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Suppose we are given a class of probability density functions $\mathcal{F}$ so that for every $f \in \mathcal{F}$ we have $\alpha \leq f \leq \beta$ for some positive $\alpha, \beta \in \mathbb{R}_+$ and the density have the same support set $B$.

Here it is proved that in this setting the $L_2$ norm between density functions, i.e., $$ \|f-g\|^2_2 = \int_B (f(x) - g(x))^2 \mu(dx) $$ is equivalent to the KL divergence between $f$ and $g$, i.e., for $$ D_{\operatorname{KL}}(f \| g) = \int_B f(x) \log \frac{f(x)}{g(x)} \mu(dx) $$ we have $$ D_{\operatorname{KL}}(f \| g) \asymp \|f-g\|^2_2 $$ where $\asymp$ denotes equiality up to constant factors (which may depend on $\alpha, \beta$ but we consider those fixed).

Here a question was asked regarding the comparison of $L_1$ and $L_2$, and was argued that for such $\mathcal{F}$ classes of functions, it is possible that $$ \|f-g\|_1 = \int_B |f(x) - g(x)| \mu(dx) \ll \|f-g\|_2 $$ in general.

By Pinsker's inequality we know that $$ \|f-g\|_1 \lesssim \sqrt{D_{\operatorname{KL}}(f \| g)} \asymp \|f-g\|_2 $$ where $\lesssim$ indicates inequality up to absolute constant factors.

My question is as follows. Suppose we impose smoothness assumptions on $\mathcal{F}$. For instance suppose $\mathcal{F}$ consists of Lipschitz continuous functions with Lipschitz constant $L$. Does a reverse Pinsker's inequality hold in this case i.e. is it possible that $$ \|f-g\|_1 \gtrsim \sqrt{D_{\operatorname{KL}}(f \| g)} \asymp \|f-g\|_2\;? $$

The counterexample given here fails to work in the Lipschtiz $\mathcal{F}$ case, and is not clear how it can be recreated because the functions need to be smooth.

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  • $\begingroup$ I've generally seen reverse pinsker inequalities only under assumptions that $c \leq f(x)/g(x)\leq c'$ are bounded for all $x$, see for example this. One can get something sort-of like a reverse pinsker inequality via bounds $D_{\mathsf{KL}(T(f)||T(g))} \leq C_T \lVert f -g\rVert_1^2$, where $T(f)$ is the markov kernel $T$ acting on the random variable with density $f$, and $C_T$ is a constant that depends on $T$, but not $f,g$. See for example this. Not relevant to your exact question, but hopefully useful. $\endgroup$ Mar 25 at 22:13

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The answer is no.

E.g., suppose that both $f$ and $g$ are supported on $B=[0,1]$, with $f=1$ on $B$ and $$g(x)=(1-h+x)\,1(0\le x<h)+1(h\le x<1-h)+(x+h)\,1(1-h\le x\le1)$$ for $h\in(0,1/2)$ and $x\in B$.

Then $f$ and $g$ are $1$-Lipschitz on $B$, $\|f-g\|_1=h^2$, and $$D_{\operatorname{KL}}(f\|g) =-\int_0^h dx\,\ln(1-h+x)-\int_{1-h}^1 dx\,\ln(x+h) \\ =-\int_0^h dx\,\ln(1-(h-x)^2) \ge\int_0^h dx\,(h-x)^2=h^3/3,$$ so that $$\|f-g\|_1\not\gtrsim\sqrt{D_{\operatorname{KL}}(f \| g)}$$ if $h$ is small enough.

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  • $\begingroup$ Awesome counterexample thanks! Do you have any thoughts on whether the exponents of $h$ can be made even more discrepant? In this example for instance $\|f-g\|_1 = h^2$ while $\|f-g\|_2 = h$. $\endgroup$ Mar 25 at 20:32
  • $\begingroup$ @spacetimewarp : Thank you for your appreciation. My guess/feeling is this is the largest discrepancy one can get given the Lipschitz condition, but I don't have a proof of this at this point. $\endgroup$ Mar 25 at 23:57

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