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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.

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    $\begingroup$ Hint: Use the "uniform boundedness principle" $\endgroup$
    – Simon
    Jun 2, 2021 at 21:55
  • $\begingroup$ @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. $\endgroup$ Jun 2, 2021 at 23:40
  • $\begingroup$ @JustDroppedIn Just did. Hope its correct, havent done functional analysis in a long time^^ $\endgroup$
    – Simon
    Jun 4, 2021 at 15:09

2 Answers 2

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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 $$ \infty > \sup_{f\in F, |y|=1} |f(y)| = \sup_{|x|=|y|=1} T(x)(y)$$
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  • $\begingroup$ Nice answer, thanks for adding it:) +1 from me! $\endgroup$ Jun 4, 2021 at 15:30
  • $\begingroup$ Thank you for your time! $\endgroup$
    – J.Spi
    Jun 4, 2021 at 17:17
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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.

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