formulas for $ \lVert A\rVert_{(1,\infty)} \text{ and } \lVert A\rVert_{(2,\infty)} $ Question: Find the formulas for $
   \lVert A\rVert_{(1,\infty)} \text{ and } \lVert A\rVert_{(2,\infty)} \text{ and sketch the proof} \text{ ( A is a Matrix Norm)}
$
Comments: I know the proof of the formula for $
   \lVert A\rVert_{\infty} = \max_i \lVert a_{(i,.)} \rVert$ 
Aslo know that $\lVert A\rVert_{(1,\infty)} = \sup\frac{\lVert Ax\rVert_{\infty}}{\lVert x\rVert_{1}} = \max_i \lVert a_{(i,.)} \rVert_1$
 And $\lVert A\rVert_{(2,\infty)} = \sup\frac{\lVert Ax\rVert_{\infty}}{\lVert x\rVert_{2}} = \max_i \lVert a_{(i,.)} \rVert_2$
I am interested to know the proofs of those formulas.

For every $p\ge1$,
$$
\|A\|_{(p,\infty)}=\max_{\|x\|_p=1}\|Ax\|_\infty=\max_{\|x\|_p=1}\max_i|a_{i\ast}x|=\max_i\max_{\|x\|_p=1}|a_{i\ast}x|.\tag{1}
$$
Since $a_{i\ast}x$ is a scalar, its absolute value $|a_{i\ast}x|$ is equal to its $p$-norm $\|a_{i\ast}x\|_p$. Hence $(1)$ gives $\|A\|_{(p,\infty)}=\max_i\max_{\|x\|_p=1}\|a_{i\ast}x\|_p=\max_i\|a_{i\ast}\|_p$. Put $p=1$ or $2$, the results follow.
