TITLE: Lower Bound of KL-Divergence Between Two Gibbs Measures
QUESTION [8 upvotes]: Suppose we have two Gibbs measures with densities 
$$
p_f(x) \propto \exp(f(x)),\quad  q_g(x)\propto \exp(g(x)).
$$
Consider the KL-divergence between $p_f$ and $q_g$, as a functional of $f$ and $g$, that is,
$$
D(f, g) :=  \text{KL}(p_f \| q_g).
$$
Question: Do we have the following lower bound:
$$
D(f, g) \geq \|f - g\|^2,
$$
where we are interested in, for example, the $L_2$-norm of $f-g$.

REPLY [5 votes]: The answer is no. Indeed, your question can be restated as follows: Is it true that for some constant $c>0$ we have 
$$\int P\ln\frac PQ\,d\mu\ge c\int(\ln P-\ln Q)^2\,d\mu, \tag{1}
$$
where $P$ and $Q$ are probability densities with respect to a measure $\mu$? 
Let $\mu$ be the counting measure on the set $\{0,1\}$, and let $P(0)=p\in(0,1)$ and $Q(0)=q\in(0,1)$. Then (1) will become 
$$p\ln\frac pq+(1-p)\ln\frac{1-p}{1-q}\ge 
c\Big[\Big(\ln\frac pq\Big)^2+\Big(\ln\frac{1-p}{1-q}\Big)^2\Big]. \tag{2}
$$
Letting now $p\to0$, we get a contradiction, because the left-hand side of (2) will go to $\ln\frac1{1-q}<\infty$, but the right-hand side of (2) will go to $\infty$.