From: ikastan@sol.uucp (ilias kastanas 08-14-90) Subject: Re: When is a function a derivative? Date: 13 Mar 1999 20:58:15 GMT Newsgroups: sci.math In article <7ce5i9$n0j@mcmail.cis.mcmaster.ca>, Zdislav V. Kovarik wrote: @In article <36EA280D.46C@rochester.rr.com>, @Jonathan W. Hoyle wrote: @>Could someone tell me what properties a function f must have so that it @>can have the property: there exists a function g(x) such that g'(x) = @>f(x)? @> @>This question is very similar to: what properties must an integrable @>function f(x) have such that if g(x) is the integral of f(x), and h(x) @>is the derivative of g(x), then f(x) = h(x). @> @[...] @ @There are two necessary conditions that f, defined on an open interval @(a,b), be the derivative of another function: @ @(1) that f be of first Baire class (the pointwise limit of a sequence of @continuous functions), @ @(2) that f have the Intermediate Value property (the image of every @subinterval of (a,b) be an interval). @ @This is the easy part. I think I remember that conditions (1) and (2) @together constitute a sufficient condition. Well... up to homeomorphism, at least. For f with domain [0,1], let B = f's of Baire class 1, IB = Baire class 1 & Int. Value property, and D = derivatives (f = g'... one-sided at 0 and 1). Then D is a proper subset of IB; but for any f in IB there exists a homeomorphism g of [0,1] such that f(g(x)) is in D. In a certain sense D is not too far from B either: any function in B can be expressed as f + g*h, with f, g, h in D. Ilias ============================================================================== From: mcmcclur@bulldog.unca.edu (Mark C McClure) Subject: Re: When is a function a derivative? Date: 17 Mar 1999 14:51:04 GMT Newsgroups: sci.math Jonathan W. Hoyle (jhoyle1@rochester.rr.com) wrote: : Could someone tell me what properties a function f must have so that it : can have the property: there exists a function g(x) such that g'(x) = : f(x)? This question is fully explored in the book Differentiation of Real Functions by Andrew Bruckner. This was published in Springer-Verlag's CRM Monongraph series Volume 5, 1994. : the Weierstrass everywhere-continuous nowhere-differentiable function is : not the derivative of any function. Any everywhere continuous function is a derivative by the fundamental theorem of calculus. In general, derivatives may be discontinuous on any set of first category. The must satisfy the intermediate value property. Hope that helps, -- Mark McClure Department of Mathematics University of North Carolina at Asheville http://www.unca.edu/~mcmcclur/ mcmcclur@bulldog.unca.edu ============================================================================== From: rld@math.ohio-state.edu (Randall Dougherty) Subject: Re: Differentiable functions with non-continuous derivatives Date: 25 Feb 1999 23:45:19 GMT Newsgroups: sci.math In article <7b3ger$3m9$1@pegasus.csx.cam.ac.uk>, Adrian Cable wrote: >Does anyone know if there exist functions which are differentiable >everywhere (in a given interval, say [-1,1]) but whose derivative is >continuous nowhere in that interval, or can offer a proof that such a >function cannot exist? No, such a function cannot exist. If F is differentiable everywhere in an interval, with derivative f, then f is a Baire class 1 function (i.e., the pointwise limit of a sequence of continuous functions), because f(x) is the limit of f_n(x) = (F(x + 1/n) - F(x))/(1/n) as n -> infinity. And it is a standard result that a Baire class 1 function must be continuous at many points (a comeager set of them). > (Or, better, functions whose first derivative is >defined everywhere in a given interval, but whose second derivative is >defined nowhere in that interval). This, on the other hand, is easy; just integrate a continuous nowhere-differentiable function. Randall Dougherty rld@math.ohio-state.edu Department of Mathematics, Ohio State University, Columbus, OH 43210 USA "I have yet to see any problem, however complicated, that when looked at in the right way didn't become still more complicated." Poul Anderson, "Call Me Joe"