In a Hilbert Space $V$, for function $f:V\to\mathbb{R}$, if $f$ is Fréchet differentiable at $x_0$, the Fréchet Derivative $\nabla f(x_0)$ is $v$ such that $$ \lim_{x\to x_0} \frac{|f(x) – f(x_0) – \langle v, x-x_0 \rangle|}{\|x-x_0\|} = 0$$
As an example, for $f(x) = \|x\|$, the Fréchet derivative is $\nabla f(x) = \frac{x}{\|x\|}$, as shown in this Wikipedia proof.
Now my question is, what is the Fréchet derivative of $f(x) = \frac{1}{\|x\|}$. My gut says something like $-\frac{x}{\|x\|^3}$ following normal derivative rules, but I can't seem to be able to solve it.
Best Answer
You can apply the product rule on the relation $$1 = \|\cdot\| \cdot \frac1{\|\cdot\|}$$
to obtain
$$D1(x_0) = \frac1{\|x_0\|}D(\|\cdot\|)(x_0) + \|x_0\|D\left(\frac1{\|\cdot\|}\right)(x_0)$$ $$0 = \frac{x_0}{\|x_0\|^2} + \|x\|D\left(\frac1{\|\cdot\|}\right)(x_0) \implies D\left(\frac1{\|\cdot\|}\right)(x_0) = -\frac{x_0}{\|x_0\|^3}.$$