Show that $T$ is continuous I have a question about how to response to:

Given a Banach space X and $T: X \rightarrow X^{*}$ a linear operator such that $\langle Tx,x\rangle \geq 0$ for all $x \in X$. Show that $T$ is continuous.

I know that this operator sends points in linear continuous functionals, I think that it is a positive form but, I actually don't know how to start my proof.
Thanks for any hint.
 A: One can do it using the Closed Graph Theorem, as follows.
Let $(x_n)$ be a sequence in $X$ such that $x_n\to 0$ and $Tx_n$ has a limit $\phi\in X^*$. We have to show that $\phi=0$.
Observe first that for any $x\in X$ and all $n\in\mathbb N$, we have 
$$\langle T(x_n+x),x_n+x\rangle\geq 0\, . $$
Now, expand the "product" (using the linearity of $T$): this gives
$$\langle T(x_n+x),x_n+x\rangle =\langle Tx_n,x_n\rangle+\langle Tx_n, x\rangle +\langle Tx,x_n\rangle+\langle Tx,x\rangle \, .$$
By assumption on the sequence $(x_n)$, it follows that 
$$ \langle T(x_n+x),x_n+x\rangle\to \langle \phi,x\rangle+\langle Tx, x\rangle\, .$$
So we have 
$$\forall x\in X\;:\; \langle \phi,x\rangle\geq -\langle Tx,x\rangle\, .$$
Replacing $x$ by $-x$ and since $\langle T(-x),-x\rangle =\langle Tx,x\rangle$, we also get $-\langle \phi,x\rangle\geq -\langle Tx,x\rangle$, i.e. $\langle \phi,x\rangle\leq \langle Tx,x\rangle$. Hence:
$$\forall x\in X\;:\; \vert\langle \phi,x\rangle\vert\leq \langle Tx,x\rangle\, . $$
Now, replace $x$ by $tx$, $t>0$: since $\langle T(tx),tx\rangle=t^2\langle Tx,x\rangle$, this gives
$$\vert \langle \phi,x\rangle\vert\leq t\,\langle Tx,x\rangle\, .$$
Letting $t\to 0^+$, it follows that $\langle \phi,x\rangle=0$ for all $x\in X$, as required.
