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Fall 2013 - Problem 9

Banach-Steinhaus, operator theory

Let $X$ be a Banach space, $Y$ be a normed linear space, and $\func{B}{X \times Y}{\R}$ be a bilinear function. Suppose that for each $x \in X$ there exists a constant $C_x \geq 0$ such that $\abs{B\p{x, y}} \leq C_x\norm{y}$ for all $y \in Y$ , and for each $y \in Y$ there exists a constant $C_y \geq 0$ such that $\abs{B\p{x, y}} \leq C_y\norm{x}$ for all $x \in X$ .

Show that then there exists a constant $C \geq 0$ such that $\abs{B\p{x, y}} \leq C\norm{x}\norm{y}$ for all $x \in X$ and all $y \in Y$ .

Solution.

First, if we fix $y \in Y$ , then $x \mapsto B\p{x, y}$ is a linear functional with operator norm at most $C_y$ , so we are working with a family of bounded linear functionals. Notice that our assumptions tell us that for a given $x \in X$ ,

\abs{B\p{x, y}} \leq C_x\norm{y} \implies \sup_{\norm{y} = 1}\,\abs{B\p{x, y}} \leq C_x.

Thus, by Banach-Steinhaus, there exists $C > 0$ such that

\sup_{\norm{y} = 1}\,\norm{x \mapsto B\p{x, y}}_{X\to\R} \leq C < \infty.

And so $\abs{B\p{x, y}} \leq C\norm{x}$ for any $\norm{y} = 1$ . If we fix $x$ , then $y \mapsto B\p{x, y}$ is a bounded linear function with operator norm smaller than $C\norm{x}$ , and so $\abs{B\p{x, y}} \leq C\norm{x}\norm{y}$ , as required.