Inequality on infinite matrices

Question

Take the square of sum of elements in every column for first finite rows (say m). Then take their sum for all the columns (say n). If $m,n \rightarrow \infty$, can we show that it is bounded by some constant times the sum of squares of all the elements of the infinite matrix.

Answer 1

If I understand you correctly you're asking whether every infinite matrix $A$ satisfies $$ \sum_{k=1}^n\Big(\sum_{j=1}^mA_{jk}\Big)^2\leq C\sum_{j,k=1}^\infty |A_{jk}|^2 $$ for some constant $C$ and all $m,n\in\mathbb N$. This, however, is not true. Take $$ A=\begin{pmatrix} 1&0&0&0&\cdots\\ \frac12&0&0&0&\cdots\\ \frac13&0&0&0&\cdots\\ \frac14&0&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots&\ddots \end{pmatrix}\,; $$ then $ \sum_{k=1}^n(\sum_{j=1}^mA_{jk})^2=(\sum_{j=1}^m\frac1j)^2 $, whereas $ \sum_{j,k=1}^\infty |A_{jk}|^2=\sum_{j=1}^\infty \frac1{j^2}=\frac{\pi^2}6 $. As $\lim_{m\to\infty}\sum_{j=1}^m\frac1j=\infty$, there for all constants $C$ exists $m\in\mathbb N$ such that $\sum_{j=1}^m\frac1j>\pi\sqrt{\frac{C}6}$. Therefore $$ \sum_{k=1}^n\Big(\sum_{j=1}^mA_{jk}\Big)^2= \Big(\sum_{j=1}^m\frac1j\Big)^2 >\frac{C\pi^2}6 =C\sum_{j,k=1}^\infty |A_{jk}|^2 \,, $$ which is the desired contradiction.