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.  
 A: 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.
