Why is such an operator continuous?

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.

• $(1)$ 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.