Bourbaki

Remark 2

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

Topological Vector Spaces, III, p. 26, § 4. The Banach-Steinhaus theorem

Let E and F be two Banach spaces, and (un)(u_n) a sequence of continuous linear mappings from E into F such that sup⁡∥un∥=+∞\sup \|u_n\| = + \infty. Then the set X of all x∈Ex \in E such that sup⁡∥un(x)∥=+∞\sup \|u_n(x)\| = + \infty is dense in E and is the intersection of a sequence of open sets in E. For, let XkX_k denote the set of all x∈Ex \in E such that sup⁡∥un(x)∥>k\sup \|u_n(x)\| > k (for k integer >0> 0). Each XkX_k is open and X is the intersection of the XkX_k. Since E is a Baire space, it is enough to show that each XkX_k is dense in E. But, if the complement of XkX_k contains a non-empty open set U, we would have ∥un(x)∥≤2k\|u_n(x)\| \leq 2k for x∈U−Ux \in U - U and, since U−UU - U is a neighbourhood of 0, we would have sup⁡∥un∥<+∞\sup \|u_n\| < + \infty.

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

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Transcribed from the printed volume. Not translated.