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Fall 2019 - Problem 5

Hilbert spaces

Prove the following claim: Let $\mathcal{H}$ be a Hilbert space with the scalar product if $x$ and $y$ denoted by $\inner{x, y}$ and let $\func{A,B}{\mathcal{H}}{\mathcal{H}}$ be (everywhere-defined) linear operators with

\forall x, y \in \mathcal{H}, \quad \inner{Bx, y} = \inner{x, Ay}.

Then $A$ and $B$ are both bounded (and thus continuous).

Solution.

Observe that for any $y \in \mathcal{H}$ ,

\sup_{\norm{x} = 1} \,\abs{\inner{Bx, y}} = \sup_{\norm{x} = 1} \,\abs{\inner{x, Ay}} \leq \sup_{\norm{x} = 1} \,\norm{x}\norm{Ay} = \norm{Ay}.

Hence, by the uniform boundedness principle,

C = \sup_{\norm{x} = 1} \norm{\inner{Bx, \:\cdot\:}} < \infty.

It follows that

\norm{Bx}^2 = \abs{\inner{Bx, Bx}} \leq C\norm{Bx} \implies \sup_{\norm{x} = 1} \,\norm{Bx} \leq C < \infty,

so by definition, $\norm{B} < \infty$ . For $A$ , observe that $\conj{\inner{Ay, x}} = \conj{\inner{y, Bx}}$ , so by symmetry, we see that $A$ is also bounded.