The Frobenius norm is submultiplicative 
Let $M_{d \times d} (\Bbb R)$ be the space of real $d \times d$ matrices. For $A = (a_{ij} ) \in M^{d×d}$, define $$ \| A \|_2 := \left( {\sum_{i,j=1}^{d}|a_{ij}|^2} \right)^\frac{1}{2} $$ Show that $$\|AB\|_2 \leq \|A\|_2 \|B\|_2$$


I don't know where to start this step. How do I show that $I+A+A^2+\cdots$ converges to a limit if $\|A\|_2 \leq 1$?  If anyone could please help me, I would appreciate it.
 A: In my humble opinion brute-force calculation should suffice.
Denote $a_{ij}$ to be the element on the $i$-th row and $j$-th column of matrix $\mathbf{A}$; similar for $b_{ij}$. Then
$$\mathbf{A B} = 
\begin{bmatrix}
\sum a_{1j} b_{j1} & \sum a_{1j} b_{j2} & \cdots & \sum a_{1j} b_{jd} \\
\sum a_{2j} b_{j1} & \sum a_{2j} b_{j2} & \cdots & \sum a_{2j} b_{jd} \\
\vdots             & \vdots             & \ddots & \vdots             \\
\sum a_{dj} b_{j1} & \sum a_{dj} b_{j2} & \cdots & \sum a_{dj} b_{jd}
\end{bmatrix} \, .
$$
Then
$$
\begin{split}
\left\Vert \mathbf{A B} \right\Vert_2^2 & = \sum_i \sum_k \left\vert \sum_j a_{ij} b_{jk} \right\vert ^ 2 \\
& \leq \sum_i \sum_k \left[ \left( \sum_j \left\vert a_{ij} \right\vert ^ 2 \right) \left( \sum_j \left\vert b_{jk} \right\vert ^ 2 \right) \right] \ (*)\\
& = \sum_i \sum_k \left[ \left( \sum_j \left\vert a_{ij} \right\vert ^ 2 \right) \left( \sum_{\color{red}{l}} \left\vert b_{\color{red}{l} k} \right\vert ^ 2 \right) \right] \\
& = \sum_i \sum_k \sum_j \sum_l \left\vert a_{ij} \right\vert ^ 2 \left\vert b_{lk} \right\vert ^ 2 \\
& = \left( \sum_i \sum_j \left\vert a_{ij} \right\vert ^ 2 \right) \left( \sum_k \sum_l \left\vert b_{lk} \right\vert ^ 2 \right) \\
& = \left\Vert \mathbf{A} \right\Vert_2^2 \left\Vert \mathbf{B} \right\Vert_2^2 \, .
\end{split} 
$$
$\left(*\right)$ follows from Cauchy-Schawarz inequality. So there is $$ \left\Vert \mathbf{A B} \right\Vert_2 \leq \left\Vert \mathbf{A} \right\Vert_2 \left\Vert \mathbf{B} \right\Vert_2 \, . $$
To show that the equality sign can hold, just take this example:
\begin{align}
\mathbf{A} & = \begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 
\end{bmatrix} \, ,
\quad \mathbf{B} = \begin{bmatrix}
2 & 2 & 2 \\
2 & 2 & 2 \\
2 & 2 & 2 
\end{bmatrix} \\
\mathbf{A B} & = \begin{bmatrix}
6 & 6 & 6 \\
6 & 6 & 6 \\
6 & 6 & 6 
\end{bmatrix} \, .
\end{align}
Apparently $\left\Vert \mathbf{A} \right\Vert_2 = 3$, $\left\Vert \mathbf{B} \right\Vert_2 = 6$, $\left\Vert \mathbf{A B} \right\Vert_2 = 18 = \left\Vert \mathbf{A} \right\Vert_2 \left\Vert \mathbf{B} \right\Vert_2$.
