# $L^p$-space

## Primary tabs

Defines:
$p$-integrable function, $L^\infty$, essentially bounded, $L^p$-norm
Synonym:
$L^p$ space, essentially bounded function
Type of Math Object:
Definition
Major Section:
Reference

## Mathematics Subject Classification

### L^\infty special?

I have a question... can't you just define L^\infty as a limit (in p)?
-apk

### Re: L^\infty special?

Won't work for spaces of infinite measure, that
is mu(X)=\infty

For such a space the constant functions are in
L^\infty, but not in any of the L^p.

Doesn't work for spaces of finite measure either

Consider X=[0,1] and f(x) = x

We have ||f||_p = (p+1)^(-p) -> 0
However ||f||_oo = 1

### Re: L^\infty special?

Your example fails : For X=[0,1] and f(x)=x, we have ||f||_p=(p+1)^(-1/p) -> 1 (not (p+1)^(-p)). In fact, the following is true :

THM : For any measure space X and function f in L^\r for some r<\infty, we have lim_{p->\infty}||f||_p=||f||_\infty.

The proof is a simple exercise with Chebyshev's inequality; see "Lebesgue Integration on Euclidean Space" by Frank Jones, p. 239 for more details.

### Re: L^\infty special?

Thanks. My example does fail - I erred. Thanks also for the THM.

### Re: L^\infty special?

I'm not sure that theorem is entirely true either. Take X to be the
degenerate measure space i.e. any subset including itself has measure 0.

and take f=0 which is in L^r for r<00, but

lim_{p->\infty}||0||_p = 0, and ||0||_\infty = -infinity, by the way

the essential supremum is defined.