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

*To*: mathgroup at smc.vnet.net*Subject*: [mg67381] Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)*From*: "jbaker75 at gmail.com" <jbaker75 at gmail.com>*Date*: Wed, 21 Jun 2006 02:12:42 -0400 (EDT)*References*: <e7589k$l5d$1@smc.vnet.net><e784ee$foq$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi Ronnen - Yes, this is always true. It follows by Holders Inequality, or the Scwartz inequality: Integral[ f(x) dx, {x, a, b} ] <= (Integral[ f(x)^2 dx, {x, a, b} ])^1/2 (Integral[ dx, {x, a, b} ])^1/2 so that Integral[ f(x) dx, {x, a, b} ]^2 <= Integral[ f(x)^2 dx, {x, a, b} ] (b-a) => r >= 1 Sincerely, Jeff Baker ronnen.levinson at gmail.com wrote: > Hi folks. > > Sorry, I omitted a trailing exponent in my definition of r: > > r = (b-a) Integral[ f(x)^2 dx, {x, a, b} ] / > Integral[ f(x) dx, {x, a, b} ]^2 > > I hope this correction makes my question clearer. > > Thanks, > > Ronnen. > > ronnen.levinson at gmail.com wrote: > > Hi. > > > > I'm trying to determine whether the following ratio > > > > r = (b-a) Integral[ f(x)^2 dx, {x, a, b} ] / > > Integral[ f(x) dx, {x, a, b} ] > > > > is always greater than or equal to one for 0 < f(x) <= 1. All values > > all real. > > > > I've obtained r>=1 for all tested choices of f(x), but seek guidance to > > find the general answer. > > > > Yours truly, > > > > Ronnen Levinson. > > > > P.S. E-mailed CC:s of posted replies appreciated.