Uniform boundedness principle and closed graph Theorem

analysisclosed-graphfunctional-analysisoperator-theory

In a functional analysis class, my professor gave a problem to prove the uniform boundedness principle from the Closed graph theorem.

The problem goes thus:

Let $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ be real Banach spaces. Let $\{T_{\alpha}\}_{\alpha \in \Delta}$ be a collection of bounded linear maps from $X$ to $Y$. Consider a new norm defined on $X$ $$\|x\|^{'} :=\|x\|_{X} + \sup_{\alpha \in \Delta}\| T_{\alpha} x\|_{Y} $$ where $\Delta$ is an arbitrary index set . Deduce from the closed graph theorem the uniform boundedness principle.

I have the following questions :

  1. The standard proof of the uniform boundedness principle uses the Baires category theorem which I am very well conversant with. In that proof, we only require that $X$ be Banach contrary to the assumption of the closed graph theorem where both $X$ and $Y$ should be Banach. Can someone explain how the closed graph theorem and the uniform boundedness theorem would then relate.

  2. What function can one define on $X$ with the new norm that would enable me apply the closed graph theorem. I am basically new to the course and would like to know what would motivate one to define such function.

Though I have searched online and saw this.

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I understand the proof but it has no affiliation with the new norm as given by my problem. Thanks

Best Answer

  1. Suppose that for every $x\in X, \sup_{\alpha\in \Delta}\|T(x)\|_Y$ is bounded, show that $(X,\|\cdot\|')$ is Banach.

  2. Consider $Id_X:(X,\|\cdot\|)\rightarrow (X,\|\cdot\|')$. Its graph is closed, so it is a bounded map. You can deduce that there exists $C>0$ such that $\|x\|_X+ \sup_{\alpha\in \Delta}\|T_{\alpha}(x)\|_Y<C\|x\|_X$.

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