I'm reading Brezis's Functional Analysis, Sobolev Spaces and Partial Differential Equations, and trying to generalize Theorem 4.14 from $\mathbb R$ to a Hilbert space $H$. Let $(X, \Sigma, \mu)$ be a $\sigma$-finite complete measure space. Let $(H, \langle \cdot, \cdot \rangle_H)$ be a Hilbert space (over the field $\mathbb R$) and $|\cdot|$ its induced norm.
Theorem: Let $\phi \in (L_1)^*$. There exists a unique $u \in L_\infty$ such that $$ \langle \phi, f \rangle = \int \langle u, f \rangle_H \quad \forall f \in L_1. $$ Moreover, $\|u\|_\infty = \|\phi\|_{(L_1)^*}$.
Could you have a check on my attempt?
My attempt: By $\sigma$-finiteness, there is an increasing (w.r.t. inclusion) sequence $(X_n) \subset \mathcal A$ such that $\mu (X_n) < \infty$ and $\bigcup_n X_n =X$.
- Uniqueness.
Assume $u, u' \in L_\infty$ such that $$ \langle \phi, f \rangle = \int \langle u, f \rangle_H = \int \langle u', f \rangle_H \quad \forall f \in L_1. $$
Then $$ \int \langle u-u', f \rangle_H = 0 \quad \forall f \in L_1. $$
We pick $f_0 = (u-u')1_{X_n}$. Because $u-u'$ is essentially bounded on $X_n$ and $\mu (X_n) < \infty$, we get $f_0 \in L_1$. So $$ \int \langle u-u', f_0 \rangle_H = 0 \iff \int |(u-u')1_{X_n}|^2 = 0. $$
It follows that $u=u'$ $\mu$-a.e. on $X_n$. Hence $u=u'$ $\mu$-a.e.
- Existence.
There exist a sequence $(\varepsilon_n) \subset \mathbb R_{>0}$ and a map $\theta \in L_2 (X, \mu, \mathbb R)$ such that $\theta (x) = \varepsilon_n$ for all $x \in X_n$. By Hölder inequality, $\theta f \in L_1$ for all $f \in L_2$. The map $L_2 \to \mathbb R, f \mapsto \langle \phi, \theta f\rangle$ is linear continuous. By this result, we have $(L_2)^* = L_2$. So there is $v \in L_2$ such that $$ \langle \phi, \theta f\rangle = \int \langle v, f \rangle_H \quad \forall f \in L_2. $$
Let $u := v /\theta$. Then $u$ is $\mu$-measurable. For all $g \in L_\infty$, we have
- $\frac{g1_{X_n}}{\theta} \in L_2$. This is because $\theta$ is constant on $X_n$, $g$ is essentially bounded on $X_n$, and $\mu(X_n) < \infty$.
- $g1_{X_n} \in L_1$. This is because $g$ is essentially bounded on $X_n$ and $\mu(X_n) < \infty$.
Hence for all $g \in L_\infty$ and $n \in \mathbb N$, we have $$ \langle \phi, g1_{X_n} \rangle = \left \langle \phi, \theta \left ( \frac{g1_{X_n}}{\theta} \right )\right \rangle = \int \left \langle v, \frac{g1_{X_n}}{\theta} \right \rangle_H = \int \left \langle \frac{v}{\theta}, g1_{X_n} \right \rangle_H = \int \langle u, g1_{X_n} \rangle_H. $$
Next we prove that $\|u\|_\infty \le \|\phi\|_{(L_1)^*}$. Fix any $C> \|\phi\|_{(L_1)^*}$ and let $A := \{x \in X | C < |u(x)|\} \in \mathcal A$. Notice that $u > 0$ on $A$. Let $g_0 := 1_{A} \frac{u}{|u|} \in L_\infty$. Then
- $$ |\langle \phi, g_0 1_{X_n} \rangle| \le \|\phi\|_{(L_1)^*} \| g_0 1_{X_n}\|_1 = \|\phi\|_{(L_1)^*} \left \| \frac{u}{|u|} 1_{A\cap X_n} \right \|_1 = \|\phi\|_{(L_1)^*} \mu (A \cap X_n). $$
- $$ \int \langle u, g_0 1_{X_n} \rangle_H = \int \left \langle u, \frac{u}{|u|}1_{A \cap X_n} \right \rangle_H = \int |u|1_{A \cap X_n} \ge C \mu (A \cap X_n). $$
It follows that $$ C \mu (A \cap X_n) \le \|\phi\|_{(L_1)^*} \mu (A \cap X_n). $$
Hence $\mu (A \cap X_n) = 0$ and thus $\mu (A) = \lim_n \mu (A \cap X_n) = 0$. So $A$ is a $\mu$-null set. We claim that $$ \langle \phi, h\rangle = \int \langle u, h\rangle_H \quad \forall h \in L_1. $$
Let $T_n$ be the truncation operator, i.e., $(T_n h) (x) = h (x)$ if $|h (x)| \le n$ and $(T_n h) (x) = n h(x)/ |h(x)|$ otherwise. Then $(T_n h)1_{X_n} \to h$ in $L_1$ by dominated convergence theorem. On the other hand, $T_n h \le n$ and thus $T_n h \in L_\infty$. As such, $$ \langle \phi, (T_nh)1_{X_n}\rangle = \int \langle u, (T_nh)1_{X_n} \rangle_H. $$
Notice that the map $L_1 \to \mathbb R, h \mapsto \int \langle u, h \rangle$ is linear continuous. Hence $$ \langle \phi, h \rangle = \lim_n \langle \phi, (T_nh)1_{X_n}\rangle = \lim_n \int \langle u, (T_nh)1_{X_n} \rangle_H = \int \left \langle u, \lim_n (T_nh)1_{X_n} \right \rangle_H = \int \langle u, h \rangle_H. $$
Finally, we have $$ |\langle \phi, h \rangle| \le \int |u| |h| \le \|u\|_\infty \int |h| = \|u\|_\infty \|h\|_1. $$
Hence $\|\phi\|_{(L_1)^*} \le \|u\|_\infty$. This completes the proof.