Lemma 16.1.3.label Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality over $K$, then the dual of $(E, \sigma(E, F))$ is $F$. In other words, for any $\phi \in L(E, \sigma(E, F); K)$, there exists a unique $y \in F$ such that $\dpn{x, \phi}{E}= \dpn{x, y}{\lambda}$ for all $x \in E$.
Proof, [IV.1.2, SW99]. Since $\phi$ is continuous, there exists $\seqf{y_k}\subset F$ such that for all $x \in E$,
\[|\dpn{x, \phi}{\lambda}| \le \sum_{k = 1}^{n} |\dpn{x, y_k}{\lambda}|\]
Assume without loss of generality that $\seqf{y_k}$ is linearly independent, then by the First Isomorphism Theorem, there exists $\Phi \in L(K^{n}; K)$ such that the following diagram commutes
\[\xymatrix{ E \ar@{->}[rd]_{\phi} \ar@{->}[r]^{{(y_1, \cdots, y_n)}} & K^n \ar@{->}[d]^{\Phi} \\ & K }\]
For each $1 \le k \le n$, let $e_{k}$ be the $k$-th standard basis vector in $K^{n}$, then for any $x \in E$,
\[\dpn{x, \phi}{E}= \sum_{k = 1}^{n} \Phi(e_{k}) \dpn{x, y_k}{\lambda}\]
$\square$