I'm reading Brezis's Functional Analysis, Sobolev Spaces and Partial Differential Equations, and trying to generalize Theorem 4.10 from $\mathbb R$ to 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. Then $L_p := L_p(X, \mu, E)$ is reflexive for all $p \in (1, \infty)$.
My attempt: The proof consists of three steps.
a. $$\left \| \frac{f+g}{2} \right \|^p_p + \left \| \frac{f-g}{2} \right \|^p_p \le \frac{\|f\|_p^p + \|g\|_p^p}{2} \quad \forall f,g \in L_p, \forall p \in [2, \infty).$$
Clearly, it suffices to prove that $$\left | \frac{a+b}{2} \right |^p + \left | \frac{a-b}{2} \right |^p \le \frac{|a|^p + |b|^p}{2} \quad \forall a,b \in E.$$
Notice that we have below inequality $$ \bigg ( \sum_{i=1}^m (x_{i})^{p} \bigg)^{q} \ge \bigg( \sum_{i=1}^m (x_{i})^{q} \bigg)^{p}, \quad \forall (x_1, \ldots,x_m \ge 0), \forall (q\ge p \ge 0). $$
Hence $$ \left | \frac{a+b}{2} \right |^p + \left | \frac{a-b}{2} \right |^p \le \left [ \left | \frac{a+b}{2} \right |^2 + \left | \frac{a-b}{2} \right |^2 \right ]^{p/2} = \left [ \frac{|a|^2+|b|^2}{2} \right ]^{p/2} \le \frac{|a|^p + |b|^p}{2}. $$
The last inequality is due to the convexity of the map $\varphi:\mathbb R_{\ge 0} \to \mathbb R_{\ge 0}, x \mapsto x^{p/2}$.
b. $L_p$ is uniformly convex and thus reflexive for all $p \in (2, \infty)$.
Fix $f,g\in L_p$ such that $\|f\|_p, \|g\|_p \le 1$ and $\|f-g\|_p \ge \varepsilon >0$. By (a), we get $$ \left \| \frac{f+g}{2} \right \|_p \le \left [ 1- \left ( \frac{\varepsilon}{2} \right)^p \right ]^{1/p} = 1- \delta \quad \text{with} \quad \delta := \left \{1-\left [ 1- \left ( \frac{\varepsilon}{2} \right)^p \right ]^{1/p} \right \} >0. $$
c. $L_p$ is reflexive for all $p \in (1, 2]$.
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.
- Because $E$ is complete, $L_p$ is complete. So the image $T(L_p)$ is closed in $(L_{p'})^*$.
- Notice that $p \in (1, 2]$ implies $p' \in [2, \infty)$. By (b), $L_{p'}$ is reflexive.
- A Banach space is reflexive if and only if its dual is reflexive. Hence $(L_{p'})^*$ is reflexive.
- A closed linear subspace of a reflexive space is reflexive. Hence $T(L_p)$ and thus $L_p$ are reflexive. This completes the proof.
