Fréchet derivative of evaluation-like functional (multivariate) I'm fairly new to functional calculus but and posting here since the question seems more appropriate than for MSE.  When coming across this post I could not help but wonder the following.
Let $H$ be the reproducing-kernel Hilbert space obtained by completing the set of all $C^1(\mathbb{R}^n,\mathbb{R}^m)$ with finite norm finite:
$$
\|f(x)\|_H:= \|f(0)\|_{\mathbb{R}^m} + \int_{x \in \mathbb{R}^n} \|(\nabla f)(x)\|_{\mathbb{R}^m} e^{-\|x\|} dx.
$$
If $w:C^1(\mathbb{R}^n,\mathbb{R}^m)\times \mathbb{R}^n\rightarrow [0,\infty)$ is the functional
$$
(f,x) \mapsto \|f(x)\|_{\mathbb{R}^m},
$$
what is its Fréchet derivative?  Thinking analogously to the linked post and appealing to the chain-rule for Fréchet derivatives, I would guess it is
$$
Dw(f,x) (g,y)= \frac1{\|g(x)+J_f(y)\|}\left(g(x) + (J_f)(y)\right).
$$
However, I don't know how to show more than this (if even it is a correct ansatz).
 A: The general procedure for the identification of a Fréchet derivative is the following

*

*Calculate the functional derivative of the given functional, then

*verify its linearity and

*verify its continuity respect to the topology that is considered on the domain of the given functional i.e., for a Banach or Hilbert space topology, verify that the norm of the derivative does not depend on the structure of the variation but only on its size (norm).

The functional $w$ is defined on a vector space structure defined on $C^1(\Bbb R^n, \Bbb R^m)\times \Bbb R^n$, since we should be able to give a meaning to the word "linear", and the topology considered on this vector space is the product topology between the Banach space topology on $\Bbb R^n$ and the topology by $H$ on $C^1(\Bbb R^n,\Bbb R^m)$: following the above list we have
$$
\begin{split}
Dw\big[(f,x)\big]\big((g,y)\big) & \triangleq \frac{\mathrm{d}}{\mathrm{d}\varepsilon}w\big[(f,x)+\varepsilon(g,y)\big]\bigg{|}_{\varepsilon = 0}\\
&=\frac{\mathrm{d}}{\mathrm{d}\varepsilon}w\big[(f+\varepsilon g, x+\varepsilon y)\big]\bigg|_{\varepsilon = 0}\\
&=\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\big\|f(x+\varepsilon y)+\varepsilon g(x+\varepsilon y)\big\|\bigg|_{\varepsilon = 0}\\
&=\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\bigg[\sum_{i=1}^m \Big(f_i(x+\varepsilon y)+\varepsilon g_i(x+\varepsilon y)\Big)^2\bigg]^\frac{1}{2}\right|_{\varepsilon = 0}\\
&=\frac{1}{2}{\big\|f(x+\varepsilon y)+\varepsilon g(x+\varepsilon y)\big\|}^{-1} \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\bigg[\sum_{i=1}^m \Big(f_i(x+\varepsilon y)+\varepsilon g_i(x+\varepsilon y)\Big)^2\bigg]\right|_{\varepsilon = 0}\\
&=\frac{1}{2}{\big\|f(x+\varepsilon y)+\varepsilon g(x+\varepsilon y)\big\|}^{-1}\\
&\qquad\cdot \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\bigg[\sum_{i=1}^m  f_i^2(x+\varepsilon y)+ 2\varepsilon f_i(x+\varepsilon y)g_i(x+\varepsilon y) +\varepsilon^2 g_i^2(x+\varepsilon y)\bigg]\right|_{\varepsilon = 0}\\
&={\big\|f(x+\varepsilon y)+\varepsilon g(x+\varepsilon y)\big\|}^{-1}\\
&\qquad\cdot \bigg[\sum_{i=1}^m  \langle\nabla f_i(x+\varepsilon y),y\rangle+  f_i(x+\varepsilon y)g_i(x+\varepsilon y) \\ 
&\qquad\qquad +\varepsilon\langle\nabla f_i(x+\varepsilon y),y\rangle g(x+\varepsilon y) +\varepsilon f_i(x+\varepsilon y) \langle \nabla g_i(x+\varepsilon y),y\rangle \\
&\qquad\qquad\qquad  + \varepsilon g_i^2(x+\varepsilon y)
+ \left.\varepsilon^2 g_i(x+\varepsilon y)\langle \nabla g_i(x+\varepsilon y),y\rangle\bigg]\right|_{\varepsilon = 0}\\
&={\big\|f(x)\big\|}^{-1}\bigg[\sum_{i=1}^m  \langle\nabla f_i(x),y\rangle
+f_i(x)g_i(x)\bigg]=\frac{\langle 1, \mathbf{J}_f(x)y\rangle+\langle f(x),g(x)\rangle }{\big\|f(x)\big\|}
\end{split}
$$
Thus, apart from errors, we have done step 1 and checked the linearity as required by step 2. Regarding step 3, we see that that if
$$
\|f(x)\|>0 \iff f(x)\neq 0
$$
for the given $x\in\Bbb R^n$, then the functional derivative norm depend only on the value $\|g(x)\|_H+\|y\|_{\Bbb R^n}$ and not on the structure of the element $(g,y)$. Thus the functional derivative of $w$ is a Fréchet derivative.
