From: tao@sonia.math.ucla.edu Newsgroups: sci.math.research Subject: Re: Bounded functions Date: Fri, 06 Mar 1998 00:35:58 -0600 In article , "Raghu Gompa" wrote: > > > Is there a characterization available > for all bounded complex valued functions f > on an interval of real numbers such that > f (x) * exp(iax) has a bounded antiderivative for all > but a finite number of real numbers a > (i is the square root of -1, x is the variable) > > exp(iax), cos(ax), sin(ax) are examples. > Are there any other (interesting) examples? Yes; in fact, you'd be hard pressed to find a smooth bounded function which did NOT have this property. Van der Corput's lemma (for a proof, see "Harmonic Analysis", by E.M. Stein) states that if f(x) = \psi(x) e^{i \phi(x)}, where psi(x) has bounded variation and \phi has derivative bounded away from zero (except on a compact set), then the primitive of f is always bounded. So in particular, any function of the form f = \psi(x) e^{i \phi(x)}, where \psi has bounded variation and \phi is convex (or has finitely many inflection points) is going to be in your class. e^{i x^2} is a good example. Of course the class is also closed under finite linear combinations. An easy integration by parts shows that it is also closed under multiplication by functions of bounded variation. Terry -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/ Now offering spam-free web-based newsreading