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Let $(\Omega, \Sigma, \mu)$ a measure space, $1 \leq p < \infty$ and $X$, $Y$ Banach spaces. Let $F: \Omega \to L(X,Y)$ a function such that $\omega \mapsto \varphi(F(\omega)x)$ is in $L^p(\Omega)$ for all $x \in X$ and $\varphi \in Y^*$. I want to show that there exists a $c \geq 0$ with $$\left(\int_\Omega \vert \varphi(F(\omega)x) \vert^p \, d\mu(\omega)\right)^{1/p} \leq c \Vert x \Vert_{X} \Vert \varphi \Vert_{Y^*}$$ for all $x \in X$ and $\varphi \in Y^*$.

I got the hint that I would need the closed graph theorem. But I don't see how to apply that. I tried the following things:

  • Using the closed graph theorem I get that the sets $\{(x, F(\omega) x) : x \in X\}$ are closed for all $\omega \in \Omega$. But I have no idea how that might help me.
  • I easily can get $$ \left(\int_\Omega \vert \varphi(F(\omega)x) \vert^p \, d\mu(\omega)\right)^{1/p} \leq \left(\int_\Omega \Vert F(\omega)x \Vert_Y^p \, d\mu(\omega)\right)^{1/p} \Vert \varphi \Vert_{Y^*}.$$ If I could show that the operator $T:X \to L^p(\Omega, Y), x \mapsto (\omega \mapsto F(\omega) x)$ is well defined and continuous (by utilizing the closed graph theorem) I would have solved the problem. But I don't even see how its well defined honestly...
  • I tried to come up with other useful operators or functionals. But I really have no good ideas...

I would appreciate some hints or solutions to the problem :)

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Hint: Consider the mapping $$\mathbb F : X \times Y^* \to L^p(\Omega), (x, \varphi) \mapsto (\omega \mapsto \varphi(F(\omega)x)).$$

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