# Compactness of an operator on $c_0$ in terms of its infinite matrix representation

Let $A\in {\cal B}(c_0)$ (${\cal B}(c_0)$ is linear bounded operators on $c_0$) and for $n\geq 1$, define $e_n \in c_0$ by $e_n(m)=\delta_{nm}$. Put $\alpha_{nm}=(Ae_n)(m)$ for $n,m\geq 1$. we have $M = \sup_m\sum_{n=1}^\infty |\alpha_{nm}|< \infty$ and for every n, $\lim_m \alpha_{mn}=0$. Also we can show $(Ax)(m) = \sum_{n=1}^\infty \alpha_{mn}x(n)$. Give necessary and sufficient condition on $(\alpha_{mn})$ for $A$ to be compact.

My attempt: for every $k>0$ define $A_k:c_0\to c_0$ such that $(A_k x)(m):= \sum_{n=1}^\infty\alpha_{mn}x(n)$ for $m\leq k$ and $(A_kx)(m)=0$ for $m>k$. Clearly $A_k$ is finite rank and $A=\lim A_n$ which shows A is compact for every $(\alpha_{mn})$.

Please check my attempt. If it's not correct, Please hint me. Thanks in advance.

## 1 Answer

If $\alpha_{nm}=\delta_{nm}$, the operator $A$ is the identity, which is not compact. So, there is a problem with your attempt... here:

$A=\lim A_n$

This is true in the sense of pointwise convergence (strong operator topology). But to conclude compactness, you need norm convergence: $\|A-A_n\|\to 0$. This is not necessarily the case, as the example $\alpha_{nm}=\delta_{nm}$ demonstrates.

As you probably found out, the norm of $A$ is equal to $\sup_m\sum_{n=1}^\infty |\alpha_{nm}|$. Thus, if you want finite-rank approximation in the norm, the goal should be to truncate $A$ by removing finitely many rows and/or columns, so that the rest of the matrix has small $\sup_m\sum_{n=1}^\infty |\alpha_{nm}|$. Formally, the condition $$\sup_{m\ge N}\sum_{n=N}^\infty |\alpha_{nm}| \to 0,\qquad N\to\infty \tag1$$ is sufficient. I think it's also necessary... indeed, if (1) fails, you should be able to find an infinite sequence of unit vectors $v_n$ with disjoint supports such that $\|A v_n\|\ge c>0$ for all $n$.

• Do you define $(A_k x)(m):=\sum^k _{ n=1} \alpha_{ mn} x(n)$ for m≤k and (A_k x)(m)=0 for m>k? In this case I think we should put the condition $$\max\{\sup_{m\leq k}\sum_{n=k}^\infty |\alpha_{mn}|, \sup_{m> k}\sum_{n=1}^\infty |\alpha_{mn}|\}\to 0$$.Is it correct?
– niki
Commented Sep 5, 2014 at 7:34
• No, I was thinking of $A_N(x) = \sum_{\min (m,n)\le M} a_{mn} x(n)$. This operator has finite rank, because it's a sum of two matrices, one of finite width, another of finite height.
– user147263
Commented Sep 5, 2014 at 10:52
• Do you mean $(A_Nx)(m)=\sum_{\min{m,n}\leq N}\alpha_{mn}x(n)$ ? (I changed M to N). So I think our definitions of $A_N$ are the same.
– niki
Commented Sep 5, 2014 at 12:51
• Yes, I did mean that. Yes, now I see these are the same things. But I don't see why you have two suprema there. My thought process: we have an infinite matrix. Chop off two bands of finite width on both sides. What's left (which is also an infinite matrix) should have small norm.
– user147263
Commented Sep 5, 2014 at 13:02
• @niki The block with finitely many columns has finite rank, because its kernel is a space of finite codimension.
– user147263
Commented Sep 5, 2014 at 19:40