}},
\end{equation}
which satisfies the equality condition in H\"older's inequality.
\end{proof}

Before proving Theorem \ref{Po}, we first show the following proposition.
\begin{prop}\label{Poprop}\cite[Proposition 5.2]{RSY16}
Let $\Omega \subset \mathbb{G}$ be an open set. If $1<p<\infty, \;f\in \widehat{\mathfrak{L}}_{0}^{1,p}(\Omega)$ and $\mathbb{E}f\in L^{p}(\Omega)$, then we have
\begin{equation} \label{Po1}
\|f\|_{L^{p}(\Omega)}\leq \frac{p}{Q}\|\mathbb{E}f\|_{L^{p}(\Omega)}.
\end{equation}
\end{prop}
\begin{proof}[Proof of Proposition \ref{Poprop}]
Let us consider the function $\zeta:\mathbb{R}\rightarrow\mathbb{R}$, which is an even and smooth function, satisfying
\begin{itemize}
\item
$0\leq\zeta\leq1,$
\item
$\zeta(r)=1 \;\;{\rm if}\;\; |r|\leq1,$
\item
$\zeta(r)=0 \;\;{\rm if}\;\; |r|\geq2.$
\end{itemize}
We set $\zeta_{\lambda}(x):=\zeta(\lambda|x|)$ for $\lambda>0$.
We have the inequality \eqref{Po1} for $f\in C_{0}^{\infty}(\mathbb{G}\backslash\{0\})$ by \eqref{L_p_weighted} when $\theta=0$. There is $\{f_{\ell}\}_{\ell=1}^{\infty} \in C_{0}^{\infty}(\Omega\backslash\{0\})$ such that $f_{\ell}\rightarrow f$ in $\widehat{\mathfrak{L}}_{0}^{1,p}(\Omega)$ as $\ell\rightarrow\infty$. Let $\lambda>0$. From \eqref{L_p_weighted} when $\theta=0$ one gets
$$\|\zeta_{\lambda}f_{\ell}\|_{L^{p}(\Omega)}\leq\frac{p}{Q}\left(\|(\mathbb{E}\zeta_{\lambda})f_{\ell}\|_{L^{p}(\Omega)}+
\|\zeta_{\lambda}(\mathbb{E}f_{\ell})\|_{L^{p}(\Omega)}\right)$$
for any $\ell\geq1$. Then, we can immediately see that
$$\lim_{\ell\rightarrow\infty}\zeta_{\lambda}f_{\ell}=\zeta_{\lambda}f,$$
$$\lim_{\ell\rightarrow\infty}(\mathbb{E}\zeta_{\lambda})f_{\ell}=(\mathbb{E}\zeta_{\lambda})f,$$
$$\lim_{\ell\rightarrow\infty}\zeta_{\lambda}(\mathbb{E}f_{\ell})=\zeta_{\lambda}(\mathbb{E}f)$$
in $L^{p}(\Omega)$. By these properties we obtain
$$\|\zeta_{\lambda}f\|_{L^{p}(\Omega)}\leq\frac{p}{Q}\left\{\|(\mathbb{E}\zeta_{\lambda})f\|_{L^{p}(\Omega)}+
\|\zeta_{\lambda}(\mathbb{E}f)\|_{L^{p}(\Omega)}\right\}.$$
Since
$$|(\mathbb{E}\zeta_{\lambda})(x)|\leq \begin{cases}
\sup|\mathbb{E}\zeta|, \;\;{\rm if}\;\; \lambda^{-1}<|x|<2\lambda^{-1};\\
0, \;\; {\rm otherwise},
\end{cases}$$
one obtains \eqref{Po1} in the limit as $\lambda\rightarrow 0$.
\end{proof}
\begin{proof}[Proof of Theorem \ref{Po}] Since $R=\underset{x\in \Omega}{\rm sup}|x|$ by Proposition \ref{Poprop} one has
\begin{equation*}\|f\|_{L^{p}(\Omega)}\leq \frac{p}{Q}\|\mathbb{E}f\|_{L^{p}(\Omega)}\\ \leq
\frac{R p}{Q}\|\mathcal{R}f\|_{L^{p}(\Omega)}=\frac{R p}{Q}\left\|\frac{1}{|x|}\mathbb{E}f\right\|_{L^{p}(\Omega)},
\end{equation*}
which implies \eqref{Po01}.
\end{proof}
\begin{proof}[Proof of Theorem \ref{1}] Since for $Q=p\theta_{4}+p-\theta_{2}\theta_{3}$ there is nothing
to prove, let us only consider the case $Q\neq p\theta_{4}+p-\theta_{2}\theta_{3}$. Using polar coordinates $(r,y)=(|x|, \frac{x}{\mid x\mid})\in (0,\infty)\times\wp$ on $\mathbb{G}$, where $\wp$ is the unit quasi-sphere
\begin{equation}\label{EQ:sphere}
\wp:=\{x\in \mathbb{G}:\,|x|=1\},
\end{equation} and by polar decomposition on $\mathbb{G}$ (see, for example, \cite{FS-Hardy} or \cite{FR}) and using the integration by parts, one calculates
\begin{equation}\label{eqp}
\int_{\mathbb{G}}
\frac{(a+b|x|^{\theta_{2}})^{\theta_{3}}}{|x|^{p\theta_{4}+p}}|f(x)|^{p}dx
=\int_{0}^{\infty}\int_{\wp}
\frac{(a+br^{\theta_{2}})^{\theta_{3}}}{r^{p\theta_{4}+p}}|f(ry)|^{p} r^{Q-1}d\sigma(y)dr.
\end{equation}
Since $\theta_{2} \theta_{3}<0$ and $p\theta_{4}-\theta_{2}\theta_{3}<Q-p$ we get
$$
\int_{\mathbb{G}}
\frac{(a+b|x|^{\theta_{2}})^{\theta_{3}}}{|x|^{p\theta_{4}+p}}|f(x)|^{p}dx$$
$$\leq\int_{0}^{\infty}\int_{\wp}
(a+br^{\theta_{2}})^{\theta_{3}}r^{Q-1-p\theta_{4}-p}\left(\frac{b r^{\theta_{2}}}{a+br^{\theta_{2}}}+\frac{a}{a+br^{\theta_{2}}}
\cdot\frac{Q-p\theta_{4}-p}{Q-p\theta_{4}-p+\theta_{2}\theta_{3}}\right)|f(ry)|^{p} d\sigma(y)dr
$$
$$=\int_{0}^{\infty}\int_{\wp}
\frac{(a+br^{\theta_{2}})^{\theta_{3}}r^{Q-1-p\theta_{4}-p}}{Q-p\theta_{4}-p+\theta_{2}\theta_{3}}
\left(\frac{\theta_{2} \theta_{3} b r^{\theta_{2}}}{a+br^{\theta_{2}}}+Q-p\theta_{4}-p\right)|f(ry)|^{p} d\sigma(y)dr
$$
$$=\int_{0}^{\infty}\int_{\wp}
\frac{d}{dr}\left(\frac{(a+br^{\theta_{2}})^{\theta_{3}}
	r^{Q-p\theta_{4}-p}}{Q-p\theta_{4}-p+\theta_{2}\theta_{3}}\right)|f(ry)|^{p}
d\sigma(y)dr$$
$$
=-\frac{p}{Q-p\theta_{4}-p+\theta_{2}\theta_{3}}\int_{0}^{\infty}(a+br^{\theta_{2}})^{\theta_{3}}r^{Q-p\theta_{4}-p}  \,{\rm Re}\int_{\wp}
|f(ry)|^{p-2} f(ry) \overline{\frac{df(ry)}{dr}}d\sigma(y)dr
$$
$$\leq \left|\frac{p}{Q-p\theta_{4}-p+\theta_{2}\theta_{3}}\right|\int_{\mathbb{G}}\frac{(a+b|x|^{\theta_{2}})^{\theta_{3}}|\mathcal{R}f(x)||f(x)|^{p-1}}{|x|^{p\theta_{4}
+p-1}}dx
$$
$$=\frac{p}{Q-p\theta_{4}-p+\theta_{2}\theta_{3}}
\int_{\mathbb{G}}\frac{(a+b|x|^{\theta_{2}})^
{\frac{\theta_{3}(p-1)}{p}}|f(x)|^{p-1}}{|x|^{(\theta_{4}+1)(p-1)}}
\frac{(a+b|x|^{\theta_{2}})^
{\frac{\theta_{3}}{p}}}{|x|^{\theta_{4}}}|\mathcal{R}f(x)|dx.$$
Here the H\"{o}lder's inequality gives
$$
\int_{\mathbb{G}}
\frac{(a+b|x|^{\theta_{2}})^{\theta_{3}}}{|x|^{p\theta_{4}+p}}|f(x)|^{p}dx$$
$$\leq\frac{p}{Q-p\theta_{4}-p+\theta_{2}\theta_{3}}\left(\int_{\mathbb{G}}\frac{(a+b|x|^{\theta_{2}})^{\theta_{3}}}{|x|^{p\theta_{4}
+p}}|f(x)|^{p}dx\right)^\frac{p-1}{p}
\left(\int_{\mathbb{G}}\frac{(a+b|x|^{\theta_{2}})^{\theta_{3}}}{|x|^{p\theta_{4}}}|\mathcal{R}f(x)|^{p}dx\right)^\frac{1}{p},
$$
which implies \eqref{Lpweighted2}.

In order to show the sharpness of the constant, we check the equality condition in above H\"older's inequality.
We consider the function
$$g_{3}(x)=|x|^{C}, $$
where $C\in\mathbb{R}, C\neq 0$ and $Q\neq p\theta_{4}+p-\theta_{2}\theta_{3}$. Then, a direct calculation gives
\begin{equation}\label{Holder_eq2}
\left|\frac{1}{C}\right|^{p}\left(\frac{(a+b|x|^{\theta_{2}})^
{\frac{\theta_{3}}{p}}|\mathcal{R}g_{3}(x)|}{|x|^{\theta_{4}}}\right)^{p}=\left(\frac{(a+b|x|^{\theta_{2}})^
{\frac{\theta_{3}(p-1)}{p}}|g_{3}(x)|^{p-1}}
{|x|^{(\theta_{4}+1) (p-1)}}\right)^{\frac{p}{p-1}},
\end{equation}
which satisfies the equality condition in H\"older's inequality.
This shows the sharpness of the constant $\frac{Q-p\theta_{4}-p+\theta_{2}\theta_{3}}{p}$ in \eqref{Lpweighted2}.
\end{proof}

We are now ready to prove Theorem \ref{twist_thm_1}.
\begin{proof}[Proof of Theorem \ref{twist_thm_1}] Let $\kappa(|z|)\neq0$ be a radial function. By polar coordinates for the $z$-plane $x=r\cos{\phi}$, $y=r\sin{\phi}$, we have
$$\int_{\mathbb{C}}\frac{|\widetilde{\nabla_{\mathcal{L}}}f|^{2}}{\kappa(|z|)}dz=\int_{\mathbb{C}}\left(\left|\left(i\partial_{x}+
y\psi(|z|)\right)f\right|^{2}+\left|\left(i\partial_{y}-
x\psi(|z|)\right)f\right|^{2}\right)\frac{dz}{\kappa(|z|)}$$
$$=\int_{0}^{\infty}\int_{0}^{2\pi}\left(\left|\left(i\cos{\phi}\partial_{r}-\frac{i\sin{\phi}}{r}\partial_{\phi}
+
\psi(r)r\sin{\phi}\right)f\right|^{2}\right)rd\phi \frac{dr}{\kappa(r)}
$$$$
+\int_{0}^{\infty}\int_{0}^{2\pi}\left(\left|\left(i\sin{\phi}\partial_{r}+\frac{i\cos{\phi}}{r}\partial_{\phi}-
\psi(r)r\cos{\phi}\right)f\right|^{2}\right)rd\phi\frac{dr}{\kappa(r)}$$
$$=\int_{0}^{\infty}\int_{0}^{2\pi}\left(|\partial_{r}f|^{2}+\frac{|\partial_{\phi}f|^{2}}{r^{2}}+\psi(r)^{2}r^{2}
|f|^{2}\right)rd\phi \frac{dr}{\kappa(r)}
-2{\rm Re}\int_{0}^{\infty}\int_{0}^{2\pi}i\psi(r)\partial_{\phi}f\cdot \overline{f}rd\phi \frac{dr}{\kappa(r)}.$$
Integrating the second integral in this equality by parts we see that
$$2{\rm Re}\int_{0}^{\infty}\int_{0}^{2\pi}i\psi(r)\partial_{\phi}f\cdot \overline{f}rd\phi \frac{dr}{\kappa(r)}=
2{\rm Re}\int_{0}^{\infty}\int_{0}^{2\pi}i\psi(r)\partial_{\phi}(|f|^{2})rd\phi \frac{dr}{\kappa(r)}=0.$$
Let us represent $f$ via its Fourier series
$$f(r,\phi)=\sum_{k=-\infty}^{\infty}f_{k}(r)e^{ik\phi}.$$
Then we obtain
$$\frac{1}{r^{2}}\int_{0}^{2\pi}|\partial_{\phi}f|^{2}d\phi=\frac{2\pi}{r^{2}}\sum_{k}k^{2}|f_{k}(r)|^{2}=
\frac{2\pi}{r^{2}}\sum_{k\neq0}k^{2}|f_{k}(r)|^{2}$$
$$\geq\frac{2\pi}{r^{2}}\min_{k\neq0}(k^{2})\sum_{k\neq0}|f_{k}(r)|^{2}=\frac{1}{r^{2}}\int_{0}^{2\pi}|f|^{2}d\phi
-\frac{2\pi}{r^{2}}|f_{0}(r)|^{2}=\frac{1}{r^{2}}\int_{0}^{2\pi}(|f|^{2}-|f_{0}(r)|^{2})d\phi,$$
where $f_{0}(|z|)=\frac{1}{2\pi}\int_{0}^{2\pi}f(|z|,\phi)d\phi$.

Thus, we arrive at
\begin{multline}\label{twist3_2}\int_{\mathbb{C}}\frac{|\widetilde{\nabla_{\mathcal{L}}}f|^{2}}{\kappa(|z|)}dz \\
\geq
\int_{\mathbb{C}}\frac{1}{\kappa(|z|)}\left|\frac{d}{d|z|}f\right|^{2}dz+
\int_{\mathbb{C}}\frac{|z|^{2}(\psi(|z|))^{2}}{\kappa(|z|)}|f|^{2}dz+
\int_{\mathbb{C}}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2}\kappa(|z|)}dz.
\end{multline}
Putting $\kappa(|z|)=|z|^{2\theta_{1}}$, it follows that
\begin{multline}\label{twist3_2_1}\int_{\mathbb{C}}\frac{|\widetilde{\nabla_{\mathcal{L}}}f|^{2}}{|z|^{2\theta_{1}}}dz \\ \geq
\int_{\mathbb{C}}\frac{1}{|z|^{2\theta_{1}}}\left|\frac{d}{d|z|}f\right|^{2}dz+
\int_{\mathbb{C}}\frac{(\psi(|z|))^{2}}{|z|^{2\theta_{1}-2}}|f|^{2}dz+
\int_{\mathbb{C}}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2\theta_{1}+2}}dz.
\end{multline}
In the abelian case $\mathbb{G}=(\mathbb{R}^{2},+)$, $p=2$ and $\theta=\theta_{1}+1$ with $\theta_{1}\neq0$, \eqref{L_p_weighted} implies
$$\int_{\mathbb{C}}\frac{1}{|z|^{2\theta_{1}}}\left|\frac{d}{d|z|}f\right|^{2}dz\geq \theta_{1}^{2}\int_{\mathbb{C}}\frac{|f|^{2}}{|z|^{2\theta_{1}+2}}dz.$$
Putting this in \eqref{twist3_2_1}, we obtain \eqref{twist3_1}.

Putting $\kappa(|z|)=|\log|z||^{-2}$ in \eqref{twist3_2}, one has
$$\int_{\mathbb{C}}|\log|z||^{2}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}dz\geq
\int_{\mathbb{C}}|\log|z||^{2}\left|\frac{d}{d|z|}f\right|^{2}dz+
\int_{\mathbb{C}}(\psi(|z|))^{2}|z|^{2}|\log|z||^{2}|f|^{2}dz$$
\begin{equation}\label{twist3_3}
+\int_{\mathbb{C}}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2}}|\log|z||^{2}dz.
\end{equation}
In the abelian case $\mathbb{G}=(\mathbb{R}^{2},+)$ and $p=2$, \eqref{L_p_weighted_log} implies
$$\int_{\mathbb{C}}|\log|z||^{2}\left|\frac{d}{d|z|}f\right|^{2}dz\geq \frac{1}{4}\int_{\mathbb{C}}|f|^{2}dz.$$
Putting this in \eqref{twist3_3}, we obtain \eqref{twist3_1_1}.

Putting $\kappa(|z|)=1$ in \eqref{twist3_2}, we get
$$\int_{\mathbb{C}}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}dz\geq
\int_{\mathbb{C}}\left|\frac{d}{d|z|}f\right|^{2}dz+
\int_{\mathbb{C}}(\psi(|z|))^{2}|z|^{2}|f|^{2}dz$$
\begin{equation}\label{twist3_3_Po}
+\int_{\mathbb{C}}\frac{|f|^{2}-|f_{0}(|z|)|^{2}}{|z|^{2}}dz.
\end{equation}
In the abelian case, when $\Omega$ is a bounded domain in $(\mathbb{R}^{2},+)$, and $p=2$, \eqref{Po01} implies
$$\int_{\Omega}\left|\frac{d}{d|z|}f\right|^{2}dz\geq \frac{1}{R^{2}}\int_{\Omega}|f|^{2}dz.$$
Putting this in \eqref{twist3_3_Po}, we obtain \eqref{twist3_1_1_Po}.

Putting $\kappa(|z|)=\frac{(a+b|z|^{\theta_{2}})^{-\theta_{3}}}{|z|^{-2\theta_{4}}}$ in \eqref{twist3_2}, we have
$$\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}}}|\widetilde{\nabla_{\mathcal{L}}}f|^{2}dz\geq
\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}}}\left|\frac{d}{d|z|}f\right|^{2}dz+
\int_{\mathbb{C}}\frac{\psi^{2}(|z|)(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}-2}}|f|^{2}dz$$
\begin{equation}\label{twist3_3_sup}
+\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}(|f|^{2}-|f_{0}(|z|)|^{2})}{|z|^{2\theta_{4}+2}}dz.
\end{equation}
Again, in the abelian case $\mathbb{G}=(\mathbb{R}^{2},+)$ and $p=2$, \eqref{Lpweighted2} gives
$$\frac{\theta_{2}\theta_{3}-2\theta_{4}}{2}
\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}+2}}|f|^{2}dz
\leq\int_{\mathbb{C}}\frac{(a+b|z|^{\theta_{2}})^{\theta_{3}}}{|z|^{2\theta_{4}}}\left|\frac{d}{d|z|}f\right|^{2}dz,$$
for $a,b>0$, $\theta_{2} \theta_{3}<0$ and $2\theta_{4}\leq \theta_{2}\theta_{3}$.
Putting this in \eqref{twist3_3_sup}, we obtain \eqref{twist3_1_1_sup}.
\end{proof}

Now we give some inequalities for real-valued functions to show the best estimates one can expect. While estimates for real-valued functions have less physical meaning than those for complex-valued functions in questions of the spectral theory, they also find their use in applications to the existence of real (or positive) solutions to some nonlinear equations.

\begin{rem}\label{twist_rem}
By a direct calculation we have
$$\int_{\mathbb{C}^{n}}|\nabla_{\mathcal{L}}f|^{2}dz=\sum_{j=1}^{n}\int_{\mathbb{C}^{n}}\left(\left|\left(i\partial_{x_{j}}+
\frac{y_{j}}{2}\right)f\right|^{2}+\left|\left(i\partial_{y_{j}}-
\frac{x_{j}}{2}\right)f\right|^{2}\right)dz$$
\begin{equation}\label{twist8}=\int_{\mathbb{C}^{n}}|\nabla f|^{2}dz+\int_{\mathbb{C}^{n}}\frac{|z|^{2}}{4}|f|^{2}dz.
\end{equation}
Using the well-known Hardy inequality for $n\geq1$ for any $f\in W_{\mathcal{L}}^{1,2}(\mathbb{C}^{n})$, we obtain
\begin{equation}\label{twist3}
\int_{\mathbb{C}^{n}}|\nabla_{\mathcal{L}}f|^{2}dz\geq\left(n-1\right)^{2}\int_{\mathbb{C}^{n}}\frac{|f|^{2}}{|z|^{2}}dz
+\int_{\mathbb{C}^{n}}\frac{|z|^{2}}{4}|f|^{2}dz,
\end{equation}
for all real-valued functions $f\in W_{\mathcal{L}}^{1,2}(\mathbb{C}^{n})$ and $n\geq1$. Here $W_{\mathcal{L}}^{1,2}$
is defined in \eqref{twist_def}.

Let $\Omega$ is a bounded domain in $\mathbb{C}$ with $0\in\Omega$ and $R\geq e\cdot \underset{z\in\Omega}{\rm sup}\{|z|\}$. Then, in the case $n=1$ we can use the critical Hardy inequality with sharp constant for $f\in W_{\mathcal{L}}^{1,2}(\Omega)$ (see for example \cite{AR01}, \cite{AS02}):
\begin{equation}\label{twist4}
\int_{\Omega}|\nabla_{\mathcal{L}}f|^{2}dz\geq \frac{1}{4}
\int_{\Omega}\frac{|f|^{2}}{|z|^{2}\left(\log\frac{R}{|z|}\right)^{2}}dz
+\int_{\Omega}\frac{|z|^{2}}{4}|f|^{2}dz,
\end{equation}
for all real-valued functions $f\in W_{\mathcal{L}}^{1,2}(\Omega)$ and $n=1$.
Since the constants in the Hardy and critical Hardy inequalities are sharp, then the constants in the obtained inequalities \eqref{twist3} and \eqref{twist4} are sharp.
\end{rem}
\begin{cor}[{\rm Uncertainty type principle}]\label{uncer_twist}
Let $\Omega$ be a bounded domain in $\mathbb{C}$ with $0\in\Omega$ and $R\geq e\cdot \underset{z\in\Omega}{\rm sup}\{|z|\}$. Then we have
\begin{equation}\label{twist5}
\|\nabla_{\mathcal{L}}f\|_{L^{2}(\mathbb{C}^{n})}\|f\|_{L^{2}(\mathbb{C}^{n})}\geq
\int_{\mathbb{C}^{n}}\left(\sqrt{\frac{(n-1)^{2}}{|z|^{2}}+\frac{|z|^{2}}{4}}\right)\left|f\right|^{2}dz
\end{equation}
for $n\geq1$ and all real-valued functions $f\in W_{\mathcal{L}}^{1,2}(\mathbb{C}^{n})$, and
\begin{equation}\label{twist6}
\|\nabla_{\mathcal{L}}f\|_{L^{2}(\Omega)}\|f\|_{L^{2}(\Omega)}\geq
\int_{\Omega}\left(\sqrt{\frac{1}{4|z|^{2}\left(\log\frac{R}{|z|}\right)^{2}}+\frac{|z|^{2}}{4}}\right)\left|f\right|^{2}dz
\end{equation}
for $n=1$ and all real-valued functions $f\in W_{\mathcal{L}}^{1,2}(\Omega)$.
\end{cor}
\begin{proof}[Proof of Corollary \ref{uncer_twist}] By \eqref{twist3} we get for $n\geq1$
$$\|\nabla_{\mathcal{L}}f\|_{L^{2}(\mathbb{C}^{n})}\|f\|_{L^{2}(\mathbb{C}^{n})}\geq
\left\|\left(\sqrt{\frac{(n-1)^{2}}{|z|^{2}}+\frac{|z|^{2}}{4}}\right)f\right\|
_{L^{2}(\mathbb{C}^{n})}\|f\|_{L^{2}(\mathbb{C}^{n})}$$
$$\geq \int_{\mathbb{C}^{n}}\left(\sqrt{\frac{(n-1)^{2}}{|z|^{2}}+\frac{|z|^{2}}{4}}\right)\left|f\right|^{2}dz,$$
which gives \eqref{twist5}.
Similarly, using \eqref{twist4} we obtain \eqref{twist6}.
The proof is complete.
\end{proof}
\begin{rem}
In the case $n=1$ of the Remark \ref{twist_rem}, we also can use the another type of the critical Hardy inequality (see for example Solomyak \cite{Solomyak94}) in \eqref{twist8}:
$$\int_{\mathbb{C}}|\nabla_{\mathcal{L}}f|^{2}dz\geq C
\int_{\mathbb{C}}\frac{|f|^{2}}{|z|^{2}(1+\log^{2}|z|)}dz
+\int_{\mathbb{C}}\frac{|z|^{2}}{4}|f|^{2}dz,$$
where $C$ is a positive constant.
\end{rem}
\begin{rem}\label{Grushin_constant_mag} Let $\mathcal{L}_{G}$ be the magnetic Baouendi-Grushin operator on $\mathbb{C}^{n}$ with the constant magnetic field
$$\mathcal{L}_{G}=\sum_{j=1}^{n}((i\partial_{x_{j}}+\psi_{1,j}(y_{j}))^{2}+(i|x|^{\gamma}\partial_{y_{j}}+\psi_{2,j}(x_{j}))^{2}),$$
where $|x|=\sqrt{|x_{1}|^{2}+\ldots |x_{n}|^{2}}$, $\psi_{1,j},\psi_{2,j}\in L_{loc}^{2}(\mathbb{R}\backslash\{0\})$. Setting $\widehat{X}_{j}=i\partial_{x_{j}}+\psi_{1,j}(y_{j})$ and $\widehat{Y}_{j}=i|x|^{\gamma}\partial_{y_{j}}+\psi_{2,j}(x_{j})$, we write
$$\nabla_{G\mathcal{L}}f=(\widehat{X}_{1}f, \ldots, \widehat{X}_{n}f, \widehat{Y}_{1}f, \ldots, \widehat{Y}_{n}f).$$
Then, a direct calculation gives for all real-valued functions $f\in C_{0}^{\infty}(\mathbb{R}^{2n}\backslash\{0\})$
$$\int_{\mathbb{C}^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}
|\nabla_{G\mathcal{L}}f|^{2}dz$$
$$=\sum_{j=1}^{n}\int_{\mathbb{C}^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}
\left(\left|\left(i\partial_{x_{j}}+
\psi_{1,j}(y_{j})\right)f\right|^{2}+\left|\left(i|x|^{\gamma}\partial_{y_{j}}+
\psi_{2,j}(x_{j})\right)f\right|^{2}\right)dz$$
$$=\int_{\mathbb{C}^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}|\nabla_{\gamma}f|^{2}dz+
\sum_{j=1}^{n}\int_{\mathbb{C}^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}(|\psi_{2,j}(x_{j})|^{2}
+|\psi_{1,j}(y_{j})|^{2})|f|^{2}dz.$$
Then putting $m=k=n$ in \eqref{Hardy3}, and hence $Q=n(2+\gamma)$, we obtain the following Hardy inequality for magnetic Baouendi-Grushin operator on $\mathbb{C}^{n}$ with the constant magnetic field and for any real-valued function $f\in C_{0}^{\infty}(\mathbb{R}^{2n}\backslash\{0\})$ with sharp constant
$$\int_{\mathbb{C}^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}
|\nabla_{G\mathcal{L}}f|^{2}dz$$
$$\geq\left(\frac{n(2+\gamma)+\alpha_{1}-2}{2}\right)\int_{\mathbb{C}^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}
\frac{|x|^{2\gamma}}{\rho^{2\gamma+2}}|f|^{2}dz$$$$+
\sum_{j=1}^{n}\int_{\mathbb{C}^{n}}\rho^{\alpha_{1}}|\nabla_{\gamma}\rho|^{\alpha_{2}}
(|\psi_{2,j}(x_{j})|^{2}+|\psi_{1,j}(y_{j})|^{2})|f|^{2}dz,$$
where $n(2+\gamma)+\alpha_{1}-2>0$ and $n+\alpha_{2}\gamma>0$.
\end{rem}

redpajama_set_name: RedPajamaArXiv


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text:
Hermagoras Karion (,  ur. ok. 40 p.n.e., zm. ok. 30 n.e) – grecki retor, uczeń Theodorosa z Gadary, działający w starożytnym Rzymie.

Najprawdopodobniej studiował u Theodorosa z Gadary na Rodos. Znany z licznych cytatów u Seneki Starszego, Kwintyliana oraz hasła w Księdze Suda (w której połączone zostały w jednym haśle dzieła Hermagorasa z Amfipolis i życiorys Hermagorasa Kariona).

Bibliografia

Pisarze greckojęzyczni starożytnego Rzymu
Pisarze I wieku p.n.e.
Pisarze I wieku
Urodzeni w I wieku p.n.e.
Zmarli w I wieku

redpajama_set_name: RedPajamaWikipedia


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focus on process. What are processes in our company, how they work?
make changes; word around us is changing, so we need to be open to change.
There is a plenty process improvement methodology: PDCA, GQM+Strategy, APIM, and so on. Some of them will be describe in future posts.


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text:
Launch of JPSS-1 scrubbed again, high upper-level winds to blame
Jason Rhian
NOAA's JPSS-1 satellite. Photo Credit: Ball Aerospace
The flight of a United Launch Alliance (ULA) Delta II rocket with its payload of the National Oceanic and Atmospheric Administration's Joint Polar Satellite System 1 (JPSS-1) spacecraft has been delayed for a second time. However, whereas yesterday's (Nov. 14) scrub was caused by wayward boaters, issues with the atmosphere were to blame for today's delay.
According to a ULA statement:
The launch of a United Launch Alliance Delta II rocket carrying the JPSS-1 mission for NASA and NOAA was scrubbed today due to a range safety hold and high upper-level winds. The team is working to establish a new launch opportunity.
Representatives with the Colorado-based launch service provider had predicted that the weather would give about an 80 percent chance of favorable conditions for launch with the flight slated to get underway at 1:47 a.m. PST.
Tagged: Delta II JPSS-1 The Range United Launch Alliance Vandenberg Air Force Base
Jason Rhian spent several years honing his skills with internships at NASA, the National Space Society and other organizations. He has provided content for outlets such as: Aviation Week & Space Technology, Space.com, The Mars Society and Universe Today.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: How can I paint random values under a curve Python? I want to paint the random numbers that are shown under the curve of a color, and those that are above another color, I tried putting my values in arrays and putting an if, I have been trying but I have not obtained result, this is my code:
import numpy as np
import matplotlib.pyplot as plt

def func(x):
    return (np.exp(-x))*x**2
f = np.linspace(0, 10, 1000)
y = func(f)
yw=np.array(y)
plt.plot(f, y)
plt.axhline(y=0.6, xmin=0, xmax=10)
h=np.random.randn(1000)
h2=np.array(h)
#if(h2>yw):
 #   plt.plot(f,h,'mo')
#else:
plt.plot(f, y)
plt.plot(f,h,'co')
plt.xlim(0,10)
plt.ylim(0,0.6012)

A: You can try looping over each value, check if its corresponding value is greater or less that the function, and plotting each point with a different color.
I tried this:
for i in range(len(f)):
    if h[i] > func(f[i]):
        plt.plot(f[i],h[i],'co')
    else:
        plt.plot(f[i],h[i],'ro')

This is the image I get:

redpajama_set_name: RedPajamaStackExchange


text:
LeanDNA Raises $15M for Software That Helps Factories Run More Efficiently
by Ellen Glover
Courtesy of LeanDNA
LeanDNA, a platform that helps factories all over the world improve their performance, has secured $15 million in Series B funding led by S3 Ventures.
They're Hiring | View 9 JobsLeanDNA is Hiring | View 9 Jobs
Richard Lebovitz, the company's CEO and founder, has almost three decades of experience in supply chain and manufacturing. Over that time, he says he noticed two big trends that are complicating the industry.
One, product customization is really important nowadays.
"If you look at aircraft, for example, 10 or 15 years ago it was all about flying safe and efficient," Lebovitz said. "Now, it's about what seat color do you want? What cabin interior colors do you want? What sort of electronics do you want?"
Two, factories have moved away from their original model of doing everything from buying and forming raw materials to shipping a finished product. Now, they're focused mainly on final assembly and using global network of suppliers to do the rest.
But, Lebovitz believes that, in all this innovation in the manufacturing industry at large, factories themselves have been left behind. They're still using Excel and other simpler methods to keep track of their inventory and optimize their business.
"Ten or 15 years ago you could do this in Excel, but with the trends and the fast pace of factories and the complexity of the industry, spreadsheets just don't work anymore," Lebovitz said. "We felt that there's gotta be a better way."
LeanDNA was created to better help factories improve on-time delivery, optimize their inventory and reduce their operating costs. Its purpose-built SaaS analytics software uses artificial intelligence to suggest important things the analyst needs to do to efficiently manage their factory's inventory and improve profitability. It also provides a workflow that lets leaders and managers monitor their factories' performance.
What makes LeanDNA special, Lebovitz explains, is its ability to get to the bottom of problems its software exposes.
"We have the ability to do root cause analysis — what caused the issue and why," Lebovitz said. "By understanding not only the prioritization of the action but also the root cause, that combination is what allows our factories to really drive significant results."
Lebovitz, a Dallas-native who attended University of Texas-Austin, chose this city as a home base for LeanDNA after returning from a stint in California.
"I think Austin is really equal in terms of matching the capabilities that you'd find in the Bay Area now," Lebovitz said. "There's a lot of companies that have grown here, so it's easier to find people with startup experience. Even in the supply chain manufacturing area, you see more and more companies that are starting to appear here in Austin."
Now, LeanDNA is in 12 countries, working with global manufacturers like Spirit AeroSystems, PSG Dover and Integra LifeSciences. The company plans to use this most recent funding to expand its customer base.
LeanDNA
Jobs at LeanDNA8 open jobs
Data + Analytics
Data Integration Engineer
Senior Technical Product Manager
Enterprise Account Executive - Mid-Atlantic
Enterprise Account Executive -- Midwest
Inside Sales Representative (ISR)
Marketing Campaigns Manager
Content Marketing Specialist


text:
Is Leonie Page Gay?
I can see that you are Looking for the truth about Leonie Page Orientation, however, allow me to answer all of your questions. Read on, and you'll find out what about it.
Leonie Page Gay-o-Meter
Leonie Page Photos
Leonie Page friends and family support his announcement, also they Don't question his sexual tastes. It is difficult to tell if there's a reality to it. We are in need of just a little bit more proof than a statements that are fabricated.
People from entourage stand by what he stated, and Since they say there is nothing to tell they don't wish to disclose any info. Whether there is truth to this or not, I'll leave it up to you. But I say we want a bit longer than that.
Close friends and family of Leonie Page state that there is no Fact to what folks are saying concerning his sexual orientation. I can not honestly say that I think them. From where I stand, I want some more evidence.
Members of Leonie Page close friends deny any rumor that he Would be homosexual. They would, wouldn't they? I don't know whether they are telling the truth or not, but what I do understand is I want more evidence than a networking statements that are social.
Sometimes you can tell a great deal about a person just by looking At the people he surrounds himself with. They like to surround themselves, although not all folks hang out with other people who have exactly the identical preferences. There's a possibility that the person who you think to be gay told the group.
If they invest a lot of time together at one another's homes, you might be right.
There show Friends are, and I will tell you that you are. Just look at his friends if you suspect someone might be gay. Tend to stick together since they can express themselves with different individuals, although which might not be the situation. Odds are that he has already told his group about his sexual orientation. In addition, they might be spending plenty of time which may just confirm your feelings.
You can tell a lot about a person judging from the group he's A component of. Just pay attention to his buddies should you suspect that somebody is homosexual. Most of the times it's going to be a lot more easy for a person to surround himself with all individuals of the identical tastes because he might get the sympathy he wants to say himself. It is likely he came out into them, something that brings comfort to him. Another sign may be the simple fact that the individual in question crashes at his new friends than normal.
Only look at if You'd like to see the nature of a man His buddies. Pay attention with whom he surrounds himself times. Men and Women tend to stick for their own, although it's not always the case, Instead of being part of groups which don't understand them. They're more Likely to come out of the cupboard facing gay people than in front of Straight ones. Furthermore, if the person spends a Whole Lot of Time at one of his gay friend's home, chances are that he's gay.
On the flip side, there are actors. When a famous Person reveals the fact that he's homosexual, people tend to react. They would consider it a act and will encourage that particular celebrity. If a person famous reveals his sexual orientation that was, it's considered a Public Relations stunt. The press will divert its attention and it'll improve his career. The example is Caitlyn Jenner. She's a TV show after she revealed that she describes as a girl.
With folks, things are entirely different. When Their new-found sexual orientation is disclosed by them, everybody praises and encourages them like it were a gesture. A shift in the preference of a star means more attention. Among the best examples I can offer you will be Kristen Stewart. She received lots of characters, both after she'd told everyone she's, in fact, a lesbian. What do you predict that?
Matters are different for celebrities. When a celebrity comes out As gay, individuals are supporting, as though it were a sort of action and extremely encouraging. Because there is a good deal it means a whole lot. The power of media is good. Take a look at what occurred to Kaitlyn Jenner. Bruce became Caitlyn, and Caitlyn obtained a brand new TV series She was not well worth it if she was Bruce, so you see where I am going for this.
Famous people have it easy. They could afford a PR disaster, But they don't get that most of the times. They receive support from each of their fans and they're praised for their courage of coming out as homosexual. Each of the press turns its attention on such subject. From Keeping Up with all the Kardashians, can you recall Bruce Jenner? He turned into Caitlyn Jenner and got a whole TV series. What about this career boost?
Is Leonie Page gay? Conclusion
I'd love it if folks left their prejudice behind. There Are kind and nice folks in the world who show their support for the community. However, there are some who don't, and they are completely. Mentality is a tough situation to change.