 Solution

For a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$, define the operator norm byIt is finite. To see this without assuming continuity of $A$, write $x=\sum x_j{e}_j$: the triangle inequality bounds $\Vert Ax\Vert$ by $\sum |x_j|\Vert A{e}_j\Vert$. Every norm in finite dimensions is equivalent to the Euclidean norm. One proof first uses this same coordinate bound to establish continuity of the norm on the Euclidean unit sphere; compactness then gives a strictly positive minimum there. Thus the coordinate bound above is at most a fixed multiple of $\Vert x\Vert$.

Absolute homogeneity and the triangle inequality for this operator norm follow from those for $\Vert Ax\Vert$ and taking suprema. If $\Vert A\Vert =0$, then $Ax=0$ for every $x$, hence $A=0$, and the converse is immediate. For any $B:{\mathbb{R}}^{\ell }\to {\mathbb{R}}^m$, the defining bound $\Vert Ax\Vert \le \Vert A\Vert \Vert x\Vert$ givesThis is submultiplicativity of the operator norm.