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

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?

• $g=1$ gives one on both sides, no?
– Dirk
Apr 6, 2020 at 21:34
• Since you changed the question, I should change my comment: the constant functions show that $C_1=1$ is the smallest such constant.
– Dirk
Apr 7, 2020 at 10:18

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 $$p>q$$).
• I should mention that log-concavity plays absolutely no role here (as should be clear from my answer above). Unless you're working in a finite space $L^p$ and $L^q$ norms cannot be equivalent. Ever. May 7, 2020 at 14:57