Theorem 12.1.10 | Jerry's Digital Garden

Theorem 12.1.10. Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then the map $L^{1}(X; K) \td{\otimes}_{\mu} E \to L^{1}(X; E)$ defined by extending

\[L^{1}(X; K) \times E \to L^{1}(X; E) \quad f \otimes x \mapsto x \cdot f\]

is an isometric isomorphism.

Proof. By (U) of the tensor product, the given map admits a unique extension

\[M: L^{1}(X; K) \otimes E \to L^{1}(X; E) \quad \sum_{j = 1}^{n} f_{j} \otimes x_{j} \mapsto \sum_{j = 1}^{n} x_{j} \cdot f_{j}\]

Restricting $M$ to the simple functions yields a linear isomorphism

\[M: [L^{1}(X; K) \cap \Sigma(X; K)] \otimes E \to L^{1}(X; E) \cap \Sigma(X; E)\]

For any $\phi \in L^{1}(X; E) \cap \Sigma(X; E)$, write

\[\phi = \sum_{y \in \phi(X) \setminus \bracs{0}}y \cdot \one_{\bracs{\phi = y}}= M\braks{\sum_{y \in \phi(X) \setminus \bracs{0}}\one_{\bracs{\phi = y}} \otimes y}\]

then

\[\normn{M^{-1}\phi}_{L^1(X; K) \otimes E}\le \sum_{y \in \phi(X) \setminus \bracs{0}}\norm{y}_{E} \cdot \mu\bracs{\phi = y}= \int \norm{\phi}_{E} d\mu = \norm{\phi}_{L^1(X; E)}\]

On the other hand, for any representation $M^{-1}\phi = \sum_{j = 1}^{n} a_{j} \one_{A_j}$,

\[\normn{\phi}_{L^1(X; E)}\le \sum_{j = 1}^{n} \norm{a_j}_{E} \mu(A_{j}) = \sum_{j = 1}^{n} \norm{a_j}_{E} \normn{\one_{A_j}}_{L^1(X; K)}\]

As this holds for all such representations, $\normn{\phi}_{L^1(X; E)}= \normn{M^{-1}\phi}_{L^1(X; K) \otimes E}$. Therefore $M$ restricted to $[L^{1}(X; K) \cap \Sigma(X; K)] \otimes E$ is an isometry. By Proposition 12.1.9, $[L^{1}(X; K) \cap \Sigma(X; K)] \otimes E$ is dense in $L^{1}(X; K) \widehat{\otimes}_{\pi} E$, and $L^{1}(X; E) \cap \Sigma(X; E)$ is dense in $E$. By the Linear Extension Theorem, $M$ extends uniquely into the given map on $L^{1}(X; K) \otimes E$, which then extends into an isometry $L^{1}(X; K) \otimes E \to L^{1}(X; E)$.$\square$

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