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