Let $V,H,V^*$ be a Gelfand-Triple, $\phi\colon V \to \mathbb{R}$ convex, lower semicontinuous and proper. There exists a so called Moreau-Enveloppe $\phi_j$, which is Gateâux-differentialable. It's derivative $\phi_j'\colon V \to V^*$ is demicontinuous and maps bounded sets to bounded sets in $V^*$. Now my question: Is the mapping $\tilde{\phi_j'} \colon L^2(0,T,V) \to L^2(0,T,V^*)$ defined by $\tilde{\phi_j'}(u)(t) \colon = \phi_j'(u(t))$ well-defined? And are in $L^2(0,T,V)$-bounded sets mapped to bounded sets in $L^2(0,T,V^*)$?
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$\begingroup$ Under mild conditions on $V$, $L^2(0,T;V^*)$ is the dual of $L^2(0,T;V)$, so the abstract theory should apply? $\endgroup$– Christian ClasonCommented Dec 2, 2016 at 17:10
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$\begingroup$ I don't understand what you mean with "weak theory" in this context. Why is the Moreau-Enveloppe a well-defined mapping from $L^2(0,T;V)$ to it's dual? How can we apply the boundedness of it as a mapping from $V$ to $V^*$? Thank you for your reply! $\endgroup$– malwinCommented Dec 2, 2016 at 17:58
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$\begingroup$ I mean: why can't you consider it as a mapping from $L^2(0,T;V)$ to $L^2(0,T;V^*)$? I assume that's where it is defined, otherwise the question of boundedness does not make sense. (You didn't tell us the definition of the mapping you're considering, so it's hard to understand what you mean here.) $\endgroup$– Christian ClasonCommented Dec 2, 2016 at 18:00
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$\begingroup$ I reformulated my question. Can you now understand my question? Thanks. $\endgroup$– malwinCommented Dec 2, 2016 at 18:31
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Your assumptions are not sufficient to consider this mapping property.
Consider $V = \mathbb{R}$ and the mapping $f(x) = x^3$. It maps bounded sets to bounded sets, but its Nemytskii operator $y \mapsto y^3$ is not well defined from $L^2(0,T)$ into itself, only from $L^{3p}(0,T)$ into $L^p(0,T)$, $p \ge 1$.
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$\begingroup$ You need additionally a growth condition, e.g., $|\phi_j(u)| \le |u|$. Such things you can find in the book "Appell J., Zabrejko P.P.: Nonlinear Superposition Operators". $\endgroup$– gerwCommented Dec 6, 2016 at 20:57
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$\begingroup$ Thank you for your answer. I edited some typos in my question. I wanted to ask for $\phi_j' := L^2(0,T,V) \to L^2(0,T,V^*)$. Your answer of course answers this question, too. Thanks. But could you give me a hint on how to prove this in general? $\endgroup$– malwinCommented Dec 6, 2016 at 21:02