```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 2013Date: 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) = 1integral(0,1) cos x = sin 1;  integral(0,1) sin x dx = cos 1If 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)/aA(a) = 1/a^2
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