Legendre-Fenchel dual of convex functional

Question

I found in the literature the following transformation. Let $Q$ be a fixed probability measure over the space $X$. Let $\pi$ be a finite measure absolutely continuous with respect to $Q$. Let $\gamma\geq 1$. Consider the functional $E_\gamma(\cdot\|Q) : \mathcal{M}(X) \to \mathbb{R}$ such that $$E_\gamma(\pi\|Q) = \int_\mathcal{X} \left(\frac{d\pi}{dQ}-\gamma\right)^+ dQ.$$ This is a convex functional. In the work they then state that convex conjugate of $E_\gamma$, let us call it $\phi^\star(f)$ is equal to $$ \phi^\star(f) = \begin{cases} \gamma\int f dQ,& \text{if } 0\leq f \leq 1\\ +\infty, & \text{otherwise} \end{cases}$$ I tried computing it but even knowing the answer I can't manage to show it properly. Formally we know that $$\phi^\star(f) = \sup_{\pi} \left\{ \int f d\pi - \int_\mathcal{X} \left(\frac{d\pi}{dQ}-\gamma\right)^+ dQ \right\} = \sup_\pi\left\{ \int dQ\left( \frac{d\pi}{dQ} f - \left(\frac{d\pi}{dQ}-\gamma\right)^+\right) \right\}.$$ Even assuming that $\frac{d\pi}{dQ}\geq \gamma$ so that one gets rid of the $^+$ I get something like $\int d\pi(f-1)+\gamma$ and clearly, if $\exists x : f(x)>1$ (or $f<0$) then I can set $d\pi = M1_x$ with $M>0$ (or $M<0$) and then let $M\to+\infty$ (resp. $-\infty$) and the $\sup$ becomes then $+\infty$. Why would this not work if $0\leq f \leq 1$? It seems that the sup is attained when one sets $d\pi = \gamma dQ$ but I cannot fully justify why. Could you help?