Suppose f is a sufficiently well-behaved real-valued function (say, a smooth function with compact support) defined on the set of real numbers), and define
T(f) = \integal _{-\infty}^x y f(y) dy
What are the eigenvalues for T ?
Which eigenvalues lead to square-integrable functions?
I tried differentiating both sides of
T(f) = cf
and obtained a differential equation
cf'(x) - x f(x) = 0,
equivalently, f'/f = x/c
provided c is not 0, in which case f will have the form
Ke^{x^2/2c}
I'm not terribly sure how to obtain the eigenvalues directly, though.
"k.hofmann" <boqui...@nospam.gmail.com> writes: > Suppose f is a sufficiently well-behaved real-valued function (say, a > smooth function with compact support) defined on the set of real numbers), > and define
> T(f) = \integal _{-\infty}^x y f(y) dy
> What are the eigenvalues for T ?
None, on that domain.
> Which eigenvalues lead to square-integrable functions?
All c with Re(c) < 0. That is, these are eigenvalues for the unbounded operator on L^2(R) with domain {f in L^2: int_{-infty}^x y f(y) dy in L^2(R)}.
> I tried differentiating both sides of
> T(f) = cf
> and obtained a differential equation
> cf'(x) - x f(x) = 0,
> equivalently, f'/f = x/c
> provided c is not 0, in which case f will have the form
> Ke^{x^2/2c}
For T to be defined on this (removing the restriction of compact support), you need Re(c) < 0.
> I'm not terribly sure how to obtain the eigenvalues directly, though.
That looks pretty direct to me. Note, however, that you may also want to look out for continuous spectrum. I think the spectrum of this unbounded operator consists of the whole complex plane. -- Robert Israel isr...@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
