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As I understand it the norm $\|f\|_2$ on the set of continuous functions on $[0,1]$ is defined by $$\|f\|_2 = \sqrt{\int_0^1|f(t)|^2dt}$$ but what is the infinity norm $\|f\|_\infty$? Is it $$\int_0^1 \max|f(t)|dt$$ so in other words just $\max|f(t)|$?

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Yes. (Fill characters to get more than 15 characters...) –  user20266 Mar 31 '12 at 18:52
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up vote 2 down vote accepted

Up there I think you meant to write

$$ \|f\|_2 := \sqrt{\int_0^1|f(t)|^2dt} $$

The $\sup$-norm is usually defined as follows

$$ \|f\|_\infty := \sup_{x \in [0,1]} |f(x)|$$

Where $f: [0,1] \to \mathbb R$ is continuous, for example. So you get a normed space of functions $(C([0,1]), \|\cdot\|_\infty)$.

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Ah thank you! :) –  user26069 Mar 31 '12 at 18:58
    
@user26069 Np : ) Glad to help. I see you corrected the $x$s in your question so I'll assume that it was just a typo and not a matter of confusion : ) –  Matt N. Mar 31 '12 at 19:00
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@user26069: Also note that continuous function from $[0,1]$ attains a maximum, so this $\sup$ is really $\max$. –  Asaf Karagila Mar 31 '12 at 19:26
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