A metric derived from KL divergence
For probability densities $p_0$ and $p_1$, KL divergence of $p_1$ from $p_0$ is defined as $K(p_0, p_1) = -E_0(\log \frac{p_1(X)}{p_0(X)})$, where $E_0$ is an expectation with respect to $p_0$.
KL divergence is regarded as “a distance” between the two probability distributions. However, it is not a metric in mathematical sense, since $K(p_0, p_1) \neq K(p_1, p_0)$ in general.
What if we define a new function $D(p_0, p_1) := \frac{K(p_0, p_1) + K(p_1, p_0)}{2}$? It is easy to show from properties of KL divergence that
\[\begin{align} &\text{(1)} \:\: D \geq 0 \\ &\text{(2)} \:\: D(p_0, p_1) = D(p_1, p_0) \\ &\text{(3)} \:\: D(p_0, p_1) = 0 \iff p_0 = p_1 \\ \end{align}\]However it is still not a metric since it is not necessarily true that the triangle inequality
\[\begin{align} &\text{(4)} \:\: D(p_0, p_1) \leq D(p_0, p_2) + D(p_2, p_1) \\ \end{align}\]holds.
By setting up a reference density $M := \frac{p_0 + p_1}{2}$, and taking a square root,
\[J(p_0, p_1) := \sqrt{\frac{K(p_0, M) + K(p_1, M)}{2}}\]has been proven to satisfy triangule inequality. $J$ here is actually a square root of Jensen-Shannon divergence $JS(p_0, p_1)$.
Hence we can say that $J := \sqrt{JS}$ is truely a metric defined on a set of density functions.