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Let $l_1$ denote the space of absolutely summable sequences and $B(l_1,l_1)$ denote the space of all bounded linear operators from $l_1$ to $l_1$. I am trying to solve the following question

Let $A \in B(l_1,l_1)$ be such that $a_{m,n}=(Ae_n)(m)$ where $\{e_i\}$ are the cannonical basis of $l_1$. Give necessary and sufficient conditions on $(a_{m,n})$ for $A$ to be compact.

Can anyone give me useful directions on how to approach the problem? I know that every compact operator on $l_1$ can be approximated by finite rank operators. But how do I use this?

pikachuchameleon
  • 2,909
  • you need to make sense to "approximated by finite rank operators". consider the case $|a_{n,n}| = 1, a_{m,n} = 0$, is it compact ? (what is the rank of the $N\times N$ identity matrix by the way ?) – reuns May 04 '16 at 04:29

1 Answers1

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The norm of a bounded linear operator $A:\ell_1\to Y$, for any normed space $Y$, is simply $\sup_{n}\|Ae_n\|$. (This follows from the triangle inequality.) For $Y=\ell^1$, in terms of the coefficients $a_{m,n}$ this becomes $$\|A\| = \sup_{n}\sum_m|a_{mn}|\tag1$$ the supremum of $\ell^1$ norms of columns. For $A$ to be the limit of finite-rank operators, we must be able to reduce the norm to $<\epsilon$ by subtracting a finite rank operator.

For example, subtracting $p_M A$, where $p_M$ is the projection onto the first $M$ coordinates, we find $$\|A-p_MA\| = \sup_{n}\sum_{m>M}|a_{mn}|\tag2$$ So, a sufficient condition for compactness is $$\lim_{M\to\infty}\sup_{n}\sum_{m>M}|a_{mn}| = 0\tag3$$

It is also necessary. Indeed, if (2) does not tend to $0$ as $M\to\infty$, then we can inductively construct infinite sequences $n_k$, $M_k$ such that $$\|A e_{n_k} - p_{M_{k-1}}Ae_{n_k}\|\ge \epsilon,\quad \|A e_{n_k} - p_{M_{k}}Ae_{n_k}\|< \epsilon/2 $$ for a fixed $\epsilon>0$.

Hence, $A e_{n_k}$ has substantial norm within coordinates numbered between $M_{k-1}$ and $M_k$, which implies that the set of vectors $\{Ae_{n_k}\}$ does not lie in a small neighborhood of a finite-dimensional subspace.