Added a characterisation of L^p.

src/fa/lp/definition.tex

@@ -127,6 +127,46 @@
     Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ strongly pointwise as $n \to \infty$. By the \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^p(X; E)$.
 \end{proof}
 
+\begin{theorem}[{{\cite[III.6.5]{SchaeferWolff}}}]
+\label{theorem:l1-tensor}
+    Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $\real \in \RC$, then the map $L^1(X; \real) \td{\otimes}_\mu E \to L^1(X; E)$ defined by extending
+    \[
+    L^1(X; \real) \times E \to L^1(X; E) \quad f \otimes x \mapsto x \cdot f
+    \]
+    
+    is an isometric isomorphism. 
+\end{theorem}
+\begin{proof}
+    By (U) of the \hyperref[tensor product]{definition:tensor-product}, the given map admits a unique extension
+    \[
+    M: L^1(X; \real) \otimes E \to L^1(X; E) \quad \sum_{j = 1}^n f_j \otimes x_j \mapsto \sum_{j = 1}^n x_j \cdot f_j
+    \]
+
+    Restricting $M$ to the simple functions yields a linear isomorphism
+    \[
+    M: [L^1(X; \real) \cap \Sigma(X; \real)] \otimes E \to L^1(X; E) \cap \Sigma(X; E)
+    \]
+
+    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
+    \[
+    \normn{M^{-1}\phi}_{L^1(X; \real) \otimes E} \le \sum_{y \in \phi(X) \setminus \bracs{0}} \norm{y}_E \cdot \mu\bracs{\phi = y} = \int \norm{\phi}_E d\mu = \norm{\phi}_{L^1(X; E)}
+    \]
+
+    On the other hand, for any representation $M^{-1}\phi = \sum_{j = 1}^n a_j \one_{A_j}$, 
+    \[
+    \normn{\phi}_{L^1(X; E)} \le \sum_{j = 1}^n \norm{a_j}_E \mu(A_j) = \sum_{j = 1}^n \norm{a_j}_E \normn{\one_{A_j}}_{L^1(X; \real)}
+    \]
+
+    As this holds for all such representations, $\normn{\phi}_{L^1(X; E)} = \normn{M^{-1}\phi}_{L^1(X; \real) \otimes E}$. Therefore $M$ restricted to $[L^1(X; \real) \cap \Sigma(X; \real)] \otimes E$ is an isometry. By \autoref{proposition:lp-simple-dense}, $[L^1(X; \real) \cap \Sigma(X; \real)] \otimes E$ is dense in $L^1(X; \real) \widehat{\otimes}_\pi E$, and $L^1(X; E) \cap \Sigma(X; E)$ is dense in $E$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $M$ extends uniquely into the given map on $L^1(X; \real) \otimes E$, which then extends into an isometry $L^1(X; \real) \otimes E \to L^1(X; E)$. 
+\end{proof}
+
+
+
 \begin{theorem}[Markov's Inequality]
 \label{theorem:markov-inequality}
     Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be a Borel measurable function, then