If $f:U_1tomathcal L^p(mu;E_2)$ is Fréchet differentiable, can we say anything about the Fréchet differentiability of $umapsto f(u)(omega)$?

Let $(Omega,mathcal A,mu)$ be a $sigma$-finite measure space, $pge1$, $E_i$ be a $mathbb R$-Banach space, $U_1subseteq E_1$ be open and $f:U_1to L$ be Fréchet differentiable at $xin U_1$, 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 $$U_1to E_2;,;;;umapsto f(u)(omega)tag1$$ for $omegainOmega$ (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 $gin L^p(mu;E_2)$, 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).