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I've been struggling with this for a bit and was wondering if anyone can give me a hint:

Suppose $\{ T_n\}_1^\infty\subset \mathcal{L}\{ X,Y\}$ is a sequence of bounded operators from a banach space $X$ to a normed space $Y$, such that $\forall x\in X:\lim_n T_n x=0$. Prove that $\lim_n(\sup\{\Vert T_n x\Vert:x\in K\})=0$ on any compact set $K\subset X$.

As far as I can see, the fact that $T_n$ attains it's maximum on $K$ for any $n$ is not enough to prove this one, since it is possible that there exists an $\epsilon>0$ such that for any $N$ $\exists n>N$ such that $\Vert T_nx\Vert>\epsilon$ for some $x\in K$, and diminishes to 0 later on. Plus, this kind of solution doesn't employ the fact the $X$ is complete.

I've been trying to work this question out as a varient of the Principle of Uniform Boundedness, but I keep getting stuck. Anyway I would be very thankful if anyone could tell me if I'm on the right track with this.

Cheers

P.S This is homework

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2 Answers

up vote 3 down vote accepted

Since $T_nx\to 0$ for each $x$, $(T_nx)_n$ is bounded for each $x$ (convergent sequences are bounded), so you can apply uniform boundedness.

Suppose the limit is not zero. It either doesn't exist or is greater than $0$, but in either case there is an $\varepsilon\gt0$ and a subsequence $(T_{n_k})_k$ such that $\sup\{\|T_{n_k}x\|:x\in K\}\gt\varepsilon$ for each $k$. This implies that there is a sequence $x_1,x_2,\ldots$ in $K$ such that $\|T_{n_k}x_k\|\gt\varepsilon$. You can then use sequential compactness of $K$ and the result of uniform boundedness to derive a contradiction.

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Thanks a bundle, i think i got it –  kneidell Feb 15 '11 at 22:10
    
You're welcome. –  Jonas Meyer Feb 16 '11 at 1:01
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Hint: a) uniform boundedness principle is the right track. b) If $||x - x_0|| < \delta$ then $||T_n(x)|| \leq ||T_n(x_0)|| + ||T_n||\delta $.

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