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In <18ltblINN8oq@darkstar.UCSC.EDU> sutin@helios.ucsc.edu (Brian Sutin) writes:

>I have a conjecture which I have been unable to find in the library, prove,
>or discover a conterexample:

>There exists some constant C > 0 such that for all non-negative functions
>f(x) on [0,inf] which satisfy the inequality

>INT{ f(x) x dx } <= C INT{ f(x) dx } ,

>the inequality

>INT{ f(x) dx }^2 <= INT{ f(x)^2 dx }

>also holds.  All integrals are over [0,inf] and converge.

Without loss of generality, we can assume INT { f(x) dx } = 1, as all the
inequalities scale.

Now we can rewrite the problem as:

INT {f(x) dx} = 1
INT {f(x) x dx} = A <= C
implies
min INT {f(x)^2 dx} >= 1

As a calculus of variations problem, we can (to find what f has the min
value) write

D[INT {f(x)^2 dx} - \lambda INT {f(x) dx} - \mu INT {f(x) x dx},f] =
  INT {2 f(x) df(x) dx} - \lambda INT {df(x) dx} - \mu {df(x) x dx} =
  INT {df(x) (2 f(x) - \lambda - \mu x) dx} >= 0,

for any df which is non-negative where f is 0.

Hence, f(x) > 0 -> f(x) = 1/2 (\lambda + \mu x)

Working over any interval [0,X] where X >= 3A, (to make it likely that
things converge and have limits), we find:

f_A(x) = (3A-x) (2/9 A^-2), for 0<=x<=3A, and 0 otherwise.

should be the extreme function.

Then, INT {f(x)^2 dx} = 4/(9 A)

In this case, we can verify directly that, if INT {g(x) dx} = 1, and
INT {g(x) x dx} = A, that

INT { g(x)^2 dx} = 
INT {(f_A(x)^2 + 2 f_A(x) (f_A(x) - g(x)) + (f_A(x)-g(x))^2) dx} =
INT {f_A(x)^2 dx} + 2 INT{f_A(x) (f_A(x) - g(x)) dx} + INT {(f_A(x)-g(x))^2) dx} =
     4/(9A)       +   (>= 0 by differencial approx.) +        >= 0
 
Hence, if C>= 9/4, the inequality holds.


--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
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let C = 1/8,
let f be L1 and L2 on [0,\infty),
and let \int_0^\infty f(x) x dx < C \int_0^\infty f(x) dx.

Then
       \int_0^C f(x) (C-x) dx >= \int_C^\infty f(x) (x-C) dx
implies with a slight amount of work
       \int_0^{2 C} f(x) dx >= \int_{2 C}^\infty f(x) dx,
meaning that the median is not more than twice the mean.

Apply Cauchy-Schwartz to f and the characteristic function on [0,2C].
Then,
       \int_0^\infty f^2 >= ( \int_0^{2 C} f )^2 / (2 C)
                         >= ( 1/2 \int_0^\infty f )^2 / (2 C)
                          = ( \int_0^\infty f )^2.

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