Let $(a(i))_{i=1}^\infty$ be an absolutely summable sequence, i.e., $\sum_{i=1}^\infty |a(i)|<\infty$, and consider the infinite matrix $$A=\begin{bmatrix} a(1)&a(2)&a(3)&\cdots\\ a(2)&a(3)&a(4)&\cdots\\ a(3)&a(4)&a(5)&\cdots\\ \vdots&\vdots&\vdots&\ddots\end{bmatrix}.$$ My problem is to show that $A$ defines a bounded compact operator on $l^2=l^2(\mathbb{N})$. Namely, the operator $A$ defines is given by, for every $x=(x_j)_{j=1}^\infty\in l^2$, $A(x)=(\sum_i a(i+j-1)x_i)_{j=1}^\infty$.

At first, I thought it would be similar to this other problem: "a compact operator on $l^2$ defined by an infinite matrix", but the only part that I managed to do is to prove that the sequence $(\sum_i a(i+j-1)x_i)_{j=1}^\infty$ is well-defined: Indeed, since $(a(i))$ is absolutely summable then it is also square-summable, and it follows from Holder's inequality that the series $\sum_i a(i+j-1)x_i$ converges whenever $(x_i)\in l^2$. I can't even prove that the operator $A$ has image in $l^2$.

One of the initial problems is that the entries of the infinite matrix are not necessarily square summable (for example, take your favorite summable sequence $(a(i))$ and put lots of zeros between consecutive entries).

Any help is welcome.


One of the main tools on infinite matrices and Hilbert spaces operators is the so-called Schur's test. This is Exercise 45 in Halmos' A Hilbert Space Problem Book.

Schur's test. Let $A=[a_{ij}]_{i,j\in\mathbb{N}}$ be an infinite matrix. Suppose that there exist positive numbers $p_i>0$, $q_j>0$ ($i,j\in\mathbb{N}$), $\beta>0$ and $\gamma>0$ satisfying $$\sum_k |a_{kj}|p_k\leq\beta q_j\qquad\text{and}\qquad \sum_k |a_{ik}|q_k\leq\gamma p_i\qquad\text{ for all }i\text{ and }j$$ Then the matrix $A$ defines an operator on $\ell^2$ with $\Vert A\Vert^2\leq\beta\gamma$.

(In Halmos' book, we only consider matrices with non-negative entries, but the same proof applies in the general case.)

To solve your problem, we apply Schur's test on your matrix $A$ with $p_i=q_j=1$ and $\beta=\gamma=\sum_i|a(i)|$. In particular, the norm of the operator which $A$ defines is $\leq\sum_i|a(i)|$.

We can prove compactness of $A$ in the following way: the preceeding argument shows that the map $\mathcal{A}:\ell^1\to B(\ell^2)$ which assigns to a sequence $(a(i))_{i=1}^\infty$ the matrix as in your question is bounded, of norm $\leq 1$. Note that if a sequence $\mathbf{a}$ is finitely supported then $\mathcal{A}(\mathbf{a})$ has only finitely many nonzero entries, hence it has finite rank and is compact. Given your sequence $(a(i))$, approximate it by finitely supported sequences (in the $\ell^1$ norm), say $\mathbf{a}_n\to (a(i))$ with each $\mathbf{a}_n$ finitely supported, so $\mathcal{A}(\mathbf{a}_n)$ is a sequence of compact operators which converges to $A=\mathcal{A}((a(i)))$, which is therefore compact.

Just for the sake of completeness, I will write the proof of Schur's test as in Halmos' book.

Proof of Schur's test. Let $\mathbf{x}=(x_i)_i\in\ell^2$. Then \begin{align*} \sum_i\left|\sum_j|a_{ij}||x_j|\right|^2&=\sum_i\left|\sum_j\left(\sqrt{|a_{ij}|}\sqrt{q_j}\right)\left(\frac{\sqrt{|a_{ij}|}|x_j|}{\sqrt{q_j}}\right)\right|^2\\ &\leq\sum_i\left(\sum_j|a_ {ij}|q_j\right)\left(\sum_j\frac{|a_{ij}||x_j|^2}{q_j}\right)\qquad\text{by Cauchy-Schwarz}\\ &\leq\sum_i\gamma p_i\left(\sum_j\frac{|a_{ij}||x_j|^2}{q_j}\right)\\ &=\gamma\sum_j\frac{|x_j|^2}{q_j}\left(\sum_i|a_{ij}|p_i\right)\\ &\leq\gamma\sum_j|x_j|^2\beta=\beta\gamma\Vert\mathbf{x}\Vert^2 \end{align*} Thus for every $i$, $\sum_j|a_{ij}||x_j|<\infty$, so $\sum_ja_{ij}x_j$ converges, and the sequence $A(\mathbf{x})=(\sum_j a_{ij}x_j)$ is well-defined. Considering the $\ell^2$-norm, the inequality above implies $$\Vert A(\mathbf{x})\Vert^2=\sum_i\left|\sum_ja_{ij}x_j\right|^2\leq\beta\gamma\Vert\mathbf{x}\Vert^2,$$ so $A$ defines an operator on $\ell^2$ with $\Vert A\Vert^2\leq\beta\gamma$. QED

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