Infinite matrix operator

Question

Given a matrix of complex numbers $A=(a_{ij})_{i,j=1}^{\infty}$ and a operator $S$ on the space $l^1(\mathbb{C}) $ as $(x_i)_{i=1}^{\infty}\mapsto (y_i)_{i=1}^{\infty} $ where $y_i=\sum_{j=1}^{\infty} a_{ij}x_j$.

I need to show that the condition" $\sum_i|a_{ij}|<C$ for any $j$ (where C is constant number) " is equivalent to the fact that $S$ is bounded linear operator from $l^1$ to $l^1$.

This condition is necessary: by taking the vectors which are zero except one index where they equal to 1. How can I show that it is also sufficient?

thank you.

Answer 1

It is a straightforward estimate: \begin{align} \|Sx\|_1&=\sum_{k=1}^\infty|(Sx)_k| =\sum_{k=1}^\infty\left|\sum_{j=1}^\infty a_{kj}x_j\right|\\ \ \\ \leq& \sum_{k=1}^\infty\sum_{j=1}^\infty |a_{kj}|\,|x_j| = \sum_{j=1}^\infty|x_j|\,\sum_{k=1}^\infty |a_{kj}|\\ \ \\ &\leq C\,\sum_{j=1}^\infty|x_j| =\|x\|_1. \end{align}