1
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Voliar Jan 12 '15 at 18:58
  • $\begingroup$ Thank you again. Don't think that I'm finding fault, but that is why I'm asking for a reference - to avoid cumbersome piece of such calculations in my article, and just refer to something. (Eventually, I constructed a counterexample by myself; but your example is definitely shorter). $\endgroup$ – Voliar Jan 13 '15 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.