Let $\lVert A \rVert_{p,q}=\sup\lbrace\lVert Ax \rVert_q ~\vert~ x \in V, \lVert x \rVert_p = 1 \rbrace$ be the operator norm.
I want to show that
$$\lVert A \rVert_{\infty, \infty} = \max_{1 \leq i \leq n} \sum_{j=1}^n{a_{ij}} $$
$$\lVert A \rVert_{1, 1} = \max_{1 \leq j \leq n} \sum_{i=1}^n{a_{ij}} $$
My first tought is to use the equivalent definition of operator norm, $\lVert Ax \rVert \leq c \lVert x \rVert$, and simplify the left member with inequality but It may not works because of the $c$.
Best Answer
(1) $$|(Ax)_i|=|\sum_jA_{ij}x_i^j|\le\sum_j|A_{ij}|\|x\|_\infty$$ so it follows that $\|Ax\|_\infty\le\sum_j|A_{ij}|\|x\|_\infty$ and so $\|A\|_{\infty,\infty}\le\max_i\sum_j|A_{ij}|$.
(2) \begin{align*}\|Ax\|_1=\sum_i|\sum_jA_{ij}x^j|&\le\sum_i\sum_j|A_{ij}||x^j|\\ &=\sum_j\left(|x^j|\sum_i |A_{ij}|\right)\\ &\le\left(\max_j\sum_i|A_{ij}|\right)\sum_j|x^j|\\ &=c\|x\|_1 \end{align*} so $\|A\|_{1,1}\le\max_j\sum_i|A_{ij}|$.