Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)

*To*: mathgroup at smc.vnet.net*Subject*: [mg67399] Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)*From*: dh <dh at metrohm.ch>*Date*: Thu, 22 Jun 2006 06:21:10 -0400 (EDT)*References*: <e7589k$l5d$1@smc.vnet.net> <e7846p$fmp$1@smc.vnet.net> <e7aoeo$8t5$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Please not that there is a typo in the foregoing msg.: Plus + should read Times *. Sorry about this, Daniel dh wrote: > Hi Ronnen, > after you changed the question to (square in the denominator): > r = (b-a) Integral[ f(x)^2 dx, {x, a, b} ] / > Integral[ f(x) dx, {x, a, b} ]^2 > your guess is certanly right. The proof is done using Cauch-Schwarz > inequality. If S is a (real) scalar product, then > S[f1,f2]^2 <= S[f1,f1]+S[f2,f2] > Now consider the scalar product: Integrate[f1[x] f2[x],{x,a,b}]: > > Integrate[1 f[x],{x,a,b}]^2 = S[1,f]^2 <= S[1,1] + S[f,f] = (b-a) + > Integrate[f[x]^2,{x,a,b}] > > Daniel >