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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...)2012-03-31

1 Answers 1

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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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    @user26069: Also note that continuous function from $[0,1]$ attains a maximum, so this $\sup$ is really $\max$.2012-03-31