Bakry-Emery criterion

Question

The most common use of the Bakry-Emery criterion is for the measure $\mu(x)=e^{-u(x)} /Z$ where $u \in \mathcal{C}^2$. I would like to ask for an application to a smaller class. Consider $u(x)=|x|^2 + V(x){1}_{|x|\geq R}$ where $V\in \mathcal{C}^2$ and $\mathrm{Hess}(V)\geq \rho$ for every $|x|>R$.However, the Hessian of $u$ doesn't exist on $|x|=R$. Could I still claim an LSI on $\mathbb{R}^d$ via Bakry-Emery? If yes, could you provide a reference for Bakry-Emery with weaker than $\mathcal{C}^2$ assumption?

Answer 1

To obtain a constant of order $1$ independent of $V$, you will need an additional drift condition at $|x|=R$ for $V$, something of the type $\langle \nabla V(x) , x \rangle \geq 0$ for those $x$ might be sufficient.

Else as a counterexample to make the constant arbitrary small, consider $V^L(x) = |x-L e_1|^2$ for $e_1$ the first coordinate vector and $L>0$. The resulting potential $|x|^2 + |x-L e_1|^2 1_{|x|\geq R}$ will be nonconvex around the point $R e_1$ for $L$ large enough.

If you want a generic solution without additional assumption on $V$, you can combine the Bakry-Émergy criterion with the Holley-Strock perturbation principle and find a decomposition of the type $$ |x|^2 + V(x) 1_{|x|\geq R} = \varphi_c(x) + \varphi_b(x) , $$ where $\varphi_c(x)$ is uniformly convex and $\varphi_b$ uniformly bounded.

Then, for those measures the Barky-Émery criterion combined with Holley-Stroock perturbation principle yields a LSI with constant $$ \frac{1}{\inf_x \lambda_{\min} \operatorname{Hess} \varphi_c(x)} \exp\left(\sup_x \varphi_b(x) - \inf_x \varphi_b(x)\right) $$ upto factors of 2 depending on the exact definiton for your LSI.

For my above (counter-)example, one probably can choose $\varphi_c$ still close to $1$-convex, however the bounded perturbation $\varphi_b$ will become large for $L$ or $R$ large making the LSI-constant exponentially large in $L$ and $R$.

Just as a visual explanation. The counterexample in the setting $R=1$ and $L=4$ in 1d, the qualitative picture does not change much in higher dimension. $R=1$ and $L=4$" />

A smoothing of the example will give you a smooth double-well potential.