According to the [Pinsker inequality](https://en.wikipedia.org/wiki/Pinsker%27s_inequality), we have the following inequality:
\begin{equation}
\delta\_{TV} (p, q)^2 \leq \frac{1}{2} D\_{KL}(p,q),
\end{equation}
where $\delta\_{TV} (\cdot, \cdot)$ and $D\_{KL}(\cdot, \cdot)$ are [total variation distance](https://en.wikipedia.org/wiki/Total_variation_distance_of_probability_measures) and [Kullback–Leibler divergence](https://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence), respectively.


On the other hand, the total variation distance is related to the $L\_1$ norm by the identity:
\begin{equation}
\delta\_{TV}(p, q) = \frac{1}{2} \int\_{\mathcal{X}} |p(x)-q(x)| \, dx,
\end{equation}
and thus by using the Cauchy–Schwarz inequality we obtain that
\begin{equation}
\delta\_{TV} (p, q)^2 \leq \frac{1}{4} \int\_{\mathcal{X}} (p(x)-q(x))^2 \, dx.
\end{equation}
I denote the RHS by $L\_2(p, q)$, i.e., $L\_2(p, q) = \int\_{\mathcal{X}} (p(x)-q(x))^2 \, dx$.


My question is: does there exist some (inequality) relation between the $D\_{KL}(\cdot, \cdot)$ and $L\_2(p, q)$?



Is there an inequality relation between KL-divergence and $L_2$ norm?
Such inequality is impossible: consider $p(x)=1$, $q(x)=1/(2\sqrt{x})$, as probability densities on $(0,1)$. Then $D\_{KL}(p\parallel q)$ is finite, while $\|p-q\|\_2=\infty$, as $q\not\in L^2$.


The reverse direction is also impossible: take $p(x)=a e^{-ax}$, $q(x)=a^2e^{-a^2x}$ on $(0,\infty)$. Then $\|p-q\|\_2\to0$, while $D\_{KL}(q\parallel p)=1/a+\ln a -1\to\infty$, as $a\to0$.


