text:
\begin{document}

\title[Spectral gap]{\bf Spectral gap for spherically symmetric log-concave probability measures, and beyond}

\author{Michel Bonnefont}
\address{Michel BONNEFONT \\ Institut de Math\'ematiques de Bordeaux, Universit\'e  de Bordeaux, 351 cours de la Lib\'eration,
33405 Talence, France.
}
\thanks{MB is partially supported by the French ANR-12-BS01-0013-02 HAB project}
\email{michel.bonnefont@math.u-bordeaux.fr}
\urladdr{http://www.math.u-bordeaux.fr/~mibonnef/}

\author{Ald\'eric Joulin}
\address{Ald\'eric JOULIN (corresponding author) \\ Universit\'e de Toulouse,
Institut National des Sciences Appliqu\'ees,
Institut de Math\'ematiques de Toulouse,
F-31077 Toulouse,
France.
}
\thanks{AJ is partially supported by the French ANR-2011-BS01-007-01 GEMECOD and ANR-12-BS01-0019 STAB projects}
\email{ajoulin@insa-toulouse.fr}
\urladdr{http://www.math.univ-toulouse.fr/~joulin/}

\author{Yutao Ma}
\address{Yutao MA\\ School of Mathematical Sciences $\&$ Lab. Math. Com. Sys., Beijing Normal University, 100875 Beijing, China.}
\thanks{YM is partially supported by NSFC 11101040, 11371283, 985 Projects and the Fundamental Research Funds for the Central Universities.}
\email{mayt@bnu.edu.cn}
\urladdr{http://math.bnu.edu.cn/~mayt/}

\keywords{Spectral gap; Diffusion operator; Poincar\'e-type inequalities; Log-concave probability measure.}

\subjclass[2010]{60J60, 39B62, 37A30, 58J50.}

\maketitle

\begin{abstract}
Let $\mu$ be a probability measure on $\rr^n$ ($n \geq 2$) with Lebesgue density proportional to $e^{-V (\Vert x\Vert )}$, where $V : \rr_+ \to \rr$ is a smooth convex potential. We show that the associated spectral gap in $L^2 (\mu)$ lies between $(n-1) / \int_{\rr^n} \Vert x\Vert ^2 \mu(dx)$ and $n / \int_{\rr^n} \Vert x\Vert ^2 \mu(dx)$, improving a well-known two-sided estimate due to Bobkov. Our Markovian approach is remarkably simple and is sufficiently robust to be extended beyond the log-concave case, at the price of potentially modifying the underlying dynamics in the energy, leading to weighted Poincar\'e inequalities. All our results are illustrated by some classical and less classical examples.
\end{abstract}

\section{Introduction}

Given a complete Riemannian manifold endowed with a smooth probability measure $\mu$, we associate a canonical non-negative diffusion operator $- \LL_\mu$ that is self-adjoint in $L^2 (\mu)$. Then a natural question arises: does the Markovian operator $-\LL _\mu$ have a gap in the bottom of its spectrum ? If yes, can we estimate the size of this gap, in particular with respect to the dimension of the manifold ? Since the pioneer work of Lichnerowicz at the end of the fifties, these questions have attracted the attention of many mathematicians in the last few decades. If $\lambda _1 (- \LL _\mu) $ denotes the size of this gap, then it can be alternatively formulated as the optimal constant in the Poincar\'e inequality. Such an object, which is a key ingredient in the theory of functional inequalities, controls the variance of functions in terms of the $L^2$-norm of their gradient. From analytical and probabilistic perspectives, the existence of a spectral gap is of crucial importance since on the one hand it governs the convergence to equilibrium of the underlying Markov process and on the other hand it has fundamental consequences in the so-called concentration of measure phenomenon but also in isoperimetric problems and their deep connections to Ricci curvature in geometry. The interested reader is referred to the set of notes of Ledoux \cite{ledoux_berlin,Ledoux} and also to the recent monograph \cite{BGL} for a nice introduction to the topic, with historical references and precise credit for this large body of work.

In the case of probability measures $\mu$ on the Euclidean space $(\rr^n , \Vert \cdot \Vert )$ with Lebesgue density proportional to $e^{-V}$, where $V$ is some smooth potential, a natural Markovian operator is given for all smooth enough functions $f$ by
$$
\LL _\mu f = \Delta f - \nabla V \cdot \nabla f .
$$
Using for instance the famous Bakry-\'Emery criterion \cite{bakry_emery} on the basis of the so-called $\Gamma_2$-calculus, a positive lower bound $K$ on the Hessian of $V$ ensures the existence of the spectral gap and moreover it is at least $K$. However, if $V$ is assumed to be only convex or if it exhibits some convexity only at infinity (allowing concavity in a localized region of the space), then there might be a spectral gap whereas the Bakry-\'Emery criterion fails. In the last thirty years, there has been many rooms of improvements of such a result and in many different directions. Recently, Bobkov \cite{bobkov_spherical} obtained a nice estimate on the spectral gap for a class of log-concave probability measures. More precisely, such measures $\mu$ are spherically symmetric and log-concave, meaning that the potential $V$ is radial and convex: it can be written as $V(x) = V(\Vert x \Vert)$ with $V : (0,+\infty) \to (0,+\infty)$ a convex function (by abuse of notation, we shall use all along the paper the same notation for the functions defined on $\rr^n$ and their radial version defined on $(0,+\infty)$). Then the operator above rewrites as
\[
\LL _\mu f( x)  = \Delta f (x) - \frac{V' (\Vert x\Vert )}{\Vert x\Vert } \, x \cdot \nabla f( x) , \quad x\in \rr^n ,
\]
and then his famous theorem can be stated as follows.
\bthm[Bobkov's theorem \cite{bobkov_spherical}]
\label{theo:bobkov}
Let $\mu$ be a spherically symmetric log-concave probability measure on $\rr^n$ ($n \geq 2$). Then the spectral gap $\lambda_1 (-\LL_\mu)$ satisfies
$$
\frac{n}{13 \int_{\rr ^n} \Vert x\Vert ^2 \mu (dx)} \leq \lambda_1 (-\LL_\mu) \leq \frac{n}{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx)}  .
$$
\nthm

A consequence of Bobkov's theorem is that the class of spherically symmetric log-concave probability measures satisfies the KLS conjecture: if the covariance of $\mu$ is the identity, meaning that for all $\theta \in \rr ^n$,
$$
\int _{\rr ^n} \vert x \cdot \theta \vert ^2 \mu (dx) = \Vert \theta \Vert ^2,
$$
then the spectral gap $\lambda_1 (-\LL_\mu)$ is bounded from below by an universal constant, independent from the dimension $n$. Recall that the KLS conjecture was originally introduced in \cite{KLS} for the Cheeger inequality but it is actually equivalent to state it for the Poincar\'e inequality according to a result of Ledoux \cite{Ledoux}, generalized recently by Milman \cite{milman}. \vspace{0.1cm}

The next theorem is the main result of the paper and is an improvement of Bobkov's theorem. In particular it gives the exact asymptotics of the spectral gap as the dimension of the space goes to infinity.
\bthm[Bobkov's theorem revisited]
\label{theo:main}
Under the assumptions and notation of Bobkov's theorem, the spectral gap $\lambda_1 (-\LL_\mu)$ satisfies
$$
\frac{n-1}{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx)} \leq \lambda_1 (-\LL_\mu) \leq \frac{n}{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx)}  .
$$
\nthm
Certainly, such a result can be adapted without additional effort to the case of a measure $\mu$ whose support is a centered Euclidean ball provided Neumann's boundary conditions are assumed for the underlying dynamics (we will investigate such an example later). \vspace{0.1cm}

As we will see in the sequel, there are two main ingredients in the proof of Theorem \ref{theo:main} which reveals to be remarkably simple. The first one is a comparison between the spectral gap of the full dynamics $-\LL_\mu$ and that of its one-dimensional radial part. Such a comparison is conceivable since the probability measure $\mu$ is close to the product measure whose marginals are its spherical and radial parts, and because of the well-known tensorization property of the Poincar\'e inequality. The second main point of the proof of Theorem \ref{theo:main} focuses on a careful estimation of the spectral gap of the one-dimensional radial part, based on a simple, but somewhat very useful, result given recently by two of the three authors \cite{bj}. \vspace{0.1cm}

Before entering into more details, let us provide the organization of the paper. In Section \ref{sect:prelim}, we briefly introduce some basic material on Markovian diffusion operators, their spectral gap and the links with the Poincar\'e inequality. Section \ref{sect:proof} is devoted to the proof of our main result, Theorem \ref{theo:main}, which is illustrated in Section \ref{sect:ex} by some examples such as probability measures $\mu$ with Lebesgue densities proportional to $e^{- \Vert x\Vert ^\alpha /\alpha}$, where $\alpha$ is a parameter greater than or equal to 1 to ensure the log-concavity of the measure $\mu$. For $\alpha = 1$ it corresponds to the Euclidean version of the exponential distribution and for $\alpha = 2$ to the classical Gaussian setting, whereas we obtain the uniform measure on the Euclidean unit ball as $\alpha \to +\infty$. Finally, we extend in Section \ref{sect:beyond} this technology to non-necessarily log-concave distributions by modifying in a convenient way the original dynamics, leading to weighted Poincar\'e inequalities. To observe the relevance of our approach beyond the log-concave case, we investigate in detail the example of heavy-tailed probability measures such as the generalized Cauchy distributions and we finish this work by revisiting the Gaussian case.

\section{Preliminaries}
\label{sect:prelim}

\subsection{Generalities on Markovian diffusion operators}
Let $(\MM,g)$ be a smooth complete connected $n$-dimensional Riemannian manifold with metric $g$. Denote $\mu$ the probability measure with density with respect to the volume measure of $(\MM,g)$ proportional to $e^{-V}$, where $V$ a smooth function on $(\MM,g)$. Let $\Delta_\MM$ be the Laplace-Beltrami operator on $(\MM,g)$ and denote $\LL _\mu$ the Markovian diffusion operator initially defined on the space $\CC ^\infty(\MM)$ of infinitely differentiable real-valued functions on $(\MM,g)$ by
$$
\LL_\mu f = \Delta _\MM f - \nabla _\MM V \cdot \nabla _\MM f ,
$$
where $\nabla_\MM $ stands for the Riemannian gradient and the underlying scalar product $\cdot$ is given with respect to the metric $g$ (denote $\vert \cdot \vert $ the associated norm). Let $\CC_0 ^\infty(\MM)$ be the subspace of $\CC ^\infty(\MM)$ consisting of compactly supported functions. The operator $\LL_\mu$ is symmetric on $\CC_0 ^\infty(\MM)$ with respect to the measure $\mu$, i.e. for every $f_1,f_2 \in \CC _0 ^\infty(\MM)$,
\[
\EE _\mu (f_1,f_2) := \int_\MM f_1 (-\LL_\mu f_2) d\mu = \int_\MM (-\LL _\mu f_1) f_2 d\mu  = \int_\MM \nabla _\MM f_1 \cdot \nabla _\MM f_2 d\mu,
\]
hence $\LL_\mu$ is non-positive on $\CC_0 ^\infty(\MM)$. Since the manifold is complete, a straightforward adaptation of the argument in \cite{strichartz} (see for example \cite{grigoryan} and the references therein) shows that $(\LL_\mu , \CC_0 ^\infty(\MM))$ is essentially self-adjoint in $L^2 (\mu)$, that is, it admits a unique self-adjoint extension (still denoted $\LL_\mu$) with domain $\DD (\LL_\mu) \subset L^2(\mu)$ in which the space $\CC_0 ^\infty(\MM)$ is dense for the norm
$$
\Vert f\Vert _{\LL_\mu} := \left( \Vert f \Vert ^2 _{L^2 (\mu)} + \Vert \LL_\mu f \Vert ^2 _{L^2 (\mu)} \right) ^{1/2} .
$$
In other words the space $\CC_0 ^\infty(\MM)$ is a core of the domain $\DD (\LL_\mu)$. In general, it is not easy to describe its domain, however it is easier to describe that of the associated Dirichlet form. The closure $(\EE _\mu , \DD (\EE _\mu))$ of the bilinear form $(\EE _\mu , \CC_0 ^\infty(\MM))$ is a Dirichlet form on $L^2(\mu)$ and we have the dense inclusion $\DD (\LL_\mu) \subset \DD (\EE _\mu)$ for the norm
$$
\Vert f\Vert _{\EE _\mu} := \left( \Vert f \Vert ^2 _{L^2 (\mu)} + \EE _\mu (f,f) \right) ^{1/2} ,
$$
where the domain $\DD (\EE _\mu)$ is the Sobolev space
\[
W^{1,2}(\mu) = \left\{ f\in L^2(\mu) : \vert \nabla _\MM f \vert \in  L^2(\mu) \right\} .
\]
If the manifold $(\MM,g)$ admits a smooth boundary, one can still consider the operator $\LL_\mu$ provided Neumann boundary conditions are assumed (in this case, the operator has to be defined initially on the space of smooth functions with vanishing normal derivative at the boundary). We refer to \cite{davies} for more details.

\subsection{Spectral gap} Let us consider the spectrum in $L^2(\mu)$ of the operator $-\LL_\mu$. First, it is a non-negative self-adjoint operator and thus its spectrum is included in $[0,+\infty)$. Since $\mu$ is assumed to be a probability measure, the constant functions belong to the domain $\DD (\LL_\mu)$ and the identity $\LL_\mu 1 = 0$ means that 0 is an eigenvalue, the constant functions being the associated eigenfunctions (by eigenvalue and eigenfunction, we mean the numbers $\lambda$ and the non-null functions $f\in \DD (\LL_\mu)$ satisfying $-\LL_\mu f = \lambda f$). In particular, the ergodicity property holds, i.e., the only functions $f$ in the domain $\DD (\LL_\mu)$ satisfying $\LL_\mu f = 0$ are the constant functions. Sometimes, it may happen that the operator $- \LL_\mu$ exhibits a gap in the bottom of its spectrum. More precisely, the spectral gap of the operator $- \LL_\mu$, usually denoted $\lambda_1 (- \LL _\mu)$, is the largest real $\lambda$ such that its spectrum lies in $\{0\} \cup [\lambda,+\infty)$. In other words, it is characterized by the variational identity
\begin{equation}
\label{eq:variational}
\lambda_1 (- \LL _\mu) = \inf \left \{ \int_\MM f(-\LL_\mu f) d\mu : f \in \DD (\EE_\mu) , \, f \perp  \mathbf 1, \, \int_\MM f^2 d\mu = 1 \right \} ,
\end{equation}
where $f \perp \mathbf 1$ means that $f$ is orthogonal to the constants, that is, $\int_\MM f d\mu =0$. Among the potential cases of interest we have in mind, one of the most famous example is the Laplace-Beltrami operator on the unit sphere $\s_{n-1} \subset \rr ^n$ ($n \geq 2$), endowed with the normalized Hausdorff measure $\sigma_{n-1}$ ($V\equiv 0$ in this case) and for which we have
$$
\lambda_1 (-\Delta _{\s ^{n-1}}) = n-1 .
$$
Writing \eqref{eq:variational} slightly differently, the spectral gap is characterized by the best (i.e., largest) constant $\lambda$ in the following Poincar\'e inequality,
\begin{equation}
\label{eq:poincare}
\lambda \Var _\mu (f) \leq \int _\MM  f (-\LL _\mu f) d\mu = \int_\MM \vert \nabla _\MM f \vert ^2 d\mu ,
\end{equation}
for all $f\in \DD(\EE_\mu)$ or, equivalently, for all $f\in \CC_0^\infty(\MM)$ by self-adjointness. Above $\Var _\mu (f)$ stands for the variance of $f$ under $\mu$, that is,
$$
\Var _\mu (f) := \int_\MM f^2 d\mu - \left( \int_\MM f d\mu \right) ^2 .
$$
As mentioned in the Introduction, such an inequality reveals to be very useful for probabilistic issues such as the $L^2$-convergence to equilibrium for the underlying Markov process, but also in analysis through measure concentration, isoperimetry and their links to geometry via the key role of the Ricci curvature.
\section{Preparation and proof of Theorem \ref{theo:main}}
\label{sect:proof} This part is devoted to the preparation and the proof of our main contribution Theorem \ref{theo:main} stated in the Introduction. Recall that the underlying Markovian operator of interest acts on functions $f \in \CC_0^\infty(\R^n)$ as
\begin{equation}
\label{eq:operator_lmu}
\LL _\mu f( x): = \Delta f (x) - \frac{V' (\Vert x\Vert )}{\Vert x\Vert } \, x \cdot \nabla f( x) , \quad x\in \rr^n ,
\end{equation}
where $\Delta $ and $\nabla$ denote the Euclidean Laplacian and gradient, respectively. The potential $V$ is assumed to be convex and the symmetric and non-positive operator $(\LL_\mu , \CC_0^\infty(\R^n))$ with respect to the measure $\mu$ is essentially self-adjoint in $L^2(\mu)$. \vspace{0.1cm}

The proof of Theorem \ref{theo:main} is based on two complementary lemmas, which follow a specific strategy:
\begin{itemize}
\item[$\circ$] to understand the competition between the spectral gaps of the full dynamics and its one-dimensional radial part;
\item[$\circ$] to analyse and estimate the spectral gap of the radial part.
\end{itemize}

\subsection{Spectral gap estimates for Sturm-Liouville dynamics}
We begin by the second point and consider the one-dimensional case. Let $\nu$ be a one-dimensional probability measure whose support is some interval $I$, possibly finite. We assume that its Lebesgue density on $I$ is proportional to $e^{-U}$, where $U$ is a smooth convex potential on $I$. Then the log-concave measure $\nu$ is invariant for the following Sturm-Liouville dynamics
\begin{equation}
\label{eq:operator_lnu}
\LL _\nu f(r) : = f''(r) - U'(r) f'(r) , \quad r \in \inte (I) ,
\end{equation}
where $\inte (I)$ denotes the interior of $I$. Obviously, if $I$ admits a boundary then we assume Neumann's conditions. Denote $\lambda_1 (-\LL_\nu)$ the associated (Neumann) spectral gap corresponding to the optimal constant in the following Poincar\'e inequality: for any $f \in W^{1,2}(\nu)$,
\begin{equation}
\label{eq:poincare1d}
\lambda \Var _\nu (f) \leq \int _I  f (-\LL _\nu f) d\nu = \int_I ( f' )^2 d\nu .
\end{equation}
Among the various criteria available to estimate the spectral gap in dimension 1 (not only in the log-concave case), one has recently been put forward in \cite{bj} by two of the three authors and can be reformulated as follows: if $\CC^\infty (I)$ stands for the space of real-valued smooth functions on $I$, then we have
\begin{equation}
\label{e:chen}
\lambda_1 (-\LL_\nu) \geq \sup_{f\in \mathcal F} \inf_{x\in I}  V_f (x) ,
\end{equation}
where
\[
 V_f := - \frac{(\LL _\nu f)'} {f'} \quad \textrm{ with } \quad \mathcal F := \left\{ f\in \CC^\infty (I) : f' > 0 \mbox{ on } \inte (I) \right\} .
\]
Moreover the equality holds in \eqref{e:chen} if $\lambda_1 (-\LL_\nu)$ is an eigenvalue. Such a result already appeared via another approach in the work of Chen and Wang \cite{chen_wang} (see also \cite{chen} for an alternative form of this variational formula and \cite{djellout} which turns the form into the present one) and has been rediscovered in \cite{bj} through intertwining relations between gradients and semigroups. In general, it provides ``easy-to-verify" conditions ensuring the existence of a spectral gap for the dynamics. However, this formula can be difficult to use in order to obtain a good estimate since one has to guess what the best function $f$ is, and this is nothing but the eigenfunction (when it exists) associated to the spectral gap. \vspace{0.1cm}

To overcome this difficulty, at least in the log-concave case, the following result derived from \cite{bj} is well-adapted.
\blem[Bonnefont-Joulin]
\label{lemme:bj}
Assume that the second derivative of $U$ is positive on $\inte (I)$. Then the spectral gap $\lambda_1 (-\LL_\nu)$ of the operator $- \LL_\nu$ defined in \eqref{eq:operator_lnu} satisfies
\begin{equation}
\label{eq:bj_gen}
\lambda_1 (-\LL_\nu) \geq \frac{1}{\int_I \frac{1}{U''} d\nu } .
\end{equation}
In particular if $\nu$ has Lebesgue density on $\rr_+$ proportional to $r^{n-1} e^{- V(r)}$ ($n\geq 2$) and $V$ is a smooth convex function, then we have
\begin{equation}
\label{eq:bj_radial}
\lambda_1 (-\LL_\nu) \geq \frac{n-1}{\int_0 ^{+\infty} r^2 \nu (dr) } .
\end{equation}
\nlem
The first estimate above is an integrated version of the one-dimensional Bakry-\'Emery criterion. Moreover, it can be seen as a one-dimensional version of a nice result of Veysseire \cite{veysseire} given in the context of compact Riemannian manifolds, our second derivative $U''$ being replaced by the Ricci curvature lower bound. Although the inequality \eqref{eq:bj_gen} is sharp in some cases like for the standard Gaussian distribution associated to the Ornstein-Uhlenbeck process on $\rr$, it can sometimes be non-optimal. For instance if $1/U''$ is not $\nu$-integrable, then the latter lower bound does not provide any information on the spectral gap. However, it reveals to be convenient at least when $U''$ tends to 0 sufficiently slowly at the boundary of $I$ (at infinity if $I$ is unbounded), since the Bakry-\'Emery criterion is not available in this context (we would have $\lambda_1 (-\LL_\nu) \geq \inf U'' = 0$). \vspace{0.1cm}

The second inequality \eqref{eq:bj_radial} is the one we are interested in. The measure $\nu$, which belongs to the class of the so-called log-concave distributions of order $n$ according to Bobkov's definition \cite{bobkov_gaussian}, is the radial part of the spherically symmetric measure $\mu$ introduced in Theorem \ref{theo:main}. In particular, due to the presence of the important prefactor $r^{n-1}$, a log-concave distribution of order $n$ is ``more" log-concave than a classical log-concave measure. Hence we expect some interesting consequences of the general inequality \eqref{eq:bj_gen} applied to this class of log-concave measures, and this is indeed the case: setting
$$
U(r) := V(r) - (n-1) \log (r), \quad r >0,
$$
one has
$$
U''(r) = V''(r) + \frac{ n-1}{r^2} \geq \frac{ n-1}{r^2}  , \quad r >0,
$$
so that the inequality \eqref{eq:bj_radial} is a straightforward consequence of \eqref{eq:bj_gen}.

\subsection{Spectral comparison with the radial part}
As announced, we turn now to the comparison between the spectral gaps of the full dynamics and its one-dimensional radial part. We mention that the next result holds for general spherically symmetric probability measure (not necessarily log-concave) admitting a finite second moment.
\blem[Spectral comparison]
\label{lemme:gen-spherically}
Let $\mu$ be a spherically symmetric probability measure on $\R^n$ ($n \geq 2$) with Lebesgue density proportional to $e^{-V(\Vert x\Vert )}$ and let $ \nu$ be its radial part, that is, the one-dimensional probability measure on $(0,+\infty)$ whose Lebesgue density is  proportional to $r^{n-1} e^{- V(r)}$. Then the spectral gaps $\lambda_1 (-\LL_\mu)$ and $\lambda_1 (-\LL_\nu)$ of the operators $-\LL_\mu$ and $-\LL_\nu$ defined in \eqref{eq:operator_lmu} and \eqref{eq:operator_lnu} respectively, satisfy
\[
\min \left \{ \lambda_1 (-\LL_\nu) , \frac{n-1}{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx) }\right \}  \leq \lambda_1 (-\LL_\mu) \leq \min \left \{ \lambda_1 (-\LL_\nu) , \frac{n}{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx) }\right \}.
\]
\nlem
\bprf
The upper bound in the right-hand-side of the desired estimates is straightforward. Indeed, considering only radial functions, the variational characterization \eqref{eq:variational} of the spectral gap gives
\[
\lambda_1 (-\LL_\mu) \leq  \lambda_1 (-\LL_\nu),
\]
whereas the second part is obtained by choosing in the Poincar\'e inequality \eqref{eq:poincare} the linear function $f(x) = \sum_{k=1} ^n x_k$. \vspace{0.1cm}

Let us concentrate our attention on the challenging lower bound. The proof given below is a refinement of Bobkov's argument used to derive his famous theorem.
The principle is to benefit from the spherically invariance of $\mu$ and then to use the tensorization property of the Poincar\'e inequality
with respect to the radial and spherical measures. More precisely, the measure $\mu$ is the image measure
of the product measure $\nu \otimes \sigma_{n-1}$ by the mapping $(r,\theta) \in (0,+\infty) \times \s ^{n-1} \to r \theta$,
where $\nu$ is the one-dimensional probability measure on $\rr_+$ corresponding to the radial part, that is, with Lebesgue density proportional to $r^{n-1} e^{-V(r)}$.
As mentioned earlier, such a measure is invariant for the Sturm-Liouville dynamics
\begin{equation}
\label{eq:SL}
\LL _\nu f (r) := f''(r) - \left( V'(r) - \frac{n-1}{r} \right) f'(r) , \quad r >0,
\end{equation}
with additional Neumann's condition at the boundary $r=0$. Let $f \in \CC _0 ^\infty (\rr^n)$ and denote $g$ the bivariate function defined on $(0,+\infty) \times \s ^{n-1}$ by  $g(r,\theta) := f(r\theta)$. First using the Poincar\'e inequality \eqref{eq:poincare1d} for the radial operator $-\LL _\nu$, we have
\begin{eqnarray}
\label{eq:poincare_nu}
\nonumber \int _{\rr^n} f ^2 d\mu & = & \int _{\s ^{n-1}} \left( \int_0 ^{+\infty} g(r,\theta) ^2 \nu (dr) \right) \sigma_{n-1}(d\theta) \\
\nonumber & \leq & \int _{\s ^{n-1}} \left( \int_0 ^{+\infty} g(r,\theta) \nu (dr) \right) ^2 \sigma_{n-1}(d\theta) \\
& & + \frac{1}{\lambda_1 (-\LL_\nu)} \int _{\s ^{n-1}} \int_0 ^{+\infty} \left \vert \partial_r g(r,\theta) \right \vert ^2 \nu (dr) \sigma_{n-1}(d\theta) \\
\nonumber & = & \int _{\s ^{n-1}} \left( \int_0 ^{+\infty} g(r,\theta) \nu (dr) \right) ^2 \sigma_{n-1}(d\theta) \\
\nonumber & & + \frac{1}{\lambda_1 (-\LL_\nu)} \int _{\s ^{n-1}} \int_0 ^{+\infty} \left \vert \theta \cdot \nabla f (r\theta) \right \vert ^2 \nu (dr) \sigma_{n-1}(d\theta).
\end{eqnarray}
Now if we set
$$
h(\theta) := \int_0 ^{+\infty} g(r,\theta) \nu (dr) ,
$$
then the integral of $h$ under the measure $\sigma_{n-1}$ is nothing but the integral of $f$ under $\mu$. Moreover the length of the intrinsic gradient on the sphere applied to $h$ gives
\begin{eqnarray}
\label{eq:CS}
\nonumber \Vert \nabla _{\s ^{n-1}} h (\theta) \Vert ^2 & = & \left \Vert \int_0 ^{+\infty} r \Pi _{\theta ^\bot} (\nabla f) (r\theta) \nu (dr) \right \Vert ^2 \\
& \leq & \int_0 ^{+\infty} r^2 \nu (dr) \int_0 ^{+\infty} \left \Vert \Pi _{\theta ^\bot} (\nabla f) (r\theta) \right \Vert ^2 \nu (dr) \\
\nonumber & = & \int_{\rr ^n} \Vert x\Vert ^2 \mu (dx) \int_0 ^{+\infty} \left \Vert \Pi _{\theta ^\bot} (\nabla f) (r\theta) \right \Vert ^2 \nu (dr),
\end{eqnarray}
where to obtain \eqref{eq:CS} we used Cauchy-Schwarz inequality, the notation $\Pi _{\theta ^\bot}$ standing for the orthogonal projection on $\theta ^\bot$.
Since the Poincar\'e inequality \eqref{eq:poincare} is satisfied for the spherical Laplacian $- \Delta_{\s ^{n-1}}$, we have
\begin{eqnarray*}
\int _{\s ^{n-1}} h(\theta) ^2 \sigma_{n-1}(d\theta) & \leq & \left( \int _{\rr ^n } f d\mu \right) ^2 + \frac{1}{\lambda_1 (-\Delta_{\s ^{n-1}})} \int _{\s ^{n-1}} \Vert \nabla _{\s ^{n-1}} h (\theta) \Vert ^2 \sigma_{n-1}(d\theta) \\
& \leq & \left( \int _{\rr ^n } f d\mu \right) ^2 + \frac{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx) }{\lambda_1 (-\Delta_{\s ^{n-1}})} \int _{\s ^{n-1}} \int_0 ^{+\infty} \left \Vert \Pi _{\theta ^\bot} (\nabla f) (r\theta) \right \Vert ^2 \nu (dr) \sigma_{n-1}(d\theta) .
\end{eqnarray*}
Together with the previous estimate we obtain
\begin{eqnarray*}
\Var _\mu (f) & \leq & \frac{1}{\lambda_1 (-\LL_\nu)} \int _{\s ^{n-1}} \int_0 ^{+\infty} \left \vert \theta \cdot \nabla f (r\theta) \right \vert ^2 \nu (dr) \sigma_{n-1}(d\theta) \\
& & + \frac{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx) }{\lambda_1 (-\Delta_{\s ^{n-1}})} \int _{\s ^{n-1}} \int_0 ^{+\infty} \left \Vert \Pi _{\theta ^\bot} (\nabla f) (r\theta) \right \Vert ^2 \nu (dr) \sigma_{n-1}(d\theta) \\
& \leq & \max \left \{ \frac{1}{\lambda_1 (-\LL_\nu)} , \frac{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx) }{\lambda_1 (-\Delta_{\s ^{n-1}})} \right \} \int _{\s ^{n-1}} \int_0 ^{+\infty} \left \Vert \nabla f (r\theta) \right \Vert ^2 \nu (dr) \sigma_{n-1}(d\theta) ,
\end{eqnarray*}
since by the Pythagorean theorem,
\[
\Vert \nabla f \Vert ^2 = \left \vert \theta \cdot \nabla f \right \vert ^2 + \left \Vert \Pi _{\theta ^\bot} (\nabla f) \right \Vert ^2 .
\]
Therefore we get the inequality
\[
\lambda_1 (-\LL_\mu) \geq \min \left \{ \lambda_1 (-\LL_\nu) , \frac{\lambda_1 (-\Delta_{\s ^{n-1}})}{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx) }\right \} ,
\]
and finally using that $\lambda_1 (-\Delta_{\s ^{n-1}}) = n-1$ entails the desired lower bound. The proof is achieved.
\nprf

\subsection{Proof of Theorem \ref{theo:main}}
Once we have in mind the two previous lemmas, we can turn to the (brief) proof of our main result, Theorem~\ref{theo:main}.
\bprf[Proof of Theorem \ref{theo:main}]
As mentioned previously, the conclusion is a straightforward consequence of Lemmas \ref{lemme:bj} and \ref{lemme:gen-spherically} since we have
\[
\lambda_1 (-\LL_\nu) \geq \frac{n-1}{\int_{0}^{+\infty} r^2 \nu(dr)} = \frac{n-1}{\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx) }.
\]
\nprf

\section{Examples}
\label{sect:ex}
In order to illustrate our Theorem \ref{theo:main}, let us revisit some classical models for which we are able to estimate the associated spectral gap.

\subsection{Uniform measure on the Euclidean unit ball}
First we consider the basic example of the Euclidean unit ball $B_n$ in $\rr^n$ ($n \geq 2$). In this case we have $V \equiv 0$ and $\mu$ is the normalized Lebesgue measure on $B_n$, i.e., its Lebesgue density on $B_n$ is given by $1/\omega_n$, with $\omega _n$ the volume of $B_n$,
$$
\omega _n = \frac{\pi ^{n/2}}{\Gamma \left( \frac{n}{2} +1 \right)},
$$
and where $\Gamma$ is the standard Gamma function
$$
\Gamma (z) := \int_0 ^{+\infty} x^{z-1} e^{-x} dx, \quad z>0.
$$
In terms of dynamics, the corresponding Markov process is the standard Brownian motion in $B_n$ reflected at the boundary (encoding Neumann's boundary conditions for the Euclidean Laplacian $\Delta$).
Then we obtain the following corollary of Theorem~\ref{theo:main}.
\begin{cor}[Euclidean unit ball]
Let $\lambda_1 (-\Delta)$ be the Neumann spectral gap associated to the Laplacian $\Delta$ and the normalized Lebesgue measure on the Euclidean unit ball $B_n$ in $\rr^n$ ($n\geq 2$). Then we have
$$
\frac{(n-1)(n+2)}{n} \leq \lambda_1 (-\Delta) \leq n+2.
$$
In particular $\lambda_1 (-\Delta) \approx n$ as the dimension $n$ is large.
\end{cor}

\subsection{Exponential power distributions}
Let us turn to another example containing the last one as a limiting case. We consider the potential $V(r) = r^\alpha /\alpha$ on $\rr_+$ which is convex provided $\alpha \geq 1$, so that the associated measure of interest $\mu$ has Lebesgue density on $\rr^n$ ($n\geq 2$) proportional to $e^{- \Vert x\Vert ^\alpha /\alpha}$. Although the terminology is not entirely stabilized in the literature, such probability measures are sometimes called exponential power distributions. The underlying dynamics is then the following:
\begin{equation}
\label{eq:exp_power}
\LL _\mu f (x) = \Delta f (x) - \Vert x \Vert ^{\alpha -2} x \cdot \nabla f (x) , \quad x\in \rr ^n .
\end{equation}
Among all these models, some particular cases reveal to be very interesting: \vspace{0.1cm}

$\circ$ When $\alpha = 2$, the associated Markov process is the Ornstein-Uhlenbeck process and $\mu$ is the centered Gaussian probability measure on $\rr^n$ with the identity matrix $I$ as covariance matrix. \vspace{0.1cm}

$\circ$ When $\alpha = 1$, the measure $\mu$ is an exponential-type distribution. In other words it is comparable to the product measure for which all the coordinates are i.i.d. and exponentially distributed with common parameter 1, since the $\ell ^1$ and $\ell ^2$ norms are equivalent in $\rr^n$. \vspace{0.1cm}

$\circ$ When $\alpha$ tends to infinity, we recover the uniform measure on the Euclidean unit ball $B_n$ studied previously. \vspace{0.1cm}

We still use the abuse of notation $V(x) = V(\Vert x\Vert )$. Then the Hessian matrix of $V$ at point $x \in \rr ^n $ is
$$
\mbox{Hess } V (x) = \Vert x\Vert ^{\alpha -2} I + (\alpha -2) \Vert x\Vert ^{\alpha -4} x x^T ,
$$
where $x^T$ stands for the transpose of the column vector $x$. The eigenvalues are $(\alpha -1) \Vert x\Vert ^{\alpha -2}$ with $x$ as corresponding eigenvector, and also $\Vert x\Vert ^{\alpha -2}$ associated to the eigenvectors orthogonal to $x$. Therefore we observe that Bakry-\'Emery criterion, which is reduced in our setting to the existence of some positive number $K$, independent from $x$, such that
$$
\mbox{Hess } V \geq K I,
$$
where the inequality has to be understood in the sense of positive matrices, cannot be applied except in the Gaussian case $\alpha = 2$ (for which it is sharp since we have the well-known result $\lambda_1 (-\LL_\mu) = 1$). Indeed, if $\alpha \in [1,2)$ then the smallest eigenvalue is $(\alpha -1) \Vert x\Vert ^{\alpha -2}$ which tends to 0 at infinity, whereas in the case $\alpha >2$, the smallest eigenvalue is now $\Vert x\Vert ^{\alpha -2}$ which vanishes at the origin. \vspace{0.1cm}

Coming back to the consequences of Theorem \ref{theo:main}, we obtain for this model the following result.
\begin{cor}[Exponential power distributions]
Let $\lambda_1 (-\LL_\mu)$ be the spectral gap of the operator $-\LL_\mu$ given in \eqref{eq:exp_power}. Then we have in any dimension $n \geq 2$,
\beq
\label{power-law1}
\frac{(n-1) \Gamma \left(\frac{n}{\alpha}\right)}{\alpha ^{2/\alpha} \Gamma \left( \frac{n+2}{\alpha}\right)} \leq \lambda_1 (-\LL_\mu) \leq \frac{n \Gamma \left(\frac{n}{\alpha}\right)}{\alpha ^{2/\alpha} \Gamma \left(\frac{n+2}{\alpha}\right)}.
\nneq
More explicitly, the two following estimates hold:
\beq
\label{power-law2}
\frac{n-1}{n+1} \times n^{1-2/\alpha} \leq \lambda_1(-\LL_{\mu}) \leq \frac{n+2}{n} \times n^{1-2/\alpha} .
\nneq
In particular we have $\lambda_1(-\LL_{\mu}) \approx n^{1-2/\alpha}$ as the dimension $n$ is large.
\end{cor}

\bprf
To obtain the inequalities \eqref{power-law1}, we use Theorem \ref{theo:main} and compute both upper and lower bounds. To establish the two last estimates, our idea is to invoke the log-convexity of the Gamma function. More precisely, if $a,b$ are two parameters such that $a>0$ and $b\in [0,1]$, then the the log-convexity of $\Gamma$ entails the two inequalities:
$$
\Gamma (a+b) \leq \Gamma (a) ^{1-b} \, \Gamma (a+1) ^b = a^b \, \Gamma (a) ,
$$
and
$$
\Gamma (a+1) \leq \Gamma (a+b) ^{b} \, \Gamma (a+b+1) ^{1-b} = (a+b)^{1-b} \, \Gamma (a+b) ,
$$
which rewrite in a condensed form as
\begin{equation}
\label{eq:gamma1}
1 \leq \frac{\Gamma (a) \, a^b}{\Gamma (a+b)} \leq \left( \frac{a+b}{a} \right) ^{1-b} .
\end{equation}
Moreover, if $b\in [1,2]$, then
$$
\frac{\Gamma (a) \, a^b}{\Gamma (a+b)} = \frac{a}{a+b-1} \times \frac{\Gamma (a) \, a^{b-1}}{\Gamma (a+b-1)} ,
$$
and since $b-1 \in [0,1]$, we can use the inequalities \eqref{eq:gamma1} to get
\begin{equation}
\label{eq:gamma2}
\frac{a}{a+b-1} \leq \frac{\Gamma (a) \, a^b}{\Gamma (a+b)} \leq \left( \frac{a+b-1}{a} \right) ^{2-b} .
\end{equation}
Now we use the previous estimates with $a = n/ \alpha >0$ and $b = 2/ \alpha \in (0,2]$. Then two cases occur: \vspace{0.1cm}

$\circ$ on the one hand, if $\alpha \geq 2$, that is $b \in (0,1]$, then the inequalities \eqref{eq:gamma1} entail the estimates
$$
1 \leq \frac{\Gamma \left( \frac{n}{\alpha} \right) \left( \frac{n}{\alpha}\right) ^{2/\alpha}}{\Gamma \left( \frac{n}{\alpha} + \frac{2}{\alpha}\right)} \leq \left( \frac{n+2}{n}\right) ^{1- 2/\alpha} \leq \frac{n+2}{n} .
$$

$\circ$ on the other hand, if $\alpha \in [1,2]$, i.e., $b\in [1,2]$, then we obtain from \eqref{eq:gamma2} the inequalities

$$
\frac{n}{n+2 - \alpha} \leq \frac{\Gamma \left( \frac{n}{\alpha} \right) \left( \frac{n}{\alpha}\right) ^{2/\alpha}}{\Gamma \left( \frac{n}{\alpha} + \frac{2}{\alpha}\right)} \leq \left( \frac{n+2 - \alpha }{n}\right) ^{2- 2/\alpha} .
$$
Finally, since in both cases the lower and upper bounds are respectively greater than or equal to $n/(n+1)$ and less than or equal to $(n+2)/n$, the desired estimates \eqref{power-law2} hold.
\nprf
As expected, the spectral gap is dimension-free only in the Gaussian case, i.e., $\alpha = 2$, reflecting that $\mu$ is a product measure.  When $\alpha = 1$ and the measure $\mu$ is of exponential-type, we obtain from \eqref{power-law1} the nice estimates
$$
\frac{n-1}{n(n+1)} \leq \lambda_1 (-\LL_\mu) \leq \frac{1}{n+1} .
$$

\subsection{Spectral gap of the one-dimensional radial part}
Let us briefly discuss the spectral gap $\lambda_1 (-\LL_\nu)$ of the associated one-dimensional radial operator $-\LL _\nu$. Recall that this generator is defined in \eqref{eq:SL} and for exponential power distributions, we have
$$
\LL _\nu f (r) := f''(r) - \left( r^{\alpha-1} - \frac{n-1}{r} \right) f'(r) , \quad r >0,
$$
with additional Neumann's condition at the boundary $r=0$. In contrast to the Cauchy-like case which will be analyzed in the next section, the exact value of the spectral gap does not seem to be accessible. However we are able to give its behaviour in terms of the underlying dimension $n$. Some simple computations in the inequality \eqref{eq:bj_radial} together with the choice of the function $f(r ) = r^\alpha$ in the variational inequality of the spectral gap show that $\lambda_1 (-\LL_\nu)$ is of order $n^{1-2/\alpha}$ for large $n$, as in the case of the full dynamics. For the exponential distribution, that is $\alpha =1$, such a result might be recovered easily. Indeed observing that the measure $\nu$, which is the Gamma distribution with $n$ degrees of freedom, is the $n$-fold convolution product of the standard exponential distribution, we obtain immediately that $\lambda_1 (-\LL_\nu)$ is at least $1/n$ times the spectral gap for the exponential measure, which is $1/4$. On the other hand, a curious phenomenon appears in the limiting case $\alpha \to +\infty$. In this situation, the measure $\nu$ converges weakly to the probability measure with density $n r^{n-1}$ on $(0,1)$ and the (Neumann) generator of interest is
$$
\LL _\nu f (r) := f''(r) + \frac{n-1}{r} \, f'(r) , \quad r \in (0,1).
$$
Then the inequality \eqref{eq:bj_radial} gives a lower bound of order $n$ on the spectral gap. However, taking the function $f' (r) = r^{(n-1)/2}$ in the inequality \eqref{e:chen}, one obtains
$$
V_f (r) = - \frac{(\LL _\nu f)'(r) }{f'(r) } = \frac{(n-1)^2}{4r^2} + \frac{n-1}{2r^2} \geq \frac{n^2 - 1}{4}, \quad r\in(0,1) ,
$$
and together with the choice of the identity function in the variational formula of the spectral gap show that the order of magnitude is $n^2$, which is different from the one of the full dynamics.

\section{Beyond the log-concave case}
\label{sect:beyond} The approach emphasized in the proof of Theorem~\ref{theo:main} allows us to obtain interesting results beyond the case of log-concave measures,
at the price of potentially modifying the energy term in the Poincar\'e inequality, leading henceforth to the so-called weighted Poincar\'e inequalities.
Let us observe how this phenomenon appears in the general setting before turning to two interesting examples: the first one is related to heavy-tailed measures, the generalized Cauchy distribution, whereas the second one revisits the Gaussian case.

\subsection{General setting}
As in the previous sections, the main protagonist of this part is still the probability measure $\mu$ on $\rr ^n$ ($n \geq 2$) with Lebesgue density proportional to $e^{-V (\Vert x\Vert)}$, but the notable difference relies in the smooth potential $V : \rr _+ \to \rr$ which is no longer convex.
If we consider as before the generator acting on smooth enough functions $f$,
$$
\LL _\mu f( x) := \Delta f (x) - \frac{V' (\Vert x\Vert )}{\Vert x\Vert } \, x  \cdot  \nabla f( x) , \quad x\in \rr^n ,
$$
then it may happen that $\lambda_1 (- \LL_\mu) = 0$, i.e., there is no spectral gap. In other words, there is no constant $\lambda >0$ such that the Poincar\'e inequality \eqref{eq:poincare} is satisfied. Among the potential reasons for which such a functional inequality can be false, a classical one is that the Euclidean norm might not be exponentially integrable with respect to $\mu$. However the story becomes different if the energy term is modified in a convenient way, that is to say, if we introduce a multiplicative (unbounded) weight $\sigma^2$ in the integral of the right-hand-side of \eqref{eq:poincare} to reach an inequality of the type
\begin{equation}
\label{eq:weigh_poinc}
\lambda \Var _\mu (h) \leq \int_{\rr^n} \sigma^2 \Vert \nabla h \Vert ^2 d\mu .
\end{equation}
Now, from our Markovian point of view, the natural question is the following: can we express the optimal constant in the latter weighted Poincar\'e inequality as the spectral gap of some alternative Markovian dynamics ? Actually the answer to this question is affirmative as we can see from now on. We introduce a new Markovian operator $\LL _\mu ^{\sigma}$ acting on functions $h\in \CC_0^\infty(\R^n)$ as follows,
\begin{equation}
\label{e:Lmusigma}
\LL ^\sigma _\mu h( x) = \sigma^2 (x) \Delta h (x) + \left( \nabla (\sigma^2) (x) - \sigma ^2(x) \frac{V' (\Vert x\Vert )}{\Vert x\Vert } x \right)  \cdot \nabla h( x) , \quad x\in \rr^n .
\end{equation}
Then the operator $\LL ^\sigma _\mu$ is symmetric and non-positive on $\CC_0^\infty(\R^n)$ with respect to the same measure probability $\mu$. Moreover, the main point resides in the fact that the weighted Poincar\'e inequality (\ref{eq:weigh_poinc}) rewrites by integration by parts as the following inequality: for smooth enough functions $h$,
$$
\lambda \Var _\mu (h) \leq \int _{\rr^n} h (-\LL ^\sigma _\mu h) d\mu ,
$$
henceforth leading to the study of the spectral gap $\lambda_1 (- \LL _\mu ^\sigma)$. \vspace{0.1cm}

Once we have realized that weighted Poincar\'e inequalities can be reduced to classical Poincar\'e inequalities for an alternative dynamics, the routine is more or less the same as in the previous section. First, let us introduce the following set of hypothesis on the weight function $\sigma$ that we will refer to as assumption $\AA$ in the sequel: \vspace{0.1cm}

$\circ$ Smoothness: the function $\sigma$ belongs to $\CC ^\infty (\R ^n)$. \vspace{0.1cm}

$\circ$ Ellipticity: it means that $\sigma$ is non-degenerate, i.e. $\sigma (x) > 0$ for every $x\in \rr ^n$. \vspace{0.1cm}

$\circ$ Completeness: the metric space $(\rr ^n, d_\sigma )$ is complete, where $d_\sigma $ is the distance
$$
d_\sigma (x,y) := \sup \, \left \{ \vert f(x) - f(y) \vert : f\in \CC ^\infty (\rr ^n) , \, \sigma \Vert \nabla f\Vert \leq 1 \right\} , \quad x,y\in \rr ^n .
$$
Then the operator $(\LL _\mu ^\sigma , \CC _0 ^\infty (\rr ^n))$ is essentially self-adjoint in $L^2(\mu)$. In this context, the Dirichlet form reads as
\[
 \EE_\mu^\sigma(f_1,f_2):= \int_{\rr ^n} \sigma ^2 \nabla f_1  \cdot  \nabla f_2 d\mu ,
\]
with domain $\DD (\EE _\mu ^\sigma)$ given by the weighted Sobolev space
\[
W_\sigma ^{1,2}(\mu) = \left\{ f\in L^2(\mu) : \sigma \Vert \nabla f \Vert \in  L^2(\mu) \right\} .
\]
As before, the self-adjointness allows us to check the Poincar\'e inequality \eqref{eq:weigh_poinc} only for functions $h\in \CC_0^\infty(\rr ^n)$. \vspace{0.1cm}

The second main object we have to deal with is the associated one-dimensional radial operator. To do so, we need a last assumption on the weight $\sigma$ which is the following: the function $\sigma$ is radial, i.e., it depends only on the length of the space variable. Recall that for radial functions, we use the same notation for the functions defined on $\rr^n$ and $(0,+\infty)$. Then the associated (Neumann) generator $\LL ^\sigma _\nu$ acts on smooth real-valued functions $g$ on $(0,+\infty)$ as
\begin{equation}
\label{e:Lnusigma}
\LL ^\sigma _\nu g (r) := \sigma ^2 (r) g''(r) + b(r)  g'( r) , \quad r >0,
\end{equation}
the drift $b$ being given by
$$
b(r) := (\sigma^2)' (r) - \sigma ^2 (r) \left( V'(r) - \frac{n-1}{r} \right) , \quad r >0.
$$

In contrast to the log-concave case studied before, there is no universal analogue of the one-dimensional estimate \eqref{eq:bj_radial}, so that we are not able to state an elegant result somewhat similar to Theorem~\ref{theo:main}. However in the spirit of Lemma~\ref{lemme:gen-spherically}, the spectral comparison between the full dynamics and its one-dimensional radial part is still possible and this is the content of the next result.
\bthm
\label{theo:gen-spherically-sigma}
Let $\mu$ be a spherically symmetric probability measure on $\R^n$ ($n \geq 2$) with Lebesgue density proportional to $e^{-V(\Vert x\Vert )}$ and let $\sigma$ be some radial function on $\rr ^n$ satisfying Assumption $\AA$. Let $\nu$ be the probability measure on $(0,+\infty)$ whose Lebesgue density is proportional to $r^{n-1} e^{- V(r)}$. Then provided the various integrals below make sense, the spectral gaps $\lambda_1 (-\LL_\mu)$ and $\lambda_1 (-\LL_\nu)$ of the operators $-\LL_\mu ^\sigma$ and $-\LL_\nu ^\sigma$ defined in \eqref{e:Lmusigma} and \eqref{e:Lnusigma} respectively, satisfy
\begin{equation}\label{e:min-maj-lambda_1-sigma}
\min \left \{ \lambda_1 (-\LL_\nu^\sigma) , \frac{n-1}{\int_{\rr ^n} \frac{\Vert x\Vert ^2 }{\sigma ^2(x) } \mu (dx)  }\right \}
\leq \lambda_1 (-\LL_\mu^\sigma)
\leq \min \left \{ \lambda_1 (-\LL_\nu^\sigma) , \frac{n \int_{\rr ^n} \sigma^2(x)  \mu (dx)} {\int_{\rr ^n} \Vert x\Vert ^2  \mu (dx)} \right \} .
\end{equation}
\nthm
\bprf
The proof is similar to the one of Lemma~\ref{lemme:gen-spherically}, with the following slight changes. First, the upper bound is obtained by choosing on the one hand radial functions $f$ in the variational characterization of the spectral gap $\lambda_1 (-\LL_\mu^\sigma)$ and on the other hand the linear function $h(x) = \sum_{k=1} ^n x_k$ in the weighted Poincar\'e inequality \eqref{eq:weigh_poinc}. \vspace{0.1cm}

For the lower bound, we follow the same computations as in the proof of Lemma~\ref{lemme:gen-spherically}, except in the Cauchy-Schwarz inequalities  \eqref{eq:poincare_nu}, \eqref{eq:CS} for which we exploit the presence of the weight $\sigma$, using $r = r /\sigma(r) \times \sigma(r)$. We arrive at the end at the estimate
\[
\lambda_1 (-\LL ^\sigma _\mu) \geq \min \left \{ \lambda_1 (-\LL ^\sigma _\nu) , \frac{\lambda_1 (-\Delta_{\sigma_{n-1}})}{\int_{\rr ^n} \frac{\Vert x\Vert ^2 }{\sigma ^2 (x) } \mu (dx) }\right \} ,
\]
which is the desired lower bound. The proof is now complete.
\nprf

Now, it is time to turn to the two examples we have in mind, which illustrate Theorem~\ref{theo:gen-spherically-sigma}: the heavy-tailed situation through generalized Cauchy distributions and then a new look at the Gaussian case.

\subsection{Generalized Cauchy distributions} First, let us focus our attention on the generalized Cauchy distribution. Let $\mu$ be the probability measure with Lebesgue density on $\rr^n$ ($n\geq 2$) proportional to $(1+\Vert x\Vert ^2) ^{-\beta}$, where $\beta >n/2$. In terms of the potential $V$, this density rewrites as $e^{-V(\Vert x\Vert )} /Z$ with
$$
V(r) := \beta \log (1+r^2), \quad r \geq 0,
$$
and where $Z$ stands for the normalization constant, that is
$$
Z := \frac{n \omega _n \, \Gamma (n/2) \, \Gamma(\beta - n/2)}{2 \Gamma(\beta)} .
$$
The measure $\mu$ is called the generalized Cauchy distribution of parameter $\beta$ on $\R^n$. \vspace{0.1cm}

Recently, Bobkov and Ledoux \cite{bob_ledoux2} used Brascamp-Lieb inequalities to establish, under the additional constraint $\beta \geq n$, the weighted Poincar\'e inequality \eqref{eq:weigh_poinc} with the radial weight
$$
\sigma^2 (x) := 1+ \Vert x\Vert ^2 ,
$$
and the constant $\lambda$ they obtained is
$$
\lambda = \frac{2(\beta -1)}{\left(\sqrt{1+\frac{2}{\beta -1}} + \sqrt{\frac{2}{\beta -1}} \right) ^2} .
$$
They also point out that a similar inequality might be established in the region $n/2 < \beta <n$, the constant $\lambda$ depending essentially on the dimension $n$. Since the weight $\sigma$ satisfies our Assumption $\AA$ and the generator $\LL_\mu ^\sigma$ rewrites as
\begin{equation}
\label{eq:sigma_mu_cauchy}
\LL_\mu ^\sigma h( x) = \left( 1+ \Vert x\Vert ^2 \right) \Delta h (x) + 2(1-\beta) x  \cdot \nabla h (x) , \quad x \in \rr ^n ,
\end{equation}
then their result means, in our Markovian language, that we have the spectral gap inequality:
$$
\lambda_1 (-\LL ^\sigma _\mu) \geq \frac{2(\beta -1)}{\left(\sqrt{1+\frac{2}{\beta -1}} + \sqrt{\frac{2}{\beta -1}} \right) ^2}.
$$
With this choice of weight function $\sigma$, we obtain after some computations involving the beta functions,
\[
\int_{\rr ^n} \Vert x\Vert ^2 \mu (dx) = \int_0^{+\infty} r^2 \, \nu(dr) = \frac{n}{2\beta - 2 - n},
\]
when $\beta > n/2 +1$ and also,
\[
\int_{\rr ^n} \frac{\Vert x\Vert ^2 }{\sigma ^2(x) } \mu (dx) = \int_0^{+\infty} \frac{r^2}{1+r^2}   \, \nu(dr) = \frac{n}{2\beta} ,
\]
so that applying Theorem~\ref{theo:gen-spherically-sigma} for this model yields, at least in the region $\beta > n/2 +1$,
\begin{equation}
\label{eq:thm_cauchy}
\min \left \{ \lambda_1 (-\LL_\nu^\sigma) , \frac{2\beta (n-1)}{n} \right \}
\leq \lambda_1 (-\LL_\mu^\sigma) \leq \min \left \{ \lambda_1 (-\LL_\nu^\sigma) , 2(\beta -1) \right \} .
\end{equation}
Recall that the associated one-dimensional (Neumann) radial generator $\LL_\nu$ is given by
\begin{equation}
\label{eq:sigma_nu_cauchy}
\LL_\nu ^\sigma g( r) = \left( 1+ r^2 \right) g''(r) + \left( (n+1 - 2\beta)r + \frac{n-1}{r} \right) g'(r) , \quad r >0.
\end{equation}
Hence it remains to estimate the spectral gap $\lambda_1 (-\LL ^\sigma _\nu)$. Surprisingly, its exact value can be computed for all $\beta > n/2$. On the one hand, consider the function
$$
g(r) := r^2 - \frac{n}{2\beta - 2 -n} , \quad r> 0.
$$
Such a function is nothing but an increasing eigenfunction of the operator $-\LL ^\sigma _\nu$ when $\beta > n/2 + 2$ and thus the corresponding eigenvalue is the spectral gap, that is,
$$
\lambda_1 (-\LL ^\sigma _\nu) = 4 \left( \beta - \frac{n}{2} - 1 \right) .
$$
On the other hand, for $\beta$ in the region $n/2 < \beta \leq n/2+2$, we claim that
\beq
\label{keybeta}
\lambda_1(-\LL_{\nu}^{\sigma}) = \left( \beta-\frac{n}{2} \right) ^2.
\nneq
In particular the spectral gap is continuous at the critical value $\beta = n/2+2$. To get the upper bound in \eqref{keybeta}, we choose $f(r):=(1+r^2)^{\vep}$ with $0<2\vep<\beta-n/2$ so that by the variational formula of the spectral gap, we have
\begin{eqnarray*}
\lambda_1(-\LL_{\nu}^{\sigma})^{-1} & \ge & \frac{\Var_{\nu}(f)}{\EE_{\nu}^{\sigma}(f, f)} \\
& =& \frac{\beta-2\vep}{2n\vep^2}\bigg(1-\frac{\Gamma (\beta-\vep-\frac n2)^2 \, \Gamma(\beta) \, \Gamma(\beta-2\vep)}{\Gamma(\beta-\frac n2) \, \Gamma(\beta-\vep)^2 \, \Gamma(\beta-2\vep-\frac n2)}\bigg),
\end{eqnarray*}
and then we take the limit as $\vep\to \beta /2 - n/4 $. To get the lower bound in \eqref{keybeta}, we choose $f(r):=(1+r^2)^{\frac{\beta}2-\frac n4}$. Using the formula \eqref{e:chen} for the dynamics $\LL_\nu ^\sigma$, we get after careful calculations,
\begin{eqnarray*}
V_f(r) & = & \frac{(-\LL_{\nu}^{\sigma} f)'(r)}{f'(r)} \\
& = & \frac{\left(\beta -\frac{n}{2}\right)^2 r^2 + 2\beta + n - n\beta + \frac{n^2}{2}}{1+r^2} ,
\end{eqnarray*}
leading to the estimate
\begin{eqnarray*}
\lambda_1(-\LL_{\nu}^{\sigma}) & \geq & \inf_{r>0} V_f(r) \\
& = & \left(\beta -\frac{n}{2}\right)^2 .
\end{eqnarray*}
Therefore combining with \eqref{eq:thm_cauchy} entails the following result for the generalized Cauchy distribution.

\begin{cor}[Generalized Cauchy distribution, case $n\geq 3$]
\label{cor:cauchy3}
Let $\mu$ be the generalized Cauchy distribution of parameter $\beta$ on $\R^n$ ($n\geq 3$). Consider the radial weight $\sigma ^2 (x):= 1+ \Vert x\Vert ^2$ and let $\LL_\mu ^\sigma$ and $\LL_\nu ^\sigma$ be the operators defined by \eqref{eq:sigma_mu_cauchy} and \eqref{eq:sigma_nu_cauchy} respectively. Then the spectral gaps $\lambda_1 (-\LL_\mu ^\sigma)$ and $\lambda_1 (-\LL_\mu ^\sigma)$ satisfy: \vspace{0.1cm}

$\circ$ for $n/2< \beta \leq n/2+2 $,
$$
\lambda_1 (-\LL ^\sigma _\mu)= \lambda_1 (-\LL ^\sigma _\nu)=\left(\beta-\frac{n}{2}\right)^2;
$$

$\circ$ for $2+n/2 < \beta \leq n(n+2)/(n+1)$,
$$
\lambda_1 (-\LL ^\sigma _\mu) = \lambda_1 (-\LL ^\sigma _\nu) = 4 \left(\beta - \frac{n}{2}-1 \right);
$$

$\circ$ for $n(n+2)/(n+1) < \beta \leq n+1$,
$$
\frac{2 \beta (n-1)}{n} \leq \lambda_1 (-\LL ^\sigma _\mu) \leq   \lambda_1 (-\LL ^\sigma _\nu) = 4 \left(\beta - \frac{n}{2}-1 \right);
$$

$\circ$ for $\beta > n+1$,
$$
\frac{2 \beta (n-1)}{n} \leq \lambda_1 (-\LL ^\sigma _\mu) \leq 2(\beta-1) <  \lambda_1 (-\LL ^\sigma _\nu) = 4 \left(\beta - \frac{n}{2}-1 \right).
$$
\end{cor}
Let us say some words about this result. \vspace{0.1cm}

$\circ$ We ignore if we still have
$$
\lambda_1 (-\LL ^\sigma _\mu) = 4 \left(\beta - \frac{n}{2}-1 \right) ,
$$
for $\beta$ in the region $(n(n+2)/(n+1), n+1]$. However when $\beta > n+1$, we know that it is no longer the case since the upper bound $2(\beta-1)$ is slightly better, meaning that the value $\beta = n+1$ is critical for the spectral comparison between the full and the one-dimensional radial dynamics. \vspace{0.1cm}

$\circ$ Our lower bound $2\beta (n-1)/n$ is not numerically comparable to that of Bobkov-Ledoux (it depends on the value of $\beta$ with respect to the dimension $n$), but both exhibit more or less the same behaviour as $\beta$ is large (they are proportional to $\beta$). \vspace{0.1cm}

$\circ$ The reader may wonder whether or not the weight function $\sigma$ is optimal. As a partial answer, let us investigate the family of weights $\sigma_a ^2 (r) := (1+ r^2) ^a$ for $a\in (0,1)$. Choosing as above the function $f(r):=(1+r^2)^{\vep}$ with $0<2\vep<\beta-n/2$, we get by the variational formula of the spectral gap,
\begin{eqnarray*}
\lambda_1(-\LL_{\nu}^{\sigma_a})^{-1} & \ge & \frac{\Var_{\nu}(f)}{\EE_{\nu}^{\sigma_a}(f, f)} \\
& =& \bigg(\frac{\Gamma(\beta-2\vep-\frac n2)}{\Gamma(\beta-2\vep)}-\frac{\Gamma (\beta-\vep-\frac n2)^2 \, \Gamma(\beta)}{\Gamma(\beta-\frac n2) \, \Gamma(\beta-\vep)^2}\bigg) \frac{\Gamma(\beta-2\vep+2-a)}{2n\vep^2\Gamma(\beta-2\vep-\frac n2+1-a)} ,
\end{eqnarray*}
and then passing through the limit as $\vep\to \beta /2 - n/4 $, the right-hand-side tends to $+\infty$ or, in other words, $\lambda_1(-\LL_{\nu}^{\sigma_a}) = 0$. Hence one deduces that there is no spectral gap for those family of weights when $a \in (0,1)$. \vspace{0.1cm}

$\circ$ In the multi-dimensional case, we observe a rather different phenomenon from the one-dimensional case. As we have already seen in the one-dimensional case, the spectral gap has an associated eigenfunction (when it exists) which is strictly monotone. Reciprocally, if there exists a strictly monotone eigenfunction, then the associated eigenvalue is nothing but the spectral gap. In the Cauchy setting, it is worth noticing that for $\beta$ in the region $( 2+n/2 ,  n(n+2)/(n+1)]$, the spectral gap $4(\beta - n/2 -1)$ is associated to the eigenfunction, $h(x) = \Vert x\Vert ^2 - n/(2\beta -2 - n)$, which does not have monotonic coordinates. Moreover, in the case $n/2+1 < \beta < n+1$, the eigenvalue $2(\beta-1)$ is strictly greater than the spectral gap whereas the underlying eigenfunction, the linear function $h(x) = \sum_{k=1} ^n x_k$, has monotonic coordinates. \vspace{0.1cm}

$\circ$ It is interesting to observe that there is no eigenfunction associated to the spectral gap of the operator $-\LL ^\sigma _\nu$ in the case $n/2 <\beta < n/2 + 2$. Indeed, using the formula \eqref{e:chen} for the function $f(r)=(1+r^2)^{\frac{\beta }2-\frac{n}{4}}$, one has
\[
V_f(r)=  \frac{(-\LL_{\nu}^{\sigma} f)'(r)}{f'(r)} > \left( \beta-\frac n2 \right) ^2.
\]
Then the Brascamp-Lieb inequality established in \cite{bj} gives for all $h\in  W_\sigma ^{1,2}(\nu)$,
\[
 \Var_\mu(h) \leq \int_0^{+\infty} \frac{(h' )^2}{V_f } \sigma^2 d\nu < \frac{1}{\left(\beta-\frac{n}{2}\right)^2}\int_0^{+\infty} (h' ) ^2 \sigma^2 d\nu .
\]
This strict inequality prevents the existence of an eigenfunction with eigenvalue $ \left(\beta- n/2 \right)^2$. \vspace{0.1cm}

When $n=2,$ the above argument for the estimates of $\lambda_1(-\LL_{\mu}^{\sigma})$ and $\lambda_1(-\LL_{\nu}^{\sigma})$ is still valid, except the comparison between $\lambda_1(-\LL_{\nu}^{\sigma})$ and $2\beta (n-1)/n$. We thus obtain the following result.
\begin{cor}[Generalized Cauchy distribution, case $n = 2$]  When $n=2$, with the same notation as in Corollary \ref{cor:cauchy3}, we have: \vspace{0.1cm}

$\circ$ for $1<\beta\le \frac{3+\sqrt{5}}{2},$  \; $\lambda_1(-\LL_{\mu}^{\sigma})=\lambda_1(-\LL_{\nu}^{\sigma})=(\beta-1)^2;$\\

$\circ$ for $\frac{3+\sqrt{5}}{2}<\beta\le 3,$ \; $\beta\le \lambda_1(-\LL_{\mu}^{\sigma}) \le \lambda_1(-\LL_{\nu}^{\sigma})= (\beta-1)^2;$\\

$\circ$ for $\beta>3,$  \; $\beta\le\lambda_1(-\LL_{\mu}^{\sigma})\le 2(\beta-1)< \lambda_1(-\LL_{\nu}^{\sigma}) =4(\beta-2).$
\end{cor}

\subsection{The Gaussian case revisited}
As announced, we turn to our second example of interest, the case of weighted Poincar\'e inequalities for the Gaussian distribution. Let $\mu$ be the centered Gaussian probability measure on $\rr^n$ ($n\geq 2$) with the identity as covariance matrix. Then the question is the following: does the technology emphasized above allow us to obtain a weighted Poincar\'e inequality of the type \eqref{eq:weigh_poinc} for the Gaussian measure ? According to the previous discussion, our objective is to estimate the spectral gap $\lambda_1 (- \LL_\mu ^\sigma)$ of the modified dynamics $\LL_\mu ^\sigma$ defined in \eqref{e:Lmusigma}, which rewrites in the Gaussian context as
\begin{equation}
\label{e:sigma_mu_gauss}
\LL ^\sigma _\mu h( x) = \sigma^2 (x) \Delta h (x) + \left( \nabla (\sigma^2) (x) - \sigma (x) ^2 x \right)  \cdot \nabla h( x) , \quad x\in \rr^n ,
\end{equation}
the potential $V$ being now $V(r) := r^2/2$. \vspace{0.1cm}

Let us investigate first the case of a similar weight $\sigma$ as for the generalized Cauchy distributions above, that is
$$
\sigma^2 (x):= 1+ \Vert x\Vert ^2 ,
$$
so that the desired weighted Poincar\'e inequality is a weak version of the classical Poincar\'e inequality (different from the so-called weak Poincar\'e inequality). For the upper bound, we choose as above the linear function $h(x):= \sum_{k=1} ^n x_k$ in the weighted Poincar\'e inequality (\ref{eq:weigh_poinc}) to obtain the simple estimate,
\begin{eqnarray*}
\lambda_1 (- \LL_\mu ^\sigma) & \leq & n \times \frac{\int_{\rr^n} \left( 1 + \Vert x\Vert ^2\right) \mu (dx)}{\int_{\rr^n} \Vert x\Vert ^2 \mu (dx)} \\
& = & n+1.
\end{eqnarray*}
For the lower bound, observe that the lower bound in \eqref{e:min-maj-lambda_1-sigma} is still available in the weighted Gaussian case,
so that once again we are led to estimate the spectral gap $\lambda_1 (-\LL ^\sigma _\nu) $ of the one-dimensional radial operator $-\LL ^\sigma _\nu$. Since it exact value is not accessible, in contrast to the Cauchy case above, we need to revisit Lemma~\ref{lemme:bj} by taking in consideration the weight $\sigma$. We are thus led to the following result, which can be stated in a more general situation than in the Gaussian case (in particular, we decide to keep the notation involving $\sigma$ since it will also be useful when dealing with another choice of weight $\sigma$). Recall that in the Gaussian situation, the (Neumann) radial operator $\LL ^\sigma _\nu$ is given by
$$
\LL ^\sigma _\nu g (r) := \sigma ^2(r) g''(r) + b(r)  g'(r), \quad r >0,
$$
where the drift $b$ is defined by
$$
b(r) := (\sigma^2)' (r) - \sigma (r) ^2 \left( r - \frac{n-1}{r} \right) .
$$
\blem[Bonnefont-Joulin's lemma revisited \cite{bj}]
\label{lemme:bj_weight}
Assume that the function $\VV_\nu ^\sigma$ defined on $(0,+\infty)$ by
$$
\VV_\nu ^\sigma := \frac{\LL _\nu ^\sigma \sigma }{\sigma } - b' ,
$$
is positive. Then the spectral gap $\lambda_1 (-\LL_\nu ^\sigma)$ satisfies the inequality
\[
\lambda_1 (-\LL_\nu ^\sigma) \geq \frac{1}{\int_0 ^{+\infty} \frac{1}{\VV_\nu ^\sigma} d\nu } .
\]
\nlem

Coming back to our Gaussian setting, we have with this choice of weight $\sigma ^2 (r) = 1+ r^2$,
\begin{eqnarray*}
\VV_\nu ^\sigma (r) & = & \frac{4r^4 + 6r^2 +1}{1+r^2} + \frac{n-1}{r^2} \\
& \geq & 4r^2 ,
\end{eqnarray*}
so that applying Lemma~\ref{lemme:bj_weight} entails the following inequality, available for any $n \geq 3$,
\begin{eqnarray*}
\lambda_1 (-\LL _\nu ^\sigma) & \geq & \frac{\int_0 ^{+\infty} r^{n-1} e^{-r^2 /2} dr }{\int_0 ^{+\infty} \frac{r^{n-1}}{4r^2} e^{-r^2 /2} dr} \\
& = &  4(n-2) .
\end{eqnarray*}
For $n=2$, we have
$$
\VV_\nu ^\sigma (r) \geq 4r^2 + \frac{1}{r^2} \geq 4 .
$$
Hence by the lower bound in \eqref{e:min-maj-lambda_1-sigma}, we obtain for any $n\geq 2$ the estimate
\begin{eqnarray*}
\lambda_1 (-\LL ^\sigma _\mu) & \geq & \min \left \{ \lambda_1 (-\LL _\nu ^\sigma) , (n-1) \times \frac{ \int_0^{+\infty} r^{n-1} e^{-r^2 /2}dr}{\int_0 ^{+\infty} \frac{r^2}{ 1+r^2} r^{n-1} e^{-r^2 /2}dr }\right \} \\
& \geq & n-1,
\end{eqnarray*}
since $r^2 /(1+r^2) \leq 1$. Let us summarize our result in the weighted Gaussian case.
\begin{cor}[Weighted Gaussian case 1]
Let $\mu$ be the standard Gaussian distribution on $\rr^n$ ($n \geq 2$) and consider the radial weight $\sigma ^2 (x) := 1 + \Vert x \Vert ^2$. Let $\lambda_1 (-\LL_\mu ^\sigma)$ be the spectral gap of the operator $-\LL_\mu ^\sigma$ defined in \eqref{e:sigma_mu_gauss}. Then we have
$$
n-1 \leq \lambda_1 (-\LL ^\sigma _\mu) \leq n+1.
$$
In particular $\lambda_1 (-\LL ^\sigma _\mu)  \approx n$ in large dimension.
\end{cor}

A more challenging problem is to obtain a weighted Poincar\'e inequality for the Gaussian measure that is a reinforcement of the classical Poincar\'e inequality. Motivated by an open question raised in \cite{bob_ledoux2}, our second choice of radial weight will be
$$
\sigma ^2 (x) := \frac{1}{1+\Vert x\Vert ^2} ,
$$
which satisfies also Assumption $\AA$. For bounding from above the spectral gap $\lambda_1 (- \LL_\mu ^\sigma)$, we choose as before the linear function $h(x) = \sum_{k=1} ^n x_k$ in the weighted Poincar\'e inequality (\ref{eq:weigh_poinc}) to get the following upper bound available for any $n \geq 3$,
\begin{eqnarray*}
\lambda_1 (- \LL_\mu ^\sigma) & \leq & n \times \frac{\int_{\rr^n} \left( 1 + \Vert x\Vert ^2\right) ^{-1} \mu (dx)}{\int_{\rr^n} \Vert x\Vert ^2 \mu (dx)} \\
& \leq & n \times \frac{\int_{\rr^n} \Vert x\Vert ^{-2} \mu (dx)}{\int_{\rr^n} \Vert x\Vert ^2 \mu (dx)} \\
& = & \frac{1}{n-2}.
\end{eqnarray*}
For $n=2$ we use the trivial inequality $\left( 1 + \Vert x\Vert ^2\right) ^{-1} \leq 1$ to get $\lambda_1 (- \LL_\mu ^\sigma) \leq 1$. \vspace{0.1cm}

For the lower bound in the general case $n\geq 2$, we apply Lemma~\ref{lemme:bj_weight} in our setting with the positive potential
$$
\VV_\nu ^\sigma (r) = \frac{(2n-3)r^4 + (3n-1)r^2 +n-1}{\left( 1+r^2 \right) ^3 r^2} , \quad r>0,
$$
and using the inequality
$$
(2n-3)r^4 + (3n-1)r^2 +n-1 \geq (n-1)(1+r^2)^2 , \quad r >0,
$$
we obtain by the lower bound in \eqref{e:min-maj-lambda_1-sigma} the estimate
\begin{eqnarray*}
\lambda_1 (-\LL ^\sigma _\mu) & \geq & \min \left \{ \frac{\int_0^{+\infty} r^{n-1} e^{-r^2 /2}dr}{\int_0 ^{+\infty} \frac{\left( 1+r^2 \right) ^3 r^{n+1}}{(2n-3)r^4 + (3n-1)r^2 +n-1} e^{-r^2 /2}dr} , (n-1)  \times \frac{\int_0^{+\infty} r^{n-1} e^{-r^2 /2}dr}{\int_0 ^{+\infty} ( 1+r^2 ) r^{n+1} e^{-r^2 /2}dr }\right \} \\
& = & (n-1) \times \frac{ \int_0^{+\infty} r^{n-1} e^{-r^2 /2}dr}{\int_0 ^{+\infty} ( 1+r^2 ) r^{n+1} e^{-r^2 /2}dr } \\
& = & \frac{n-1}{n(n+3)}.
\end{eqnarray*}
Finally we get the following estimates, exhibiting a different behaviour with respect to the dimension.
\begin{cor}[Weighted Gaussian case 2]
Let $\mu$ be the standard Gaussian distribution on $\rr^n$ ($n \geq 2$) and consider the radial weight $\sigma ^2 (x) := 1/(1 + \Vert x \Vert ^2)$. Let $\lambda_1 (-\LL_\mu ^\sigma)$ be the spectral gap of the operator $-\LL_\mu ^\sigma$ defined in \eqref{e:sigma_mu_gauss}. Then we have
$$
\frac{n-1}{n(n+3)} \leq \lambda_1 (-\LL ^\sigma _\mu) \leq \frac{1}{n-2} \wedge 1 ,
$$
so that $ \lambda_1 (-\LL ^\sigma _\mu) \approx 1/n$ for large $n$.
\end{cor}

config: arxiv


TITLE: Find the spectral basis of the following endomorphism
QUESTION [0 upvotes]: I have the following problem, I have to find the spectral basis ( I hope I am translating this correctly) for the following endomorphism ( homomorphism ):
$\ R^3 -> R^3 f(x,y,z)=( (x,6y,2z),(2y),(x,4y,2z)) $
I created the associated matrix and found the eigenvalues
$$
   \begin{bmatrix}
   1 & 6 & 2 \\
   0 & 2 & 0 \\
   1 & 4 & 2 \\
   \end{bmatrix}
$$
I tried to find det(A-$\lambda$I)
    $$
   \begin{bmatrix}
   1-\lambda & 6 & 2 \\
   0 & 2-\lambda & 0 \\
   1 & 4 & 2-\lambda \\
   \end{bmatrix}
$$
And from this I found the roots of the eigenvalues as $\lambda=0,2,3$
Now I tried finding the basis by re-writing the matrices with every $\lambda$ value
$$\lambda = 0$$
$$
   \begin{bmatrix}
   1 & 6 & 2 \\
   0 & 2 & 0 \\
   1 & 4 & 2 \\
   \end{bmatrix}$$
The equivalent linear solution is :
$$
\begin{cases}
x+6y+2z=0\\
2y=0\\
x+4y+2z=0\\
\end{cases}
$$
Since $ y=0 $ I have attributed to z a parametric solution, and my result is:
$$
\begin{cases}
x=-2s\\
y=0\\
z=s\\
\end{cases}
$$
If I create a vector from that I get :
$$ s*\begin{pmatrix}
-2\\
0\\1\\ \end{pmatrix} $$
Doing the same calculations for 2 and 3 I get:
$$\lambda=3$$
$
\begin{bmatrix}
   -2 & 6 & 2 \\
   0 & -1 & 0 \\
   1 & 4 & -1 \\
   \end{bmatrix}$
$ \begin{cases}
-2x+2z=0\\
y=0\\
z=s\\
\end{cases}$
Vector:
$ s*\begin{pmatrix}
1\\
0\\1\\ \end{pmatrix} $
$$\lambda=2$$
$
\begin{bmatrix}
   -1 & 6 & 2 \\
   0 & 0 & 0 \\
   1 & 4 & 0 \\
   \end{bmatrix}$
$ \begin{cases}
x=-4s\\
y=s\\
z=-5s\\
\end{cases}$
Vector:
$ s*\begin{pmatrix}
-4\\
1\\-5\\ \end{pmatrix} $
And so I find $ B_1=\{-2,0,1\}$ $ B_3=\{1,0,1\}$ that are correct basis for the endomorphism, however for $\lambda = 2$ I get the basis
$ B_2=\{-4,1,-5\}$  that is exactly half of what I should get, I am unsure weather I have proceeded correctly, but seeing as I have 2/3 correct results, I don't know if it's something I'm doing wrong or I'm just going insane, I would appreciate some insight, thank you !

REPLY [0 votes]: You answer is fine. If $\mathbf v$ is an eigenvector with eigenvalue $\lambda$, then so is $c\mathbf v$ for $c\ne0$. Unless there’s some convention about which eigenvectors to choose out of all of the possibilities, your answer is just as correct as the one you’re comparing against.


TITLE: Confused about Erdos-Gallai theorem
QUESTION [9 upvotes]: Consider the sorted degree sequence $1,3,3,3$.
Erdos-Gallai states that for $k$ in $1 \leq k \leq n$:
$$\sum_{i=1}^{k} d_i \leq k(k-1) + \sum_{j=k+1}^n min(d_j,k)$$.
And that $\Sigma d_i$ must be even, which it is.
$k=1$ gives the inequality $1 \leq 1(0) + (1 + 1 + 1)$, or $1 \leq 3$, which passes.
$k=2$ gives the inequality $1 + 3 \leq 2(1) + (2 + 2)$, or $4 \leq 6$, which passes.
$k=3$ gives the inequality $1 + 3 + 3 \leq 3(2) + (3)$, or $7 \leq 9$, which passes.
$k=4$ gives the inequality $1 + 3 + 3 + 3 \leq 4(3)$, or $10 \leq 12$, which passes.
But it's easy to see that the degree sequence $1, 3, 3, 3$ is agraphical. What am I missing here?

REPLY [6 votes]: As hardmath commented, my ordering was backwards. Erdos-Gallai states that the degree sequence must be ordered largest degree first; that is, the sequence must be $3,3,3,1$.
So we have:
$k=1$ gives $3 \leq 1(0) + (1 + 1 + 1)$, or $3 \leq 4$, which passes.
$k=2$ gives $3 + 3 \leq 2(1) + (2 + 1)$, or $6 \leq 5$, which fails.
Then we're done, but for good measure:
$k=3$ gives $3 + 3 + 3 \leq 3(2) + 1$, or $9 \leq 7$, which fails.
$k=4$ gives $3 + 3 + 3 + 1 \leq 4(3)$, or $10 \leq 12$, which passes.
Thanks hardmath!


TITLE: Plotting Complex Sets in a Research Paper
QUESTION [1 upvotes]: I am meant to "sketch" the set $|z - 1 + i| \leq 3$ for my Complex Analysis homework. I am able to draw this set by hand, so my question isn't about that! Rather, I would like to know how a professional mathematician would generate such an image if he or she were submitting the image as part of a journal article. What program/software/code would he or she use?...all I need is a direction to head in!
(For context: I get extra credit in the class for submitting my homework typed out in Latex. And for this question, I'm sure I could just leave a blank space and hand draw the diagram. I also found this answer, which is kind of a hacker way to generate such diagrams. But my aspiration is to become a mathematician, and so I wanna know how a real mathematician would make such a diagram! Thanks in advance!)

REPLY [0 votes]: Another tool that can be incorporated into LaTeX documents: Asymptote. It's a programming language, so there's a learning curve there. On the plus side, it comes standard with TeX distributions, and the usual editors can easily be configured to run Asymptote code in the same .tex file as everything else.
Drawing circles is in the base capabilities of Asymptote even before importing any of the standard libraries.


TITLE: Finding an explicit formula for $a_n$ defined recursively by $a_{n+1}=a_n^2+6$.
QUESTION [5 upvotes]: Let $\{a_n\}$ be a sequence defined recursively by $a_1=1,a_{n+1}=a_n^2+6$, find an explicit formula for $a_n$.
I tried some of the usual mathods but none of them led to a solution. It seems that there just isn't such a formula (with only elementary functions). So is there? If there isn't, how do you prove that there isn't such a formula?

REPLY [5 votes]: This is much too long to fit into the comment section beneath Yves Daoust's answer:

Define $\{c_n\}$ by $c_n^{2^{n-1}}=a_n$. Then we have $c=\lim_{n\to\infty}c_n$ and
$$
c_{n+1}^{2^n}=a_{n+1}=a_n^2+6=a_n^2\left(1+\tfrac 6{a_n^2}\right)
$$
and so
$$
\begin{align}
2^n\log(c_{n+1})&=2\log(a_n)+\log(1+6/a_n^2)\\
&=2^n\log(c_n)+\log(1+6/a_n^2)
\end{align}
$$
therefore
$$
\log(c_{n+1})=\log(c_n)+\frac1{2^n}\cdot\log(1+6/a_n^2)\leq\log(c_n)+\frac{6}{2^n\cdot a_n^2}
$$
and inductively we see that
$$
\begin{align}
\log(c)&\leq\log(c_n)+\sum_{k=n}^\infty\frac{6}{2^k\cdot a_k^2}\\
&\leq\log(c_n)+\frac6{a_n^2}\sum_{k=n}^\infty\frac{1}{2^k}\\
&=\log(c_n)+\frac{6}{2^{n-1}\cdot a_n^2}
\end{align}
$$
and we know that $a_n=\left\lfloor c^{2^{n-1}}\right\rfloor$ will work if and only if
$$
a_n\leq c^{2^{n-1}}< a_n+1
$$
We already know that $c>c_n$ so the first inequality holds. The second is equivalent to $2^{n-1}\log(c)<\log(a_n+1)$. Since $a_n+1=a_n(1+1/a_n)$ and $c_n^{2^{n-1}}=a_n$ the latter can be rewritten as
$$
\log(a_n+1)=2^{n-1}\log(c_n)+\log(1+1/a_n)
$$
and after dividing by $2^{n-1}$ the inequality becomes
$$
\log(c)<\log(c_n)+\frac{1}{2^{n-1}}\cdot\log(1+1/a_n)
$$
Furthermore we have the first order Taylor approximation $\log(1+x)\approx x$ which is quite accurate for small values of $x$ and already for $x\leq1/a_2=1/7$ we get $\log(1+x)>\frac67x$ and thus for $a_n\geq 7$
$$
\begin{align}
\log(c)&\leq\log(c_n)+\frac{6}{2^{n-1}\cdot a_n^2}\\
&\leq\log(c_n)+\frac{6}{2^{n-1}\cdot 7a_n}\\
&<\log(c_n)+\frac{1}{2^{n-1}}\cdot\log(1+1/a_n)
\end{align}
$$
as desired.

BTW this analysis is easily generalized to $a_{n+1}=a_n^2+b$ for positive numbers other than $b=6$, namely using the same definition of $c_n,c$ then for $a_n\geq b+1$ we have $a_n=\left\lfloor c^{2^{n-1}}\right\rfloor$.
One huge drawback of this method in it's current form, is that it requires one to already compute $a_n$ in order to approximate $c$ well enough to be able to compute $a_n$ with it. That could potentially be fixed by using some analysis like the one I gave above to speed up the convergence $c_n\to c$.


TITLE: Find all values of $k$ for which the given augmented matrix corresponds to a consistent system
QUESTION [0 upvotes]: Consider the augmented matrix
\begin{bmatrix}1&k&-1\\4&8&-4\end{bmatrix}
I want to find all the values of $k$ for which the corresponding linear system is consistent. Upon dividing the second row by $\frac{1}{4}$ we have
\begin{bmatrix}1&k&-1\\1&2&-1\end{bmatrix}
Upon subtracting the second row from the first row, and then assigning the results to the second row, we have
\begin{bmatrix}1&k&-1\\0&k-2&0\end{bmatrix}
In order to avoid a contradiction, it follows that $k$ can only equal $2$. However, the solution to this problem says that every $k\in \mathbb{R}$ causes the system to be consistent. What is the problem?

REPLY [1 votes]: Well, I think what I failed to understand here is the algebra that allows me to deduce the unrestrictive or arbitrary nature of $k$, for my last system in my post converts to
$$\begin{align} x_1 + kx_2 &= -1 \\
      (k-2)x_2 &= 0 \end{align}$$
We immediately see from the second equation that either $x_2 = 0$, which means that any value for $k$ will not pose any problems to our system, so it is consistent for all values of $k$ in this scenario; and if $k=2$, we also don't have any inconsistency.
Therefore, for all $k$ the system is consistent.


TITLE: What formula to give small values more weight?
QUESTION [2 upvotes]: I have the following image which contain images ordered and scaled according a rank value, the biggest having a rank of 1 and the smallest a rank of 77:

I would like to give smaller ranks more 'weight', so they become a little bigger.
The logarithm seems quite interesting but the problem is that I don't know how to 'contain' it in a 0.0 to 1.0 space:

Do you know what formula could be used to give smaller values more weight ?

REPLY [3 votes]: If your purpose is to shift the values from $[1,77]$ to $[0,1]$ while increasing the relative weights of the smaller ones comparing with the larger ones, then you can try either one of the following options:

$\dfrac{\log(x)}{\log(77)}$
$\sqrt{\dfrac{x}{77}}$


TITLE: The Dyadic Partition is finer when m>n
QUESTION [0 upvotes]: Given: For each $n\in \Bbb N$, $P_n = \{j/2^n : j = 0,1,...2^n\}$ is a partition of the interval [0,1].
Prove: $P_m$ is finer than $P_n$ when $m>n$
Attempt at proof: Since $m>n$, we have $2^m>2^n$. This implies that $P_m = \{j/2^m : j = 0,1,...2^n,..,2^m\}$.Thus $\forall$ $x\in P_n$, we have $x \in P_m$.
Thus $P_n \subset P_m$. Thus $P_m$ is finer than $P_n$.
Is my proof correct? Thanks!

REPLY [1 votes]: It is not very clear why $x\in P_n\implies x\in P_m$.
Try putting $j=k\cdot 2^{m-n}$ for $k=0,1,\ldots,2^n$ in the second partition.
Do you have to prove that $P_m$ is strictly finer? If you have to, consider $1/2^m$.


TITLE: Property of surjective submersions?
QUESTION [1 upvotes]: Let $M$, $N$ and $P$ be smooth manifolds and $f:M\longrightarrow N$ and $g:N\longrightarrow P$ be two smooth maps. Let us write $h:=g\circ f:M\longrightarrow P$.
Suppose $h$ is a surjective submersion and $f$ is surjective. Does this imply $g$ is a surjective submersion?
Thanks.

REPLY [1 votes]: Since $h$ is surjective, $g$ is surjective. This is true for any maps of sets satisfying the hypotheses. By the chain rule,
$$
dh_x = dg_{f(x)}\circ df_x,
$$
and $dh_x$ is surjective. Thus we glean that $dg_{f(x)}$ is surjective, so that $g$ is a submersion at $f(x)$. Since any point $y\in N$ may be written $f(x)$ for some $x\in M$, $dg_y$ is surjective for all $y\in N$, so $g$ is a submersion. Putting it all together, $g$ is a surjective submersion.


text:
\section{Introduction}
\IEEEPARstart{M}{odel} predictive control (MPC) is a popular control strategy that computes control inputs by solving a numerical optimization problem. A mathematical model is used to predict the future behavior of the system and an optimal sequence of  control inputs is computed by solving an optimization problem that minimizes a given objective function subject to constraints. The main reasons for its success are the possibility of handling systematically multiple-input multiple-output systems, nonlinearities as well as constraints.
The main challenge of MPC is that it requires the solution of an optimization problem at each sampling time of the controller. For this reason, traditional applications focused on slow systems such as chemical processes~\cite{qin2003} \cite{rawlings2009}.

During the past two decades, a large research effort has been devoted to extending the application of MPC algorithms to fast embedded systems. To achieve this goal, two different approaches have been followed. The first approach included the development of fast solvers and tailored implementations  \cite{mattingley2012cvxgen} that can solve the required optimization problems in real time for fast systems. Different variations of the Nesterov's fast gradient method (see e.g. \cite{richter2012} \cite{giselsson2013}, \cite{koegel2011}) and of the alternating directions method of multipliers (ADMM)  \cite{boyd2011distributed} have been very successful for embedded optimization and model predictive control.  Different versions of these algorithms have been used to obtain MPC implementations on low-cost microcontrollers  \cite{zometa2013}, \cite{lucia2016_ARC} or high-performance FPGAs \cite{jerez2014} \cite{lucia2018_TII}.

The second approach to extend the application of MPC to fast and embedded systems is usually called Explicit MPC. The MPC problem for linear time invariant systems is a parametric quadratic program whose solution is a piecewise affine function defined on polytopes and only depends on the current state of the system \cite{bemporad2002}. Explicit MPC exploits this idea by precomputing and storing the piecewise affine function that completely defines the MPC feedback law. The online evaluation of the explicit MPC law reduces to finding the polytopic region in which the system is currently located and applying the corresponding affine law.
The main drawback of explicit MPC is that the number of regions on which the control law is defined grows exponentially with the prediction horizon and the number of constraints.  Therefore its application is limited to very small systems and small prediction horizons, especially in the case of embedded systems with limited storage capabilities.

To counteract the problems of explicit MPC, some approaches try to simplify the representation of the control law by eliminating redundant regions \cite{geyer2008} or by using different number representations \cite{ingole2017}. Other approaches try to approximate the exact explicit MPC solution to further reduce the memory requirements of the approach (see a review in \cite{alessio2009}). Approximate explicit MPC schemes include the use of simplicial regions \cite{Bemporad2011}, neural networks \cite{parisini1995}, radial basis functions \cite{cseko2015}, or using a smaller number of regions to describe the MPC law \cite{holaza2013}.

The main contribution of this paper is to present a new approach for the exact representation as well as for the efficient approximation of explicit MPC laws based on deep learning. We make use of new advances in the theoretical description of the representation capabilities of deep neural networks \cite{silver2016}, \cite{safran2017}, which show that deep neural networks (with several hidden layers) can represent exponentially many more linear regions than shallow networks (with only one hidden layer). We establish bounds on the size (width and depth) that a network should have to be able to achieve an exact representation of the MPC law and we show that an approximate explicit MPC based on deep learning achieves better accuracy with less memory requirements when compared to other approximation techniques. The contributions developed in this paper give a theoretical background and support our preliminary results presented in \cite{karg2018} for mixed-integer quadratic programs.


TITLE: Model churn when churn rate diminishes over time
QUESTION [0 upvotes]: If I'm calculating a consistent rate, it's easy to model that as exponential decay. For example, if I lose 10% of customers each year, starting with A customers.
$$
y = A 0.9^x
$$
However, what if that churn is less with each year. Rather than $ A * 0.9 * 0.9 * 0.9 ... $ I need $ A * 0.9 * 0.88 * 0.86 ... $ or something along those lines.
Is there a standard way to model something like that?

REPLY [2 votes]: Adding on to Ertxiem's answer here, maybe the rate of decrease isn't necessarily linear. Then the base would no longer be $r-kx$ raised to some exponent $cx$. We can derive this by assuming a few starting points:
$$f(0)=1$$
$$f(1)=0.9$$
$$f(2)=0.9\cdot(0.9*0.98)=0.9\cdot0.882=0.9^2\cdot0.98$$
$$f(3)=0.9\cdot(0.9*0.98)\cdot(0.9*0.98^2)=0.9\cdot0.882\cdot0.86436=0.9^3\cdot0.98^3$$
We can see a pattern emerging in the exponents if we keep going. The exponent for $0.9$ will be increasing by 1, but the exponent for $0.98$ will actually be the triangular numbers!
Essentially $f(x) = (0.9)^x\cdot (0.98)^{x(x-1)/2}$
Just some more food for thought.


TITLE: if $\sum_{n=1}^{\infty} a_nx^n$ has a convergence radius $R<1$ then $\lim_{n\to\infty} a_n \neq 0$
QUESTION [0 upvotes]: I'm supposed to prove  the Title.
I have tried to prove by contradiction and using the limit definition but with not much luck.
thanks for the help

REPLY [0 votes]: If $R<1$ , then $ \lim \sup |a_n|^{1/n}>1.$ Hence $|a_n|^{1/n}>1$ for infinitely many $n$, thus $|a_n|>1$ for infinitely many $n$. This gives the result.


TITLE: Show y is odd in the equation $y^3 =x^2 +2$
QUESTION [4 upvotes]: Found a question which asks you to find all integer solutions $y^3 =x^2 +2$ .    In order to do so, it first asks you to show that y is odd.  How would one go about showing this?
Many Thanks

REPLY [1 votes]: If $y$ is even, $y^3\equiv0\pmod 8\equiv0\pmod 4$
So, $x^2=y^3-2\equiv-2\pmod 4\equiv2$
Now, $0^2\equiv0\pmod 4,(\pm1)^2\equiv1, (\pm2)^2\equiv0$


TITLE: Modular Forms in Pari/GP
QUESTION [2 upvotes]: I'm trying out Pari's new modular forms package, and I've run into a small issue that I couldn't resolve.
I want to use the modular parameterization of an elliptic curve $E$ given by the elltaniyama(E) function and then use some of the new features in the modular forms package like mfslashexpansion among other things.
However, the output of elltaniyama(E) is a series, and so I need to make a conversion to the modular form type before this can work. Is there any way this can be done?
If not, I can generate the coefficients for the modular parameterization through my own function, but is there a way to construct a modular form object from just a list of coefficients of the q expansion?
I've tried looking through Pari's user guide for both elliptic curves and modular forms, but haven't found anything on the subject.
I tried posting this on stack-overflow, but they told me it was too specialized of a question and to ask on Math StackExchange.
Thanks in advance!

REPLY [1 votes]: I think you can do what you want using, for example:
? print(q*Ser(ellan(E=ellinit([0, -1, 1, 0, 0]),7),q))
q - 2*q^2 - q^3 + 2*q^4 + q^5 + 2*q^6 - 2*q^7 + O(q^8)

You can check this with LMFDB curve 11.a3. An alternative is
? default(seriesprecision, 6)
? xy = subst(elltaniyama(E), 'x, q);
? print(q*deriv(xy[1])/(2*xy[2] + E.a1*xy[1] + E.a3))
q - 2*q^2 - q^3 + 2*q^4 + q^5 + 2*q^6 - 2*q^7 + O(q^8)

To get a modular form object do:
? M = mffromell(E);

For your question about constructing modular forms from coefficients, the function mfsearch() will probably help you. For more help read Tutorial for Modular Forms in Pari/GP.
As another example, consider the LMFDB curve 64.a4:
? print(q*Ser(ellan(E=ellinit([0, 0, 0, 1, 0]),29),q))
q + 2*q^5 - 3*q^9 - 6*q^13 + 2*q^17 - q^25 + 10*q^29 + O(q^30)

The "modular parameterization" using elltaniyama() is:
? xy = subst(elltaniyama(E), 'x, q);
? print( 1/xy[1]); print( -1/xy[2]);
q^2 + q^6 - q^14 - q^18 - q^22 + 2*q^30 + O(q^33)
q^3 + q^7 - q^11 - 2*q^15 + q^19 + 2*q^23 - 2*q^27 - 2*q^31 + O(q^34)

Note that the first q-series is a generating function of
OEIS sequence A092869 and the second
q-series is a generating function of OEIS sequence A226559. The first is a modular function for
$\, \Gamma_1(16) \,$ of weight $0$.


TITLE: Evaluation of the 'Odd Harmonic' Continued Fraction
QUESTION [1 upvotes]: How to prove that this continued fraction $1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\ddots}}}}$ evaluates to
$\displaystyle\frac{2}{\displaystyle 1+\frac{I_1(\frac14)}{I_0(\frac14)}}$?
The recurrence relation of these Besselfunctions is a different one than the recurrence associated with this Continued Fraction.

REPLY [4 votes]: The continued fraction $\cfrac{1}{1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\ddots}}}}}$ can be evaluated as follows:
Consider $w_n=\cfrac{1}{1+\cfrac{w_{n+1}}{(2n+1)}}$
Assume $w_n=\cfrac{I_{n+1}}{I_n}$
This results in the recurrence relation ${(1+2n)I_n}={(1+2n)I_{n+1}}+{I_{n+2}}$
Euler's Differential method can be used to find an integral which satisfies this recurrence. Applying Euler's method with $a=1, \alpha=2, b=1, \beta=2, c=1, \gamma=0$ results in a ODE. See Sergey Khruschev's Orthogonal Polynomials and Continued Fractions: From Euler's Point of View for technical details of Euler's Differential Method.
The choice R(x)=x satisfies and results in:
$$w_0=\cfrac{I_1}{I_0}=\cfrac{1}{1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\ddots}}}}=\cfrac{\int_{0}^{1} e^{x/2}\cfrac{x^{1/2}}{(x-1)^{1/2}}dx}{\int_{0}^{1} e^{x/2}\cfrac{1}{x^{1/2}(x-1)^{1/2}}dx}$$
Using the substitution $x=\cfrac{1+cos(\phi)}{2}$ it follows easily that this quotient of integrals equals $\frac{\displaystyle I_0(\frac14)+I_1(\frac14)}{\displaystyle2I_0(\frac14)}$
Taken reciprocals proves that $1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\ddots}}}} = \frac{\displaystyle 2}{\displaystyle 1+\frac{\displaystyle I_1(\frac14)}{I_0(\frac14)}}$


TITLE: Can a positive definite second order elliptic operator over an unbounded domain give rise to a compact semigroup?
QUESTION [4 upvotes]: Let $A$ denote some positive definite second order elliptic operator which is defined over $L^2(\Omega)$ with domain $D(A) = H^{2m}(\Omega) \cap H^{m}_{0}(\Omega)$. Here $\Omega$ is a bounded domain in $R^m$. Assume that the coefficient functions of $A$ are nice enough, then it can be shown that the semigroup generated by $A$ is compact.
Now what if $\Omega$ is $[0, \infty)$? Is it possible that $A$ can still generate compact semigroup, if replacing the function spaces in the bounded case by some weighted function spaces? If so, are there any references please?

REPLY [2 votes]: Let $\Omega\subset \mathbb R^n$ be any domain, bounded or unbounded. If $(e^{-tA})$ is ultracontractive, i.e., $e^{-tA}$ maps $L^2(\Omega)$ into $L^\infty(\Omega)$ for $t>0$ and $V\in L^1_{\mathrm{loc}}(\Omega)$ is nonnegative and $\{x\in \Omega|V(x)\leq M\}$ has finite measure for every $M>0$, then $(e^{-t(A+V)})$ is compact in $L^2(\Omega)$. This is Theorem 1 in Simon. Schrödinger Operators with Purely Discrete Spectrum.
For example, the Dirichlet Laplacian on any domain generates an ultracontractive semigroup. On $\mathbb R^n$, this can be seen from the explicit formula for $(e^{t\Delta})$, and for general domains this follows from a comparison principle. Also, if $A$ satisfies the Gagliardo-Nirenberg-Sobolev inequality
$$
\|u\|_{p^\ast}\leq C\langle Au,u\rangle^{1/2}
$$
for some $p^\ast>2$, then $(e^{-tA})$ is ultracontractive. In particular, if $A$ is an elliptic operator in divergence form
$$
Au=\sum_{j,k}\partial_j(a_{jk}\partial_ku)
$$
with coefficients $a_{j,k}\in L^1_{\mathrm{loc}}(\Omega)$ with $a_{jk}=a_{kj}$ and $\sum_{j,k}a_{jk}\xi_j\xi_k\geq \lambda\sum_{j}\xi_j^2$ for some $\lambda>0$, then $(e^{-tA})$ is ultracontractive.
A note on the domains: The domain of such operators is best described by the form method. In the case of the Laplacian the quadratic form associated with $-\Delta+V$ is given by
\begin{align*}
D(q_{-\Delta+V})&=\left\{u\in H^1_0(\Omega):\int_\Omega V|u|^2\,dx<\infty\right\}\\
q_{-\Delta+V}(u,v)&=\int _\Omega(\nabla u\cdot\nabla v+Vuv)\,dx,\;u,v\in D(q_{-\Delta+V}).
\end{align*}
Then
$$
D(-\Delta+V)=\{u\in D(q_{-\Delta+V}):\exists v\in L^2(\Omega)\,\forall w\in D(q_{-\Delta+V})\colon q_{-\Delta+V}(u,w)=\langle v,w\rangle_2\}.
$$
In the case of general elliptic operators in divergence form as described above, one has to replace $H^1_0(\Omega)$ by the form domain of $A$ (with Dirichlet boundary conditions), everything else stays the same. I won't go into details here, suffice it to say that if the coefficient matrix is uniformly bounded, then the form domain of $A$ is simply $H^1_0(\Omega)$.


TITLE: Solve differential equation $\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$
QUESTION [2 upvotes]: I want to solve the following first- order nonlinear ordinary differential equation:
$\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$
where a,b and c are constants. I rewrote the equation:
$\leftrightarrow 1=\frac{1}{cx^2-x(b+c)+a}\frac{dx}{dt}\\
\leftrightarrow \int 1dt=\int \frac{1}{cx^2-x(b+c)+a} dx\\
\leftrightarrow t+k= \int \frac{1}{cx^2-x(b+c)+a} dx\\
\leftrightarrow t+k= \int \frac{1}{c(x-\frac{b+c}{2c})^2+a-\frac{(b+c)^2}{4c}} dx
$
for some arbitrary number k. How do I solve the last integral? Wolfram-Alpha tells me that it is
$\frac{2tan^{-1}(\frac{-c-b+2cx}{\sqrt{-c^2-b^2-2cb+4ca}})}{\sqrt{-c^2-b^2-2cb+4ca}}$
But I don't know how to calculate that on my own.

REPLY [0 votes]: The main step is converting the fraction
$$\frac{1}{cx^2+x(b+c)+a}$$
into the form, expected from the integral tables:
$$\int\frac{1}{t^2+q^2}dt=\frac{1}{q}\arctan \frac{t}{q}$$
You take out the extra $c$, complete the square and change variables:
$$\frac{1}{c}\frac{1}{\color{red}{(x+\frac{b+c}{2c})}^2-(\frac{b+c}{2c})^2+\frac{a}{c}}$$
Now you have $t=x+\frac{b+c}{2c}$ and $q^2=\frac{a}{c}+(\frac{b+c}{2c})^2$.


TITLE: What is the sum of only half the exponential terms that give the Dirac comb?
QUESTION [5 upvotes]: The following infinite sum of exponential terms gives a Dirac comb:
$$ \sum_{n=-\infty}^\infty e^{i n x} = 2 \pi \sum_{n=-\infty}^\infty \delta(x - 2 \pi n) $$
Of course the sum doesn't strictly converge, but only in the same sense in which the Dirac delta-function is defined.
What is the result of a semi-infinite sum of such terms?
$$ \sum_{n=1}^\infty e^{i n x} =~? $$

REPLY [3 votes]: Hint:
(1) write the semi-infinite sum in the following way
$$\sum_{n=1}^\infty e^{i n x} = \frac{1}{2} \sum_{n=-\infty}^\infty e^{i n x}
+ \frac{1}{2} \sum_{n=-\infty}^\infty \mathop{\rm sgn}(n) e^{i n x} - \frac{1}{2}.$$
The first summand is the Dirac-comb. So your question reduces to figuring out what the second term equals to...
(2) Lagrange's trigonometric identities might help there...