Is the following function Gâteaux-differentiable Like I said in the header, I want to know, if this function has a Gâteaux-derivative (or at least a Newton-derivative):
\begin{align*}
f\colon L^2(\Omega) &\to L^2(\Omega) \\
v &\mapsto \begin{cases}
v(x), \text{ if } |v(x)| < 1 \\
\frac{v(x)}{|v(x)|}, \text{else} 
\end{cases}
\end{align*}
Here $\Omega \subset \mathbb{R}^n$ is a bounded and Lipschitz. I have no idea how to start...
 A: You can define your map for any Banach space, and it will never be differentiable. Let $\nu$ be a norm one vector, consider the smooth path $\gamma(t) = t\,\nu$. Then $f(\gamma(t))= \nu\cdot\min(1,t) $ for $t≥0$. At $t=1$ you would have:
$$d_{\gamma}f\lvert_{t=1} =\nu\lim_{h\to0} \frac{\min(1,1+h)-1}{h}$$
which does not exist!
A: You have $f(v)(x) = \sigma(v(x))$ where $\sigma(y) = {1 \over \max(1,\|y\| )} y$.
Then $f$ has a Gâteaux derivative at $v$ iff the set $E=\{ x \in \Omega | \|v(x)\| = 1 \}$ has measure zero. 
Note that $\sigma(y)$ is the projection of the point $y$ onto the
closed unit ball and is Lipschitz with rank one. In particular, we
have $\| {f(v+th)(x)-f(v)(x) \over t} \| \le \|h(x)\|$
Note that if $\|y\| <1$ then $D\sigma(y)\delta = \delta$ and if
$\|y\| >1$ then
$D \sigma(y)\delta = ({1 \over \|y\|}I-{1 \over \|y\|^3} y y^T)\delta$.
If $E$ has measure zero, then choose some direction $h$. We see that
${f(v+th)(x)-f(v)(x) \over t} \to D \sigma(u(x)) h(x)$ for all $x \notin E$. A little dominated convergence shows that
$d f(u, h) (x)= D \sigma(u(x)) h(x)$ for ae. $x$.
If $E$ has strictly positive measure, then choose $h$ defined by
$h(x) = 1_E(x) u(x)$. For $t >0$ we have ${f(v+th)(x)-f(v)(x) \over t} = 0$ and for $t <0$ (and $t <-1$) we have
${f(v+th)(x)-f(v)(x) \over t} = h(x)$. Hence
$\lim_{t \uparrow 0} {f(v+th)-f(v) \over t} = h$ and
$\lim_{t \downarrow 0} {f(v+th)-f(v) \over t} = 0$ from which it follows that the Gâteaux derivative does not exist.
