# Proving that a linear operator is bounded

Let $$X$$ be a real Banach space and $$T: X \rightarrow X^*$$ a linear operator such that $$[T(x)](y) = [T(y)](x)$$ for all $$x,y \in X$$. Prove that $$T$$ is bounded.

So, I need to prove that there exists an $$M \geq 0$$ such that $$\|T(x)\| \leq M \|x\|$$, for all $$x \in X$$, or, equvalently, that $$T$$ is continuous.
I have shown that, if $$[T(x)](x) \geq 0$$ for all $$x \in X$$, then $$T$$ is bounded. I don't know if this result is related to what I'm trying to show though.

I can't really see where to start with this. Any help would be much appreciated. Thanks.

• Hint: Use the "uniform boundedness principle" Jun 2, 2021 at 21:55
• @Simon Could you please demonstrate what you have in mind? I gave an answer using the closed graph theorem but I can't see how one would use the principle of uniform boundedness instead. Jun 2, 2021 at 23:40
• @JustDroppedIn Just did. Hope its correct, havent done functional analysis in a long time^^ Jun 4, 2021 at 15:09

The answer by @JustDroppedIn is fine. For completeness, here is one using uniform boundedness:

• Consider the set of operators $$F=\{ T(x)\ |\ x\in X, |x|=1 \}$$
• For any fixed $$y\in X$$ it is $$\sup_{f\in F} f(y) = \sup_{|x|=1} T(x)(y) = \sup_{|x|=1}T(y)(x)=||T(y)||<\infty$$
• By the uniform boundedness principle, it follows that $$\infty > \sup_{f\in F, |y|=1} |f(y)| = \sup_{|x|=|y|=1} T(x)(y)$$
• Nice answer, thanks for adding it:) +1 from me! Jun 4, 2021 at 15:30
• Thank you for your time! Jun 4, 2021 at 17:17

I think the standard idea is to use the closed graph theorem. So $$T:X\to X^*$$ is a linear operator between Banach spaces. In order to show that it is bounded we can employ the closed graph theorem. Therefore, we can assume that we are given a sequence $$(x_n)\subset X$$ so that $$x_n\to 0$$ and we know that $$T(x_n)\to\phi$$ for some $$\phi\in X^*$$. If we manage to show that $$\phi=0$$ (the zero functional), then we will be able to deduce that $$T$$ is bounded, thanks to the closed graph theorem.

Indeed, suppose that we have the above sequence $$(x_n)\subset X$$ satisfying $$x_n\to0$$ and $$T(x_n)\to\phi$$. If $$z\in X$$, then $$\phi(z)=\lim_{n\to\infty}T(x_n)(z)=\lim_{n\to\infty}T(z)(x_n)=T(z)(0)=0$$ where in the second last equation we use the fact that $$T(z)\in X^*$$, so $$T(z)$$ is a continuous functional and $$x_n\to0$$. This shows that $$\phi(z)=0$$ for all $$z\in X$$, so $$\phi=0$$ and as we explained this proves the claim.