From: israel@math.ubc.ca (Robert Israel) Subject: Re: Riemann-Lebesgue operators Date: 14 Apr 2000 17:23:49 GMT Newsgroups: sci.math Summary: [missing] In article , Allan Adler writes: > A self-adjoint not-necessarily-bounded operator A on a Hilbert space H > is said to be a Riemann-Lebesgue operator if the operator exp(itA) > converges to 0 in the weak operator topology as t goes to plus or > minus infinity. If A has this property, so does -A, but 0 doesn't, > so the Riemann-Lebesgue operators are not closed under addition. > Suppose A,B are Riemann-Lebesgue operators on H, suppose that A,B commute > and suppose that A+B is invertible. Is it true that A+B is also a > Riemann-Lebesgue operator? If so, where can I find a detailed proof? You have to be a bit careful with the definition of "commute" when unbounded operators are involved. See e.g. Reed and Simon, "Functional Analysis", sec. VIII.5. The "good" definition is that A and B commute if all the projections in their associated projection-valued measures commute, or equivalently exp(isA) and exp(itB) commute for all real s and t. Then there is a projection valued measure P on \R^2 such that exp(isA) exp(itB) = int_{\R^2} exp(i [s,t].[x,y]) dP(x,y). The "marginals" P_A(E) = P(E x \R) and P_B(E) = P(\R x E) are projection-valued measures for A and B. Now the Riemann-Lebesgue property for A says that for any u and v in H, the Fourier transform of the measure goes to 0 at +infinity and -infinity. This will certainly be true if A has only absolutely continuous spectrum, and is certainly false if A has an eigenvalue, but may or may not be true in the case of singular continuous spectrum. Anyway, the answer to the original question is no. It is possible for A+B to be invertible and have an eigenvalue when A and B individually have only absolutely continuous spectrum. For example, let H be L^2[0,1] with Lebesgue measure, A multiplication by x and B multiplication by 1-x. The Riemann-Lebesgue properties for A and B follow directly from the Riemann-Lebesgue Lemma, but of course A+B=I does not have the property. But what you can say is that if the joint projection-valued measure P is absolutely continuous with respect to Lebesgue measure, A+B (or any nonzero linear combination of A and B) will be Riemann-Lebesgue. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2