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I am sitting through a course on "Operators in Hilbert Spaces". The instructor has asked us to look at the following problem:

Let $H$ be a hilbert space and $E \subset H$. E is called weakly bounded if $$\forall y \in H, \exists \alpha_y > 0 \ni |\langle x, y \rangle| \leq \alpha_y \forall x \in E$$ To show that $E$ is weakly bounded $\iff$ $E$ is bounded.

The hint given to us by the instructor is Uniform Boundedness Principle.

Now, my knowledge in functional analysis is through self-reading so it is pretty patchy. I would appreciate if someone could guide me in the right direction with this problem.

Uniform Boundedness Principle: Let $X$ be a Banach space, $Y$ be a normed space and $\mathcal{F}$ be a subset of $BL(X, Y)$ such that for each $x \in X$, the set $\{F(x) : F \in \mathcal{F}\}$ is bounded in $Y$. Then for each bounded subset $E$ of $X$ the set $\{F(x) : x \in E, F \in \mathcal{F}\}$ is bounded in $Y$, that is, $\mathcal{F}$ is uniformly bounded on $E$. In particular, $\sup \{\|F\|\ : F \in \mathcal{F}\} < \infty$.

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possible duplicate of Weak Convergence implies boundedness and componentwise convergence – Norbert Jan 24 '13 at 9:39
Could you give the precise formulation of the uniform boundedness principle you were given in class? Some formulations give $\Longrightarrow$ directly from the statement, while $\Longleftarrow$ is a direct consequence of Cauchy-Schwarz. – Martin Jan 24 '13 at 9:50
@Norbert: Not really a duplicate, I think. The question you suggest is about bounded sequences in $\ell^2$, not bounded subsets of an arbitrary Hilbert space. – Martin Jan 24 '13 at 9:51
@Martin I have added the uniform boundedness principle to the question. – TenaliRaman Jan 24 '13 at 10:02
@Martin, the same arguments applies to arbitrary sets. I don't think this small modification had to be taken into account – Norbert Jan 24 '13 at 10:20
up vote 1 down vote accepted

I assume that the scalar product is linear in the second variable. Each $x \in H$ gives an element $\varphi_x \in H^\ast = BL(H,\mathbb{C})$ by setting $\varphi_x(y) = \langle x,y\rangle$.

The map $H \to H^\ast, x \mapsto \varphi_x$ is norm-decreasing by Cauchy-Schwarz: $\lvert \varphi_x(y)\rvert = \lvert \langle x, y\rangle \rvert \leq \lVert x \rVert \lVert y\rVert$, so $\lVert \varphi_x\rVert \leq \lVert x\rVert$ and it is isometric since $\lvert\varphi_x(x)\rvert = \lVert x\rVert^2$, so $\lVert \varphi_x\rVert \geq \lVert x\rVert$. (By the Riesz representation theorem, $x \mapsto \varphi_x$ is onto $H^\ast$, but we won't need this.)

Let $E$ be a weakly bounded subset of $H$. We will apply the uniform boundedness principle with $X = H$, $Y = \mathbb{C}$ to the set $\mathcal{F} = \{\varphi_x \mid x \in E\} \subset H^\ast$. By hypothesis, for each $y \in H$ there is $\alpha_y$ such that $\lvert\varphi_x(y)\rvert \leq \alpha_y$ for all $x \in E$. By the "in particular" of your formulation of the uniform boundedness principle, $\sup_{x \in E} \lVert \varphi_x\rVert \lt \infty$, but $\lVert\varphi_x\rVert = \lVert x\rVert$ by the second paragraph of this answer, so $\sup_{x \in E} \lVert x\rVert \lt \infty$, i.e., $E$ is bounded.

Conversely, suppose $E$ is bounded. Then $C = \sup_{x \in E} \lVert x \rVert \lt \infty$ and therefore Cauchy-Schwarz gives $\lvert \langle x,y\rangle\rvert \leq \lVert x \rVert \lVert y\rVert \leq C \lVert y\rVert$, so we may take $\alpha_y = C\lVert y\rVert$.

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Thank you! This answer has gone a long way into making me understand several different things! :-) – TenaliRaman Jan 24 '13 at 13:58
You're welcome :-) I fixed an unfortunate typo in the third paragraph of my answer. – Martin Jan 24 '13 at 14:02

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