Proposition 8

Tag 07R5, permanent. Cite it as [Bourbaki, Tag 07R5].

Topological Vector Spaces, chapter IV, § 1. Duality

— (i) Let $E$ be a normed space, with dual $E'$ . For every $x \in E$ , we have

\|x\| = \sup_{x' \in E', \|x'\| \leq 1} |\langle x, x' \rangle|.

(ii) Let $E_1$ and $E_2$ be two normed spaces and $u$ a continuous linear mapping from $E_1$ into $E_2$ . We have

\|u\| = \|t u\|.

Let $x \in E$ . For every $x' \in E'$ such that $\|x'\| \leq 1$ , we have

|\langle x, x' \rangle| \leq \|x\| \cdot \|x'\| \leq \|x\|.

By Hahn-Banach theorem (II, p. 23, cor. 2), there exists an element $x'$ in $E'$ such that $\|x'\| \leq 1$ and $\langle x, x' \rangle = \|x\|$ . This proves (i).

Let us now prove (ii). By formula (3) and the definition of the transpose, we have

\|^{t}u\| = \sup_{\|y'\| \leq 1} \|^{t}u(y')\| = \sup_{\|y'\| \leq 1, \|x\| \leq 1} |\langle x, ^{t}u(y') \rangle| = \sup_{\|x\| \leq 1, \|y'\| \leq 1} |\langle u(x), y' \rangle| = \sup_{\|x\| \leq 1} \|u(x)\| = \|u\|.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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