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For Functionals, Uniform Boundedness Principle can be rephrased as the following :

Let ${X}$ be a Banach Space, $K$ be the field($\mathbb{R}$ or $\mathbb{C}$). Let $\mathcal{F}$ be the subset of $BL(X,K)$ such that for each $x \in X$, the set $\{F(x): F \in \mathcal{F}\}$ is bounded in $K$. Then $\{||F||:F \in \mathcal{F}\}$ is bounded i.e uniformly bounded on the unitball of $X$.

The closed graph theorem states that:

Let $X$ and $Y$ be Banach Spaces. Let $F: X \to Y$ be a closed Linear map. Then $F$ is continuous.

Does Closed Graph Theorem imply Uniform Boundedness Principle?

I don't know if it is possible or not. But to make it possible, all I need to do is find a map from $X$ to $K$ which is linear and closed. The first thing that comes to my mind is : $\sup\{F(x)|F \in \mathcal{F}\}$. But this is not a linear map. So it doesn't work. I can take a slightly detour from here and use Zabreiko's Theorem to prove (since $\sup\{F(x)|F \in \mathcal{F}\}$ is a seminorm, which is countably subadditive). But that deviates from what I want to prove here.

tattwamasi amrutam
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1 Answers1

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Yes, it does. Consider the map $$ \Phi:X\to\ell^{\infty}\left(\mathcal{F}\right),x\mapsto\left(f\left(x\right)\right)_{f\in\mathcal{F}}. $$ By assumption, $\Phi$ is well-defined and $X,\ell^{\infty}\left(\mathcal{F}\right)$ are Banach spaces. Since each $f\in\mathcal{F}$ is a continuous linear functional, it is easy to see that $\Phi$ has closed graph.

Hence, it is continuous and thus bounded, so that $$ \sup_{\left\Vert x\right\Vert \leq1}\left\Vert \Phi\left(x\right)\right\Vert _{\ell^{\infty}\left(\mathcal{F}\right)}=\sup_{\left\Vert x\right\Vert \leq1}\sup_{f\in\mathcal{F}}\left|f\left(x\right)\right|=\sup_{f\in\mathcal{F}}\sup_{\left\Vert x\right\Vert \leq1}\left|f\left(x\right)\right|=\sup_{f\in\mathcal{F}}\left\Vert f\right\Vert $$ is finite.

In fact, an easy modification of the above proof shows that the uniform boundedness principle is a consequence of the closed graph theorem.

Very nice question, by the way!

EDIT: The principle which I applied here is that one can often "hide" a nonlinearity (in this case the sup of the absolute value) in a norm (in this case the $\ell^\infty (\mathcal{F})$ norm). This is a nice trick (or technique) to remember.

PhoemueX
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  • ..Oh nice!!! You considered the whole sequence of it instead of taking the supremum. . thanks. liked it. By the way your proof shows that CGT implies UBT. – tattwamasi amrutam Oct 10 '15 at 15:33
  • You meant UBT implies CGT?? – tattwamasi amrutam Oct 10 '15 at 15:39
  • @tattwamasiamrutam: No, I meant that the UBP follows from (is a consequence of) the CGT. – PhoemueX Oct 10 '15 at 17:32
  • What goes wrong if we do the same for linear maps... I think we can do the same.. $l^{\infty}(Y)$ is still complete... –  Nov 15 '15 at 17:23
  • @PraphullaKoushik: Yes, we can do the same, but the OP only asked for linear functionals. – PhoemueX Nov 15 '15 at 18:59
  • Yes Yes... One more question... Isn't it necessary for $\mathcal{F}$ to be countable... I am aware of $l^{\infty}$ as a collection of sequences... Countable... @PhoemueX –  Nov 15 '15 at 19:20
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    @PraphullaKoushik: No, you can just consider the space $\ell^\infty (I)$ of sequences $(x_i){i \in I} \in \Bbb{C}^I$ (or $Y^I$ for a Banach space $Y$) which are bounded, i.e. for which $\sup{i \in I} |x_i|$ (or $\sup_{i \in I} \Vert x_i \Vert$) is finite. It is not hard to see that this is a Banach space. – PhoemueX Nov 15 '15 at 19:31
  • Oh.. Thank you.. :) @PhoemueX –  Nov 15 '15 at 19:39
  • After thinking about this again and again i think this is the most obvious function to be considered after supremum.... Thanks again... This makes much sense... –  Nov 16 '15 at 04:57
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    In Gert Pedersen's book "Analysis Now", the Uniform Boundedness Principle is proven by deduction from the Closed Graph Theorem. (This is Theorem 2.2.9 in the book.) – M. Mueger Apr 19 '20 at 16:05