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Let $E$ be a Banach space and let $T:E\to E^{\star}$ be a linear operator satisfying $\langle Tx,x\rangle\geq 0$ $\forall x\in E$. Prove that $T$ is a bounded operator.

My Solution (but I have trouble to conclude):

With the closed graph theorem we need to show that the graph of $T$ is closed. For this let $x_n\in E$ be a sequence converging to $x$. Furthermore we have $(x_n,Tx_n)\in \text{graph}(T)$ that converges to $(x,f)$. Goal: Show that $f=Tx$. We have that

\begin{equation}\langle Tx-Ty,x-y\rangle\geq 0\end{equation} passing to the limit we get $\langle f-ty,x-y\rangle\geq 0$. But here I'm get stuck... Question: how can I conclude?

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We can assume that $x=0$ hence we have to show that $f=0$. Here is an approach: for a fixed $u\in X$, we have $$ 0\leqslant\langle T\left(x_n -x\right),x_n -x\rangle.$$ Expanding and letting $n$ going to infinity, we get that
$$\langle f,u\rangle\leqslant \langle Tu,u\rangle.$$ Replacing $u$ by $Ru$, we get by linearity
$$\langle f,u\rangle\leqslant \frac 1R\langle Tu,u\rangle$$ and since $R$ is arbitrary, $\langle f,u\rangle\leqslant 0$. Replace $u$ by $-u$ and use Hahn-Banch theorem to conclude that $f=0$.

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