# $\int_\Omega |\nabla u^+|^2 \, dx$ is not differentiable with respect to $u$ in $W_0^{1,2}(\Omega)$

Let $u \in W_0^{1,2}(\Omega)$, where $\Omega$ is some domain in $\mathbb{R}^N$, $N \geq 1$.

Denote $u^+ := \max\{u, 0\}$. (It is know that $u^+$ also belongs to $W_0^{1,2}(\Omega)$ (see, e.g., Theorem A.1 in Kinderlehrer-Stampacchia).

Consider now the functional $$u \mapsto \int_\Omega |\nabla u^+|^2 \, dx.$$ It is noted in the article of M. Clapp & T. Weth, p.4, that this functional is not differentiable (in the Fréchet sense) in $W_0^{1,2}(\Omega)$. Unfortunately, they don't provide any reference for a counterexample.

I reinvented the wheel, and constructed a simple 1D counterexample, using the function $u(x) \approx x^\alpha$ near $0$, where $\alpha \in (0,1)$. But, obviously, somewhere should be a published result concerning this non-differentiability, which I can cite.

Maybe somebody met the same question and can provide the reference?

Thanks!

P.S. This functional is differentiable for $u \in W_0^{1,2}(\Omega) \cap W^{2,2}(\Omega)$ (see T. Bartsch, T. Weth, Lemma 3.1, p.7).

Take $\Omega=(0,1)$, define $u_t(x) = x-t$. Set $$f(t) := \int_\Omega |\nabla u_t^+|^2 dx.$$ Then it holds $f(t) =1$ for all $t\le 0$, but $f(t) = 1-t$ for $t>0$. If the functional would have been differentiable, so would have been $f$. Thus, the functional is not differentiable.
Edit: To obtain a counter-example in $W-^{1,2}$, my proposal would be: $\Omega=(-2,2)$. Set $u(x)=|x|$ and $h(x) =1$ for $x\in(-1,1)$. Then extend $u$ and $h$ such that $u,h\in W^{1,2}_0$, $u,h\in C^2(\Omega \setminus[-1,1])$, $u\ge0$ and $h\ge0$, $u'(-2) >0$, $u'(2)<0$, $h(-2)=h'(2)=0$.
Then $u+th$ should be still non-negative on $\Omega \setminus[-1,1]$ for $|t|$ small. And the integral behaves non-smooth as in the $W^{1,2}$ example above.
• Thanks for your example! However, it is not in $W_0^{1,2}(\Omega)$ space. Is there a simple extension of $u_t(x)$ to $W_0^{1,2}(\Omega)$, which keeps non-differentiability? – Voliar Jan 12 '15 at 18:58