$f$-divergence: $D_f(P || Q) = 0 \iff P = Q$ Let $f$ be a function such that:
(1) $f$ is convex. 
(2) $f(1) = 0$
(3): $f : \mathbb{R}_{> 0} \mapsto \mathbb{R}$
(4): $f$ is strictly convex around $1$ which means $$f(kx + (1 - k)y) < k f(x) + (1 - k) f(y) \, \,\, \,\, \,\, \, $$
$$\forall \, k \in [0, 1] \, \, \, \forall x, y \in \mathbb{R}_{> 0} \text{ such that } kx + (1 - k)y = 1$$
Now let $P, Q$ be probability measures on $\mathcal X$ and other necessary conditions 
 found on wikipedia . 
\begin{align*}
D_f(P || Q) &= \mathbb{E}_Q f \left(\frac{dP}{dQ} \right) \\
&\geq f \left( \mathbb{E}_Q \frac{dP}{dQ} \right)\\
&= f \left( \int_\mathcal{X}  dP \right)\\
&= f(1)\\
&= 0   
\end{align*}
I want to prove $D_f(P || Q) = 0 \iff P = Q$
$P = Q$ $\implies $ $D_f(P || Q) = 0 $ is easy.
Assume $D_f(P || Q) = 0$. Then, from above we know $$\mathbb{E}_Q f \left(\frac{dP}{dQ} \right) = f \left( \mathbb{E}_Q \frac{dP}{dQ} \right)  $$
How to complete the proof to show $P = Q$? 
 A: NOTE: Not sure what etiquette is here. The answer below (beyond the first dashed line) was given prior to the addition of condition (4) above, suggested by @stochasticboy321. For my own sake, I wrote out his reasoning more fully beyond the second dashed line. I added below the third dashed line a slightly more complicated counterexample which may be used to sketch the necessity of strict convexity at 1.

This is not true in general. As a very simple example, take $f = 0$, which gives you that the divergence between any two measures is 0.
It would be interesting to know if there are simple conditions on $f$ which would guarantee this property. For KL divergence, Hellinger distance, and total variation (all examples of f-divergences), this equivalence holds and you can find proofs just by tracing through Wikipedia.

As suggested, if we have strict convexity of $f$ at 1, then we have that $f(y) > a(y-1)$ for some $a \in \mathbb{R}$. We then note: 
$$\mathbb{E}_Q\left[f\left(\frac{dP}{dQ}\right)\right] = \mathbb{E}_Q\left[f\left(\frac{dP}{dQ}\right) \mathbb{1}_{\{dP/dQ \neq 1\}}\right] > \mathbb{E}_Q\left[a\left(\frac{dP}{dQ} - 1 \right)\mathbb{1}_{\{dP/dQ \neq 1\}}\right].$$
The second strict inequality only holds if $Q(\{dP/dQ \neq 1\}) > 0$, which is equivalent to $P \neq Q$. This last term is equal to 0:
$$\mathbb{E}_Q\left[a\left(\frac{dP}{dQ} - 1 \right)\mathbb{1}_{\{dP/dQ \neq 1\}}\right]=a\mathbb{E}_Q\left[\left(\frac{dP}{dQ} - 1 \right)\right]=a(1-1)=0.$$
Thanks for the suggestion @stochasticboy321!

It also seems to me that strict convexity at 1 is not only sufficient, but necessary. Lack of strict convexity means that $f$ is linear in some neighborhood of 1, and it is not difficult to construct counterexamples then.
For an explicit one, consider $f(y) = y-1$, and $dQ = dx$ and $dP = 2x \, dx$ on the unit interval. Then we have that:
$$\mathbb{E}_Q\left[f\left(\frac{dP}{dQ}\right)\right] = \int^1_0 2x-1 \, dx = 0,$$
despite the fact that $dP \neq dQ$. It's not too difficult to see how some version of this simple example can be "plugged in" locally for any $f$ that is locally linear near 1.
