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I am trying to consider hypothesis on $g$ such that the operator $$ H_0^1 (\Omega) \to L^2(\Omega), \qquad v \mapsto g(v) $$ is $\mathcal C^1$. As additional hypothesis $\Omega$ is bounded and $g(0) = 0$.

I believe that $g \in W^{2,\infty} (\mathbb R)$ is sufficient, but I am quite stuck on the proof.

Thanks in advance,

D

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  • $\begingroup$ It seems like you're on the right track. It's just the chain rule, no? $\endgroup$
    – Deane Yang
    Sep 21, 2014 at 18:38
  • $\begingroup$ Basically, yes. But you have to proof the convergence $$ \frac{g(v + w) - g(v)}{|w|} \overset{ L^2 }\to g'(v) w, \quad \text{as } w \overset{H^1} \to 0 $$ which presents some technical difficulties. $\endgroup$
    – D G
    Sep 21, 2014 at 18:47

2 Answers 2

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In the realm of Sobolev spaces, if $k>\frac{\dim(F)}2$, for the composition mapping $H^{k+l}(F,\mathbb R) \times H^k(F,F) \to H^k(F,\mathbb R)$, left translations are $C^l$ and right translations are smooth; i.e., composition is $C^l$ in the right hand side variable, and is smooth in the left hand side variable. This is folklore; for a detailed recent proof see

  • H. Inci,T. Kappeler and P. Topalov, On the Regularity of the Composition of Diffeomorphisms, Memoirs of the American Mathematical Society, vol. 226 (American Mathematical Society, 2013).

In your case we have a left translation, but you are not above the Sobolev threshold with $v$, in general.

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  • $\begingroup$ Thank you for your answer. I will go on looking for translations instead of Nemitskii operators and I am sure the reference you gave me (which I have already shifted through) will be very useful in the future. However, I need far less regularity on the result, since $g \circ v$ need only be $L^2$. Ultimately my aim is to use $\langle g(v), \cdot \rangle_{L^2}$ as an element in $H^{-1}$ for a weak formulation of a PDE so conditions of the type $s > \frac n 2$ are not of much use, as $s$ should be $2$. I would also like Lipschitz-like conditions on $g$. $\endgroup$
    – D G
    Sep 21, 2014 at 19:19
  • $\begingroup$ Hello can you please help on the question I asked here mathoverflow.net/q/337841/108824 $\endgroup$
    – Red shoes
    Aug 16, 2019 at 15:08
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${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$ Part 1 of Nemytskii operator differentiability Part 2 of Nemytskii operator differentiability

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  • $\begingroup$ In the proof of Lemma 2.1 in place of the exponent $\frac 12\,(q-2)$ there should be $q^{-1}(q-2)\,$, and in the proof of Theorem 2.2 in place of the zero vector $\boldsymbol 0_E$ of $E$ there should be the representative $\boldsymbol 0=\Omega\times\{\kern.3mm 0\kern.3mm\} \,$. $\endgroup$
    – TaQ
    Jul 25, 2016 at 19:04
  • $\begingroup$ I believe globally bounded and Lipschitzian are quite strong assumptions, and can be relaxed. $\endgroup$
    – Saj_Eda
    Sep 27, 2018 at 18:52
  • $\begingroup$ Do you have any idea for a proof with the "relaxed" assumptions, and what would they be? The assumptions in Theorem 2.2 are just a bit weaker than being in $W^{2,+\infty}(\mathbb R)$ that the OP believed sufficient. $\endgroup$
    – TaQ
    Dec 18, 2018 at 10:33

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