Bounding $\|Ax\|_2$ in terms of $\|A\|_1$ and $\|x\|_2$

Question

Given a matrix $A$ of size $m\times n$ and a vector $x$ of size $n\times 1$, how can be bouund the $l_2$ norm $\|Ax\|_2$ in terms of $\|A\|_1$ and $\|x\|_2$, where $\|A\|_1$ is the sum of the absolute values of all entries of the matrix $A$.

Answer 1

One useful and simple bound is that $$||Ax||_2 \leq \sqrt{m}||A||_1 ||x||_2.$$ To see this, the $n$-th entry of $Ax$ is bounded by $$|(Ax)_n| = \left| \sum_{k} A_{nk} x_k \right| \leq \sum_k |A_{nk}| |x_k| \leq ||A||_1 \sum_k |x_k|,$$ so that $$||Ax||_2 = \sqrt{\sum_{j=1}^m (Ax)_n^2 } \leq ||A||_1 \sqrt{\sum_{j=1}^m \sum_{k=1}^n |x_j|^2} = ||A||_1 \sqrt{m \sum_{k=1 }^n |x_k|^2} = \sqrt{m}||A||_1 ||x||_2.$$