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Let $(\Omega,\mathcal A,\mu)$ be a $\sigma$-finite measure space, $p\ge1$, $E_i$ be a $\mathbb R$-Banach space, $U_1\subseteq E_1$ be open and $f:U_1\to L$ be Fréchet differentiable at $x\in U_1$, where $L=\mathcal L^p(\mu;E)$ or $L=L^p(\mu;E)$.

For a function $f$ of this form, can we say anything about the Fréchet differentiability of $$U_1\to E_2\;,\;\;\;u\mapsto f(u)(\omega)\tag1$$ for $\omega\in\Omega$ (outside a $\mu$-null set)?

Strictly speaking, the notion of Fréchet differentiability is not defined for $L=\mathcal L^p(\mu;E_2)$, since $\mathcal L^p(\mu;E_2)$ is only a semi-normed space and hence limits are not unique (only up to a $\mu$-null set). However, the assumption could be understood in the sense that the convergence in the definition of Fréchet differentiability holds.

On the other hand, if $L=L^p(\mu;E)$, we've got the problem that the elements of $L^p(\mu;E_2)$ are only equivalence classes and hence the expression $g(\omega)$ is not well-defined for $g\in L^p(\mu;E_2)$, unless we pick a particular representative. Maybe we need the existence of continuous representatives. So, maybe the desired conclusion is possible if $L^p(\mu;E)$ is replaced by a suitable closed subspace (such as a Sobolev space).

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  • $\begingroup$ There are two dollar sign missing in the text below formula (1). $\endgroup$
    – Daniele Tampieri
    Commented Jul 24, 2020 at 11:03
  • $\begingroup$ @DanieleTampieri Thanks for noting. Didn't saw that for some reason. $\endgroup$
    – 0xbadf00d
    Commented Jul 24, 2020 at 12:25
  • $\begingroup$ This is very similar to "convergence in $L^1$ $\implies$ convergence almost everywhere?", and it seems you can use this to produce counterexamples. FWIW, one situation where I think the implication holds, is when $\Omega$ has a nice topology, $E_1 = \mathbb C$, $f$ is jointly continuous and $L^p$-holomorphic. When you then write $f(u, \omega) - f(0, \omega) = u f'(0, \omega) + u R(u, \omega)$; you can use Cauchy's integral formula to prove that $f'$ and $R$ are jointly continuous, and then $R \to 0$ in $L^p$ implies $R \to 0$ everywhere; i.e. $f$ is pointwise holomorphic (=for fixed $\omega$). $\endgroup$
    – Bart Michels
    Commented Jul 26, 2020 at 11:53
  • $\begingroup$ @BartMichels Convergence in $L^1$ at least implies convergence in measure and hence convergence almost everywhere along a subsequence. $\endgroup$
    – 0xbadf00d
    Commented Jul 26, 2020 at 13:57

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