$L^p$ norm inequalities with respect to strongly-log-concave densities

Question

Let $\pi(x)=\frac{e^{-f(x)}}{\int_{\mathbb{R}^d}e^{-f(u)}du}$ be a strongly-log-concave distribution, i.e., $f(x):\mathbb{R}^d\rightarrow R$ is an $m$-strongly convex function. Also, $f(x)$ has $L$-Lipschitz gradient. Define the norm of a function $g:\mathbb{R}^d\rightarrow R$:

$$\|g\|_p=\left(\int_{\mathbb{R}^d} |g(x)|^p \pi(x)dx\right)^\frac{1}{p}$$.

Question: For $p\geq q$, is there any inequality like the following: $$C_0\|g\|_p\leq \|g\|_q\leq C_1\|g\|_p$$

where $C_0,C_1$ are constants. By Jensen's inequality, $C_1=1$ satisfies the upper bound. Is it possible to get $C_0,C_1$ which uses the fact that the density is strongly log-concave and $f(x)$ has $L$-Lipschitz gradient?

Answer 1

No. Only the upper bound holds, and as Dirk pointed out the best constant is $C_1=1$. The lower bound cannot hold for any $C_0$, since otherwise the $L^p$ and $L^q$ norms would be equivalent. This is well-known to fail. To see this, observe that with your assumptions your reference measure $\pi(x)dx$ is locally equivalent to the Lebesgue measure $dx$ (i-e $c_0 dx\leq \pi(x)dx\leq c_1 dx$ on any ball $B_R$). The comparison between $L^p$ and $L^q$ norms on $\mathbb R^d$ is well known for the Lebesgue measure: Taking a radial function $g(x)=\chi_{B_R}(x) |x|^\alpha$ and tuning $\alpha<0$ (denpending on $p,q$) easily gives a counterexample to $C_0\|g\|_{L^p}\leq \|g\|_{L^q}$, i-e such that $\|g\|_{L^p}=+\infty$ but $\|g\|_{L^q}<\infty$. (Here I'm using a cutoff function to localize on a ball $B_R$ so that $\pi(x)dx$ and $dx$ are equivalent). More explicitly, take $\alpha=-\frac{d}{q}+\epsilon$ (for very small $\epsilon>0$) so that $|g(x)|^qdx\sim r^{\alpha q}r^{d-1}dr=r^{\alpha q+d-1}=r^{-1+\epsilon}dr$ is bordeline integrable at the origin, while $|g(x)|^pdx\sim r^{\alpha p+d-1}dr$ is not (due to the exponent $\alpha p+d-1<-1$ since $p>q$).