Positive operator is bounded 
For a real Banach space $X$ let $A:X\rightarrow X^*$ be a positive operator in the sense that $(Ax)(x)\geq 0$ for all $x\in X$. Show that $A$ is bounded.

I don't know how to do that, maybe it's an application of the closed graph theorem?
 A: It's indeed an application of closed graph theorem. First, we take $\{x_n\}\subset X$ a sequence which converges to $x$ and such that $Ax_n\to l$, where $l\in X^*$. As the sequence $\{x_n\}$ is bounded, we have $\langle T(x_n),x_n\rangle\to l(x)$. For $y\in X$, we have 
$$0\leq \langle Tx_n-Ty,x_n-y\rangle=\langle Tx_n,x_n\rangle-\langle Ty,x_n\rangle-\langle Tx_n,y\rangle+\langle Ty,y\rangle.$$
Taking the limit $\lim_{n\to +\infty}$, we get 
$$0\leq l(x)-\langle T(y),x\rangle-l(y)+\langle T(y),y\rangle,$$
which gives 
$$l(y-x)\leq \langle T(y),y-x\rangle.$$
Let $y=z+x$. Then 
$$l(z)\leq \langle Tz,z\rangle+\langle T(x),z\rangle.$$
Replacing $z$ by $az$, where $0<a$ and taking $a\to 0$, we get $l(z)\leq \langle T(x),z\rangle$. Replacing $z$ by $-z$, we get $l=Tx$.
As $X$ and $X^*$ are Banach spaces, we conclude by closed graph theorem.
A: We prove it by closed graph theorem. Assume $ x_n \rightarrow x $ in $ E $ and $ Tx_n \rightarrow t $ in $ E^* $, we show that $ t = Tx $, which is the same as $ \left\langle Tx-t, z\right\rangle, \forall z\in E $.
We use $ \lim $ to mean that we take limit with respect to $ n $. 
Introducing $ P(x,y):=\left\langle Tx, y\right\rangle +\left\langle x,Ty\right\rangle  $ and using the fact $ P(x-ty,x-ty)\geq 0, \forall t \in \mathbb{R} $, it is not hard to show:
     $$ \left| P(x,y)\right| \leq\sqrt{P(x,x)\,P(y,y)} $$ 
The limit $ \lim P(x_n - x, x_n -x) $ exists and equals to zero for the assumption that we take and linearity of brackets. Therefore, $ \lim P(x_n-x, z) = 0, \forall z\in E $.
However, we can have the following calculation:
     $$
  \begin{aligned}
     \lim P(x_n-x,z)&=\lim \left\langle Tx_n-Tx,e\right\rangle +\lim \left\langle x_n - x, Tz\right\rangle  \\
   &=\lim \left\langle Tx_n-Tx,e\right\rangle = \left\langle t-Tx, z\right\rangle 
  \end{aligned}
 $$
     As so far, we have shown $\left\langle t-Tx, z\right\rangle =0  $ and hence finished the proof.
