14.2 Duality of $L^{p}$ Spaces

Proof. Since $f(X)$ is separable, assume without loss of generality that $E$ is separable. By Proposition 12.4.1, there exists $\seq{z_n}\subset \bracsn{z \in F|\ \norm{z}_F \le 1}$ such that for each $y \in F$, $\norm{y}_{F} = \sup_{n \in \natp}\dpn{y, z_n}{\lambda}$.

For each $N \in \natp$ and $x \in X$, let $F_{N}(x) = 0 \vee \bigvee_{n = 1}^{N} |\dpn{f(x), z_n}{E}|$, then $0 \le F_{N} \le \norm{F}_{E}$ and $F_{N} \upto \norm{f}_{E}$ pointwise as $N \to \infty$.

For every $1 \le n \le N$, inductively define

Let $\phi_{N} = \sum_{n = 1}^{N} z_{n} \one_{A_{N, n}}$, then $|\dpn{f, \phi_N}{\lambda}| = F_{N}$, and

1. (1)

   Since $\norm{z_n}_{F} \le 1$ for all $n \in \natp$, $\norm{\phi_N}_{F} \le 1$ for each $N \in \natp$.

2. (2)

   For every $N \in \natp$, $|\dpn{f, \phi_N}{\lambda}| = F_{N}$, so $|\dpn{f, \phi_N}{\lambda}| \le \norm{f}_{E}$.

3. (3)

   As $F_{N} \upto \norm{f}_{E}$ pointwise as $N \to \infty$, $|\dpn{f, \phi_N}{\lambda}| \upto \norm{f}_{E}$ pointwise as $N \to \infty$.

$\square$

Proof. For each $A \in \cm$, define

Let $f, g \in L^{p}(X; E)$ and $A, B \in \cm$ such that

1. (1)

   $X = A \sqcup B$.

2. (2)

   $f|_{B} = 0$, and $g|_{A} = 0$.

3. (3)

   $\norm{f}_{L^p(X; E)}= \norm{g}_{L^p(X; E)}= 1$.

Suppose that $p \in (1, \infty)$, then for each $t \in \real$,

Since $p > 1$, for each $t \in (0, 1)$, $(1 - t)^{p - 1}, t^{p - 1}< 1$, so $\norm{(1 - t)f + tg}_{L^p(X; E)}^{p} < 1$.

On the other hand, if $p = \infty$, then

Therefore for any $A, B \in \cm$ with $A \cap B = \emptyset$, $\norm{\phi_A}_{L^p(X; E)^*}> 0$, and $\norm{\phi_B}_{L^p(X; E)^*}> 0$,

Now, by density of simple functions,

Let $\seq{A_n}\subset \cm$ such that $\mu(A_{n}) < \infty$ for all $n \in \natp$, and $\norm{\phi_{A_n}}_{L^p(X; E)^*}\upto \norm{\phi}_{L^p(X; E)^*}$ as $n \to \infty$. Let $A = \bigcup_{n \in \natp}A_{n}$, then $A$ is $\sigma$-finite and $\norm{\phi_A}_{L^p(X; E)^*}= \norm{\phi}_{L^p(X; E)^*}$. By maximality, there exists no $B \in \cm$ with $B \cap A = \emptyset$ and $\norm{\phi_B}_{L^p(X; E)^*}> 0$, so $\phi = \phi_{A}$.$\square$

Proof, [Proposition 6.13, Theorem 6.14, Fol99]. (1): By Lemma 14.2.2, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^{p}(X; E)$,

Therefore $g|_{A^c}= 0$ almost everywhere, and $\bracs{g \ne 0}$ is $\sigma$-finite.

(2, truncated): First suppose that $g \in L^{q}(X; F)$. Assume without loss of generality that $\norm{g}_{L^q(X; F)}= 1$.

By Lemma 14.2.1, there exists $\seq{\phi_n}\subset \Sigma(X, \cm; E)$ such that:

1. (1)

   For each $n \in \natp$, $\norm{\phi_n}_{E}\le 1$.

2. (2)

   For every $n \in \natp$, $0 \le |\dpn{g, \phi_n}{\lambda}| \le \norm{g}_{F}$.

3. (3)

   $|\dpn{g, \phi_n}{\lambda}| \upto \norm{g}_{F}$ pointwise as $n \to \infty$.

(2, a, truncated): Let $\Phi(x) = \norm{g(x)}_{F}^{q - 1}$^[1] for each $x \in X$. If $p < \infty$, then by Lemma 14.1.4,

Otherwise, $\Phi = 1$, and $\norm{\Phi}_{L^\infty(X; \real)}= 1$.

Let $\seq{\Phi_n}\subset \Sigma(X, \cm; E) \cap L^{p}(X; E)$ be non-negative such that $\Phi_{n} \upto \one_{\bracs{g \ne 0}}\cdot \Phi$. For each $n \in \natp$, $\Phi_{n} \phi_{n} \in \Sigma(X, \cm; E) \cap L^{p}(X; E)$ with $\norm{\Phi_n \phi_n}_{L^p(X; E)}\le 1$. By assumption,

Let $f_{n} = \Phi_{n} \phi_{n} \cdot \ol{\sgn \dpn{\phi_n, g}{H}}$, then by the Monotone Convergence Theorem,

(2, b, truncated): Let $\alpha \in (0, 1)$, then $\mu\bracs{\norm{g}_F \ge \alpha}> 0$. Since $\mu$ is semifinite, there exists $A \in \cm$ with $A \subset \bracs{\norm{g}_F \ge \alpha}$ and $0 < \mu(A) < \infty$. For each $x \in X$, let $\Phi(x) = \one_{A}/\mu(A)$, then $\norm{\Phi}_{L^1(X; \real)}= 1$.

For every $n \in \natp$, let $f_{n} = \Phi \phi_{n} \cdot \ol{\sgn \dpn{\phi_n, g}{H}}$, then $\norm{f_n}_{L^1(X; E)}\le 1$, and by the Monotone Convergence Theorem,

As the above holds for all $\alpha \in (0, 1)$, $\norm{\phi_g}_{L^1(X; E)^*}= 1$.

(2, general): If (a) holds, then $\bracs{g \ne 0}$ is $\sigma$-finite. In both cases, there exists $\seq{g_n}\subset L^{q}(X; F)$ such that

1. (1)

   $\norm{g_n}_{F}\upto \norm{g}_{F}$ as $n \to \infty$.

2. (2)

   For each $n \in \natp$, $\norm{\phi_{g_n}}_{L^p(X; E)^*}\le \norm{\phi_g}_{L^p(X; E)^*}$.

By the truncated case, for each $n \in \natp$,

If $q < \infty$, then the Monotone Convergence Theorem implies that $\norm{g}_{L^q(X; F)}\le \norm{\phi_g}_{L^p(X; E)^*}$. Otherwise,

The above argument shows that the truncation argument was technically not required. By applying the truncated case again, $\norm{g}_{L^q(X; F)}= \norm{\phi_g}_{L^p(X; E)^*}$.$\square$

Proof, [Theorem 6.15, Fol99]. By Theorem 14.2.3, the given map is isometric. Thus it is sufficient to show that it is surjective. Let $\phi \in L^{p}(X; H)^{*}$.

(Finite): First suppose that $\mu$ is finite, then $\Sigma(X, \cm; H) \subset L^{p}(X; H)$, and $\phi$ induces an $H$-valued measure on $(X, \cm)$, absolutely continuous with respect to $\mu$. By the Radon-Nikodym Theorem, there exists $g \in L^{1}(X; H)$ such that for each $f \in \Sigma(X, \cm; H)$,

By Theorem 14.2.3, $g \in L^{q}(X; H)$.

(Arbitrary): In the case of (a), by Lemma 14.2.2, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^{p}(X; H)$, $\dpn{f, \phi}{L^p(X; H)}= \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a $\sigma$-finite set satisfying the same restriction condition.

Let $\seq{A_n}\subset \cm$ such that $\mu(A_{n}) < \infty$ for all $n \in \natp$, and $A = \bigsqcup_{n \in \natp}A_{n}$. By the finite case, there exists $\seq{g_n}\subset L^{q}(X; H)$ such that for each $n \in \natp$ and $f \in L^{p}(X; H)$,

Let $g = \sum_{n = 1}^{\infty} g_{n}$. If $q < \infty$, then $g \in L^{q}(X; H)$ by the Monotone Convergence Theorem. Otherwise,

For every $f \in L^{p}(X; H)$,

by the Dominated Convergence Theorem.

Therefore the mapping is surjective, and hence an isomorphism.$\square$