text:
\section{Introduction}

A classical topic in geometry is to understand the structure of spaces that maximize a given geometric quantity, within a suitable class of spaces. An emblematic example that motivates some of our work here is the Cheeger-Gromoll splitting theorem, which states that manifold with nonnegative Ricci curvature that contain an infinite-length geodesic must split off a line. There are many examples of results with a similar flavor, involving quantities such as the diameter, the spectral gap, isoperimetric-type properties, etc... We shall discuss relevant examples later in this introduction.

Once such a result is known, it is natural to investigate the structure of spaces that \emph{almost} maximize the geometric quantity, and determine whether they are close in some sense to having the same structure as an extremal space. When investigating classes of compact spaces, a non-explicit stability estimate can often be derived by a compactness argument, via Gromov's precompactness theorem, and showing that a sequence of spaces that saturate in the limit the geometric constraint must converge to an extremal space. See \cite[Theorem 1.5]{CM17a} for an example where this strategy is used. Our work here will deal with a class of non-compact spaces, for which this strategy is unavailable. Moreover, our focus will be on deriving estimates that have a fully explicit dependence on the various parameters (and in particular, making sure the estimates are dimension-free).

In this work, we shall consider metric measures spaces $(X,d,m)$ which belong to the class of $RCD(1,\infty)$ spaces, normalized so that $m(X)=1$. These spaces are a natural extension of Riemannian spaces satisfying a curvature condition. It is stronger than the Lott-Sturm-Villani curvature dimension condition CD$(1,\infty)$, but more appropriate for the type of questions we investigate here, since it is for example the right setting for the non-smooth generalization of the Cheeger-Gromoll splitting theorem \cite{Gig14}. For some of the results we obtain here, we shall further assume the existence of a $CD(1,\infty)$ disintegration of the metric measure space. Such spaces are studied -for general $K$ and $N$- in \cite{CaMi21} and denoted by $CD^1(K,N)$. These notions will be fully defined in Section \ref{sect_prelim}.

Examples of metric measures spaces admitting a CD$(1,\infty)$ disintegration are weighted Riemannian manifolds whose Bakry-Emery curvature is greater or equal to one, as proved by Klartag \cite{Kla17}, and $RCD(1,N)$ m.m.s (for any $1<N<+\infty$) as proved by Cavalletti and Mondino in \cite{CM17a}. It seems that an obvious adaptation of the proof \cite[Theorem 3.8]{CM17a} yields that metric measure spaces which are both RCD$(1,\infty)$ and $CD_{loc}(K,N)$, for $K \in \mathbb{R}$ and $1<N <+\infty$, also meet the $CD^1(1,\infty)$ condition.

In an RCD$(1,\infty)$ space, there is a spectral gap $\lambda_1 \geq 1$ for the first eigenvalue of the Laplacian; note that $\lambda_1=1$ holds true for the Gauss space. In \cite{GKKO}, Gigli, Ketterer, Kuwada, and Ohta proved the following rigidity result:

\begin{thm}\label{GKKO}
Let $(X,d,m)$ be an RCD$(1,\infty)$-space, and assume that $\lambda_1=1$. Then there exists an RCD$(1,\infty)$-space $(Y,d_Y,m_Y)$ such that:

\begin{itemize}

\item The metric space $(X,d)$ is isometric to the product space $(\mathbb{R},|\cdot|)\times (Y,d_Y)$ with the product metric.

\item Through the isometry above, the  measure $m$ coincides with the product measure $e^{-x^2/2}dx\otimes m_Y$.
\end{itemize}
\end{thm}

Other rigidity results for RCD$(1,\infty)$ spaces were obtained in \cite{OT20, Han21}. For smooth Riemannian manifolds endowed with their volume measure, the analogous inequality is due to Lichnerowicz and the case of equality -which characterizes the unit sphere- is due to Obata. Obata's theorem has been generalized to RCD$(N-1,N)$-spaces by Ketterer (with $N<\infty)$.

Recall that RCD$(N-1,N)$-spaces have diameter bounded from above by $\pi$ and thus, by Gromov's precompactness theorem, form a compact class of metric measure spaces for the measured Gromov Hausdorff distance. Therefore Ketterer's result also leads to stability results for  RCD$(N-1,N)$-spaces with almost minimal first eigenvalue (or equivalently, almost maximal diameter). In the setting of Riemannian manifolds, the equivalence follows from work of Cheng \cite{Cheng} and Croke \cite{Croke}. No such compactness property is available for the RCD$(1,\infty)$ spaces, which leaves the properties of such spaces with almost minimal first eigenvalue unknown.

Part of our work here will be to study RCD spaces with almost maximal spectral gap, that is RCD$(1,\infty)$ spaces with $\lambda_1 \leq 1 + \epsilon$. Let us recall that an RCD$(1,\infty)$-space has a discrete spectrum \cite[Proposition 6.7]{GMS15}. One of our main results is the following about higher eigenvalues:

\begin{thm} \label{main_thm_rcd}
Consider an RCD$(1, \infty)$-space, and assume its spectral gap $\lambda_1$ is smaller than $1+\epsilon$ with $\epsilon$ less than some fixed $\epsilon_0$. Then for any $n \geq 1$, $\theta < 1/2$, there is an eigenvalue $-\lambda_n$ of the Laplacian $\Delta$ such that $|\lambda_n - n\lambda_1| \leq C(n, \theta, \epsilon_0)\epsilon^{1/2 - \theta}.$
\end{thm}

Since by assumption $\lambda_1$ is close to one, we hence prove that positive integers are close to being eigenvalues of the Lpalacian when the spectral gap is almost minimal. The integers appear here as the eigenvalues of the Ornstein-Uhlenbeck operator, which is the Laplacian on the Gaussian space.

In this setting, higher eigenvalues are stable under measured Gromov-Hausdorff convergence \cite{GMS15}. The argument in that work uses a compactness argument, so it is not explicitly quantitative. Moreover, since the eigenfunctions are not necessarily globally Lipschitz, it does not seem like this type of statement can be immediately deduced from a statement on closeness to a Gaussian factor, such as Theorem \ref{thm_gaussian_component_eigen} below.

In the Euclidean setting, it is known that the sequence of ordered eigenvalues is bounded from below by the ordered eigenvalues of the Ornstein-Uhlenbeck operator on the same space \cite{Mil18}, as a consequence of Caffarelli's contraction theorem. To our knowledge, an analogous result in the full geometric setting is an open problem at the time of writing.

In the smooth finite-dimensional setting, Petersen \cite{Pet99} proved that an $n$-dimen\-sional manifold with Ricci curvature bounded from below by $n-1$, whose $(n+1)$-th eigenvalue is close to $n$, is close in the Gromov-Hausdorff sense to a sphere. Aubry \cite{Au05} later showed that the statement still holds for the $n$-th eigenvalue, with quantitative bounds, and that a control on the $(n-1)$-th eigenvalue is not enough. More recently, Takatsu \cite{Tak21} proved convergence of the spectral structure of low-dimensional projections of spheres of high-dimension to Gaussian spaces, which is a particular example to which all the results presented here apply.

We shall also show that the pushforward of the reference measure by a normalized eigenfunction is close to Gaussian, with dimension-free quantitative bounds, partially extending results of \cite{DPF17,CF20} from the Euclidean setting to the full setting of RCD spaces. Similar questions for manifolds with small deficit in the Bakry-Ledoux isoperimetric inequality were raised in \cite[Remark 7.6]{MO19}, and can be positively answered with the tools we use here.

\begin{thm} \label{thm_gaussian_component_eigen}
Let $(M,d,\mu)$ be an RCD$(1,\infty)$ probability space, and $f$ be a normalized eigenfunction of $-\Delta$ associated to the eigenvalue $\lambda$. Let $\nu=f_{\sharp} \mu$, and $\gamma$ be the standard one-dimensional Gaussian measure. Then,
$$W_1(\nu, \gamma) \leq  4\times 2^{\lambda/2}\sqrt{\lambda -1}.$$
\end{thm}

Here $W_1$ stands for the $L^1$ Wasserstein distance. Asymptotic normality for eigenfunctions on manifolds was studied by E. Meckes \cite{Mec09}, and (a variant of) Theorem \ref{thm_gaussian_component_eigen} can be directly deduced from her abstract theorem via our main Lemma \ref{keylemma} below. When the eigenspace associated with the spectral gap is of dimension higher than one, or if there are several eigenvalues close to one, then a multivariate version of \ref{thm_gaussian_component_eigen} holds, and can be proved with the arguments we develop here, as in the Euclidean setting \cite{CF20}.

A problem left open by this work is a complete stability estimate for Theorem \ref{GKKO}. We would expect a space with almost minimal spectral  gap to be close in some sense to a product space with Gaussian factor, but we have not been able to prove a satisfactory statement showing an approximate splitting at the level of the metric.

A second part of our work will deal with stability of the observable diameter, motivated by Ketterer's work on Obata's theorem in RCD spaces, as well as \cite{MO19}. While seeking for upper bound on the diameter of an RCD$(1,\infty)$-space is meaningless, Gromov's observable diameter is a good candidate to replace standard diameter. Let us recall its definition.

\begin{defn}[Observable diameter]
The observable diameter $D_{obs}$ of the metric-measure space $(M, d, \mu)$ is a function of $\kappa \in (0, 1)$ defined by
$$D_{obs}((M,d,\mu); \kappa) := \sup_{f:M \rightarrow \mathbb{R} }\Big\{ \inf_{E \operatorname{Borel}} \left\{ \operatorname{diam}(E); f_{\sharp}\mu(E) \geq 1-\kappa\right\}; f \; 1\operatorname{-Lipschitz}\Big\}.$$
For ease of notation, the observable diameter is also denoted by $D_{obs}(\mu; \kappa)$.
\end{defn}

As proved in Theorem \ref{ObsCompa} below, the observable diameter of an RCD$(1,\infty)$-space is pointwise bounded from above by that of the Gauss space.

We shall prove some results connecting RCD$(1,\infty)$ spaces with almost minimal spectral gap to those with almost maximal observable diameter. More precisely, we prove

\begin{thm} \label{sg_to_diam}
Let $(X,d,\mu)$ be an RCD$(1,\infty)$ space admitting a $CD(1,\infty)$ disintegration in the sense of Assumption \ref{assumpt_needle}, and whose spectral gap satisfies $\lambda_1 \leq 1 + \epsilon$. Then for any $\theta>0$ small enough and $\kappa\geq C(\theta) \epsilon^{1/20-\theta}$, the following inequalities hold
$$ D_{obs}( \gamma;\kappa-C(\theta)\epsilon^{1/20 - \theta}) \geq D_{obs}( \mu;\kappa - C(\theta)\epsilon^{1/20 - \theta}) \geq D_{obs}( \gamma;\kappa).$$

\end{thm}

The assumption on the $CD(1,\infty)$ disintegration in this result could be relaxed, in that we only need a disintegration with respect to one particular test function, namely an eigenfunction associated with the spectral gap.

This theorem can be compared with \cite[Theorem 4.4]{CMS19}, which proved that under a CD$(N-1,N)$ condition if the spectral gap is almost minimal then the diameter is almost maximal, with dimension-dependent exponent and constants.

Finally, we also obtain a converse result, deducing closeness of the spectral gap from closeness of the observable diameter:

\begin{thm}\label{thm_dobs_to_sg}
Let $(X,d,\mu)$ be an RCD$(1,\infty)$ space admitting a $CD(1,\infty)$ disintegration in the sense of Assumption \ref{assumpt_needle}, and assume that
$$D_{obs}((M,d,\mu); \kappa) \geq D_{obs}((\mathbb{R}, |\cdot|, \gamma); \kappa)     -\epsilon$$
for a given $0<\kappa<1$ and $\epsilon>0$ small enough. Then there exists $C = C(\kappa)$ such that the spectral gap $\lambda_1$ satisfies
$$\lambda_1 \leq 1 + C\epsilon^{1/22}.$$

\end{thm}

As a corollary, this implies rigidity of the observable diameter, under the extra Assumption \ref{assumpt_needle}, via the main Theorem of \cite{GKKO}. This statement seems to be new.

The remainder of the article is as follows: in Section 2, we present some necessary preliminaries on RCD spaces and needle decompositions. In Section 3, we prove Theorems \ref{main_thm_rcd} and \ref{thm_gaussian_component_eigen}, and Section 4 contains the proofs of Theorems \ref{sg_to_diam} and \ref{thm_dobs_to_sg}.

\section{Preliminaries}
\label{sect_prelim}

\subsection{RCD spaces}

\subsubsection{Definitions}

In this work, we shall always work with separable, complete metric spaces, which we shall denote by $(M, d)$. We shall endow it with a Borel probability measure $\mu$. Additionally, we shall assume the metric space is \emph{geodesic}: every pair of points $x, y$ are connected by a minimal geodesic $\gamma : [0, 1] \longrightarrow M$ such that $\gamma(0) = x$, $\gamma(1) = y$ and $d(\gamma(s), \gamma(t)) = |t-s|d(x,y)$ for all $s, t \in [0, 1]$.

We can endow the space of Borel probability measures on $M$ with finite second order moment, denoted by $\mathcal{P}_2(M)$, with the $L^2$ Wasserstein distance from optimal transport. We refer to the monograph \cite{Vil03} for the definition of $L^p$ Wasserstein distances.

The relative entropy functional with respect to $\mu$ on $\mathcal{P}_2(M)$ is defined as
$$\Ent_{\mu}(\nu) := \int{\rho \log \rho d\mu}$$
if $\nu = \rho \mu$ is an absolutely continuous probability measure, and $+ \infty$ otherwise.

With these objects in mind, we can define the curvature-dimension condition CD$(K, \infty)$:

\begin{defn}
The space $(M, d, \mu)$ is said to satisfy the curvature-dimension condition CD$(K, \infty)$ for $K \in \mathbb{R}$ if the relative entropy is $K$-convex along $W_2$-geodesics on $\mathcal{P}_2(M)$, that is for any pair of probability measures $\nu_1$, $\nu_2$ with finite relative entropy with respect to $\mu$, there is a minimal $W_2$ geodesic $(\nu_t)_{t \in [0,1]}$ connecting them, and such that for all $t \in [0, 1]$
$$\Ent_{\mu}(\nu_t) \leq (1-t)\Ent_{\mu}(\nu_0) + t \Ent_{\mu}(\nu_1) -\frac{Kt(1-t)}{2}W_2(\nu_0, \nu_1)^2.$$
\end{defn}

The parameter $K$ plays the role of a Ricci curvature lower bound. Indeed, when the space is a smooth manifold endowed with its (normalized) volume measure, it satisfies this condition iff the Ricci curvature tensor is bounded from below by $K$ times the metric tensor. This definition is a particular instance of the more general curvature dimension CD$(K, N)$, that takes into account the dimension. Here we shall mostly deal with the infinite-dimensional setting, so we omit the full definition, and refer to \cite{Vi09} for the more general setting.

Given the metric-measure structure, we can also define the Cheeger energy
$$\Ch(f) := \frac{1}{2} \inf_{(f_i)} \liminf \int{|\Lip f_i|^2d\mu}$$
where the infimum runs over the set of all sequences of locally-Lipschitz functions that converge to $f$ in $L^2(\mu)$, and where $\Lip (f)(x)$ is the local Lipschitz constant.

Given an $L^2$ function with finite Cheeger energy, there exists a minimal weak upper gradient $|\nabla f| \in L^2(\mu)$ such that
$$\Ch(f) = \frac{1}{2}\int{|\nabla f|^2 d\mu}.$$
The Sobolev space $W^{1,2}$ is the space of $L^2$ functions with finite Cheeger energy. We refer to \cite{AGS14b} for more details about these notions.

\begin{defn}
A metric-measure space $(M, d , \mu)$ that satisfies the CD$(K, \infty)$ condition is said to satisfy the Riemannian curvature-dimension condition RCD$(K, \infty)$ if the Cheeger energy is a quadratic form, that is
$$\Ch(f+ g) + \Ch(f-g) = 2\Ch(f) + 2\Ch(g)$$
for all $f, g \in W^{1,2}$.
\end{defn}

The most basic example of an RCD$(K, \infty)$ with positive $K$ is a Gaussian measure on $\mathbb{R}$  with variance $K^{-1}$. More generally, a probability measure on $\mathbb{R}^d$ with density $e^{-V}$ w.r.t. the Lebesgue measure satisfies the RCD$(K, \infty)$ condition iff $\Hess V \geq K$.

On a smooth manifold endowed with its volume measure, the RCD$(K, \infty)$ condition is equivalent to requiring the Ricci curvature tensor to be bounded from below by $K$. More generally, if the measure has density $e^{-V}$ w.r.t. the volume measure, the RCD$(K, \infty)$ condition is equivalent to the Bakry-Emery condition
$$\operatorname{Ric} + \Hess V \geq K\tilde{g}$$
where $\tilde{g}$ is the Riemannian metric tensor of the manifold. However, there are also non-smooth spaces that satisfy the RCD$(K, \infty)$ condition, including Alexandrov spaces \cite{Pet} and some stratified spaces \cite{Ber}.

On an RCD space, we can then define by polarization the scalar product of two elements of $W^{1,2}$ and the Dirichlet form
$$\langle \nabla f, \nabla g \rangle := \frac{1}{4}(|\nabla(f+g)|^2 - |\nabla (f-g)|^2) \in L^1(\mu); \hspace{2mm} \mathcal{E}(f,g) := \int{\langle \nabla f, \nabla g \rangle d\mu}.$$
The natural analog of the Laplacian on $(M, d, g)$ is the operator $\Delta : D(\Delta) \longrightarrow L^2$ such that
$$\mathcal{E}(f, g) = - \int{g(\Delta f) d\mu} \hspace{2mm} \forall g \in W^{1,2}.$$
The domain $D(\Delta)$ is dense in $W^{1,2}$. We refer to \cite{AGS14} for details of the construction. Note that even when the space is a smooth manifold, $\Delta$ is not simply the Laplace-Beltrami operator, since it takes into account the reference measure $\mu$, that is not necessarily the volume measure. For example, for the Euclidean space endowed with a measure with density $e^{-V}$, the natural Laplace operator is $\Delta_{eucl} - \nabla V \cdot \nabla$, where $\Delta_{eucl}$ is the usual Laplacian on $\mathbb{R}^d$. Finally, note that since the Cheeger energy is used as a Dirichlet form, the induced Laplacian is a local operator, which will allow us to use the diffusion property.

\subsubsection{Properties of RCD$(K, \infty)$ spaces}

An important property of RCD$(K, \infty)$ is the tensorization property, which states that this class of spaces is stable for the product:

\begin{prop}[Tensorization property]
If two spaces $(M_i, d_i, \mu_i)$ satisfy the RCD$(K, \infty)$ condition, then the product space $(M_1 \times M_2, d_1 \oplus d_2, \mu_1 \otimes \mu_2)$ also does. Here $d_1 \oplus d_2((x_1,x_2), (y_1,y_2)) = \sqrt{d_1(x_1,y_1)^2 + d_2(x_2,y_2)^2}$.
\end{prop}

In particular this property explains why rigidity statements such as Theorem \ref{GKKO} involve a splitting: the product of an RCD$(1,\infty)$ space with minimal spectral gap and
any other RCD$(1,\infty)$ space is still an RCD$(1,\infty)$ space, and still has minimal spectral gap.

The spectral gap $\lambda_1$ can be reformulated as the sharp constant in the Poincar\'e inequality for $\mu$, that is the largest constant $C_P$ such that
$$\Var_{\mu}(f)  \leq \frac{1}{C_P}\int{|\nabla f|^2d\mu} \hspace{3mm} \forall f \in W^{1,2}.$$

As we have mentioned in the introduction, an RCD$(K, \infty)$ space with positive $K$ satisfies a Poincar\'e inequality with constant $K^{-1}$, that is $\lambda_1 \geq K$. Moreover, as proved in \cite{GKKO} (with results in the smooth setting obtained in \cite{CZ17}), this bound is rigid: if the sharp spectral gap of an RCD$(K, \infty)$ is equal to $K$, then the space splits off a Gaussian factor that has variance $K^{-1}$.

The sharp constant is indeed the inverse of the smallest positive eigenvalue $\lambda_1$ of $-\Delta$, that is there exists an eigenfunction $f$ such that
$$\Delta f = -C_P^{-1}f.$$
Existence is ensured by the validity of the logarithmic Sobolev inequality, which is strictly stronger than the Poincar\'e inequality, and provides enough compactness to ensure there is a function that achieves equality in the sharp Poincar\'e inequality, see \cite[Proposition 6.7]{GMS15}.

\subsection{Localization of the curvature condition}

Part of our work requires an additional assumption on the metric measure space, namely the existence of a suitable needle decomposition. A brief reminder on that decomposition is provided below. This notion originates in Klartag's work \cite{Kla17} in the smooth setting, and was extended to metric measure spaces by Cavalletti and Mondino \cite{CM17a}. Recently, \cite{CGS20} showed that the needle decomposition provides an equivalent way of defining curvature-dimension conditions with finite dimension (which unfortunately is not our setting here).

\begin{thm}[Needle decomposition] \label{thm_needle_dec}
Let $(M, d, \mu)$ be an RCD$(1, N)$ space ($N<+\infty$) and $f$ be a centered, $L^1$ function such that
$$\int{|f(x)|d(x, x_0)d\mu} < \infty.$$
Then the space $X$ can be written as the disjoint union of two measurable sets $Z$ and $T$, where $T$ admits a partition in the following sense. There exists a $1$-Lipschitz function $g$, a partition $(X_q)_{q \in Q}$ of $T$, a probability measure $\mathfrak{q}$ on $Q$ and a family of probability measures $(m_q)_{q \in Q}$, each supported on $X_q$, such that
\begin{enumerate}
\item For any measurable $A$, $\mu(A\cap T) = \int{m_q(A)d\mathfrak{q}(q)}$;

\item For $\mathfrak{q}$ a.e. $q$, for any $x, y$ in $X_q$, $d(x,y) = |g(x) - g(y)|$, and for any $z \notin X_q$, there exists $x \in X_q$ such that $|g(z) - g(x)| < d(x,z)$;

\item For $\mathfrak{q}$ a.e. $q$, if $X_q$ is not a singleton then $(X_q, d, m_q)$ is RCD$(1, N)$;

\item For $\mathfrak{q}$ a.e. $q$, $\int_{X_q}{fdm_q} = 0$ and $f=0$ $\mu$-a.e. in $Z$.
\end{enumerate}
\end{thm}

Note that the properties of the needle decomposition require that $\nu$-almost every needle is not reduced to a singleton, and that $\operatorname{Lip} (g) = 1$ almost everywhere.

The $1$-Lipschitz function $g$ is called a guiding function, and is obtained as an optimizer in the variational problem

$$\sup_{u 1-lip} \int{f u d\mu}.$$

It can be interpreted as the potential in the Kantorovitch-Rubinstein dual formulation of the $L^1$ optimal transport problem between the positive measures $f_+ \mu$ and $f_-\mu$, which have same mass when $f$ is centered.

Since we consider infinite-dimensional spaces, the current state of knowledge on needle decompositions does not guarantee existence of the decomposition, as the proof uses compactness of finite-dimensional positively curved spaces. However, it may nonetheless exist (indeed, its study in the Euclidean context goes back to work of Payne and Weinberger \cite{PW60}). So we make the following assumption for Theorems \ref{sg_to_diam} and \ref{thm_dobs_to_sg}.

\begin{assumption} \label{assumpt_needle}
The space $(M, d, \mu)$ satisfies the conclusion of the above theorem with $N = \infty$.
\end{assumption}

Note that since the second moment is finite and the eigenfunction is $L^2$, the integrability condition $\int{|f|d(x,x_0)d\mu} < \infty$ is automatically satisfied.

\begin{prop}[Structure of needles] \label{prop_struct_needle}
Without loss of generality, for a.e. $q$ we can identify $(X_q, d, m_q)$ with a space of the form $(\mathbb{R}, |\cdot|, e^{-\psi_q}dx)$ where $\psi''_q \geq 1$ (in the weak sense) and  $\psi_q$ is finite in some interval $I$, and such that, on $X_q$, $g(t) = t + c_q$.
\end{prop}

\begin{proof}
Let $I = \{g(x), x \in X_q\}$. It is an interval of $\mathbb{R}$. From the structure of the needle, we can parameterize $X_q$ by $I$, and the measure $m_q$ is of the form $e^{-\psi_q}dx$ on $I$. We can then extend $\psi_q$ to take the value $+\infty$ outside $I$, which allows us to identify $(I, |\cdot|, e^{-\psi_q}dx)$ with $(\mathbb{R}, |\cdot|, e^{-\psi_q}dx)$.
\end{proof}

\subsection{Observable diameter}

We now define the notion of observable diameter, which plays the role of diameter for RCD$(1,\infty)$ spaces, by taking advantage of concentration inequalities. This notion was introduced by Gromov in \cite{Gro}. Informally, a control on the observable diameter states that, even if the diameter itself is not finite, most of the mass is concentrated inside a set of controlled diameter.

\begin{defn}
The observable diameter $D_{obs}$ of the metric-measure space $(M, d, \mu)$ is a function of $\kappa \in (0, 1)$ defined by
$$D_{obs}((M,d,\mu); \kappa) := \sup_{f:M \rightarrow \mathbb{R} }\Big\{ \inf_{E \operatorname{Borel}} \left\{ \operatorname{diam}(E); f_{\sharp}\mu(E) \geq 1-\kappa\right\}; f \; 1\operatorname{-Lipschitz}\Big\}.$$

\end{defn}

The observable diameter can be estimated by means of the separation distance whose definition (in the case of two sets with the same lower bound on their mass) is recalled below.

\begin{defn}
The separation distance is a function of $\kappa \in (0, 1)$ defined by
$$ Sep((M,d,\mu); \kappa):=\sup_{A_1,A_2\subset X \;\rm{Borel}} \left\{ \operatorname{d} (A_1,A_2); \mu(A_i) \geq \kappa, i \in\{1,2\}\right\}.$$

\end{defn}

The following inequality holds (see \cite[Proposition 2.26]{Shioya} for a proof):

\begin{prop} Let $(M, d, \mu)$ be a metric measure space such that $\mu(M)=1$. Then, for any $\kappa \in (0,1)$

\begin{equation}\label{ObsSep}
D_{obs}((M,d,\mu); \kappa) \leq Sep((M,d,\mu); \kappa/2).
\end{equation}

\end{prop}

\begin{rmq}
In the case of the Gaussian space $(\mathbb{R},|\cdot|,\gamma)$, the symmetry of the density with respect to the origin yields for any $\kappa \in (0,1)$
$$ D_{obs}((\mathbb{R},|\cdot|,\gamma); \kappa) = Sep((\mathbb{R},|\cdot|,\gamma); \kappa/2).$$
\end{rmq}

On RCD$(1,\infty)$ spaces, the Lévy-Gromov type isoperimetric inequality has been proved by Ambrosio and Mondino \cite{AmMo16}:

\begin{thm}\label{IIn} Let $(X,d,\mu)$ be an RCD$(1,\infty)$ probability space. Then, for any Borel subset $E\subset X$ it holds
$$ \mathcal{P}(E) \geq \frac{1}{\sqrt{2\pi}} e^{-a_E^2/2},$$
where $\mathcal{P}$ stands for perimeter, and $ \mu(E)= \gamma((-\infty,a_E])$.
\end{thm}

As a consequence, one gets the following comparison inequality for the observable diameter.

\begin{thm}\label{ObsCompa} Let $(M,d,\mu)$ be an RCD$(1,\infty)$ probability space. Then for any $\kappa \in (0,1)$, the following inequalities hold
\begin{equation*}
\begin{split}
 D_{obs}((M,d,\mu); \kappa) & \leq   Sep((M,d,\mu); \kappa/2) \\ & \leq  Sep((\mathbb{R},|\cdot|,\gamma); \kappa/2) = D_{obs}((\mathbb{R},|\cdot|,\gamma); \kappa).\end{split}
\end{equation*}

\end{thm}

\begin{proof}
The only property that is yet to prove  is the inequality involving the separation distances of $(\mathbb{R},|\cdot|,\gamma)$ and $(M,d,\mu)$. The proof is the same as the one of \cite[Lemma 2.32]{Shioya}, we report it since we need the argument later in this work.

Let $A_1,A_2 \subset X$ be two mutually disjoint sets such that $\mu(A_i) \geq \kappa/2$ for $i \in \{1,2\}$. Set $r:= d(A_1,A_2)$ and notice that $A_1 \cap V_r(A_2)=\emptyset$, where $V_r(A_2)$ stands for the open $r$-neighborhood of $A_2$. Let us also define $\sigma(s):= \gamma((-\infty,s])$ and note that $\sigma$ is an increasing function of $s$. Consequently, making use of the isoperimetric inequality for RCD$(1,\infty) $ spaces in integrated form, we infer for any $s>0$
$$ \mu (V_s(A_2)) \geq \gamma((-\infty, -a +\delta +s]) \geq  \gamma((-\infty, -a +s]),$$
where $-a+\delta :=\sigma^{-1}(\mu(A_2)) \geq -a:=\sigma^{-1}(\kappa/2)$. With this notation,
$$ Sep((\mathbb{R},|\cdot|,\gamma); \kappa/2)=2 |\sigma^{-1}(\kappa/2)| =  2a.$$
Now, observe that
\begin{eqnarray*}
\kappa/2 \leq \mu(A_1)  & \leq &  1 -\mu(V_r(A_2))\\
					 & \leq &  1 - \gamma((-\infty, -a +r])

\end{eqnarray*}
Therefore, by symmetry of $\gamma$,
$$ \gamma ((-\infty,-a])  +\gamma(([a - r,+\infty)) = \kappa/2 + \gamma((-\infty, -a +r])\leq  1.$$
Thus, $a-r \geq -a$ hence $2|\sigma^{-1}(\kappa/2)| \geq r$ and the result is proved.
\end{proof}

\section{Properties of RCD$(1,\infty)$ spaces with almost minimal spectral gap}

In this section, we shall investigate properties of RCD$(1,\infty)$ spaces when the spectral gap is close to one, and prove Theorems \ref{main_thm_rcd} and \ref{thm_gaussian_component_eigen}.

In the sequel, $L^p$ norms will always be with respect to the reference measure $\mu$, that is
$$||f||_p := \left(\int{|f|^pd\mu}\right)^{1/p}.$$

\subsection{Main lemma}

We shall first prove the main technical lemma that underlies the proofs of our results, Lemma \ref{keylemma}, which states that if the spectral gap is close to one, then the gradient of an associated eigenfunction is almost constant.

We shall use the following $L^p$ version of Poincar\'e inequalities:

\begin{prop} \label{prop_p_poincare}
Under the curvature condition, for any $p \geq 1$we have
$$\left|\left|f - \int{fd\mu}\right|\right|_{p} \leq 2p\left|\left||\nabla f|\right|\right|_{p}.$$
\end{prop}

This inequality is strongest for $p = 1$, where it corresponds to the classical Cheeger inequality. See for example \cite[Proposition 2.5]{Mil09} for a proof. The constant in this formulation is not sharp, since for $p=2$ this is not the best estimate on the spectral gap.

When considering eigenfunctions in positive curvature, we have nice integrability estimates in all $L^p$ spaces, which are summarized in the following proposition:

\begin{prop}[Integrability of eigenvectors] \label{integ_eigen}
Assume that curvature is bounded from below by $1$. Let $f$ be an eigenfunction of $-{\Delta}$ with eigenvalue $\lambda$ and normalized so that $||f||_2 = 1$. Then
\begin{enumerate}
\item $||f||_p \leq (p-1)^{\lambda/2}$ for any $p \geq 2$;

\item $|||\nabla f|||_p \leq 2^{\lambda }||f||_{2p}^2 \leq (4p-2)^{\lambda}$ for any $p \geq 1$.
\end{enumerate}
\end{prop}

\begin{proof}
The proof of (i) can be found in \cite[Section 5.3]{BGL14}.

To prove (ii), we use the gradient bound  $|\nabla P_tf |^2 \leq (e^{2t} -1)^{-1}P_t(f^2)$ (see \cite[Theorem 4.7.2]{BGL14}). We apply it to the eigenfunction $f$ with $t= (\ln 2)/2$ to get
$$\Gamma(f)^p \leq \left(2^{\lambda }P_t(f^2)\right)^p \leq 2^{\lambda p}P_t(f^{2p})$$
and then integrate with respect to the reference measure $\mu$ to conclude.
\end{proof}

The key lemma in our proofs is the following result, stating that the gradient of the eigenfunction is almost constant when $\lambda$ is close to one, with quantitative bounds in all $L^p$ norms. A version of this result in the Euclidean setting was derived in \cite{DPF17}, and proved using optimal transport and Caffarelli's contraction theorem. The proof we present here, in addition to being more general, is simpler even in the Euclidean setting.

\begin{lem}[Key Lemma]\label{keylemma}
Let $f$ be an eigenfunction of $-\Delta$ associated with eigenvalue $\lambda \leq 1+\epsilon$, with $\int{fd\mu} = 0$, $\int{f^2d\mu} = 1$. Then for any $p \in [1, 2[$, we have
$$|||\nabla f|^2 - \lambda||_p \leq 4p(\lambda-1)^{1/2}\lambda^{1/2}\left(\frac{6p-4}{2-p}\right)^{\lambda/2}. $$

Assume now $\lambda \leq \bar{\lambda}$ for some fixed threshold $\bar{\lambda}$. Then for any $\theta \in (0, 1/2)$ and $p \geq 2$ there exists $C = C(p, \theta, \bar{\lambda})$ such that
$$|||\nabla f|^2 - \lambda||_p \leq C(\lambda - 1)^{1/2 - \theta}.$$
\end{lem}

\begin{proof}

In the non-smooth setting, under the curvature condition we have the following pointwise estimate of \cite[Theorem 3.4]{Sav15} (generalizing a result of \cite{Bak83, Bak94}) that holds almost everywhere:
\begin{equation}
|\nabla |\nabla f|^2| \leq 2|\nabla f|\sqrt{\Gamma_2(f) - |\nabla f|^2}
\end{equation}
This is only proved for globally Lipschitz $f$ in \cite{Sav15}, but we shall only use its integrated version, so we can proceed by approximation, by taking $h_{n,t} = P_t \max(-n, \min(f, n))$ for $t, n > 0$, which is globally lipschitz (see for example \cite[Proposition 2.5]{GKKO}), and then letting $n$ go to infinity, then using the fact that $f$ is an eigenfunction so $P_t f = e^{-\lambda t}f$ to let $t$ go to zero. In the smooth setting, $|\nabla |\nabla f|^2|^2 \leq 4||\operatorname{Hes} f||_2^2 |\nabla f|^2$. A version of this estimate involving Hessians in the nonsmooth context was investigated in \cite{Gig18}.

Therefore, using Holder's inequality, ad denoting by $\Gamma_2$ the usual operator
$$\Gamma_2(h) = \frac{1}{2}L\Gamma(h) - \Gamma(h, Lh),$$
we have for $1 \leq p < 2$
\begin{align*}
|||\nabla |\nabla f|^2|||_p^p &\leq 2^p\left(\int{(\Gamma_2(f) - |\nabla f|^2)d\mu}\right)^{p/2}\left(\int{|\nabla f|^{2p/(2-p)}d\mu}\right)^{(2-p)/2} \\
&\leq 2^p(\lambda-1)^{p/2}\left(\int{|\nabla f|^2d\mu}\right)^{p/2}\left(\int{|\nabla f|^{2p/(2-p)}d\mu}\right)^{(2-p)/2} \\
&\leq 2^p(\lambda-1)^{p/2}\lambda^{p/2}\left(\frac{6p-4}{2-p}\right)^{\lambda p/2}
\end{align*}
To conclude, we just apply the $L^p$ Poincar\'e inequality to $|\nabla f|^2$.

To obtain the inequality for $p \geq 2$, using Holder's inequality we have
$$|||\nabla f|^2 - \lambda||_p \leq |||\nabla f|^2-\lambda||_1^{1-\theta}|||\nabla f|^2-\lambda||_{p'}^{\theta} \leq |||\nabla f|^2-\lambda||_1^{1-\theta}(\lambda + |||\nabla f|^2||_{p'})^{\theta}$$
with $p'$ suitably large enough, and we then apply Proposition \ref{integ_eigen}.
\end{proof}

\subsection{Spectral properties}

Let us start with the following general lemma:

\begin{lem} \label{perturbed_eigen} Given $\mathcal{L}$ an unbounded operator on $L^2(\mu)$ with discrete spectrum. Assume that there is a function $f$ such that $\mathcal{L}f = \alpha f + g$ with $||f||_2 = 1$ and $||g||_2 \leq \epsilon$. Then there is an eigenvalue of $\mathcal{L}$ at distance at most $\epsilon$ of $\alpha$.
\end{lem}

\begin{proof}
Let $\alpha_i$ be the collection of eigenvalues of $\mathcal{L}$, with associated orthonormal basis of eigenfunctions $f_i$. We can write $f = \sum \beta_i f_i$ and $g = \sum \delta_if_i$ with $\sum \beta_i^2 = 1$ and $\sum \delta_i^2 \leq \epsilon^2$. Then
$$\sum (\alpha_i -\alpha)\beta_if_i =  \sum \delta_i f_i$$
and hence for any $i$, $(\alpha_i -\alpha)\beta_i = \delta_i$, so that $\sum (\alpha_i -\alpha)^2\beta_i^2 \leq \epsilon^2$. But since $\sum \beta_i^2 = 1$, there is at least one index $i$ such that $|\alpha_i - \alpha| \leq \epsilon$.
\end{proof}

\smallskip

With all this in hand, we can now prove Theorem \ref{main_thm_rcd}:

\begin{proof}[Proof of Theorem \ref{main_thm_rcd}]
Consider $H_n$ the $n$-th Hermitte polynomial, defined via $H_0 = 1$, $H_1(x) = x$ and $H_{n+1}(x) = xH_n(x) - nH_{n-1}(x)$. Recall that $H_n''(x) - xH_n'(x) = -nH_n(x)$, and that they form an orthogonal basis of eigenfunctions for the one-dimensional Ornstein-Uhlenbeck generator.

Let $f$ be a normalized eigenfunction of $-\Delta$ with eigenvalue $\lambda$. From the diffusion property \cite[Section 2.2]{Sav15},
\begin{equation}
{\Delta}H_n(f) = H_n'(f){L}f + H_n''(f)|\nabla f|^2 = -\lambda n H_n(f) + H_n''(f)(|\nabla f|^2 - \lambda).
\end{equation}

According to Lemma \ref{keylemma},   we have a suitable $L^p$ bound on $|\nabla f|^2 - \lambda$ for all $p$, and $H_n''(f)$ has controlled $L^p$ norm via Proposition \ref{integ_eigen}, we control the $L^2$ norm of $H_n''(f)(|\nabla f|^2 - \lambda)$ by some $C(n, \theta, \bar{\lambda})(\lambda-1)^{1/2-\theta}$, and can then apply Lemma \ref{perturbed_eigen} to conclude.
\end{proof}

\subsection{Proof of Theorem \ref{thm_gaussian_component_eigen}}

We shall now prove that the pushforward of the measure by the eigenfunction is approximately Gaussian, generalizing part of the main result of \cite{CF20} to the RCD setting. The proof will be based on (a version of) Stein's lemma, a classical result in probability and statistics, which allows to control distances to a Gaussian target measure via deficits in an integration-by parts formula. This was already the strategy used in \cite{Mec09,CF20}. We refer to \cite{Ros11} for the proof, and an introduction to Stein's method. We could also proceed by directly combining the main abstract theorem of \cite{Mec09} with Lemma \ref{keylemma}.

\begin{lem}
Let $\gamma$ be the standard Gaussian measure on $\mathbb{R}$. Then for any $\mu \in \mathcal{P}(\mathbb{R})$, we have
$$\max\left( d_{TV}(\mu, \gamma), W_1(\mu, \gamma) \right)\leq \sup_{||f||_{\infty}, ||f'||_{\infty} \leq 4} \int{f' - xf d\mu}. $$
\end{lem}

Here $W_1$ stands for the $L^1$ optimal transport distance, and $d_{TV}$ for the total variation distance on $\mathcal{P}(M)$.

To keep the statement compact, we have not given the sharp required bounds on $f$ and $f'$ in the integration by parts formula. If we only wish to control $W_1$, we can additionally assume the second derivative to be bounded. Higher-dimensional versions of these estimates are available, although typically getting good estimates on the total variation distance in higher dimension is much harder than estimates on $W_1$, due to regularity issues for solutions to certain Poisson equations.

Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be a $1$-Lipschitz function. According to Stein's lemma,
$$W_1(\nu, \gamma) \leq 4\sup_{\tiny g \,1\mbox{-Lipschitz}} \int{(g' - xg)d\nu}.$$
Actually, the factor $4$ is not necessary, if one uses a version of Stein's lemma with sharper constants.
We have
\begin{align*}
\int{xg(x)d\nu} &= \int{f(x)g(f(x))d\mu} = -\lambda^{-1}\int{Lf(x) g(f(x))d\mu} \\
&= \lambda^{-1}\int{\Gamma(f, g \circ f)(x)d\mu} = \lambda^{-1}\int{g'(f(x))\Gamma(f)(x)d\mu}
\end{align*}

Hence, using Lemma \ref{keylemma}, we get
\begin{align*}
\sup_{\tiny g \,1\mbox{-Lipschitz}} \int{(g' - xg)d\nu} &\leq  \lambda^{-1}\int{|\Gamma(f)-\lambda|d\mu} \\
&\leq  4\times 2^{\lambda/2}\sqrt{\lambda -1}.
\end{align*}

\section{Spectral gap and observable diameter}

This section aims at proving that for an RCD$(1,\infty)$ space that admits $CD(1,\infty)$ disintegration, the spectral gap is close to that of the Gaussian space if and only if the observable diameter is close to that of the Gauss space.

\subsection{Proof of Theorem \ref{sg_to_diam}}

Let $f$ be a normalized eigenfunction of $-\Delta$, with eigenvalue $\lambda_1$, as in the previous section. We consider the guiding function in the needle decomposition
\begin{equation}\label{gf}
g := \underset{\tiny u \,1\mbox{-Lipschitz}}{\argmax}  \hspace{1mm} \int{uf d\mu}.
\end{equation}
Without  loss of generality, we may assume that $g$ is centered. Note that since $f$ is an eigenfunction, $g$ also maximizes $\int{\nabla u \cdot \nabla fd\mu}$

The approach follows in the spirit the same one as in \cite{MO19}. In a first lemma, we estimate the $L^2$-norm of $f$ and $|\nabla f|$ along many needles.

\begin{lem}\label{EstNeed1}
Assume $\lambda_1 \leq 1 + \epsilon$ with $\epsilon \leq 1$. There is a set of needles $Q_{\delta}$ such that $\mathfrak{q}(Q_{\delta}) \geq 1 - \delta$ and for any needle $q \in Q_{\delta}$ we have
$$1 - \frac{48\sqrt{\epsilon} + 2\epsilon}{\delta} \leq \int{f^2dm_q} \leq 1 + \frac{48\sqrt{\epsilon}}{\delta}$$
and
$$\int{f^2dm_q} \leq \int{|\nabla f|^2dm_q} \leq \int{f^2dm_q} + \frac{2\epsilon}{\delta}.$$
\end{lem}

\begin{proof}

On a.e. needle $q$, we have by the Poincar\'e inequality on the needle
$$\int{f^2dm_q} \leq \int{|f'|^2dm_q} \leq \int{|\nabla f|^2dm_q}.$$
Therefore, by Markov's inequality, we have
\begin{eqnarray*}
\mathfrak{q}\left(\left\{q; \int{|\nabla f|^2dm_q} \geq \int{f^2dm_q} + 2\epsilon/\delta\right\}\right)& \leq &\frac{\delta\big(\iint{|\nabla f|^2dm_qd\mathfrak{q}(q)} - \iint{f^2dm_qd\mathfrak{q}(q)\big)}}{2\epsilon} \\ &\leq & \delta/2.\end{eqnarray*}
So for an arbitrarily large proportion of needles, we almost have equality in Poincar\'e's inequality for $f$, up to a deficit of order $\epsilon$.

Then from Lemma \ref{keylemma} with $p=1$,
$$\int{\left|\int{(|\nabla f|^2 - 1)dm_q}\right| d\mathfrak{q}(q)} \leq \int{||\nabla f|^2 - 1|d\mu} \leq 24\sqrt{\epsilon}.$$
Hence
$$\mathfrak{q}\left(\left\{q; \left|\int{|\nabla f|^2dm_q} - 1\right| \geq 48\sqrt{\epsilon}/\delta\right\}\right) \leq \delta/2.$$
Taking the intersection, we get a set of needles of mass at least $1 - \delta$ on which
$$1-\frac{48\sqrt{\epsilon}}{\delta} \leq \int{|\nabla f|^2dm_q} \leq 1 + \frac{48\sqrt{\epsilon}}{\delta}$$
and
$$\int{f^2dm_q} \leq \int{|\nabla f|^2dm_q} \leq \int{f^2dm_q} + \frac{2\epsilon}{\delta}.$$
Reordering these bounds leads to the desired result.
\end{proof}

Next, we prove the $L^2$-closeness of $f$ and $g$ for a large set of needles.
\begin{lem}\label{fg-clo}
For any $q \in Q_{\delta}$, there exists a constant $c_q$ such that
$$\int{(f - g - c_q)^2dm_q} \leq \frac{C\sqrt{\epsilon}}{\sqrt{\delta}}.$$
\end{lem}

\begin{rmq}
Recall that for $\mathfrak{q}$-a.e. $q$, $\int f \, dm_q=0$. Since the average is an $L^2$ projection, we can therefore replace $c_q$ by $\int{gdm_q}$ in the above estimate.  Moreover, we can infer from the lemma above that
\begin{equation}\label{cq-mean}
\left| \int g \, dm_q -c_q\right| \leq  \frac{C\epsilon^{1/4}}{\delta^{1/4}}.
\end{equation}
\end{rmq}

\begin{proof}
Let $\Theta = \left(\int{f^2dm_q}\right)^{-1/2}$. Recall that we can identify each needle $X_q$ with the space $( \mathbb{R}, |\cdot|, e^{-\psi_q}dt)$, by Proposition \ref{prop_struct_needle}.  Using the results of \cite{CF20} on stability of the Bakry-Emery theorem on an Euclidean space, and since $\int{fdm_q} = 0$, we have
$$\int_{\mathbb{R}}{(\Theta f- t)^2e^{-\psi_q(t)}dt} \leq \frac{\sqrt{18\epsilon}}{\sqrt{\delta}}$$
and therefore, from the parametrization of $g$ on the needle,
$$\int_{X_q}{(\Theta f- g - c_q)^2e^{-\psi_q(t)}dt} \leq \frac{\sqrt{18\epsilon}}{\sqrt{\delta}}.$$
From Lemma \ref{EstNeed1}, and provided $\sqrt{\epsilon}/\delta\leq C_0 $, we have $|\Theta ^2 - 1| \leq C\sqrt{\epsilon}/\delta$; therefore
$$|\Theta - 1|^2 \leq C\epsilon/\delta^2.$$
Then
\begin{eqnarray*}
\int_{\mathbb{R}}{( f- g - c_q)^2e^{-\psi_q(t)}dt} &\leq & 2(\Theta -1)^2\int_{\mathbb{R}}{f^2} e^{-\psi_q(t)}dt  \\ &{} &\;\hspace{2,3cm} +     2\int_{\mathbb{R}}{(\Theta f- g - c_q)^2e^{-\psi_q(t)}dt}, \\
&\leq &  C\sqrt{\epsilon}/\sqrt{\delta}.
\end{eqnarray*}
\end{proof}

In the last technical lemma of this section, we show that $g$ is almost centered on a large set of needles.
\begin{lem}\label{teclem2}
There exists a set of needles $\bar{Q}_{\alpha}$ with $\mathfrak{q}(\bar{Q}_{\alpha}) \geq 1 - \alpha$ such that for any $q \in \bar{Q}_{\alpha}$ we have
$$\left|\int{g\,dm_q}\right|^2 \leq C\epsilon^{1/5}/\alpha$$
\end{lem}

\begin{proof}
Recall that we assumed $g$ is centered with respect to $\mu$. Therefore, since $g$ is $1$-Lipschitz, the $L^2$ version of the Poincaré inequality yields
$$1 \geq \Var_{\mu}(g) = \int{\Var_{m_q}(g)d\mathfrak{q}(q)} + \int{\left(\int{gdm_q}\right)^2d\mathfrak{q}(q)}.$$
If $q \in Q_{\delta}$ (which is still the set provided by Lemma \ref{EstNeed1}) then
$$\Var_{m_q}(g) \geq \left(\Var_{m_q}(f)^{1/2} - \Var_{m_q}(f-g)^{1/2}\right)^2 \geq 1 - C\epsilon^{1/4}/\delta^{1/4},$$
where we use Lemma \ref{fg-clo} and \eqref{cq-mean} to get the inequality.
Consequently, we obtain
$$\int{\Var_{m_q}(g)d\mathfrak{q}(q)} \geq \mathfrak{q}({Q}_{\delta})(1 - C\epsilon^{1/4}/\delta^{1/4}) \geq (1-\delta)(1 - C\epsilon^{1/4}/\delta^{1/4}),$$
so that
$$\int{\left(\int{gdm_q}\right)^2d\mathfrak{q}(q)} \leq 1 - (1-\delta)(1 - C\epsilon^{1/4}/\delta^{1/4}) \leq \delta + C\epsilon^{1/4}/\delta^{1/4}.$$
Taking $\delta$ of order $\epsilon^{1/5}$ and using Markov's inequality yields the result.
\end{proof}

We finally prove the $H^1$-closeness of $f$ and $g$ provided that the spectral gap is close to $1$.

\begin{thm} Let $(M,d,\mu)$ be an RCD$(1,\infty)$ space admitting CD$(1,\infty)$ disintegration. Assume $\lambda_1 \leq 1 + \epsilon$ with $\epsilon < 1$. For any $\theta$ small enough, there exists $C(\theta) > 0$ such that
$$\int{(f-g)^2d\mu} \leq C(\theta)\epsilon^{1/10 - \theta}$$
and
$$\int{|\nabla f - \nabla g|^2d\mu} \leq C(\theta)\epsilon^{1/20 - \theta}.$$
\end{thm}

\begin{proof}
Let us decompose $\|f-g\|_{L^2}^2$ as follows:
$$\int{(f-g)^2d\mu} = \int_{Q_{\delta}\cap\bar{Q}_{\alpha}}{\int{(f-g)^2dm_q} d\mathfrak{q}(q)} + \int_{(Q_{\delta}\cap\bar{Q}_{\alpha})^c}{\int{(f-g)^2dm_q} d\mathfrak{q}(q)}.$$
To estimate the first term on the right hand side, we build on Lemmas \ref{fg-clo} and \ref{teclem2}:
\begin{align*}
&\int_{Q_{\delta}\cap\bar{Q}_{\alpha}}{\int{(f-g)^2dm_q} d\mathfrak{q}(q)} \\
&= \int_{Q_{\delta}\cap\bar{Q}_{\alpha}}{Var_{m_q}(f-g) d\mathfrak{q}(q)} + \int_{Q_{\delta}\cap\bar{Q}_{\alpha}}{\left(\int{f-g dm_q}\right)^2 d\mathfrak{q}(q)}\\
&\leq \int_{Q_{\delta}\cap\bar{Q}_{\alpha}}{\int{(f-g - c_q)^2dm_q} d\mathfrak{q}(q)} + \int_{Q_{\delta}\cap\bar{Q}_{\alpha}}{\left(\int{g dm_q}\right)^2 d\mathfrak{q}(q)} \\
&\leq C\sqrt{\epsilon}/\sqrt{\delta} + C\epsilon^{1/5}/\alpha.
\end{align*}
In order to estimate the second term, we set $p$ and $q$ two positive numbers with $p^{-1} + q^{-1} = 1$.
\begin{align*}
&\int_{(Q_{\delta}\cap\bar{Q}_{\alpha})^c}{\int{(f-g)^2dm_q} d\mathfrak{q}(q)} \leq \mathfrak{q}((Q_{\delta}\cap\bar{Q}_{\alpha})^c)^{1/p}\left(\int{(f-g)^{2q}d\mu}\right)^{1/q} \\
&\leq 2(\alpha + \delta)^{1/p}\left(\left(\int{f^{2q}d\mu}\right)^{1/q} + \left(\int{g^{2q}d\mu}\right)^{1/q}\right)  \\
&\leq 2(\alpha + \delta)^{1/p}((2q-1)^{1 + \epsilon} + 16q^2) \\
&\leq Cq^2(\alpha + \delta)^{1/p}.
\end{align*}
Here we used Proposition \ref{integ_eigen} to control the $L^{2q}$ norm of $f$ and Proposition \ref{prop_p_poincare} to control the $L^{2q}$ norm of $g$.

Combining the two estimates yields
$$\int{(f-g)^2d\mu} \leq C(q^2(\alpha + \delta)^{(q-1)/q} + \sqrt{\epsilon}/\sqrt{\delta} + C\epsilon^{1/5}/\alpha).$$
We then take $\alpha$ of order $\epsilon^{1/10}$, $q$ large enough and $\delta$ of order say $\epsilon^{3/10}$ (the influence of $\delta$ is on lower order terms anyway).

For the second estimate, we expand the norm of the gradients as follows:
\begin{align*}
\int{|\nabla f - \nabla g|^2d\mu} &= \int{|\nabla f|^2 + |\nabla g|^2 - 2\nabla f \cdot \nabla g d\mu} \\
&\leq 2 + \epsilon + 2 \int{(Lf)gd\mu} \\
&= 2 + \epsilon -2  \lambda_1 \int{fgd\mu}  \\
&= 2 + \epsilon -2\lambda_1 \int{f^2d\mu} - 2\lambda_1 \int{f(g-f)d\mu} \\
&\leq \epsilon + 2\lambda_1||f||_{L^2}||f-g||_{L^2} \\
&\leq C(\theta)\epsilon^{1/20 - \theta}.
\end{align*}
\end{proof}

Proceeding as in the proof of Theorem \ref{thm_gaussian_component_eigen}, we also obtain

\begin{cor}\label{cor_dist_proj_lip}Let $(X,d,\mu)$ be an RCD$(1,\infty)$ space admitting $CD(1,\infty)$ disintegration, and whose spectral gap satisfies $\lambda_1 \leq 1 + \epsilon$. Let $g$ be the guiding function as in \eqref{gf}. Then,
$$W_1(g_{\sharp}\mu, \gamma) \leq C(\theta)\epsilon^{1/20-\theta}$$
$$d_{TV}(g_{\sharp}\mu, \gamma) \leq 2C(\theta)\epsilon^{1/20-\theta}$$
\end{cor}

We shall use the estimate on the total variation distance to get an estimate on the observable diameter via the following lemma:

\begin{lem}\label{lem_obs_tv}
Assume that $\mu$ and $\nu$ are two probability measures on a metric space $(X,d)$ such that $d_{TV}(\mu, \nu) \leq \epsilon$, where $\epsilon>0$. Then for any $\kappa > \epsilon$, the following inequality holds
$$D_{obs}(\mu; \kappa - \epsilon) \geq D_{obs}( \nu; \kappa).$$
\end{lem}

\begin{proof}
We shall argue by contradiction. Fix $\kappa$, and assume that $D_{obs}(\mu; \kappa - \epsilon) < D_{obs}(\nu; \kappa) - t$ for some $t > 0$. Then for any $f$ $1$-Lipschitz there exists $E \in \mathbb{R}$ such that $\mu(E) \geq 1 - \kappa + \epsilon$ and $\operatorname{diam}(f(E)) \leq D_{obs}(\nu; \kappa) - t$. But then $\nu(E) \geq 1 - \kappa + \epsilon - d_{TV}(\mu, \nu) \geq 1 - \kappa$, and therefore, taking the sup over all $f$, we get $D_{obs}(\nu; \kappa) \leq D_{obs}(\nu; \kappa)-t$ hence a contradiction.
\end{proof}

Since the function $g$ in Corollary \ref{cor_dist_proj_lip} is $1$-Lipschitz, by definition of observable diameter, we infer
 $$D_{obs}(\mu; \cdot) \geq D_{obs}( g_{\sharp}\mu, \cdot)$$
 on $(0,1)$ (see \cite{Shioya} for a proof). Therefore, by combining Theorem \ref{ObsSep}, Corollary \ref{cor_dist_proj_lip}, and Lemma \ref{lem_obs_tv}, we get the proof of Theorem \ref{sg_to_diam}

\subsection{Proof of Theorem \ref{thm_dobs_to_sg}}

We assume throughout this part that

\begin{equation}\label{amod}
D_{obs}((M,d,\mu); \kappa) \geq D_{obs}((\mathbb{R}, |\cdot|, \gamma); \kappa)     -\epsilon
\end{equation}
for a given $0<\kappa<1$ and $\epsilon>0$ small enough. Our aim is to show  Theorem \ref{thm_dobs_to_sg}, namely that such an RCD-space has almost minimal spectral gap.

Let us denote by $V_s(A)$ the $s$-open neighborhood of a subset $A$ of a given metric space ($(\mathbb{R},|\cdot|)$ in most cases).

As a consequence of Theorem \ref{ObsCompa}, we get

\begin{equation}\label{amsd}
Sep((M,d,\mu); \kappa/2) \geq Sep((\mathbb{R},|\cdot|,\gamma);\kappa/2) -\epsilon.
\end{equation}
Instead of the observable diameter, we shall mainly deal with the separation distance through \eqref{amsd}. The proof mostly relies on a careful study of the 1D case. We will then infer the bound on the spectral gap by means of the disintegration property. In what follows, $(\mathbb{R}, |\cdot|, m)$ stands for an $RCD(1,\infty)$-space, namely $m=e^{-\phi(x)}\, dx$ is a probability measure such that $\phi(x) - x^2/2$ is convex.

 \subsubsection{The $1D$ case}

In this part we assume that  $(\mathbb{R}, |\cdot|, m)$ is an RCD$(1,\infty)$-space such that for some $\theta>0$ and for $\eta>0$ sufficiently small

\begin{equation}\label{amsd1}
 D_{obs}((\mathbb{R}, |\cdot|, m); \theta) \geq D_{obs}((\mathbb{R}, |\cdot|, \gamma); \theta)     -\eta.
 \end{equation}

Let  us start with some basic properties regarding the Gauss measure. Recall that $\sigma(x) = \gamma(]-\infty, x])$.

\begin{lem}For $0<\kappa_1 \leq a<b < 1/2$ where $\kappa_1 $ is fixed and sufficiently small, the following estimates hold
 \begin{equation}\label{sigEst}
 {\sqrt{2\pi}}(b-a) \leq \sigma^{-1}(b)- \sigma^{-1}(a) \leq \frac{\sqrt{2\pi}}{\kappa_1^2}(b-a).
\end{equation}
\end{lem}

\begin{proof}
We denote by $e^{-\phi_{\gamma}}$ the density of the standard Gauss measure on $\mathbb{R}$. By definition of $\sigma^{-1}$, we have
$$ \int_{\sigma^{-1}(a)}^{\sigma^{-1}(b)} e^{-\phi_{\gamma}(s)}\,ds = b-a.$$
The first inequality follows readily from this equality. For the second one, we need to estimate $|\sigma^{-1}(\kappa_1)|$ from above. We write for $\kappa_1 < 1/2$
\begin{eqnarray*}
\kappa_1 & = & \int_{-\infty}^{\sigma^{-1}(\kappa_1)} e^{-\phi_{\gamma}(s)} \, ds \\
         & \leq & \frac{1 }{\sqrt{2\pi}|\sigma^{-1}(\kappa_1)|} e^{-(\sigma^{-1}(\kappa_1))^2/2}.
\end{eqnarray*}
From the inequality above, we infer $|\sigma^{-1} (\kappa_1)| \leq 2 \sqrt{-\ln \kappa_1}$  as soon as $\sigma^{-1}(\kappa_1) \leq -1$, which in turn implies the needed lower bound on $e^{-\phi_{\gamma}(s)}$  on $[\sigma^{-1}(a), \sigma^{-1}(b)]$ for $\kappa_1 < \gamma(]-\infty, -1])$.

\end{proof}

\begin{rque}\label{EstSep}Recall that $Sep((\mathbb{R},|\cdot|,\gamma);\theta/2) = 2|\sigma^{-1} (\theta/2)|$. Thus, the preceding lemma also gives bounds on the separation distance on the Gauss space .
\end{rque}

Easy computations assert that the interval $[\sigma^{-1} (\theta/2), -\sigma^{-1} (\theta/2)]$ is the only closed interval of given length $R:= 2|\sigma^{-1} (\theta/2)|$ whose $\gamma$-measure is maximal, namely $1-\theta$. More precisely, there exists a constant $C(R)$, depending continuously on $R>0$, such that

\begin{equation}\label{stabObsGa}
\left| \gamma([s-R/2,s+R/2])  -(1-\theta)\right| \geq C(R)\, s^2,
\end{equation}
for any $s$ sufficiently small (say, such that $  \gamma([s-R/2,s+R/2]) \geq 1/2$).

 \begin{lem}\label{closemass} Let $(\mathbb{R}, |\cdot|, m)$ be a $CD(1,\infty)$-space.  Let $A_1,A_2\subset \mathbb{R} $ be two Borel subsets.
 Then, the following inequality holds
$$ d(A_1,A_2) \leq -\sigma^{-1}(m(A_1)) - \sigma^{-1}(m(A_2)).$$
In particular, assuming $\min \{ m(A_1), m(A_2)\} \geq \theta/2>0$ and
$$d(A_1,A_2) \geq Sep((\mathbb{R},|\cdot|,\gamma);\theta/2) -2\eta>0,$$
the following inequality holds
$$ \max\{ m(A_1), m(A_2) \} \leq \theta/2 + \sqrt{2/\pi}\, \eta.$$

\end{lem}

\begin{proof}

The proof is a simple consequence of an integrated version of the isoperimetric inequality: given a Borel subset $A$ of $\mathbb{R}$ and $t \geq 0$,
\begin{equation}\label{eII}
m(V_t(A)) \geq \gamma((-\infty, \sigma^{-1}(m(A)) +t]).
\end{equation}
In particular, for $r = d(A_1,A_2)$,
 \begin{eqnarray*}
\gamma((-\infty, \sigma^{-1}(m(A_1)) +r]) &  \leq & m(V_r(A_1)) \\
               & \leq & 1 -m(A_2), \\
 			  & \leq & \gamma((-\infty,|\sigma^{-1}(m(A_2))|]),

 \end{eqnarray*}
where the second inequality follows from $A_2 \cap V_r(A_1) =\emptyset$. The inequality relating the expressions at the ends implies
$$ \sigma^{-1}(m(A_1)) +r \leq -\sigma^{-1}(m(A_2))$$
and the first inequality is proved.

To prove the second statement let us rewrite the first inequality as
\begin{equation}\label{es}d(A_1,A_2) \leq 2|\sigma^{-1} (\theta/2)| + (\sigma^{-1}(\theta/2) - \sigma^{-1}(m(A_1)) + (\sigma^{-1}(\theta/2) - \sigma^{-1}(m(A_2)).
\end{equation}
We are left with estimating $ (\sigma^{-1}(\theta/2) - \sigma^{-1}(m(A_i))$ from above. According to \eqref{sigEst}, we have
$$  \sigma^{-1}(m(A_i))  - \sigma^{-1}(\theta/2) \geq \sqrt{2\pi} (m(A_i) -\theta/2).$$
 By combining \eqref{es} together with the above inequality, we get the second claim.

\end{proof}

\begin{rque}\label{remclosemass}
The proof above applies verbatim to general RCD$(1,\infty)$-spaces since it is mainly based on the isoperimetric inequality stated in Theorem \ref{IIn}. It gives the same results.
\end{rque}
In the same vein, we shall also need the following estimate:

\begin{lem}\label{uppboundmass}Let $(\mathbb{R},|\cdot |,m)$ be a CD$(1,\infty)$ space, and let $A_1,A_2\subset \mathbb{R} $ be two Borel subsets such that $\min \{ m(A_1), m(A_2)\} \geq \theta/2>0$ and
$$r:= d(A_1,A_2) \geq Sep((\mathbb{R},|\cdot|,\gamma);\theta/2) -2\eta>0.$$

Then, for any $s \in [0, r]$ and $i \in \{1,2\}$
$$ m((V_s(A_i)) \leq  \gamma((-\infty,\sigma^{-1}(m(A_i))+s]) + \sqrt{2/\pi}\,\eta. $$
\end{lem}

\begin{proof}

We argue as in the proof of the previous lemma. Set $i=1$.
 \begin{eqnarray*}
 m(V_s(A_1))  & \leq &1 -m(V_{r-s}(A_2)) , \\
 		&\leq	& 1 - \gamma( [ -\sigma^{-1}(m(A_2))-r +s, +\infty)) \\
 			  & \leq & \gamma((-\infty, -\sigma^{-1}(m(A_2)) +2\sigma^{-1}(\theta/2) +2\eta +s]), \\
 			   & \leq & \gamma((-\infty, \sigma^{-1}(\theta/2)) +s + 2\eta]),
 			  \end{eqnarray*}
where we use the symmetry of $\gamma$, the inequality on $r=d(A_1,A_2)$ to infer the inequality on the third line, and $m(A_1) \geq \theta/2$ to get the estimate in the last one. The proof then follows from $ \gamma((c, c+\delta) )\leq \frac{1}{\sqrt{2\pi}} \delta,$ where $c$ is arbitrary.

\end{proof}

Let us fix some notation. Let $A_1$ and $A_2$ as in Lemma \ref{uppboundmass}.
Without loss of generality, one can assume that the interval $[a_-,a_+]$ is such that $(a_-,a_+,) \cap (A_1 \cup A_2) =\emptyset$ and $a_+ -a_- = d(A_1,A_2)$. Up to reversing $A_1$ and $A_2$, we can further assume that $a_- \in \overline{A_1}$ and $a_+ \in \overline{A_2}$.

Our next goal is to show that $\phi$ is uniformly close to $\phi_{\gamma}$ on $[a_-,a_+]$. To this aim, we shall use properties of the optimal map $T$ such that $T_{\sharp} \, \gamma = m$. Let us recall that $T$ is non-decreasing and $1$-Lipschitz, as a consequence of Caffarelli's contraction theorem \cite{Caf00, FGP20}.

We set
\begin{equation}\label{defalpha}
\alpha_-:= \sigma^{-1}(m(A_1)) \mbox{ and } \alpha_+:= \sigma^{-1}(1-m(A_2))= -\sigma^{-1}(m(A_2)),
\end{equation}  in other terms $T(\alpha_{\pm})= a_{\pm}$.

\begin{prop}\label{UnifEsti} Let $(\mathbb{R}, |\cdot|, m=e^{-\phi(s)}ds)$ be a CD$(1,\infty)$ space such that for $\theta>0$ and $0<\eta <\eta_0(\theta)$,
$$   d(A_1,A_2) \geq Sep((\mathbb{R},|\cdot|,\gamma);\theta/2) -2\eta>0.$$
Then, up to translating $m$ so that the median is at the origin, there exists $C=C(\theta)>0$ such that for any $s \in [a_-,a_+]$,
$$ |\phi(s) - \phi_{\gamma} (s) | \leq  C\eta^{1/2}. $$
\end{prop}
\begin{proof}
In this proof, $C(\theta)$ will be a constant that only depends on $\theta$, but which may change from line to line. Set $\delta_+:= a_+-\alpha_+$ and $\delta_-:=a_- -\alpha_-$. We first claim that
\begin{eqnarray*}
 \phi (s) &\leq & \phi_{\gamma} (s - \delta_+) \; \; \mbox{ on } [s_l, a_+]  \\
  \phi (s) & \leq & \phi_{\gamma} (s - \delta_-) \; \; \mbox{ on } [a_-, s_l],
\end{eqnarray*}
where $s_l = \rm{argmin} \, \phi$ (recall $\phi$ is $1$-convex).
We shall only prove the first inequality, the second one is proved in the same way.

Setting the median of $m$ at the origin is equivalent to enforcing $T(0) = 0$, since the median of the Gaussian measure is also at the origin. Thanks to the monotonicity of $T$ and the Lipschitz bound recalled above, we notice that $ T(s) \geq s + \delta_+$ on $(-\infty,\alpha_+].$ By combining this together with the fact
that $e^{-\phi}$ is decreasing on $[s_l,+\infty)$, we get on $[s_l-\delta_+,\alpha_+]$
$$ e^{-\phi(T(s))} \leq e^{-\phi(s + \delta_+)}.$$
By definition of $T$, we have for almost every $s$,
$$e^{-\phi_{\gamma}(s)}=e^{-\phi(T(s))} T'(s)\leq e^{-\phi(T(s))}, $$
 and the proof of the inequality is complete by continuity of $\phi $ and $\phi_{\gamma}$.

Next, we prove that $\delta_+ \simeq \delta_-\simeq T(0) = 0$. Using that $s \mapsto T(s) -s$ is non-increasing on $[\alpha_-,\alpha_+]$, we get that
$$  0\leq   (  T(\alpha_-) -\alpha_-)  - (  T(\alpha_+)) -\alpha_+   )=  \delta_- - \delta_+ . = (\alpha_+-\alpha_-) -(a_+-a_-) $$
By assumption on $d(A_1,A_2) = a_+ -a_-$ and using that $-\sigma^{-1}$ is non-increasing, we obtain
\begin{eqnarray*}
a_+ -a_- & \geq &  Sep((\mathbb{R},|\cdot|,\gamma);\theta/2) -2\eta \\
         & \geq & -2\sigma^{-1}(\theta/2) -2\eta \\
         & \geq & -\sigma^{-1} (m(A_1))  -\sigma^{-1} (m(A_2)) -2\eta \\
         & \geq & \alpha_+-\alpha_- -2\eta,
\end{eqnarray*}
where we use \eqref{defalpha} to get the last inequality. Therefore, we obtain
\begin{equation}\label{deltaEst}
\left\{  \begin{array}{rcl} & 0 & \leq  \delta_- -\delta_+ \leq 2\eta  \\
\mbox{ and }  & \delta_+ & \leq  T(0) = 0 \leq \delta_-
\end{array}\right.
\end{equation}  by monotonicity. As a consequence, $|\delta_+| + |\delta_-| \leq 2\eta$.

Now recall that by definition of $\alpha_{\pm}$,  $\min (\gamma ((-\infty,\alpha_-]), \gamma([\alpha_+,+\infty))) \geq \theta/2$. Moreover, Lemma \ref{closemass} implies
\begin{equation}\label{EstiAlph} \max (\gamma ((-\infty,\alpha_-]), \gamma([\alpha_+,+\infty))) \leq \theta/2 + \sqrt{2/\pi} \, \eta.
\end{equation}
Hence, \eqref{stabObsGa} implies the existence of $C(\theta)>0$ such that
\begin{equation}\label{EstiAlphII} |(\alpha_+  + \alpha_-)|/2 \leq C(\theta) \sqrt{\eta},
\end{equation}

By combining this together  with \eqref{deltaEst}, we infer
\begin{equation}\label{ApproMid} \left|\frac{a_++a_-}{2}\right| \leq C \sqrt{\eta}.
\end{equation}
As a consequence,
\begin{align}\label{EstiPhi}
 \phi (s) &\leq \phi_{\gamma} (s -\delta_+)  \notag \\
&= \phi_\gamma(s) -\delta_+ s + \delta_+^2/2 \notag \\
&\leq \phi_\gamma(s) + C(\theta)\eta
\end{align}
on $[a_-,a_+] \subset [-C(\theta), C(\theta)]$.

The last step of the proof relies on Markov's inequality and the fact that $\phi-\phi_{\gamma}$ is a convex function. More precisely, given $\nu >0$, we set
$$S_{\nu}:= \{ s \in [a_-,a_+]\, | \phi(s) \geq \phi_{\gamma}(s) -\nu \}.$$
Observe that $(a_-,a_+) \cap (A_1 \cup A_2)=\emptyset$ implies
\begin{equation}\label{EstiInt4.11} \int_{[a_-,a_+]} e^{-\phi(s)} \,ds  \leq  1 - m(A_1) -m(A_2)=  \int_{[\alpha_-,\alpha_+]}  e^{-\phi_{\gamma}(s)} \,ds.
 \end{equation}
 Besides,  \eqref{EstiAlph} implies the existence of $C>0$ such that
 $$ \int_{[a_-,a_+]} e^{-\phi_{\gamma} (s)} \, ds \geq  \int_{[\alpha_-,\alpha_+]} e^{-\phi_{\gamma} (s)}\, ds -C\eta.$$

Using \eqref{EstiPhi}, we have
\begin{eqnarray*}
\int_{[a_-,a_+]} e^{-\phi(s)} \,ds  & \geq & e^{\nu}\, \int_{S_{\nu}^c}  e^{-\phi_{\gamma}(s )} \,ds  + e^{-C \eta}\, \int_{S_{\nu}}   e^{-\phi_{\gamma}(s)} \,ds   \\
        & \geq & \big(e^{\nu}-1\big) \, \int_{S_{\nu}^c}  e^{-\phi_{\gamma}(s)} \,ds  + e^{-C \eta}\left( \int_{[\alpha_-,\alpha_+]}  e^{-\phi_{\gamma}(s)} \,ds  -C\, \eta\right).
\end{eqnarray*}
This estimate and \eqref{EstiInt4.11} yield
$$ \gamma(  S_{\nu}^c ) \leq C \, \frac{\eta}{e^{\nu}-1} \leq C \, \frac{\eta}{\nu},$$
which suggests to choose $\nu= {\eta}^{1/2}$.

Now, the convexity of $h:=\phi-\phi_{\gamma}$  forces $S_{\nu}^c$ to be an interval, and the Lebesgue measure of $S^c_{\eta^{1/2}}$ is bounded from above by $C(\theta) \eta^{1/2}$. Thales' theorem and the convexity of $h$ then imply  $\min h \geq -{C}(\theta) {\eta}^{1/2}$ and the proof is complete.

\end{proof}

Let us set $a_{\eta}:= |\sigma^{-1}(\sqrt{\eta})|$.  Our goal is to extend the estimate on $|\phi-\phi_{\gamma}|$ from $[a_-,a_+]$ to a much larger interval (depending on $\eta$). This is the content of the next lemma.

\begin{lem}\label{EstPhiII} Let $(\mathbb{R}, |\cdot|, m=e^{-\phi(s)}ds)$ be a CD$(1,\infty)$ space such that for $\theta>0$ and $0<\eta <\eta_0(\theta)$
$$  Sep((\mathbb{R},|\cdot|,m);\theta/2) \geq Sep((\mathbb{R},|\cdot|,\gamma);\theta/2) -2\eta>0.$$
Then, up to translating $m$ so that the median is at the origin, there exists $C=C(\theta)>0$ and $b_\eta > 0$ such that, for any $s \in [-b_\eta,b_\eta]$,
$$ |\phi(s) - \phi_{\gamma} (s) | \leq  C\eta^{1/10}, $$
and $m((-\infty, -b_\eta]) \leq \gamma((-\infty, -b_\eta]) \leq C\eta^{1/10}$.
\end{lem}

 \begin{proof}
First of all, by definition of $a_\eta$, $\gamma((-\infty, -\aet]) = \sqrt{\eta}$, and since the median of $m$ is at the origin, the Gaussian concentration inequality (which is the integrated form of the isoperimetric inequality for RCD$(1, \infty)$ spaces states that $m((-\infty, a_\eta]) \leq \gamma((-\infty, -a_\eta]) $.

 Set $h  :=\phi -\phi_{\gamma}$.  Recall that $|h|\leq C\eta^{1/2}$ on $[a_-,a_+]$ and that $h$ is convex. Markov's inequality yields the existence of $\hat{a}_+ \in [a_-,a_+] $ such that
$$ |\phi'(\hat{a}_+) -\phi_{\gamma}'(\hat{a}_+) | \leq C \eta^{1/4}   \mbox{  and  } |a_+ -\hat{a}_+| \leq \eta^{1/4}.$$
For $s \geq \hat{a}_{+}$, the $1$-convexity of $\phi$ yields
$$ \phi(s) \geq \phi(\hat{a}_{+}) + \phi'(\hat{a}_{+}) (s-\hat{a}_{+}) +  1/2(s-\hat{a}_+)^2.$$
Noticing that $\phi_{\gamma}(\hat{a}_+)+ \phi_{\gamma}'(\hat{a}_+) (s-\hat{a}_+)  +  1/2(s-\hat{a}+)^2 = \phi_{\gamma}(x),$ we infer from the previous inequality:
$$ \phi(s) \geq \phi_{\gamma}(s) + E(s),$$
where $E(s) = -C {\eta}^{1/4} (1 + (s-\hat{a}_+)).$

\noindent Recall that $\aet = |\sigma^{-1}(\sqrt{\eta})| \sim \sqrt{\ln (1/\eta)}$ for $\eta \sim 0$ thus, for $\eta$ small enough, there exists a constant $C$ such that
\begin{equation}\label{EtEsti}
\aet \leq C\sqrt{\ln (1/\eta)},
\end{equation}
and there exists a constant $C$ such that
 $$E(s) \geq -C {\eta}^{1/4} (1 + a_\eta) \geq -C \eta^{1/5}$$
on $[\hat{a}_{+}, \aet]$. Consequently,

 \begin{equation}\label{UppEstEphi}
 \phi \geq \phi_{\gamma} - C \eta^{1/5 }
\end{equation}
on $[{a}_{+}, \aet] \subset [\hat{a}_{+}, \aet]$.

The same method applies on $[-\aet, \hat{a}_-]$ (where the definition of $\hat{a}_-$ is clear from the context)
  and leads to the same estimate $ \phi \geq \phi_{\gamma} - C \eta^{1/5}$.

We then have, using the estimates on the interval $[a_-, a_+]$ and that $|a_+ - \hat{a}_+| \leq  {\eta}^{1/4}$,
\begin{align*}
\left| \int_{[-\aet,\aet]\setminus [{a}_-,{a}_+]} e^{-\phi(s)} -e^{-\phi_{\gamma}(s)} \, ds\right|&= \left|\gamma([{a}_-, {a}_+]) - m([{a}_-, {a}_+]) \right.\\
& \hspace{2cm} \left. + \gamma([-\aet, \aet]^c) - m([-\aet, \aet]^c)\right| \\
&\leq 2\gamma([\aet, + \infty)) + m((-\infty, -\aet]) \\
& \hspace{2cm} + m([\aet, + \infty)) + C\sqrt{\eta} \\
&\leq C\sqrt{\eta} .
\end{align*}

As in the proof of Proposition \ref{UnifEsti}, Markov's inequality and the estimates above imply that
$$S_{\eta}:=\{s\in  [-\aet,\aet]\setminus [{a}_-,{a}_+]| \,(\phi-\phi_{\gamma})(s) \leq \eta^{1/10}\}$$
satisfies
$$\gamma (S_{\eta}^c) \leq C \eta^{1/10}.$$

Finally, the convexity of $\phi-\phi_{\gamma}$ implies that $S_{\eta}$ is the union of two intervals; besides, thanks to Proposition \ref{UnifEsti}, $S_{\eta}$ contains a neighborhood of ${a}_-$ and ${a}_+$. Therefore $S_{\eta}^c$ is the union of two intervals $[-\aet,-\bet] \cup [\bet,\aet]$.
We also have
$$m((-\infty, -b_\eta]) \leq \gamma((-\infty, -b_\eta]) \leq \gamma((-\infty,-\aet]) + \gamma ((-\aet,-\bet]) \leq C\eta^{1/10}.$$

 \end{proof}

 We are left with estimating the variance of the identity map relative to the measure $m$.

 \begin{lem}\label{VarEsti1D} Let $(\mathbb{R}, |\cdot|, m=e^{-\phi(s)}ds)$ be a CD$(1,\infty)$ space such that for $\theta>0$ and $0<\eta <\eta_0(\theta)$,
$$  Sep((\mathbb{R},|\cdot|,m);\theta/2) \geq Sep((\mathbb{R},|\cdot|,\gamma);\theta/2) -2\eta>0.$$
 Then, up to translating $m$ so that its median is at the origin, the following estimates hold

 $$ \left| \int_{\mathbb{R}} s \, dm(s)\right| \leq C \eta^{1/11},$$
 and
 $$ \int_{\mathbb{R}} s^2 \, dm(s) \geq 1 -C \eta^{1/11}.$$
 In particular,
 $$ Var_m(s) \geq 1 -C \eta^{1/11}.$$

 \end{lem}

 \begin{proof}
Let us start with some estimates relative to the Gaussian measure. Using $$\frac{e^{-\phi_{\gamma}}(\bet)}{\bet}\sim \int_{\bet}^{+ \infty} e^{-\phi_{\gamma}(s)} \, ds \leq C\eta^{1/10},$$
we infer, for $\eta$ small enough,

$$ \int_{\bet}^{+ \infty} s\, e^{-\phi_{\gamma}(s)} \, ds = e^{-\phi_{\gamma} (\bet)} \leq  C \bet \eta^{1/10} \leq C\eta^{1/10} \sigma^{-1}(\eta^{1/10}) \leq C\eta^{1/11}.$$
Moreover, thanks to
$ \int_{\bet}^{+ \infty} s^2\, e^{-\phi_{\gamma}(s)} \, ds = \bet e^{-\phi_{\gamma}(\bet)} + \int_{\bet}^{+ \infty} \, e^{-\phi_{\gamma}(s)} \, ds$, we get
$$ \int_{\bet}^{+ \infty} s^2\, e^{-\phi_{\gamma}(s)} \, ds \leq  C\eta^{1/11}. $$

Now, we use Lemma \ref{EstPhiII} to infer similar bounds for $m$. More precisely,

\begin{eqnarray*} \left|\int_{[-\bet,\bet]} s\, e^{-\phi(s)} \, ds - \int_{[-\bet,\bet]} s\, e^{-\phi_{\gamma}(s)} \, ds \right| & \leq & \int_{[-\bet,\bet]} |s| \big(e^{|\phi(s)-\phi_{\gamma}(s)|}-1\big) e^{-\phi_{\gamma} (s)} \, ds \\
& \leq & \int_{[-\bet,\bet]}|s| \big(e^{C \eta^{1/10}}-1 \big)e^{-\phi_{\gamma} (s)} \, ds \\
  & \leq & 2 (1 -e^{-\phi_{\gamma}(\bet)}) {C \eta^{1/10}} \leq C \eta^{1/10}.
 \end{eqnarray*}
We also have
 \begin{eqnarray*} \left|\int_{[-\bet,\bet]}s^2 e^{-\phi(s)} \, ds - \int_{[-\bet,\bet]} s^2 e^{-\phi_{\gamma}(s)} \, ds \right| & \leq & \int_{[-\bet,\bet]} s^2\big( e^{C \eta^{1/10}}-1\big) e^{-\phi_{\gamma} (s)} \, ds \\
   & \leq & C \eta^{1/10}.
 \end{eqnarray*}

Last, we claim that $0 \leq T(s) \leq s $ (resp. $ 0 \geq T(s)\geq s $) on $[\bet,+\infty) \subset [\alpha_+,+\infty)$ (resp. on $(-\infty,-\bet]$), since $T$ is monotone and $1$-Lipschitz and $T(0) = 0$.

With this property, we can estimate the remaining part of the expectation of $m$:
\begin{eqnarray*} \int_{[\bet, +\infty]} s\, e^{-\phi(s)} \, ds &\leq &
  \int_{[T^{-1}(\bet), +\infty)}  T(u) \, e^{-\phi_{\gamma}(u)} \, du \\
 & \leq & \int_{[\bet, +\infty)} u e^{-\phi_{\gamma}(u)} \, du \\
 & \leq & C\eta^{1/11},
\end{eqnarray*}
where we use the upper tail bound for the Gaussian measure. The same method applies on $(-\infty,-\bet]$, and the proof is complete.
 \end{proof}

\subsubsection{The general case}
According to \eqref{amsd}, there exists sets $A_1$ and $A_2$ such that
$$ d(A_1,A_2) \geq Sep((\mathbb{R},|\cdot|,\gamma);\kappa/2) -2\epsilon,$$
and $\min (\mu(A_1), \mu(A_2)) \geq \kappa/2$.

We now introduce a needle decomposition relative to a function $f$ defined in terms of the sets $A_1$ and $A_2$ above. Precisely, we set
\begin{equation}\label{ndf}
f= \chi_{A_1} -\chi_{A_2} - \mu(A_1) + \mu(A_2).
\end{equation}
This function satisfies

$$ \int |f(x)| d(x,x_0)\, d\mu< \infty \mbox{         and       } \int f\, d\mu =0.$$

Applying the needle decomposition associated to $f$ defined on  $(M,d,\mu)$, we get that for $\mathfrak{q}$ a.e $q$, $\int_{X_q} f \, dm_q =0$. This yields for a.e $q$

$$ m_q(A_1\cap X_q) - m_q (A_2 \cap X_q) = \mu(A_1) - \mu(A_2).$$

Therefore, according to Lemma \ref{closemass} (see Remark \ref{remclosemass}), we get
\begin{equation}\label{mqE}
|m_q(A_1\cap X_q) - m_q (A_2 \cap X_q)|\leq \sqrt{2/\pi}\, \epsilon.
\end{equation}

From this estimate, we infer the following result

\begin{lem}
Let $(M,d,\mu)$ be an RCD$(1,\infty)$ space admitting CD$(1,\infty)$ disintegration and such that \eqref{amod} holds. Then, relatively to the needle decomposition associated to $f$ as in \eqref{ndf}, there exists a set of needles $Q_{\delta}$ such that $\mathfrak{q}(Q_{\delta}) \geq 1 -{\delta}$, and for any needle $q \in Q_{\delta}$, we have
\begin{equation}\label{amod1d}
\min\{ m_q (A_1), m_q(A_2)\} \geq \kappa/2 - \frac{\sqrt{8/\pi} \,\epsilon}{\delta}-  \sqrt{2/\pi}\, \epsilon.
\end{equation}
\end{lem}

\begin{proof}
Since we assume there is a needle decomposition, there exists a partition of $M$: $M= \Tau \sqcup Z$ such that $\mu|_{\Tau}$ admits a needle decomposition and $f=0$ $\mu$-a.e. on $Z$. Now, by the very definition of $f$, $Z \subset (A_1 \cup A_2)^c$ (actually $Z=\emptyset$ whenever $\mu(A_1)\neq \mu(A_2)$). Therefore, we obtain
\begin{equation}\label{mma} \mu(A_i) = \int_{\Tau} \chi_{A_i} \, d\mu = \int_Q m_q(A_i) \, d\mathfrak{q}.
\end{equation}
Recall that each needle is in particular a geodesic, thus the distance between $A_1\cap X_q$ and $A_2 \cap X_q$ along $X_q$ is greater or equal to $d(A_1,A_2)$, where
$$d(A_1,A_2) \geq Sep((\mathbb{R},|\cdot|,\gamma);\kappa/2) -2\epsilon> Sep((\mathbb{R},|\cdot|,\gamma);\kappa/2 + \sqrt{2/\pi} \epsilon)$$
according to \eqref{sigEst} and Remark \ref{EstSep}. Using Theorem \ref{ObsCompa}, we infer
$$ d(A_1,A_2) > Sep((X_q,|\cdot|,m_q);\kappa/2 +\sqrt{2/\pi} \epsilon).$$
This inequality yields
$$ \min\{ m_q (A_1), m_q(A_2)\} \leq \kappa/2 + \sqrt{2/\pi} \epsilon,$$
and we can replace $\min$ by $\max$ in the above equation thanks to \eqref{mqE}, provided that $\sqrt{2/\pi} \epsilon$ is replaced by $2\sqrt{2/\pi} \epsilon$.

The end of the proof consists in applying Markov's inequality in \eqref{mma} combined with $\max\{ m_q (A_1), m_q(A_2)\} \leq \kappa/2 + \sqrt{8/\pi} \epsilon$.

\noindent Defining $Q_{\delta}:=\{ q \in Q|\, m_q (A_1) \geq \kappa/2 -\frac{\sqrt{8/\pi} \,\epsilon}{\delta}\}$, we get
\begin{eqnarray*}
\mu(A_1)& \leq & \int_{Q_{\delta}^c} m_q(A_1)\, d\mathfrak{q} + \int_{Q_{\delta}} m_q(A_1)\, d\mathfrak{q} \\
          & < & (\kappa/2 -\frac{\sqrt{8/\pi} \epsilon}{\delta}) \mathfrak{q}(Q_{\delta}^c) + (\kappa/2 + \sqrt{8/\pi} \epsilon) ( 1 - \mathfrak{q}(Q_{\delta}^c)),
\end{eqnarray*}
from which we infer (since $\mu(A_1)\geq \kappa/2$)
$$ \mathfrak{q}(Q_{\delta}^c ) \leq \delta.$$
Applying \eqref{mqE} yields the result.

\end{proof}

 For a.e. $q \in Q_{\sqrt{\epsilon}}$, there exists a constant $\widehat{C}>0$ such that  $$\min\{m_q(A_1), m_q(A_2)\} \geq \kappa/2 -\widehat{C}\,\sqrt{\epsilon},$$
while, as explained in the proof above, for almost every $q$,
$$d(A_1\cap X_q, A_2\cap X_q) \geq d(A_1,A_2) \geq  Sep((X_q,|\cdot|,m_q);\kappa/2) -2\epsilon.$$
Besides, using  \eqref{sigEst} and Remark \ref{EstSep}, we obtain
$$  Sep((X_q,|\cdot|,m_q);\kappa/2) -2\epsilon  \geq  Sep((X_q,|\cdot|,m_q);\kappa/2 -\widehat{C}\sqrt{\epsilon}) - C \sqrt{\epsilon}.$$

According to Lemma \ref{VarEsti1D}, this implies for a.e $q \in Q_{\sqrt{\epsilon}} $
the existence of a constant $C=C(\kappa)>0$ such that
\begin{eqnarray}\label{VarEstII}
Var_{m_q} (s) &\geq & 1 -C\epsilon^{1/22}, \nonumber \\
\mbox{and } \hspace{0,8cm} \mathfrak{q}(Q_{\sqrt{\epsilon}}) &\geq & 1 -\sqrt{\epsilon}.
\end{eqnarray}

\subsubsection{Proof of Theorem \ref{thm_dobs_to_sg}}

 The proof builds upon the decomposition relative to $f$ as in \eqref{ndf}. According to Theorem \ref{thm_needle_dec}, there exists a guiding function $u$ which is $1$-Lipschitz, and such that for a.e. $q \in Q$, $u(s)=s +c_q$ on $X_q$ (with $c_q \in \mathbb{R})$. As a consequence, we have
 $$ \int |\nabla u|^2 \, d\mu \leq 1,$$
 and
 \begin{eqnarray*}
  Var_{\mu} (u) & = & \int_X u^2 -\left(\int_X u \,d\mu\right)^2 \, d\mu \\
  				& \geq & \int_Q \int_{X_q} u^2 \, dm_q \,d\mathfrak{q}(q) - \int_Q \left(\int u\, dm_q\right)^2 \,d\mathfrak{q}(q) \\
  				& = & \int_Q Var_{m_q} (u)\, d\mathfrak{q}(q),
 \end{eqnarray*}
where the inequality follows from Cauchy-Schwarz. Using \eqref{VarEstII}, we then get
$$  Var_{\mu} (u) \geq \mathfrak{q}(Q_{\sqrt{\epsilon}}) \big(1 - C\epsilon^{1/22}\big) \geq 1 - C\epsilon^{1/22} \geq (1 - C\epsilon^{1/22})\int{|\nabla u|^2d\mu},$$
which gives the upper bound on the spectral gap.

\vspace{3mm}

\underline{\textbf{Acknowledgments}}: We thank Fabio Cavalletti, Thomas Courtade, Guido De Philippis, Bo'az Klartag, Michel Ledoux and Shin-Ichi Ohta for valuable discussions. M.F. was supported by ANR-11-LABX-0040-CIMI within the program ANR-11-IDEX-0002-02, as well as ANR projects EFI (ANR-17-CE40-0030) and MESA (ANR-18-CE40-006). J.B. was supported by ANR-17-CE40-0034.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


By all accounts, this book is a favorite among toddlers and adults alike. Generations will recall their first time reading along as lovable, furry old Grover begs the reader not to turn the page... for a monster is at the end of the book! A classic, not to be missed. Images and image descriptions available.


A lovely interview with knitting goddess and one of Knitty's heroes, Barbara Walker. The interview appeared in a Columbus, Ohio, newspaper, ahead of Ms. Walker's appearance at last weekend's TNNA needlearts tradeshow. A new DVD featuring some of Ms. Walker's knitting work has just been launched.
Get on the yarn train!
Residents of the Toronto area should mark Saturday, July 9th, on their calendars: it's the annual TTC Knitalong. Torontonians can regularly be found knitting on the buses, streetcars and subways of the TTC (Kate included!), but this day is extra-special. The TTC Knitalong comprises a yarn crawl, a fundraising event, and an awards ceremony and after-party at a local pub …. Details here.
The "Visit Guernsey" tourist information organization, in cooperation with I Knit London, is hosting a competition to win a trip to Guernsey! More info on Facebook.
A cozy bench for reading.
Eugene, Oregon's Knotty Knitters hit a local pedestrian bridge.
20 trees in Leith, Scotland, have been wrapped in little wooly jumpers to mark the opening of the 2011 Leith Festival.
Our own Mandy and her collaborators were caught in the act in Vancouver.
An excellent lamp-post in Pioneer Square in Seattle, WA.
I imagine a ladder was required for this one. Photo courtesy Suzanne Tidwell.
Shall We Knit, an LYS in Waterloo, Ontario (and a favorite of several of us at Knitty), got hit while the owner was at TNNA.
Tons more pictures on the Facebook group and the blog.
I love the light post. I enjoy yarn bombing so much more when it is a coherent piece (much more of an installation than a bunch of swatches tied to things).


text:
BEYOND EIGHTEEN Release Day!
Beyond Eighteen Max Goldstein has been confronted by death, betrayal, and unwanted expectations. He must work to find the delicate balance of a love that has been tested and a family that has been broken. Returning to the warmth of Wilson's arms, he finds the only place he can be truly healed. Unfortunately, teetering on the edge of losing everything he's ever wanted, Max must make decisions that pull at his sense of responsibility and push at his desire to protect the life he's found with Wilson. When Wilson Mooney is thrust into adulthood, she never expects her childhood demons to come knocking. Struggling to balance the delicate nature of her relationship with Max and the burning sting of betrayal; Wilson realizes she must dig deep within her soul to find forgiveness. Will she allow circumstances out of her control to shape her future? Or will she find the strength and maturity to go beyond eighteen and have the life she desires with the man she loves? Will Wilson and Max ever get to have their happily ever after?
Enter for a chance to win the entire Wilson Mooney Series!
RELEASING October 2, 2013
Add it to your TBR on Goodreads
About the Author: By day Gretchen teaches computers, by night she dawns the cloak of motherhood, wifehood, and authorship. She is making her way through self-publishing and is truly learning to let go with every curve and bump in the creative process. She enjoys writing about first loves and first times, in the first person. She is a firm believer that anything is possible if you set your mind to it; and what you expect out of life, always finds a way of showing up. Gretchen happily lives in Northern California with her amazing husband, their three terrific boys, her talented mother, one goofy black lab, one crazy kitty, and eight happy chickens. Even though it sounds like it, she doesn't live on a farm. www.facebook.com/gretchendelaoauthor www.facebook.com/wilsonmooneytheseries
Amazon: http://www.amazon.com/Beyond-Eighteen-Wilson-Mooney-ebook/dp/B00FKAK9SS/ref=sr_1_3?s=digital-text&ie=UTF8&qid=1380669210&sr=1-3&keywords=gretchen+de+la+O
Smashwords... https://www.smashwords.com/books/view/363294
Get UNWRAPPED
Which NAtural are You?
Cover Reveal: FAST FORWARD by Marion Croslydon
HEAT UP THE FALL: New Adult Boxed Set
NAked with Delancey Stewart
Ninety Eight by Shannon Mayer
Little of This, Little of That, or: What's This Po...
Cover Reveal: Entwine by Rebecca Berto
The Sexy Side of Non-Contemporary New Adult & Give...
release event: The Story of You and Me by Pamela D...
Upcoming Release: The Long Game, by J.L. Fynn
cover reveal: The Best Laid Plans by Tamara Mataya...
Cover Reveal: The Dividing Line by Victoria H. Smi...
STAY WITH ME by Abigail Kerber
New Adult - it's not necessarily what you think.
Cover Reveal: CRASHING INTO YOU by B.D. Rowe

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text:
Small-scale clot detector, big value to trauma patients
Clot device-Nathan White McKenna Princing Dr. Nathan White is an emergency-medicine specialist at Harborview. Nathan White A small, plastic-and-silicon card to...
Professor helps craft U.S. bill to aid women fleeing abuse
It is not uncommon for a woman to flee, with her child, from an abusive spouse, hoping to start fresh elsewhere. Some even seek refuge in a new country. Often, though, problems aren't so easily...
Is it really an emergency? ER specialist gives guidance
It's a place no one wants to end up: the emergency room. But according to Dr. Gregory Schroedl, an emergency medicine physician and chief medical officer at Northwest Hospital & Medical Center,...
Implications for teens in trend of legal pot for adults
Kevin Haggerty graduated from high school in 1977 – among the foggiest, Cheech-and-Chongiest years ever in terms of U.S. teens smoking marijuana, government data says. For more than 25 years, he has...
Maple Valley teen is NFL's guest at Super Bowl
Zack Lystedt's long road back from a 2006 head injury included moments of bliss this past weekend. Not only did he watch his beloved Seahawks win the Super Bowl, but he and his parents were at the...
Text-while-driving risk is like 1.9% blood-alcohol level
In Washington state's first study to examine driver use of electronic devices, UW investigators saw that more drivers were engaging with such devices behind the wheel than previously estimated. Data...
Harborview patients update: Oso, Wash., mudslide
April 22 Amanda Skorjanc, 25, has been re-admitted to Harborview Medicine Center in satisfactory condition. No interviews at this time. April 21 (9 a.m.) A 38-year-old man injured in the Oso, Wash. ...
(-) Remove emergency medicine filter emergency medicine
(-) Remove legislation filter legislation


text:
Japan's Princess Mako celebrates 30th birthday ahead of marriage
Japan's Princess Mako, left, the eldest daughter of Crown Prince Akishino and Crown Princess Kiko, talks to her younger sister Princess Kako, as they stroll in the garden of their Akasaka imperial property residence in Tokyo, Japan, Oct. 6, in this photo provided by the Imperial Household Agency of Japan. AP-Yonhap
Japan's Princess Mako celebrated her 30th birthday Saturday, marking the last birthday she would have as a member of the imperial family before she marries her boyfriend next week.
Her marriage to commoner Kei Komuro, Tuesday, will mean she becomes a commoner in keeping with laws saying female members of the imperial family cannot keep their royal status after marriage.
Photos of Mako taking a walk on the imperial grounds with her younger sister were released by the Imperial Household Agency, but no comment was given by the princess to mark her birthday.
The three-year engagement between Princess Mako and Komuro was marked with scandal and controversy, which started when a tabloid reported that Komuro was embroiled in a money scandal.
The Imperial Household Agency said in a news conference held in early October that Princess Mako had developed post-traumatic stress disorder (PTSD) from the intense media scrutiny her engagement came under. (Reuters)


text:
Home / 18th November election in Anambra State / 2017 Governorship election in Anambra / 2018 FIFA World Cup in Russia / 2019 elections / DONAL TRUMP IN DANGER AS DEMOCRATS MOVE TO IMPEACH HIM.
DONAL TRUMP IN DANGER AS DEMOCRATS MOVE TO IMPEACH HIM.
Unknown Thursday, November 16, 2017 18th November election in Anambra State, 2017 Governorship election in Anambra, 2018 FIFA World Cup in Russia, 2019 elections
BY BENEDICT ODINAKA EZEOKA.
The House Democrats on Wednesday, introduced articles of impeachment against President Donald Trump, though they they believe that their efforts are exercise in futility and as such has no chance of success because Republicans control both houses of Congress.
Steve Cohen from Tennessee, introduced five articles of impeachment that include obstruction of justice for Trump's decision to fire former FBI Director James Comey, two emoluments clause violations, undermining the independence of the federal judiciary and undermining the freedom of the press. "The time has come to make clear to the American people and to this President that his train of injuries to our Constitution must be brought to an end through impeachment," he said.
Cohen, the ranking member on the House judiciary committee's Constitution subcommittee, acknowledged the limitations of his proposal saying, "I don't expect the House judiciary committee, which is operated like a branch of the administration, to take up hearings".
He told members of the press he would likely be facilitating briefings in lieu of hearings.
"There are many reasons why I think the President is an awful President, an awful person, but not all those reasons rise to the level of impeaching a sitting President. We are not seeking his impeachment because of what he did before he was President," Rep. Luis Gutiérrez, D-Illinois, one of the bill's co-sponsors, said beside Cohen.
White House Press Secretary Sarah Sanders responded in a statement, saying time spent calling for Trump's impeachment "would be better spent focusing on tax relief for American families and businesses. It's disappointing that extremists in Congress still refuse to accept the President's decisive victory in last year's election," she said.
Michael Ahrens, a spokesman for the Republican National Committee, called the impeachment effort radical saying "House Democrats lack a positive message and are completely unwilling to work across the aisle, so instead they've decided to support a baseless radical effort that the vast majority of Americans disagree with".
This isn't the first time Democrats have argued to impeach Trump. Earlier this month, Rep. Pramila Jayapal, D-Washington, said the President has committed "significant constitutional impeachable violations," adding that Democrats needed to act
DONAL TRUMP IN DANGER AS DEMOCRATS MOVE TO IMPEACH HIM. Reviewed by Unknown on Thursday, November 16, 2017 Rating: 5


text:
Q: grounding a moving robot I'm am designing an external battery pack for a robot, Pepper, a robot by Softbank (I'm am not making the battery itself but the circuit connecting two batteries). I am going to connect a 25V (LiFe-PO4) battery to Pepper's internal (Li-ion) battery and strap it to its back. Due to the delicacy of (Li-ion) batteries I included a voltage regulator. But I also want a ground in this circuit, in case of a short circuit or anything.
The problem is: Pepper is a moving robot, so a real ground isn't possible. I was wondering whether it is possible to use its (small) metal chassis as an earth. Similarly like they do in cars.
(Just wiring the - to the chassis)
Pepper is a robot which is about 1,4m high, and only has a metal chassis in it's base, which is about 40cm broad and 20 cm high
If this is not advisable, is there another way of making the circuit more safe? (I am already going to include a fuse.)

External battery specs:
-Capacity 10Ah

-Output voltage25,6V

-Maximum pulse discharge 55A

-Maximum continuous discharge 15A

Internal battery specs:
-Capacity  30Ah

-Max charge current 8A

-Max charge voltage 29,4V

P.S.
Any other tips about my project are more then welcome

A: There is no such thing as a "real ground", or perhaps I should say there's no such thing as a "fake ground". A "ground" is a reference point. It's arbitrary. You can even pick the '+' side of your battery as "ground" if you want to. The purpose of "ground" is to give you a point relative to which you measure voltage. In a closed system like your robot the negative side of your battery will serve just fine as "ground".
If you want to protect your robot against lightning strikes, then you'll need an earth ground, but that's only because earth is probably the biggest body of electrons you're going to find around here, and that's what the lightning is going to want to strike. In the case of a basic robot though, just use the chassis and connect the chassis to the battery '-'.

A: A car chassis is insulated from earth because of the rubber wheels. The chassis is 'ground', i.e. battery negative--the common reference point for the car's electronics. It's more convenient to connect to chassis than to connect a wire back to battery negative. Having the whole chassis connected to battery negative also ensures that wiring faults to chassis (as caused by wire chafing) cause large current flow, which blows fuses and breaks the circuit.
('Earth' is the dirt underneath the car. 'Earthing' a car's chassis is to connect it to the dirt.)
Your system voltages are not hazardous. Should the unit become energized to 30V, nothing will happen. Adding a chassis ground does not mitigate any shock risk because there is no shock risk in the first place.
The biggest hazard in your system is the battery. It likely contains internal protections to prevent it from overheating/exploding. You can stick a fuse in series with it to ensure that current doesn't exceed expected limits.

A: With a moving object you can not normally have a true ground. There is one exception where the wheels are conductive and the device is moving over a metal floor or rails, but I'm going to ignore that here.
With an isolated system providing safety to users involves providing sufficient isolation to ensure that both the "positive" and "negative" side of the power source are never presented to the user. If the thing were in a metal enclosure, connecting that enclosure to either the positive or the negative side of the battery would be acceptable provided nothing exposed of the opposite polarity pokes through that enclosure.
Either way, and with a partly non conductive chassis, you need to ensure that any metal parts the user may come into contact with are either double insulated or are connected to the same terminal as the metal parts of the chassis.
The voltages in question here are non-lethal, however, you also have to be concerned about a hard short caused by some inadvertent metallic connection between exposed "terminals." Such a connection will result in a mighty spark, even with a fused power supply.
If this device is intended for out-door use, you should also consider what measures you need to take to prevent water from reaching exposed and connected metal parts.
One other thing you need to consider is static build-up on the robot. With an isolated system a significant charge can build up, kilovolts, between the chassis and the floor it stands on. That charge, will give anyone that touches it a shock, and that discharge may interfere with the operation and reliability of the robot. As such, use of a high impedance discharge belt that hangs under the robot and drags on the ground may be prudent. However, that aspect is and entirely different issue and not related to the battery isolation. Since it is intermittent at best, it should never be relied upon as a safety ground.

redpajama_set_name: RedPajamaStackExchange


text:
GetUpside Hires Former Gulf Oil Exec to Lead Expansion
Posted by Fuels Market News | April 16, 2020
GetUpside, a digital platform that personalizes brick-and-mortar commerce, announced that it has acquired A-Frame Ventures and brought two new executives on to the team. Even amid the current COVID-19 environment, GetUpside will continue to hire locally in the Chicago, D.C., and Austin markets.
Meredith Sadlowski, former Senior Vice President of Branded Sales and Marketing at Gulf Oil, has joined GetUpside as President of Fuel and C-store. Additionally, Arman Ghosh, A-Frame's co-founder and managing partner, is now the President of Restaurants and Revenue Operations at GetUpside. The acquisition of A-Frame and their team will enable GetUpside to expand into new verticals to drive more value and experiences for consumers and merchants.
"The mission of GetUpside is to help all communities thrive. We started in gas and have over 19,000 amazing merchant partners and millions of consumers that engage with our app on a daily basis," said GetUpside co-founder and CEO Alex Kinnier. "But in order to truly achieve that mission, we need to bring proven personalization to every part of the physical shopping world. The addition of Meredith, Arman, and the A-Frame team will enable us to quickly expand our product to convenience stores, restaurants, grocery, and beyond."
Sadlowski, a senior business development executive with extensive experience in the downstream oil and gas industry, will manage all activity and growth related to GetUpside's fuel and convenience store business, which already drives $1 billion in commerce through the platform. Prior to GetUpside, she worked with the Carlyle Group of companies as a chief commercial officer in the infrastructure energy fund and spent more than a decade in various roles at Gulf Oil.
"The downstream oil and gas industry is more competitive than ever due to M&A-led market consolidation and reductions in fuel demand. Brands must provide innovative program offerings to attract and retain distributors and dealers in their portfolio. Likewise, distributors and dealers must find ways to drive remaining volume into their stations," said Sadlowski. "I've spent years analyzing various marketing programs; GetUpside is the only product that uses the power of personalization to drive proven incremental sales and volume for brands, distributors, and dealers alike. I am proud to join the GetUpside team as we continue to provide brand partners and merchants with a competitive advantage in the market today."
Before the acquisition, Ghosh and partner Brian Fichio co-founded and managed A-Frame Ventures, a company that built, advised, and invested in technology companies. Prior to A-Frame, Ghosh served as an executive at multiple venture backed companies. Of particular note is his tenure as Senior Vice President of Sales at Belly, where he scaled the team from five to 150, grew the network to over 7,500 paying merchants and five million users.
"We have deep experience in brick-and-mortar commerce and local sales, and we understand the challenges and opportunities that restaurants face on a daily basis," said Ghosh. "GetUpside is the silver bullet brick-and-mortar merchants have been looking for; it gives us a rare opportunity to materially change local commerce for the better. I'm excited to bring the GetUpside platform to local and enterprise restaurants across the country."
While the current environment changes day-to-day, GetUpside is fortunate to be growing and continuing the search for new talent. Sadlowski, Ghosh, and the rest of the GetUpside team are actively hiring in Chicago, IL, Washington, D.C., and Austin, TX. The GetUpside careers page has more information about GetUpside's work environment and all open positions.
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NextWEBINAR: Reset 2020! Improve Your Fuel Program While Prices are Still Low
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NATSO Analysis Confirms Commercial Rest Areas Limit Truck Parking Capacity
ATA: Truckload Turnover Rate Rises Slightly in First Quarter
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text:
Der Begriff Hanhart-Syndrom ist in der Fachliteratur nicht eindeutig definiert.

Die Namensgebung bezieht sich jeweils auf den Erstautor der Erstbeschreibungen durch den schweizerischen Internisten und Humangenetiker Ernst Hanhart (1891–1973).

Er bezeichnet folgende Syndrome:
 Keratodermie mit Lipomen
 Heredodegenerativer Zwergwuchs
 Oro-akraler-Fehlbildungskomplex
 Familiäre spastische Paraplegie

Die Syndrome sind nicht zu verwechseln mit dem Richner-Hanhart-Syndrom.

Keratodermie mit Lipomen
Synonyme sind:  Keratosis plamo-plantaris mit systematisierten Lipomen;

Die Bezeichnung bezieht sich auf eine Beschreibung aus dem Jahre 1947.

Heredodegenerativer Zwergwuchs
Synonyme sind: Hanhart-Dwarfismus;

Die Bezeichnung bezieht sich auf eine Beschreibung aus dem Jahre 1925.

Oro-akraler-Fehlbildungskomplex
Synonyme sind: Aglossie-Adaktylie-Syndrom; Hanhart-Syndrom; Jussieu-Syndrom; Hypoglossie-Hypodaktylie-Syndrom

Die Bezeichnung bezieht sich auf eine Beschreibung aus dem Jahre 1950.

Familiäre spastische Paraplegie
Synonyme sind: Hanhart-Syndrom IV;

Es handelt sich um eine autosomal-dominant vererbte familiäre spastische Paraplegie mit geistiger Behinderung.

Die Bezeichnung bezieht sich auf eine Beschreibung aus dem Jahre 1936.

Einzelnachweise

Fehlbildung
Seltene Krankheit
Erbkrankheit

redpajama_set_name: RedPajamaWikipedia