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Let $T$ be a linear operator in a Banach space $\mathbf{B}$. Suppose that for every $x \in \mathbf{B}$ there exists some real numbers $c_x>0$ and $a_x<1$ such that $||T^nx|| \leq ca^n$, for all $n \in \mathbb{N}$. Prove that $||T^n|| \to 0$ as $n \to \infty$.

I could only deduce that $T^nx \to 0$ for all $x \in \mathbf{B}$, but I don't think it's enough. It's easy to show, using the uniform boundedness principle, that $||T^n||$ is bounded, but again that's not enough.

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Is this really exercise from textbook? Why are you sure that this is true? –  userNaN Jul 4 '12 at 17:17
    
It is indeed from one of my tasks. I can only be as sure of this as I trust my professor... –  Abe Jul 4 '12 at 17:43
    
@Abe Can we assume $T$ is bounded? –  Alex Becker Jul 4 '12 at 21:21
    
@AlexBecker Yes. Sorry for not pointing it out. –  Abe Jul 4 '12 at 21:26

1 Answer 1

Apply uniform boundedness to the family of bounded operators $\{ nT^n \}$. Since $\| T^n x \|$ decays so rapidly, we still have $\| nT^n x\| \to 0$ for any fixed $x$, so $\|nT^n\|$ is bounded by some constant $c$, and so $\|T^n\| \le c/n$.

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