# Does Closed Graph Theorem imply Uniform Boundedness Principle for functionals?

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.

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.

• ..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... – user87543 Nov 15 '15 at 17:23
• @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