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](https://math.stackexchange.com/questions/2463116/frechet-derivative-of-evaluation-function) 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).



Fréchet derivative of evaluation-like functional (multivariate)
The general procedure for the identification of a Fréchet derivative is the following


1. Calculate the [functional derivative](https://mathoverflow.net/questions/349057/question-about-functional-derivatives) of the given functional, then
2. verify its linearity and
3. 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](https://math.stackexchange.com/questions/3678005/what-does-it-mean-for-something-to-be-linear)", and the topology considered on this vector space is the [product topology](https://en.wikipedia.org/wiki/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.


