Distributional derivative of the norm function

Question

Let $H$ be a real finite-dimensional Hilbert space. My question is simple: what is the distributional derivative of the norm function $\|\cdot\|: H\to[0,+\infty)$?

I know that when $H=\mathbb{R}$, then $\frac{d}{dx}|x|=2\Theta(x)-1$. But I am curious to know what happens in the general case? Especially, I am very interested to know what is the distributional derivative of $\|\cdot\|$ at zero in a general finite-dimensional Hilbert space.

Answer 1

The norm function is weakly differentiable with weak gradient $\frac{x}{\lVert x\rVert}$ (extend to $0$ by whichever value you like). Indeed, if $\Phi\in C_c^\infty(\mathbb{R}^n;\mathbb{R}^n)$, then we get from the divergence theorem $$ \int_{\|x\|>\epsilon} \|x\|\nabla\cdot\Phi(x)\,dx=-\int_{\|x\|>\epsilon}\frac{x}{\|x\|}\cdot \Phi(x)\,dx-\int_{\|x\|=\epsilon}\|x\|\frac{x}{\|x\|}\cdot\Phi(x)\,d\sigma(x). $$ Observe that $$ \left\lVert\int_{\|x\|=\epsilon}x\cdot\Phi(x)\,d\sigma(x)\right\rVert\leq \epsilon \sigma(\{\|x\|=\epsilon\})\|\Phi\|_\infty, $$ where the surface area $\sigma(\{\|x\|=\epsilon\})$ is given by $C_d\epsilon^{d-1}$ with some dimensional constant $C_d$. Hence this term goes to zero as $\epsilon\searrow 0$.

Thus we can take the limit as $\epsilon\searrow 0$ in the first equation to get $$ \int_{\mathbb{R}^n}\|x\|\nabla\cdot\Phi(x)\,dx=-\int_{\mathbb{R}^n}\frac{x}{\|x\|}\cdot\Phi(x)\,dx. $$

Answer 2

Notation: The directional derivative is $$D_v f(x)=\frac\partial {\partial t}f(x+tv)|_{t=0}.$$

Hint: Since $$D_v||x||^2=2||x||D_v||x||$$it's enough to find $D_v||x||^2$. Write $||x+tv||^2$ in terms of the inner product, optionally expand, and differentiate.