Theorem 14.2.3.label Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $\dpn{E, F}{\lambda}$ be a norming duality of normed vector spaces over $K$, $p, q \in [1, \infty]$ be Hölder conjugates such that one of the following holds:
- (a)
$p \in (1, \infty]$ and $q \in [1, \infty)$.
- (b)
$p = 1$, $q = \infty$, and $\mu$ is semifinite.
Let $g: X \to F$ be a strongly measurable function such that for every $f \in \Sigma(X, \cm; E) \cap L^{p}(X; E)$, $\dpn{f, g}{\lambda}\in L^{1}(X; E)$, and the mapping
is continuous in the $L^{p}(X; E)$ norm, then:
- (1)
If (a) holds, then $\bracs{g \ne 0}$ is $\sigma$-finite.
- (2)
$g \in L^{q}(X; F)$ with $\norm{g}_{L^q(X; F)}= \norm{\phi_g}_{L^p(X; E)^*}$.
- (3)
The mapping
\[L^{q}(X; F) \to L^{p}(X; E)^{*} \quad g \mapsto \phi_{g}\]is isometric.
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)
For each $n \in \natp$, $\norm{\phi_n}_{E}\le 1$.
- (2)
For every $n \in \natp$, $0 \le |\dpn{g, \phi_n}{\lambda}| \le \norm{g}_{F}$.
- (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)
$\norm{g_n}_{F}\upto \norm{g}_{F}$ as $n \to \infty$.
- (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, $\norm{g}_{L^q(X; F)}= \norm{\phi_g}_{L^p(X; E)^*}$.$\square$
- Under the convention that $0^{0} = 1$keyboard_return
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