Application of closed graph theorem I m working on the following problem : Let $E$ be a Banach Space and let $T:E\rightarrow E^*$ be a linear operator satisfying $\langle Tx,x\rangle \geq 0$ for all $x\in E$. Show that $T$ is bounded operator. 
So here is the proof: Let $x_n\rightarrow x\in E$ and $Tx_n\rightarrow f$, then for all $y\in E$. 
$$\langle Tx_n-Ty, x_n-y\rangle\geq 0$$ Applying the limit 
$$\langle f-Ty, x-y\rangle \geq 0$$
Then, Brezis wrote, let $y=x+tz$ for some $t\in\mathbb{R}$ and $z\in E$, then we can conclude $y=Tz$. How does the last implication work? So when you plug $y=x+tz$, you get $$\langle f-Tx-Ttz,-tz\rangle \geq 0$$, where do you go from here
 A: A correction of the question: Brezis' conclusion is $T(x)=f$.
Now, when we do the substitution we obtain:
$$\langle f-Ty,x-y\rangle=\langle f-Tx,-tz\rangle-\langle T(tz),tz\rangle\geq 0$$
for all $z\in E$. By assumption $\langle T(tz),tz\rangle\geq 0$, hence
$$\langle f-Tx,-tz\rangle=-t\langle f-Tx,z\rangle\geq 0.$$
Which implies
$$t\langle f-Tx,z\rangle\leq 0.$$
for all $t$ and $z$.If for some $z$, the value $\langle f-Tx,z\rangle$ is not zero, you can take $t$ such that this value gives a number $>0$. Hence, for all $z\in E$, $\langle f-Tx,z\rangle=0$ and hence the conclusion.
A: Plug $y=x+tz$ in and we have
$$
\langle f-Tx-tTz,-tz\rangle=-t\langle f-Tx,z\rangle+t^2\langle Tz,z\rangle\geq0.
$$
For any fixed $z\in E$, the above quadratic form $p(t)$ is positive. By standard quadratic form knowledge, we must have
$$
\frac{\langle f-Tx,z\rangle}{2\langle Tz,z\rangle}\leq 0,
$$
which implies 
$$
\langle f-Tx,z\rangle\leq 0,~~\forall z\in E.
$$
Consider $-z$ and we also have
$$
\langle f-Tx,-z\rangle=-\langle f-Tx,z\rangle\leq 0.
$$
Hence
$$
\langle f-Tx,z\rangle=0,~~\forall z\in E,
$$
which means $Tx=f$.
