Date: Apr 5, 2013 10:15 PM
Author: William Elliot
Subject: Re: Is it possible to bound these functions?
From marsh@panix.com Fri Apr 5 19:13:04 2013

Date: Fri, 5 Apr 2013 19:13:04 -0700 (PDT)

From: William Elliot <marsh@panix.com>

To: William Elliot <marsh@panix.com>

Subject: Re: Is it possible to bound these functions?

> Define A{f(x)} as a mapping from the set of functions defined on the

interval [0,1] to the Reals.

> The functions are as "nice, smooth and integrable" as you may need them

> to be.

>

> A{f(x)} = [ int cos(int f(t) t=0..x) x=0..1 ]^2 +

> [ int sin(int f(t) t=0..x) x=0..1 ]^2

>

A(f) = (integral(0,1) cos(integral(0,x) f(t) dt) dx)^2

+ (integral(0,1) sin(integral(0,x) f(t) dt) dx)^2

> Given that a <= f(x) <= b, can it be shown that A{a} => A{f(x)} => A{b} ?

No, let a = -2, b = 2, f = 0.

integral(0,x) 0 dx = 0; A(0) = 1

integral(0,1) cos x = sin 1; integral(0,1) sin x dx = cos 1

If a /= 0, then, integral(0,x) a dt = ax;

integal(0,1) cos ax = (sin a)/a; integral(0,1) sin ax dx = (cos a)/a

A(a) = 1/a^2