Re: > Eigenvalue problem for sufficiently well-behaved functions

"k.hofmann" <boquiqui@xxxxxxxxxxxxxxxx> 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 israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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