Proving that a linear operator is bounded

analysisbanach-spacesfunctional-analysisoperator-theory

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.

Best Answer

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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