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From: solrac140 on 6 Jan 2010 04:39
Why the integral operator K: L^2 (0,1) -> L^2(0,1) defined by:

Kx(t) = Integral[ (1+ts)exp(ts)x(s) ds from s=0 to s=1]

_cannot_ take constant values? i.e why the constant functions do not belong to the range of K.
From: William Elliot on 7 Jan 2010 02:22
On Wed, 6 Jan 2010 solrac140(a)hotmail.com wrote:

> Why the integral operator K: L^2 (0,1) -> L^2(0,1) defined by:
>
> Kx(t) = Integral[ (1+ts)exp(ts)x(s) ds from s=0 to s=1]

> _cannot_ take constant values? i.e why the constant functions do not
> belong to the range of K.

K0 = 0.
From: solrac140 on 7 Jan 2010 05:29
Sorry. I meant non-zero constant functions.
From: Robert Israel on 7 Jan 2010 20:19
On Wed, 06 Jan 2010 14:39:21 EST, solrac140(a)hotmail.com wrote:

> Why the integral operator K: L^2 (0,1) -> L^2(0,1) defined by:
>
> Kx(t) = Integral[ (1+ts)exp(ts)x(s) ds from s=0 to s=1]
>
> _cannot_ take constant values? i.e why the constant functions do not belong to the range of K.

(later amended to: nonzero constant functions).


Consider the Taylor series (1+z) exp(z) = sum_{j=0}^infty c_j z^j,
where c_j > 0 for all nonnegative integers j. This has radius of
convergence infinity, since (1+z) exp(z) is entire.
K(x)(t) = sum_{j=0}^infty c_j int_0^1 (ts)^j x(s) ds
= sum_{j=0}^infty c_j t^j int_0^1 s^j x(s) ds

So for K(x)(t) to be constant, we need all the moments
int_0^1 s^j x(s) ds = 0 for positive integers j. But polynomials
are dense in L^2...

--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: TCL on 8 Jan 2010 17:27
On Jan 7, 8:19 pm, Robert Israel
<isr...(a)math.MyUniversitysInitials.ca> wrote:
> On Wed, 06 Jan 2010 14:39:21 EST, solrac...(a)hotmail.com wrote:
> > Why the integral operator K: L^2 (0,1) -> L^2(0,1) defined by:
>
> > Kx(t) = Integral[ (1+ts)exp(ts)x(s) ds from s=0 to s=1]
>
> > _cannot_ take constant values? i.e why the constant functions do not belong to the range of K.
>
> (later amended to: nonzero constant functions).
>
> Consider the Taylor series (1+z) exp(z) = sum_{j=0}^infty c_j z^j,
> where c_j > 0 for all nonnegative integers j.  This has radius of
> convergence infinity, since (1+z) exp(z) is entire.
> K(x)(t) = sum_{j=0}^infty c_j int_0^1 (ts)^j x(s) ds
>         = sum_{j=0}^infty c_j t^j int_0^1 s^j x(s) ds
>
> So for K(x)(t) to be constant, we need all the moments
> int_0^1 s^j x(s) ds = 0 for positive integers j.  But polynomials
> are dense in L^2...
>
> --
> Robert Israel              isr...(a)math.MyUniversitysInitials.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia            Vancouver, BC, Canada

I agree that polynomials are dense in L^2, but here we are dealing
with polynomials of degree at least one since j>0. Is this set of
functions also dense?
-TCL
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