Norm on the Lp space

**willy0625** · May 20th 2010, 04:01 PM

In showing that $\Vert .\Vert{p}$ defines a norm on $L_{p} (p\geq1)$ , I'm not so sure about checking the axiom

$\Vert f \Vert_{p}=0$ iff $f=0$ .

A textbook mentions defining an equivalence relation on $L_{p}$ space but I don't really get what that's all about.

**Drexel28** · May 20th 2010, 04:10 PM

Originally Posted by willy0625

In showing that $\Vert .\Vert{p}$ defines a norm on $L_{p} (p\geq1)$ , I'm not so sure about checking the axiom

$\Vert f \Vert_{p}=0$ iff $f=0$ .

A textbook mentions defining an equivalence relation on $L_{p}$ space but I don't really get what that's all about.

Don't take what I am about to say too seriously, wait for another member to verify.

It defines a seminorm on the space. The common thing to do in this case is to define an equivalence relation on your space by modding out by the kernel of $\|\cdot\|_p$

**Focus** · May 20th 2010, 06:15 PM

Originally Posted by willy0625

In showing that $\Vert .\Vert{p}$ defines a norm on $L_{p} (p\geq1)$ , I'm not so sure about checking the axiom

$\Vert f \Vert_{p}=0$ iff $f=0$ .

A textbook mentions defining an equivalence relation on $L_{p}$ space but I don't really get what that's all about.

Drexel is right except that $L^p$ is already defined as the equivalence class of functions under the relation f ~ g iff $f(x)=g(x)$ almost everywhere.

The main reason for this is that if you are doing measure theory (e.g. probability) you essentially don't care what happens on null sets. You will notice that people will abuse the notation a lot, by for example saying let $f \in L^p$ and consider f(x), but there are ways around this.

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