I have a new matrix norm defined as the sum of the absolute values of the elements: for $A \in \mathbb{C}^{m\times n}$ define:
$$\|A\| = \sum_{i=1}^{m}\sum_{j=1}^{n}|\alpha_{i,j}| = \sum\left(\begin{matrix} |\alpha_{1,1}| & ... & |\alpha_{1,n}| \\ \vdots & \ddots & \vdots \\ |\alpha_{m,1}| & ... & |\alpha_{m,n}| \end{matrix}\right)$$
I'm trying to prove this norm is submultiplicative, i.e. $\|AB\| \leq \|A\|\|B\|$. Here's what I have, if somebody wouldn't mind checking it:
\begin{aligned} \|AB\| &= \left\| \begin{matrix} \tilde{a}_1^Tb_1 & ... & \tilde{a}_1^Tb_n \\ \vdots & \ddots & \vdots \\ \tilde{a}_m^Tb_1 & ... & \tilde{a}_m^Tb_n\end{matrix} \right\| \\ &= \sum_{i = 1}^m \sum_{j = 1}^n |\tilde{a}_i^T b_j| & \text{def. of norm}\\ &\leq \sum_{i = 1}^m \sum_{j = 1}^n \|\tilde{a}_i\|_2 \|b_j\|_2 & \text{Cauchy-Schwartz}\\ &= \left(\sum_{i = 1}^m \|\tilde{a}_i\|_2 \right) \left( \sum_{j = 1}^n \|b_j\|_2 \right) & \text{algebra} \\ &\leq \left(\sum_{i = 1}^m \|\tilde{a}_i\|_1 \right) \left( \sum_{j = 1}^n \|b_j\|_1 \right) & \|x\|_2 \leq \|x\|_1 \\ & = \|A\|\|B\| & \text{norm is defined as sum of absolute value of elements} \end{aligned}