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If $f:A\rightarrow B$ is continuous, A is a compact set of some metric $Y$, how dows one show $||f||$ is continuous, ...if it even is continuous. I am inducing that it is since if a real functions is continuous, then its absolute value function is as well.

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What is $\sup \|f(x)\|_Y$ defined as a function over? The space $C(A,B)$ of continuous functions from $A$ to $B$? – Alex Becker Apr 12 '12 at 6:40
the set of x's in A. – cap Apr 12 '12 at 6:41
...with the supremum taken over all $x\in A$? – Alex Becker Apr 12 '12 at 6:43
sorry, Alex, I edited the question. – cap Apr 12 '12 at 6:50
@capItan: Do you mean 'show the function $x \mapsto ||f(x)||$ is continuous'? – copper.hat Apr 12 '12 at 6:59
up vote 1 down vote accepted

Using the triangle inequality, you have $||x|| \leq ||x-y|| + ||y||$, and similarly with $x,y$ reversed. From this you can conclude that $|\;||x||-||y||\;| \leq ||x-y||$, ie, the norm is continuous. Now apply this, replacing $x$ by $f(x)$ and $y$ by $f(y)$, and use continuity of $f$.

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