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Let $\mathbb{H}$ be a hilbert space, $E \subset \mathbb{H}$. We say that $E$ is weakly bounded if for every $y \in \mathbb{H}$, there is some $\alpha_{y} \geq 0$ such that $|<x, y>| \leq \alpha_{y}$ for all $x \in E$.

Then show that a subset of a HIlbert space is weakly bounded iff it is bounded. .......................................................

I am trying to use Uniform boundedness theorem, but it seems to be not working.

mathonlymath
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  • Uniform Boundedness Principle is precisely the correct approach. Where did you fail to conclude? – Vobo Nov 20 '13 at 16:09
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    Remember points and bounded linear functionals can be more or less interchanged in Hilbert land. – copper.hat Nov 20 '13 at 16:11

2 Answers2

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It is clear that you need to prove only weakly bounded $\Rightarrow$ bounded. You are on the right path: you have to use the Banach Steinhaus theorem on the following family of operators: \begin{align} E_{y}:H &\longrightarrow H\\ x &\longmapsto \langle x, y \rangle \end{align} for all $y \in E$. Then $||E_y||$ is uniformly bounded and by reflexivity of Hilbert space, you can conclude.

Mik
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  • Can you please clarify that why reflexivity will conclude the proof? Or, we reisz representation will directly conclude the proof . – mathonlymath Nov 21 '13 at 12:53
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Let $E$ be weakly bounded. For each $y\in E$ we consider the map $T_y:\mathbb{H}\to\mathbb{C}$ defined by $T_y(x)=\langle x,y\rangle$ for every $x\in\mathbb{H}$. Then since $E$ is weakly bounded, therefore for every $x\in\mathbb{H}$, the set $\{T_y(x):y\in E\}$ is bounded. Thus, by the uniform boundedness principle, the set $\{\|T_y\|:y\in E\}$ is bounded. Also it is easy to see that each $T_y$ is a linear operator. Thus each $T_y$ is a bounded linear functional on $\mathbb{H}$, ad thus from the Riesz's theorem, we must have $\|T_y\|=\|y\|$ for each $y$ (given how the $T_y$'s are defined). Thus the set $\{\|y\|:y\in E\}$ is bounded, i.e., $E$ is bounded.

Kuldeep Guha Mazumder
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