Why is such an operator continuous? These two questions were in one question of a list of exercises.
Let $E$ be a Banach space and $T : E \longrightarrow E^*$ be linear.


*

*If $\langle T(x),x \rangle \geq 0$ holds for all $x \in E$, then $T$ is continuous. 

*If $\langle T(x),y \rangle = \langle x ,T(y) \rangle$ holds for all $x, y \in E$, then $T$ is continuous.
I tried to expand it, as in the proof of the Cauchy-Schwarz inequality, to get a polynomial of degree $2$. Any solution or hint?
 A: For $x\in S_{E}$ consider the family $F_x$ of bounded functionals given by
$$
F_x(y)=\langle Tx,y\rangle
$$
We  have $|F_x(y)|=|\langle Tx,y\rangle|=|\langle x,Ty\rangle|\leq||x||\cdot||Ty||=||Ty||$ 
Therefore, the family $\{F_x : x\in S_E\}$ is pointwise bounded, and it follows from the Uniform Boundness Principle that is also norm bounded. Note that this works for both $T$ symmetric and anti-symmetric.
Since the family is norm bounded, there exists $K$ such that  for any $x\in S_E$, we have
$$
||Tx||=\sup_{y\in S_E}|\langle y, Tx\rangle| <K
$$ 
which means that $T$ is bounded, thus showing $(2)$.
For $(1)$, if the vector spaces are over $
\mathbb{C}$, the condition implies the anti-symmetry of $T$, as Daniel Fischer noted before. The proof above works just as well for $T$ anti-symmetric. Don't know about the real case.
A: *

*$(1)$ is true also in the real case. Here is one possible proof although I'm not sure its the quickest (since its an adaption of a proof showing (possibly nonlinear) monotone operators are locally bounded). If $T$ were not bounded, then there would exist a sequence $x_n \to 0$ with $\|Tx_n\| \to \infty$. Define
$$c_n = 1 + \|Tx_n\|\|x_n\|.$$
Now let $z \in E$. Then by assumption
$$0 \le \langle T(z - x_n), z - x_n \rangle $$
which after expanding and rearranging turns into
$$\langle Tx_n, z \rangle \le \langle Tx_n, x_n - z \rangle + \langle Tz, z - x_n \rangle.$$
Since $c_n > 1$, we get
$$c_n^{-1}\langle Tx_n, z \rangle \le c_n^{-1}\langle Tx_n, x_n - z \rangle + \langle Tz, z - x_n \rangle$$
$$\le 1 + c_n^{-1}\|Tz\|\|z -  x_n\| \le M(z)$$
where $M(z)$ is some constant independent of $n$. We can repeat the same argument with $-z$ in place of $z$ to get
$$-c_n^{-1}\langle Tx_n, z \rangle \le M(-z)$$
where again $M(-z)$ is independent of $n$. Thus we can use the Banach-Steinhaus Theorem to conclude that
$$\sup c_n^{-1}\|Tx_n\| \le C < \infty.$$
Recalling the definition of $c_n$ we get
$$\|Tx_n\| \le C(1 + \|Tx_n\|)\|x_n\| $$
so
$$(1 - C\|x_n\|)\|Tx_n\| \le C$$
for all $n$. This implies $\|Tx_n\| \le 2C$ when $\|x_n\| \le \frac{1}{2C}$ contradicting the fact that $\|Tx_n\| \to \infty$ as $x_n \to 0$. So $T$ is bounded.

*Here is also an alternative to $(2)$ which mimicks the Hellinger-Toeplitz Theorem. Let $x_n \to x$ in $E$ be such that there exists $y \in E^*$ with $Tx_n \to y$. Then we have
$$\langle y, z \rangle = \lim \langle Tx_n,z \rangle = \lim \langle Tz, x_n \rangle $$
$$ = \langle Tz, x \rangle = \langle Tx, z \rangle$$
for all $z \in E$ (where we used continuity of the linear functional $Tz$ in the third equality). This means that $y = Tx$ and therefore the graph of $T$ is closed. Hence $T$ is continuous by the Closed Graph Theorem.
A: Here is an alternative proof for the first statement.
Suppose that $x_n\to x$ in $E$ and $Tx_n\to f$ in $E'$. By hypothesis, we have that $$\langle Tx_n-Ty,x_n-y\rangle\ge 0,\ \forall \ y\in E\tag{1}$$
If we pass the limit in $(1)$ we get that $$\langle f-Ty,x-y\rangle\ge 0,\ \forall\ y\in E\tag{2}$$
Now take $y=x+tv$ where $t\in \mathbb{R}$ and $v\in X$. We have that $$\langle f-Tx-tTv,-tv\rangle\geq 0,\ \forall\ t\in\mathbb{R},\ v\in E\tag{3}$$
We get from $(3)$ that $-t\langle f-Tx, v\rangle\geq-t^2\langle Tv,v\rangle$ or equivalently $$\langle f-Tx, v\rangle\leq t\langle Tv,v\rangle,\ \forall\ t> 0,\ v\in E \tag{4}$$
Now it is straightforward to conclude from $(4)$.
