0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ $g=1$ gives one on both sides, no? $\endgroup$
    – Dirk
    Apr 6 '20 at 21:34
  • $\begingroup$ Since you changed the question, I should change my comment: the constant functions show that $C_1=1$ is the smallest such constant. $\endgroup$
    – Dirk
    Apr 7 '20 at 10:18
1
$\begingroup$

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 $p>q$).

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ May 7 '20 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.