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 $d$ ?