Log-Sobolev constant

Question

Let $\nu \propto e^{-f}$ be a probability density on $\mathbb{R}^d$ with full support. We say $\nu$ satisfies the log-Sobolev inequality (LSI) with constant $\alpha$ if for every smooth function $g:\mathbb{R}^d \to \mathbb{R}$, we have $$ \mathbb{E}_\nu [g^2 \log g^2] - \mathbb{E}_\nu[g^2] \log\mathbb{E}_\nu[g^2] \leq \frac{2}{\alpha} \mathbb{E}_\nu[|\nabla g |^2] $$ If $f$ is $\mu$-strongly convex (or equivalently $\lambda_{\min} (\nabla^2 f(x)) \geq \mu $, for some constant $\mu >0$), then it is well-known that $\nu$ satisfies LSI with constant $\alpha =\mu$.

Now, suppose $$\lambda_{\min} (\nabla^2 f(x)) \geq \mu \frac{|x|}{1+|x|}.$$ What is the LSI constant $\alpha$ in this case ? Specifically, will it be polynomial or exponential in $d$, or independent of $d$ ?

Answer 1

Late answer, but since $\nu$ can be shown to be close to a measure that satisfies a dimension-free LSI, we can obtain a dimension-free constant using the Holly-Stroock perturbation lemma (eg see Proposition 5.1.6 in Bakry's book).

For example, if we define the measure $\mu \propto e^{-g}$ with $g(x) = f(x) + \frac{\mu}{4} \|x\|^2 \mathbb{1}_{\|x\| \leq 1}$, then we have that $\lambda_{\min}(\nabla^2 g) \geq \frac{\mu}{2}$ and thus, $\mu$ satisfies an LSI with constant $\frac{\mu}{2}$. Furthermore, since $|f-g| \leq \frac{\mu}{4}$, it follows from the perturbation lemma that $\mu$ satisfies an LSI with constant $\frac{\mu}{2} e^{-\mu/2}$. The downside of this approach is that it is a weak estimate in the setting where $\mu$ is large – this can be partially remedied by fiddling with the function $g$.