From: "Noel Vaillant"
Subject: Re: Real Analysis question
Date: Wed, 24 Nov 1999 05:10:07 -0000
Newsgroups: sci.math
Keywords: L^p norms converge to L^\infty norm
> Could someone explain to me in a non-condescending manner how I would
> go about showing that if f is a bounded measurable function on [0,1],
> then lim (as p goes to infinity) of ||f||(sub p) = ||f||(sub infinity).
> where ||f||(sub p) is the p norm in L^p space and ||L||(sub infinity)
> is the infinity norm = ess sup|f(t)|.
Let ||f|| be the infinity norm of f, and Np(f) be its p-norm.
Then Np(f)<=||f||.
Suppose ||f||>0 and let e>0 be such that ||f||-e>0
Define A={x in [0,1]: |f(x)|>||f||-e}
Then m(A)>0, where m is the lebesgue measure on [0,1].
We have: int_[0,1].|f|^p.dm>=int_A.|f|^p.dm>=(||f||-e)^p.m(A)
So Np(f)>=(||f||-e).m(A)^(1/p)
Since m(A)>0, m(A)^(1/p) tends to 1 as p tends to infinity.
Hence, for p sufficiently large:
||f||>=Np(f)>=||f||-2e
I hope this is correct and that I wasn't condescending :-)
Regards. Noel.
-------------------------------------------
Dr Noel Vaillant
http://www.probability.net
vaillant@probability.net