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I have that $T_n$ are bounded operators on $H_n$ ($n\geq 1$) and that $\sup ||T_i||<\infty$. Define $T=\oplus T_n$ and $H=\oplus H_n$. I want to show that $T$ is compact iff $T_n$ is compact for all $n$ and $||T_n||\rightarrow 0$.

Here is what I have so far:

Assume that $T$ is compact, and let $B_n$ be the unit ball in $H_n$. Then we have that $\overline{T_n(B_n)}$ is a closed subset of $\overline{T(B)}$ (the unit ball in $H$), so we get compactness of $T_n$, and to see that $|T_n|\rightarrow 0$ just note that if the limit didn't go to zero, then for some $\epsilon>0$ there is an infinite subsequence $\{n_i\}$ such that $|T_{n_i}|>\epsilon$. Pick $m$ large, and let $h_{n_i}\in H_{n_i}$ be such that $|T_{n_i}(h_{n_i})|\geq \epsilon$ for $i=1,...,m$. Let $h\in H$ be equal to $h_{n_i}$ in the $n_i$-position and $0$ elsewehere. Then, $|h|=\sqrt{m}$ so $|T(h)/\sqrt{m}|\geq \epsilon \sqrt{m}$, so letting $m\rightarrow\infty$ we get that $T$ is unbounded, a contradiction.

For the other direction I am a little stuck, I was thinking of using a theorem that says that for a bounded operator $S$, we have that $S$ is compact iff there is a sequence $S_n$ of operators of finite rank such that $|S-S_n|\rightarrow 0$. Maybe call $S_i$ to be $T_1\oplus...\oplus T_i$, and arguing that $S_i$ has finite rank? I can see that $|T-S_n|\rightarrow 0$ for if $h$ is a unit vector, then $$|(T-S_n)(h)|=|\sum_{n+1}^\infty T_n(h_n)|\leq \sup_{i\geq n+1}|T_i|\rightarrow 0$$, but I don't know where to use the hypothesis that $\sup |T_n|<\infty$ and how to show that $S_n$ has finite rank.

Daniel Montealegre
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    I think $|T(h)| = (m \epsilon^2)^{1/2}$ so that $|T(h)|/|h| \geq \epsilon$. So you need another proof for why the $|T_n|$ go to $0$ (uniform bounded alone ensures T is bounded). But anyway, this is an old question, it just confused me for a bit :) This question was also something I was looking for :D – abe.nong Jun 15 '17 at 12:57
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    I think your argument of $\epsilon \sqrt{m}$ is wrong. Note that $\lVert Th \rVert^2 = \sum_n \lVert T_n h \rVert^2 > \epsilon^2 m$, so $\lVert T(h) \rVert = \epsilon \sqrt{m}$ and hence $\lVert T(h) \rVert / \lVert h \rVert = \epsilon$. – Minsheng Liu Oct 26 '17 at 05:22
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    The true reason seems like this. Consider the sequence ${ h_n } \in \mathcal{H}_n$ where each $\lVert T_n(h_n) \rVert > \epsilon$. Clearly, for any $i \neq j$, $\lVert T_i(h_i) - T_j(h_j) \rVert > \sqrt{2}\epsilon$, so there is no convergent subsequence. The contradiction is the same as an infinite-dimensional closed unit ball is not a compact set. – Minsheng Liu Oct 26 '17 at 05:25

1 Answers1

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Fix $\varepsilon>0$. Choose $n_0$ such that $\|T_n\|<\varepsilon$ if $n> n_0$. For each $n=1,\ldots,n_0$, there exists a finite-rank $S_n$ with $\|S_n-T_n\|<\varepsilon$. Put $S_n=0$ for $n>n_0$. Then $\bigoplus_1^{n_0}S_n$ is finite-rank and $$ \|T-S\|=\sup\{\|S_n-T_n\|,\ n\in\mathbb N\} <\varepsilon. $$ So $T$ is a limit of finite-rank operators.

Martin Argerami
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