simply using the concept of Gâteaux differentiability. When $m$ represents the density of a probability measure, the functional becomes a functional of a probability measure, but of a special kind. Suppose that $F$ extends to a general probability measure, then the concept of differentiability does not extend. For instance if $x^{j}$ is a point in $\mathbb{R}^{n}$, neither can the concept of differentiability be extended to $F(\delta_{x^{j}}),$ nor more generally to $F(\frac{\sum_{j=1}^{K}\delta_{x^{j}}}{K}).$

Nevertheless, we may have the following situation: the functional $F(m)$ is differentiable in $m$, and the function $$\Phi(x)=\Phi(x^{1},\cdots,x^{N})=F(\frac{\sum_{j=1}^{K}\delta_{x^{j}}}{K})\label{eq:6.100}$$ is differentiable in $x$. Note that differentiability refers to two different set up, one with respect to arguments in $L^{2}(\mathbb{R}^{n})$ and the other one, with respect to arguments in $\mathbb{R}^{nN}.$ We want to study the link between these two concepts.

\[prop:6.1\] Assume that $F(m)$ and $\Phi(x)$ are sufficiently differentiable in their own sense, and that expressions below are well-defined. Then we have the following relations $$\int_{\mathbb{R}^{n}}D_{\xi}\frac{\partial F(m)}{\partial m}(\xi).B(\xi)m(\xi)d\xi=\sum_{j=1}^{K}D_{x^{j}}\Phi(x).B(x^{j});\label{eq:6.101}$$ $$\int_{\mathbb{R}^{n}}\Delta_{\xi}\frac{\partial F(m)}{\partial m}(\xi)m(\xi)d\xi+\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\text{tr}D_{\xi}D_{\eta}(\frac{\partial^{2}F(m)}{\partial m^{2}}(\xi,\eta))m(\xi)m(\eta)d\xi d\eta=\sum_{j,k=1}^{K}\text{tr}(D_{x^{j}}D_{x^{k}}\Phi(x));\label{eq:6.102}$$ $$\int_{\mathbb{R}^{n}}\text{tr}(a(\xi)D_{\xi}^{2}(\frac{\partial F(m)}{\partial m}(\xi)))m(\xi)d\xi \approx \sum_{j=1}^{K}\text{tr}(a(x^{j})D_{x^{j}}^{2}\Phi(x)),\:\text{for large } K\label{eq:6.103};$$ where $m = \frac{1}{K} \sum_{j=1}^K \delta_{x^j}$.

We first comment how to understand these relations. The left hand side makes sense for sufficiently smooth functionals $F(m).$ The results are functionals of 