If $U$ is finite dimensional, then operator norm is finite Let $M:U\to V$ be a linear map between normed vector space $U$ and $V$. We know $U$ is finite dimensional (but don't know about $V$). Define $\|M\| = \sup \{\|Mv\|\;:\;\|v\| = 1\}$. I want to show that $M$ is continuous and that $\|M\|$ is bounded.
There are two difficulties, first is that I know the proof for $M:\mathbb{R}^m\to\mathbb{R}^n$. Continuity of $M$ is due to each entry of operator matrix of $M$ is linear thus continuous. I show unit sphere is compact and therefore continuous function obtains max. But now we are not in $\mathbb{R}$ any more, I don't know if unit sphere is compact or not.
The second difficulty is that $V$ might not be finite dimensional any more, so I can't represent $M$ in a matrix of finite dimension any more. Please help.  

Further, not knowing if $U$ and $V$ are finite dimensional or not, I want to prove that $M$ continuous then $\|M\|$ finite.
 A: For the first part of the question:
Assume first that the norm on $U$ is given by $\lVert u \rVert_1 = \sum_{i=1}^n \lvert u_i \rvert $ where $u = \sum_{i=1}^n u_i e_i$ with respect to the basis $\{e_1,\dots,e_n\}$. With this norm we have that $\lVert u\rVert \leq 1$ implies $\lvert u_i\rvert \leq 1$. Then your estimate shows that $\lVert Mu \rVert \leq \sum \lvert u_i \rvert \lVert Me_i\rVert \leq C \lVert u\rVert_1$, where $C = \max\{\lVert Me_i\rVert \mid i=1,\dots,n\}$, so $\lVert M\rVert \leq C$.
Since $U$ is finite-dimensional, all norms on $U$ are equivalent. Therefore there is a constant $D \gt 0$ such that $\lVert u\rVert_1 \leq D \lVert u\rVert$. But then $\lVert Mu\rVert \leq C \lVert u\rVert_1 \leq CD \lVert u\rVert$, showing that $M$ is also bounded with respect to the original norm.

For the second question, use the $\varepsilon$-$\delta$-criterion for continuity. Since $M$ is continuous, it is continuous at $0 \in U$. Therefore there is $\delta \gt 0$ such that $\lVert u - 0\rVert \leq \delta$ implies $\lVert Mu \rVert = \lVert Mu - M0\rVert \leq 1$. Thus, $\lVert u\rVert \leq 1$ implies $\lVert Mu\rVert \leq 1/\delta$ and hence $\lVert M \rVert \leq 1/\delta$.
