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Let $I=[a,b]$ and consider the vector spaces $\mathcal{C}(I)=\{f:I\mapsto\mathbb{R}:f \text{ is continuous}\}$ and $\mathcal{R}(I)=\{f:I\mapsto\mathbb{R}:f \text{ is Riemann integrable}\}$.

What I'm trying to understand is why the function defined below,

$$\Vert f\Vert_1=\int_a^b|f(x)|\,dx$$

is a norm over $\mathcal{C}(I)$, but not over $\mathcal{R}(I)$. I'm thinking that maybe a function $f$ with some discontinuities and zero otherwise would satisfy $\int_a^b|f(x)|\,dx=0$, but $f(x)\neq 0$; if $f$ had to be continuous, this would not be allowed.

user401936
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1 Answers1

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The condition $\|f\|_1 = 0\implies f = 0$ is not satisfied on $\mathcal{R}(I)$, for example $\left\|\chi_{\{a\}}\right\|_1 = 0$.

However, you can turn $(\mathcal{R}(I), \|\cdot\|_1)$ into a normed space by identifying $$f = g \stackrel{def}{\iff} \|f-g\|_1 = 0 \iff \int_a^b |f-g| = 0$$

but then you are operating with equivalence classes of functions. See here for an example.

mechanodroid
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