Bourbaki

Remark

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

Topological Vector Spaces, chapter III, § 3. Spaces of continuous linear mappings

— 3) Let A be a dense subset of the unit ball of E. In view of the continuity of uu, we also have

r(u)=sup⁡x∈Aq(u(x)). r(u) = \sup_{x \in A} q(u(x)) .

For example

r(u)=sup⁡p(x)<1q(u(x)). r(u) = \sup_{p(x) < 1} q(u(x)) .

Since we have u(tx)=tu(x)u(tx) = tu(x) for t∈Rt \in \mathbf{R}, we also have,

r(u)=sup⁡p(x)=1q(u(x))=sup⁡p(x)≠0q(u(x))p(x). r(u) = \sup_{p(x) = 1} q(u(x)) = \sup_{p(x) \neq 0} \frac{q(u(x))}{p(x)} .

whenever p≠0p \neq 0.

4) The formula (2) shows that the map u↦r(u)u \mapsto r(u) is lower semi-continuous on Ls(E;F)\mathscr{L}_s(E; F).

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

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Cites

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Transcribed from the printed volume. Not translated.