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Isn't $\{x:1=||x||\}$ always a closed set and aren't norms always continuous? If that's the case then why is the norm of a matrix defined as a supremum instead of as a maximum?

la flaca
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    You need to a compact set, which you have if you are considering matrices. However, the maximum does not generalise to infinite dimensions. – Peter Jan 18 '17 at 12:21
  • @Peter so if we are dealing with a finite dimensional normed vector space is it ok to define the norm as the maximum instead of the supremum? I didn't know there were infinite matrices, very intersting! – la flaca Jan 18 '17 at 12:27
  • Also, why is that the maximum does not generalises to infinite dimensions? Why the extreme value theorem would not apply if the space is infinite dimensional? – la flaca Jan 18 '17 at 12:30
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    @laflaca The unit closed ball is compact if and only if the dimension is finite. EVT depends on compactness not closedness. – Leon Sot Jan 18 '17 at 12:37

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If $\max$ exists, then $\sup$ also exists and is the same value. So, one can replace $\max$ by $\sup$ in any context and never use $\max$. More consistent notation is generally a good thing. The flowchart for "when to use $\sup$ and when to use $\max$" simplifies to "use $\sup$". (*)

Similar to matrices, you will also often see $\|f\|_{C[a,b]} = \sup\{|f(x)|:x\in [a,b]\}$. This could be $\max$, but (a) does not really matter; (b) writing $\sup$ is consistent with what we would do for discontinuous functions and for functions on an unbounded interval, if such a space was considered there.

For matrices, using $\sup$ is consistent with what is done for operators on infinite-dimensional spaces.

(*) In practice, it's typical to use $\max$ for finite sets, where it's easier to think about picking the maximal element out of a finite collection. Also, $\max$ remains common when it really matters that the maximum is attained — for example, when we care about where it is attained.