How to prove this linear operator is bounded? Suppose $X$ is a Banach space, $T:X\to X^*$ is a linear operator, which satisfies
$$\langle Tx,x\rangle\ge0,\quad\forall x\in X.$$
How to prove $T$ is bounded? The question will be easy if $X$ is a complex Banach space. I'm wondering whether we can prove $T$ is bounded when $X$ is real.
 A: Let $x_n \in X$ be a sequence, converging to $0$. We shall prove that $Tx_n$ is a bounded sequence (this fact, obviously, implies the boundedness of $T$). Let $\alpha_n = (1 + \|x_n\|\|Tx_n\|)^{-1}$. Let $v \in X$ and consider the following inequalities:
$$
\langle T(x_n + v), x_n + v\rangle \ge 0,\;\;\langle T(x_n - v), x_n - v\rangle \ge 0.
$$
Trivial calculations ensure that the foregoing inequalities can be rewritten as
$$
\langle Tx_n, v\rangle \ge -\langle Tx_n, x_n\rangle - \langle Tv, x_n + v\rangle,\;\;\langle Tx_n, v\rangle \le \langle Tx_n, x_n\rangle - \langle Tv, x_n - v\rangle.
$$
These inequalities trivially imply that $\langle \alpha_n T x_n, v\rangle$ is a bounded sequence for all $v \in X$. Therefore, Banach-Steinhaus theorem implies, that the sequence $\alpha_n T x_n$ is bounded. So, there exists $M > 0$ s.t.
$$
\|Tx_n\| \le M(1 + \|x_n\| \|Tx_n\|).$$
Finally, consider $m \in \mathbb N$ s.t. $M \|x_n\| \le 1/2$ for $n \ge m$ (note that $x_n \rightarrow 0$). It follows that for $n \ge m$
$$
(1 - M\|x_n\|) \|Tx_n\| \le M \Rightarrow \|Tx_n\| \le 2M.
$$
Thus, $T$ is bounded.
A: You can use the closed graph theorem: suppose that $x_n \to x$ in $X$ and $Tx_n \to x^*$ in $X^*$. To show that $x^*=Tx$ it suffices to show that
$$ \langle x^*-Tz,x-z \rangle \ge 0 ~~~ \forall z \in X . \tag{1}$$
Indeed, suppose that $(1)$ holds. Fix $y \in X$ and let $z = ay +x$ where $a>0$. Then $(1)$ implies that
$$ \langle x^*-aTy-Tx,ay \rangle \ge 0.$$
Dividing by $a$ and then letting $a \to 0$ you obtain that
$$\langle x^*,y\rangle \ge \langle Tx,y\rangle .$$
In a similar fashion, letting $z=ay+x$ where $a<0$ you obtain that
$$\langle x^*,y\rangle \le \langle Tx,y\rangle .$$
This shows that $\langle x^*-Tx,y\rangle = 0$ and since $y$ was arbitrary, $x^*=Tx$.
Showing $(1)$ is fairly simple: if $z \in X$ then
$$ \langle x^*-Tz,x-z \rangle= \lim_n \langle Tx_n-Tz,x_n-z \rangle \ge 0 $$
since $T$ is positive.
