Consider $E: \mathcal P (X) \rightarrow \mathbb R \cup \{ \infty \}$ a functional (with a convex and dense domain, $E< +\infty$) over $\mathcal P(X)$ the set of probability measures of a metric compact space $X$.
Then consider the minimisation problem $\underset{m \in \mathcal P(X) }{\inf} E[m] $
Suppose that there is a $\widetilde m \in \mathcal P(X)$ solves the minimisation problem and that $E$ is differentiable in $\widetilde m$ is the sense that there is a function $\frac{\delta E}{\delta m}[\widetilde m]$ such that for all $m$ in the domain of $E$
$$ 0 \leq \lim_{\epsilon \rightarrow 0^+} \frac{E[\widetilde m + \epsilon (m-\widetilde m)]-E[\widetilde m]}{\epsilon}\leq \int_X \frac{\delta E}{\delta m}[\widetilde m] ~d(m -\widetilde m)$$
Then we have for all $m$ in the domain of $E$ $$ \int_X \frac{\delta E}{\delta m}[\widetilde m] ~d \widetilde m \leq \int_X \frac{\delta E}{\delta m}[\widetilde m] ~dm .$$
How to conclude that $\widetilde m$ is concentrated on $(\arg \min \frac{\delta E}{\delta m}[\widetilde m]) $?
Here is my idea. Since $$ \int_X \frac{\delta E}{\delta m}[\widetilde m] ~d \widetilde m \leq \int_X \frac{\delta E}{\delta m}[\widetilde m] ~dm $$
folows for any $m$ so by taking $m(x) = \begin{cases} 1, ~\text{if } x = \arg \min \frac{\delta E}{\delta m} [\widetilde m]\\ 0, \text{otherwise}\end{cases}$
we have
$$ \int_X \frac{\delta E}{\delta m}[\widetilde m] ~d \widetilde m \leq \arg \min \frac{\delta E}{\delta m} [\widetilde m]$$
but $$ \arg \min \frac{\delta E}{\delta m} [\widetilde m]\leq \int_X \frac{\delta E}{\delta m}[\widetilde m] ~d \widetilde m \leq \arg \max \frac{\delta E}{\delta m} [\widetilde m]$$
therefore
$$ \int_X \frac{\delta E}{\delta m}[\widetilde m] ~d \widetilde m = \arg \min \frac{\delta E}{\delta m} [\widetilde m]$$
and so $\widetilde{m}(x) = \begin{cases} 1, ~\text{if } x = \arg \min > \frac{\delta E}{\delta m} [\widetilde m]\\ 0, > \text{otherwise}\end{cases}$
Many thanks