Existence of first variation

Question

I am trying to compute the first variation of the functional

$$\mathcal F(\rho) = \int_{\Omega} R(x;\rho) d\rho(x)$$

where $R$ is some function of $x$ that also depends on $\rho$. Here $\rho$ is a probability measure, i.e. $\rho \in \mathcal P(\Omega)$. I'm using the Definition 7.12 from Optimal Transport for Applied Mathematicians by Santambrogio, which defines $\frac{\delta \mathcal F}{\delta \rho}(\rho)$ as a measurable function satisfying

$$\frac{d}{d\epsilon} \mathcal F_1(\rho+\epsilon\chi) \bigg|_{\epsilon=0} = \int \frac{\delta\mathcal F_1}{\delta\rho}(\rho) d\chi$$

for all perturbations $\chi=\tilde\rho-\rho$ with $\tilde\rho\in L_c^\infty(\Omega)\cap\mathcal P(\Omega)$. Since $\mathcal F$ didn't seem to fit into the types of functionals listed in Santambrogio's book, I resorted to computing this directly, which gave me

$$\frac{d}{d\epsilon} \mathcal F_1(\rho+\epsilon\chi) = \int R(x;\rho) d\chi(x) + \frac{d}{d\epsilon} \int R(x;\rho+\epsilon\chi) d\rho(x) + \epsilon \frac{d}{d\epsilon} \int R(x;\rho+\epsilon\chi) d\chi(x).$$

Then I wrote $f(x,\delta) = R(x;\rho+\delta\chi)$ and substituted in the Taylor series $R(x;\rho+\epsilon\chi) = f(x,0) + \epsilon f_\delta(x,\xi_\epsilon)$ for some $\xi_\epsilon \in[0,\epsilon]$. This gave me

$$\frac{d}{d\epsilon} \mathcal F_1(\rho+\epsilon\chi) = \int R(x;\rho) d\chi(x) + \int f_\delta(x,\xi_\epsilon) + \epsilon f_{\delta,\delta}(x,\xi_\epsilon) d(\rho +\epsilon\chi)(x).$$

I then set $\epsilon=0$ so that

$$\frac{d}{d\epsilon} \mathcal F_1(\rho+\epsilon\chi) \bigg|_{\epsilon=0} = \int R(x;\rho) d\chi(x) + \int f_\delta(x,0) d\rho(x).$$

As per the definition, I need to write the remainder as an integral against $\chi$. This is simple for some special cases. For example, if $R(x;\rho)=\log(\rho(x))$ then $\int f_\delta(x,0)d\rho(x) = \int d\chi(x)$ and so $\frac{\delta \mathcal F}{\delta \rho}(\rho)(x) = \log(\rho(x))+1$.

However, this seems to not work as well in general. I had the idea to use the Radon-Nikodym derivative, which would give $\frac{\delta \mathcal F}{\delta \rho}(\rho)(x) = R(x;\rho) + f_\delta(x,0)\frac{d\rho}{d\chi}(x)$, but I don't think $\chi$ should appear in the first variation.

I have two questions:

1. Does this just mean that $\mathcal F$ does not have a first variation for many choices of $R$? Even for sufficiently regular $R$ where the remainder just can't be written as an integral against $\chi$?
2. Are there some references dealing with first variations of functionals like this one? Perhaps with a more general definition to handle the remainder above?

Some more context for my particular setting: I generally don't know the form of $R$, but I know $\rho(x)\nabla_x R(x,\rho) = \mathscr F^{-1}(g\cdot\mathscr F(\nabla\rho))(x)$ where $\mathscr F$ denotes the Fourier transform and $g$ is a known function, which is usually very smooth. In particular, $g=1$ gives the special case $R(x;\rho)=\log(\rho(x))$ mentioned above.

Answer 1

In general, a first variation is just the collection of all directional derivatives $\frac{d}{d\epsilon} \mathcal{F}(\rho+\epsilon\chi)|_{\epsilon = 0}$. For fixed $\rho$ one can treat them as a function of $\chi$ and in many cases this turns out to be a bounded linear operator, or even the same as the Frechet-derivative of $\mathcal{F}$. Now in the case of measures, these operators live in the dual space of the space of measures, which is not the nicest space to work with. However it includes operators of the form $\chi \mapsto \int f d\chi$, where $f$ is a bounded, measurable functions, which is more or less what I believe Santambrogio uses for his definition.

Regarding your problem, I think the example is leading you in the wrong direction. Given the identity for $R$, it should be a nonlocal operator, i.e. for fixed $x$, $R(x,\rho)$ depends on all of $\rho$ and not just on $\rho(x)$ or its behaviour in a neighborhood. It is just in the special case $g=1$ that the Fourier-transform and its inverse cancel. Thus you are dealing with functions and your Taylor-series approach is simply wrong.

What you need to do is to find (at least formally) the derivative $\frac{\partial R}{\partial \rho}$ of $R(.,\rho)$ with respect to $\rho$, which hopefully will be a linear map from the space of measures to some space of measureable functions. I would guess that differentiating your identity for $R$ with respect to $\rho$ at least gives you at least some information about that. Then the first variation at $\rho$ in direction $\chi$ is given by $$ \int \frac{\partial R}{\partial \rho}(\cdot,\rho) \chi d\rho + \int R(x;\rho) d \chi$$ as you have already calculated. If $\rho$ is sufficiently nice, it is likely that you can rewrite the first term as an integral with respect to $\chi$ as well.