Frechet derivative of $\delta_{x(t)}$ is $\delta_{x'(t)}$?

Question

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 $$\frac{\|\delta_{x(t+h)}-\delta_{x(t)}-h\delta_{x'(t)}\|_{\text{TV}}}{|h|}\to0\text{ when }h\to 0$$

Answer 1

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, $\lim_{h\rightarrow0}\mu_h=-\delta_{\mathbf{0}}\neq0$

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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 $\lim_{h\rightarrow0}\nu_{t,h}$.

Answer 2

Technicalities aside (see comments to the main question), I don't think the result is correct. Take an arbitrary continuous function $f$. You can pair it with that $\delta_{x(t)}$ and the result is $$\tag{1}\langle \delta_{x(t)}, f\rangle = f(x(t)).$$ Now, if $\delta_{x(t)}$ was differentiable, whatever that means, it MUST imply that the left-hand side of (1) is a differentiable real-valued function of $t$. But this is not the case, because the function $f$ is just a generic continuous function, it needs not be differentiable at any point.

Answer 3

To find the derivative (if it exists) we have to consider the measure $$ \frac{\delta_{x(t+h)}-\delta_{x(t)}}{h}\quad\text{given by}\quad A\mapsto \frac 1 h \cdot \begin{cases} 1 & \text{if $x(t+h)\in A$, $x(t)\notin A$}, \\ -1 & \text{if $x(t+h)\notin A$, $x(t)\in A$}, \\ 0 & \text{otherwise}. \end{cases} $$ Claiming that this would converge to $\delta_{x'(t)}$ as $h\to 0$ would mean this measure converges to $$ A\mapsto \begin{cases} 1 & \text{if $x'(t)\in A$}, \\ 0 & \text{otherwise}. \end{cases} $$ You could check this by computing the total variation of the difference, but it should be rather obvious that this isn't the case and the first measure above does not converge for $h\to 0$.

Answer 4

For a very different perspective we have the following result: the tangent vector of the curve $t \mapsto \delta_{x(t)}$ at a point $s$ is $\delta_{x'(s)}$ provided $x$ is absolutely continuous (and hence differentiable almost everywhere).

Here is what I mean by tangent vector (for a detailed treatment you have to consult Ambrosio, Savaré and Gigli's book: Gradient flows in metric spaces and in the space of probability measures or the PhD Thesis of Nicolá Gigli (On the geometry of the space of probability measures endowed with the quadratic optimal transport distance), available on his website).

Let $P_2(X)$ be the set of Radon measures on $X$ with finite second moment and $I \subset \mathbb R$ be an interval open to the right.

Definition 1 (Tangent vector). A tangent vector to a curve $\gamma \colon I \to P_2(\mathbb R^d)$ at $t \in I$ is a "velocity plan" $v_t \in T_{\gamma(t)} P_2(\mathbb R^d)$ such that for every $v_{t, h} \in \exp_{\gamma(t)}^{-1}(\gamma(t + h))$, $h > 0$ we have \begin{equation*} \lim_{h \searrow 0} W_{\gamma(t)}\left(v_t, \frac{1}{h} v_{t, h}\right) = 0, \end{equation*} that is, $v_t = \lim_{h \searrow 0} v_{t, h}$ with respect to $W_{\gamma(t)}$.

Let me give some explanations for all the notations: the space $T_{\gamma(t)} P_2(\mathbb R^d)$ is a sort of "geometric" tangent space of $P_2(\mathbb R^d)$ at $\gamma(t)$. In the case of $\gamma(t) = \delta_{x(t)}$ we have $T_{\gamma(t)} P_2(\mathbb R^d) = \{ \gamma(t) \otimes \nu: \nu \in P_2(\mathbb R^d) \}$.

Further, $W_{\gamma(t)}$ is a Wasserstein-2-type metric on measures of the form $\gamma(t) \otimes \mu$ ("fixed base measure $\gamma(t)$").

Next, for a constant $c > 0$ and a measure $\mu \in P_2(\mathbb R^d \times \mathbb R^d)$, we define $c \cdot \mu := (\pi_1, c \cdot \pi_2)_{\#} \mu$, where $\pi_j \colon \mathbb R^d \to \mathbb R^d \to \mathbb R^d$, $(x_1, x_2) \mapsto x_j$ are the projections onto the $j$-th component for $j \in \{ 1, 2 \}$ and $\#$ is the pushforward of measures.

Lastly, $\exp_{\gamma(t)}^{-1}\big( \gamma(t + h) \big)$ is the inverse of a suitably chosen exponential map (this makes sense since $P_2(\mathbb R^d)$ behaves similarly to an infinite-dimensional Riemannian manifold). In the case of a curve of Diracs, like here, it is simply equal to $ \left\{ \gamma(t) \otimes \delta_{\{ x(t + h) - x(t) \}} \right\}$, so that $$\frac{1}{h} v_{t, h} = \gamma(t) \otimes \delta_{\left\{ \frac{x(t + h) - x(t)}{h} \right\}} \xrightarrow[W_{\gamma(t)}]{h \searrow 0} \gamma(t) \otimes \delta_{\{ x'(t) \}}.$$