I don't really know anything about the Fréchet derivative but I was wondering if the Fréchet derivative of $\delta_{x(t)}$ was $\delta_{x'(t)}$. More precisely, if we consider the Banach space $(\mathcal{M}(\mathbb{R}^d), \|\cdot\|_{\text{TV}})$ of signed measures with total variation norm, $t\mapsto x(t)$ is a smooth curve in $\mathbb{R}^d$, and $\delta_{x(t)}$ the Dirac measure in the point $x(t)$ and consider the map
\begin{align}
\mathbb{R}\to&\mathcal{M}(\mathbb{R}^d)\\
t\mapsto& \delta_{x(t)}
\end{align}
I was wondering if this is Frechet differentiable and if like intuition suggest we have that it's Frechet derivative at point $t$ is $\delta_{x'(t)}$. I think that what I have to verify is (is it?) that $$\frac{\|\delta_{x(t+h)}-\delta_{x(t)}-h\delta_{x'(t)}\|_{\text{TV}}}{|h|}\to0\text{ when }h\to 0$$
Best Answer
The answer is no. Consider any smooth function $x:(-1,1)\rightarrow\mathbb{R}^n$ such that $x(t)\neq x'(t)$ and $\|x'(t)\|>0$. The measure(s) $$\mu_h:= \frac{1}{h}\Big(\delta_{x(t+h)}-\delta_{x(t)}-h\delta_{x'(t)}\Big)$$ have in principle total variation $\frac{2}{|h|}+1\not\rightarrow0$ as $h\rightarrow0$. Indeed, the variation measure $|\mu|=\mu_+ + \mu_-$ is given by $$|\mu_h|=\frac{1}{|h|}\Big(\delta_{x(t+h)}+\delta_{x(t)}+|h|\delta_{x'(t)}\Big)$$
Even in the simple case of $x_c(t)\equiv \mathbf{c}$ for some $\mathbf{c}\in\mathbb{R}^n$ the anger is not:
$$\mu_h:=\frac{1}{h}\Big(\delta_{x_c(t+h)}-\delta_{x_c(t)}-h\delta_{x_c'(t)}\Big)=-\delta_{\mathbf{0}}$$ and so, $\lim_{h\rightarrow0}\mu_h=-\delta_{\mathbf{0}}\neq0$
It seems that the appropriate setting would be (weak) convergence with respect a space of smooth functions, maybe $\mathcal{C}^1(\mathbb{R}^d)$, $\mathcal{D}(\mathbb{R}^b)$, or the Schwartz space $\mathcal{S}(\mathbb{R}^n)$.
For example, if $\phi$ is continuously differentiable and $\nu_{t,h}=\frac{1}{h}(\delta_{x(t+h)}-\delta_{x(t)})$, then $$\nu_h\phi(t):=\frac{1}{h}(\phi(x(t+h))-\phi(x(t))\xrightarrow{h\rightarrow0}\phi'(x(t))x'(t)=x'(t)\delta_{x(t)}D\phi$$ where $D$ is the derivative operator acting on $\mathcal{C}^1(\mathbb{R}^n)$. From this, one can see that the space of measures $\mathcal{M}(\mathbb{R}^d)$ of finite total variation is not big enough to hold the (weak) limit $\lim_{h\rightarrow0}\nu_{t,h}$.