Relationship between KL, chi-squared, and Hellinger There are many well-known relationships between the KL divergence, chi-squared ($\chi^2$) divergence, and the Hellinger metric. In the paper "Assouad, Fano, and Le Cam" by Bin Yu, the author appears to use the following relationship (see Example 2, p. 433):
$$
\text{KL}(f,g)
\le \int \frac{(\sqrt{f}-\sqrt{g})^2}{f}\,dx.
$$
This looks almost like the standard relationship $\text{KL}(f,g)\le \chi^2(f,g)$, but differs in the square roots taken in the numerator. I have not seen this general relationship previously.
My question is: Does this inequality hold for general densities $f,g$?
(Note: In the paper, this inequality is applied to special choices of $f$ and $g$, and so maybe the author only means it for these functions. It is not clear to me based on the writing.)
 A: This inequality is false in general.
E.g., let $f$ and $g$ be pdf's on $[0,1]$ given by the formulas $f=1$ and $g=f+t\,1_{(0,1/2)}-t\,1_{(1/2.1)}$ for $t\in(0,1)$. Then the left- and right-hand sides of your inequality are, respectively, $\sim t^2/2$ and $\sim t^2/4$ as $t\downarrow0$, so that your inequality fails to hold.
Moreover, if here $t\uparrow1$, then the left- and right-hand sides of your inequality go, respectively, to $\infty$ and $1-1/\sqrt2$. So, your inequality will fail to hold even if any extra universal constant factor is inserted on its right-hand side.

Here is the graph of the ratio the left-hand side of your inequality to its right-hand side:


The reason for such counterexamples is that $\text{KL}(f,g)=\int f\ln\frac fg$ can be arbitrarily large if $g$ is allowed to take values much smaller than the corresponding values of $f$, whereas the right-hand side of your inequality will remain bounded if $f$ is (say) bounded away from $0$ on the set where $g>0$.
A: The inequality does not hold for even simple densities which fulfill the conditions given in Section 29.3 of the paper shared by the OP.

Here is a piecewise constant counterexample,
$$ f(x) = \frac{3}{2} 1_{\{(0,1/2)\}}(x) + \frac{1}{2} 1_{\{ (1/2,1) \}}(x) $$
$$ g(x) = \frac{1}{2} 1_{\{(0,1/2)\}}(x) + \frac{3}{2} 1_{\{ (1/2,1) \}}(x) $$
These functions are (i) probabilities densities whose support is $(0,1)$; (ii) they are bounded from above/below; and (iii) piecewise constant.  However, the KL divergence between the corresponding distributions is
$$
\int_0^1 f(x) \log(f(x)/g(x)) dx = \log(3)/2 \ge 1/2
$$
whereas the claimed upper bound is
$$
\int_0^1 \frac{(\sqrt{f(x)} - \sqrt{g(x)})^2}{f(x)} dx = \frac{(\sqrt{6}-\sqrt{2})^2}{3} \le 1/2 \;. 
$$

And, here is a smooth counterexample,
$$ f \equiv 1 $$
$$ g(x) = 2 (1-x) $$
In this case, by direct integration, the KL divergence is $(2-4\sqrt{2})/3 > 1/4$ while the claimed upper bound is $2 - 4 \sqrt{2}/3<1/4$.

All this seems to point to a possible bug in Example 2 of the paper.
