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