Operator norm: submultiplicativity and comparison with Frobenius

Idea. Submultiplicativity follows from applying the definition twice. The comparison with $\|\cdot\|_F$ uses Cauchy-Schwarz in one direction and a sum over the $n$ unit vectors $e_j$ in the other.

Step 1: the basic inequality. By definition of the supremum, for $x\ne0$ we have $|Ax|=|x|\,|A(x/|x|)|\le\|A\||x|$ (and it is trivial for $x=0$).

Step 2: submultiplicativity. For every $x$, $|ABx|\le\|A\||Bx|\le\|A\|\|B\||x|$. Taking the supremum over unit $x$ gives $\|AB\|\le\|A\|\|B\|$.

Step 3: $\|A\|\le\|A\|_F$. Let $r_i$ be the rows of $A$. Then $(Ax)_i=\langle r_i,x\rangle$ and by Cauchy-Schwarz
$$|Ax|^2=\sum_i\langle r_i,x\rangle^2\le\sum_i|r_i|^2|x|^2=\|A\|_F^2|x|^2.$$
Take $|x|=1$ and the supremum.

Step 4: $\|A\|_F\le\sqrt n\|A\|$. The columns of $A$ are the vectors $Ae_j$. Hence
$$\|A\|_F^2=\sum_{j=1}^n|Ae_j|^2\le\sum_{j=1}^n\|A\|^2|e_j|^2=n\|A\|^2.$$

Remark. Both bounds are sharp: the first is an equality for rank-one matrices, the second for orthogonal matrices (see the problem on the best constants). One consequence, used constantly: $\|A\|$ is controlled by the entries, e.g. $\|A\|\le\|A\|_F\le n\max|a_{ij}|$.