Fréchet derivative of evaluation-like functional (multivariate)

Question

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).

Answer 1

The general procedure for the identification of a Fréchet derivative is the following

1. Calculate the functional derivative 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", 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.