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norm p was defined as such:

$|.|_p=(\int_0^1|f(x)|^p dx)^{1/p}$

so, the norm 1 should look like this:

$|.|_1=(\int_0^1|f(x)| dx)$

when proving homogeneity and triangle inequality i just used the integral properties of constant multiple and sum of integrals; respectively. Now, my problem its with positive definiteness. my guess is this: since every function you are going to pluck its continous over a close interval, then its riemann integrable. If you're going to take the absolute value of that function, its always gonna be positive, then the result of the integral its always positive since the result of the integral of a positive riemann integrable function its always positive. I am doubtful about the last bit.

is this alright? Is there something i am missing?

thanks, any hint will help.

Edwyn Collins
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1 Answers1

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Continuity matters. The only continuous nonnegative function with integral equal to zero on $[0,1]$ is the zero function. For continuous $f:[0,1]\to\Bbb R$, $\int_0^1|f|=0\iff\|f\|_1=0\iff f\equiv0$.

This is well known and is proved many times on this site.

The result of the integral of a positive Riemann integrable function its always positive

Well, $f:x\mapsto\begin{cases}1&x=0\\0&0<x\le1\end{cases}$ is Riemann integrable, and not identically zero, yet $\int f=0$. So I'm not quite sure what you're getting at here: continuity is important. Either you do that, or as in the $\mathscr{L}^p$-space theory, we identify functions that are almost everywhere equal to get around this issue.

FShrike
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