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For a real Banach space $X$ let $A:X\rightarrow X^*$ be a positive operator in the sense that $(Ax)(x)\geq 0$ for all $x\in X$. Show that $A$ is bounded.

I don't know how to do that, maybe it's an application of the closed graph theorem?

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It's indeed an application of closed graph theorem. First, we take $\{x_n\}\subset X$ a sequence which converges to $x$ and such that $Ax_n\to l$, where $l\in X^*$. As the sequence $\{x_n\}$ is bounded, we have $\langle T(x_n),x_n\rangle\to l(x)$. For $y\in X$, we have $$0\leq \langle Tx_n-Ty,x_n-y\rangle=\langle Tx_n,x_n\rangle-\langle Ty,x_n\rangle-\langle Tx_n,y\rangle+\langle Ty,y\rangle.$$ Taking the limit $\lim_{n\to +\infty}$, we get $$0\leq l(x)-\langle T(y),x\rangle-l(y)+\langle T(y),y\rangle,$$ which gives $$l(y-x)\leq \langle T(y),y-x\rangle.$$ Let $y=z+x$. Then $$l(z)\leq \langle Tz,z\rangle+\langle T(x),z\rangle.$$ Replacing $z$ by $az$, where $0<a$ and taking $a\to 0$, we get $l(z)\leq \langle T(x),z\rangle$. Replacing $z$ by $-z$, we get $l=Tx$.

As $X$ and $X^*$ are Banach spaces, we conclude by closed graph theorem.

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