I'm reading Brezis's Functional Analysis, Sobolev Spaces and Partial Differential Equations, and trying to generalize Theorem 4.11 from $\mathbb R$ to a Hilbert space $E$. Could you have a check on my attempt?
Theorem: Let $(X, \Sigma, \mu)$ be a $\sigma$-finite measure space. Let $(E, \langle \cdot, \cdot \rangle_H)$ be a Hilbert space (over the field $\mathbb R$) and $|\cdot|$ its induced norm. Let $L_p := L_p(X, \mu, E)$ and $L_{p'} := L_{p'}(X, \mu, E)$ with $\frac{1}{p} + \frac{1}{p'} = 1$. Then $(L_{p'})^* = L_{p}$ for all $p \in (1, \infty)$.
My attempt: We define an operator $T: L_p \to (L_{p'})^*$ with $\frac{1}{p} + \frac{1}{p'} = 1$ by $$ \langle Tf, g \rangle := \int_X \langle f, g\rangle_H \mathrm d \mu \quad \forall f \in L_p, g \in L_{p'}. $$
Fix $f \in L_p$. Clearly, $Tf$ is linear. By Cauchy-Schwarz and Hölder inequalities, $$ |\langle Tf, g \rangle | \le \int_X |f| \cdot |g| \mathrm d \mu \le \|f\|_p \|g\|_{p'} \quad \forall f \in L_p, g \in L_{p'}. $$
Therefore, $\|T f\|_{(L_{p'})^*} \le \|f\|_{p}$. Let $g_0 := |f|^{p-2} f$. Then $$ \int_X |g_0|^{p'} \mathrm d \mu = \int_X |f|^{(p-1)p'} \mathrm d \mu = \int_X |f|^{p} \mathrm d \mu < \infty. $$
Then $g_0 \in L_{p'}$. Also, $$ \langle Tf, g_0 \rangle = \int_X \langle f, |f|^{p-2} f\rangle_H \mathrm d \mu = \|f\|_p^p. $$
Hence $$ \frac{\langle Tf, g_0 \rangle}{\|g_0\|_{p'}} = \frac{\|f\|_p^p}{\|f\|_p^{p/p'}} = \|f\|_p. $$
Hence $\|T f\|_{(L_{p'})^*} = \|f\|_{p}$. This implies $T$ is a linear isometry. Let's prove that $T$ is surjective. Because $E$ is complete, $L_p$ is complete. So the image $T(L_p)$ is closed in $(L_{p'})^*$. So it suffices to prove that $T(L_p)$ is dense in $(L_{p'})^*$. Let $\varphi \in (L_{p'})^{**}$ such that $\langle \varphi, Tf \rangle =0$ for all $f \in L_p$. By Hahn-Banach theorem, it suffices to show that $\varphi \equiv 0$. We have $L_{p'}$ is reflexive, so $\varphi \in L_{p'}$. Then $$ \langle \varphi, Tf \rangle = \langle Tf, \varphi \rangle = \int_X \langle f, \varphi\rangle_H \mathrm d \mu = 0 \quad \forall f \in L_p. $$
We pick $f_0 := |\varphi|^{p'-2} \varphi \in L_p$. Then $\|\varphi\|_{p'}^{p'} = 0$ and thus $\varphi = 0$. This completes the proof.
