is it true that the infinity norm can be bounded using the $L_2$ norm the following way? Let $v \in \mathbb{R}^k$, and let $A \in \mathbb{R}^{m \times k}$ and let $B \in \mathbb{R}^{m \times n}$ such that each column of $B$, $B_i$, has $$||B_i||_2 \le 1.$$
Is it true that:


*

*$||v A^{\top} B||_{\infty} \le ||v A^{\top}||_2$ ?

*If the spectral norm of $A$ is such that $||A||_{\mathrm{spectral}} \le 1$, is it true that $||v A^{\top}||_2 \le ||v||_2$ ?
Thanks.
 A: I give it a try with your first question:
If $A\in\mathbb{R}^{m\times n}$ then
$$\frac{1}{\sqrt{n}}\|A\|_\infty\leq \|A\|_2\leq \sqrt{m}\|A\|_\infty$$
$$\frac{1}{\sqrt{m}}\|A\|_1\leq \|A\|_2\leq \sqrt{n}\|A\|_1$$
Also
$$\|AB\|_a\leq \|A\|_a\|B\|_a$$
Note that $\|B_i\|_2<1$ implies that absolute value of the every element of matrix B is less than 1, $|B_{ij}|<1$, then $\|B\|_\infty<n$ (infinity norm of a matrix is the maximum absolute row sum of the matrix). And with the same argument $\|B\|_1<m$.
Then using these properties we have
$$\|vA^TB\|_\infty\leq \sqrt{n}\|vA^TB\|_2\leq \sqrt{n} \|vA^T\|_2\|B\|_2\leq \sqrt{nm} \|vA^T\|_2\|B\|_\infty\leq n\sqrt{nm} \|vA^T\|_2$$
or
$$\|vA^TB\|_\infty\leq \sqrt{n}\|vA^TB\|_2\leq \sqrt{n} \|vA^T\|_2\|B\|_2\leq n \|vA^T\|_2\|B\|_1\leq nm \|vA^T\|_2$$
(these are conservative bounds you can probably obtain a tighter bound)
A: 1)   Cauchy-Schwarz says $$|(v A^T B)_i| = |v A^T B_i| \le \|v A^T\|_2 \|B_i\|_2 \le \|v A^T\|_2$$
2) Yes because the spectral norm is the operator norm corresponding to the $2$-norm on vectors, and $\|A\|_{\text{spectral}} = \|A^T\|_{\text{spectral}}$.
