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IIf If $f:U_1\to\mathcal L^p(\mu;E_2)$ is Fréchet differentiable, can we say anything about the Fréchet differentiability of $u\mapsto f(u)(\omega)$?

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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$$ for $\omega\in\Omega$ (outside a $\mull$$$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).

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 $\mull$-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).

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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IIf $f:U_1\to\mathcal L^p(\mu;E_2)$ is Fréchet differentiable, can we say anything about the Fréchet differentiability of $u\mapsto f(u)(\omega)$?

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 $\mull$-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).