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From: TCL on 9 Jan 2010 10:59
On Jan 8, 5:27 pm, TCL <tl...(a)cox.net> wrote:
> 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- Hide quoted text -
>
> - Show quoted text -

The answer is yes. See my other post: Characterize this set in L^2.
-TCL
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