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


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.


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