From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: convergence of the derivatives of a sequence of functions Date: 17 Sep 1999 13:14:22 -0400 Newsgroups: sci.math Keywords: f_n -> f =>? (f_n)' -> f' ? In article <7rtlmq$c1r$1@nnrp1.deja.com>, wrote: :I have a doubt about the convergence of the derivatives of a sequence :of functions and would like some opinions on this subject. Let (Fn) be :a sequence of functions from R to R, defined on an interval J, and :suppose F'n (the derivative of Fn) exists on J for any x in J. I want :to determine under what conditions the sequence (F'n) will converge on :J to a function F' that is the derivative of F, the limit of (Fn). : :According to most calculus books, (Fn) will converge on J to a function :F and (F'n) to F' if (but not only if): : :The numeric sequence (Fn(x0)) converges for some xo in J (It's not :clear if J must be closed or not) : :The sequence of functions (F'n) converges uniformly on J : :This conclusion is often presented as a theorem. It's not assumed that :the functions F'n are continuous on J. From the conditions above, it's :proved (the proof is long and not trivial at all)that (Fn) will :converge uniformly to a function F whose derivative F' is the limit of :(F'n). : :My doubt is that, like I said, it's not clear if J is closed or not :and, anyway, it's not assumed the functions F'n are continuous. But :though it's not explicitly mentioned, during the prrof it's assumed :that de derivatives F'n are bounded on J, since the prrof is based on :the concept of norm of a function on a set J, defined as supremum {|f(x) :| : x is in J}. The prrof assumed the norm J of the derivatives F'n :exists for any natural n. : :If it's not assumed the the derivatives F'n are bounded, then is the :prrof still valid? And if J is an opem interval? : :Thank you :ACS : Here is a couple of theorems from an old (perhaps forgotten) local Analysis I textbook (in Slovak), and I checked the proofs. [Bracketed remarks are added by me.] (1) Suppose {f_n'} converges uniformly on an [open bounded] interval (a,b). Suppose also that there is a point c in (a,b) for which the sequence {f_n(c)} is convergent. Then the sequence {f_n} is uniformly convergent on (a,b). (2) [A sort of closed graph statement] Suppose {f_n} converges to f uniformly on an open interval J [bounded or unbounded]. Suppose also that {f_n'} converges to a function g uniformly on J. Suppose that f_n' are all continuous. Then f is differentiable in J and f'=g. Observe that in theorem (1), the derivatives were not expected to be continuous or even bounded. In theorem (2), the derivatives are expected to be continuous, but need not be bounded. The openness of the intervals is associated with the fact that endpoint derivatives were not needed (they would be one-sided), and modifications for closed intervals are easy to make (extend linearly beyond the endpoints for a short interval). The proofs are done with the use of the Mean Value Theorem, so that connectedness of the domain is needed. Local versions are available, and they are mentioned in exercises. Cheers, ZVK(Slavek). ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: Uniform convergence Date: 15 Jun 1999 02:07:48 -0400 Newsgroups: sci.math In article <7k3kdj$13t$1@nnrp1.deja.com>, wrote: >A uniform limit of complex analytic functions is complex analytic. One >may prove this using Morera's theorem. > >Is it also true that a uniform limit of real differentiable functions >f_n(x) -> f(x) is real differentiable? [...] The distinction between the behaviour of real differentiable functions and complex ones is dramatic: Suppose, for simplicity, that the domain is a closed bounded interval J. Then there exists a function f continuous on J and nowhere differentiable, and there exists a sequence of polynomials (can't ask for smoother functions) convergent uniformly to f on J. The example and the approximation theorem (for any function continuous on J) are both due to Karl Weierstrass. Later it was found that in a way, an "overwhelming majority" of continuous functions on J are nowhere differentiable. (If you want to violate differentiability at a single point, consider J=[-1, 1] and f_n(x) = sqrt(x^2+1/n), convergent uniformly to abs(x).) A remedy: Integration behaves nicely under uniform convergence (it is a continuous operation on the space of continuous functions with maximum norm). From this (for continuous differentiability), or from the Mean Value Theorem (for differentiability not necessarily continuously), we can prove an amended and valid test (related to a "closed graph statement" for the operator of differentiation): Suppose f_n are differentiable on J, and {f'_n} converges uniformly to g, and there exists c in J such that the sequence of numbers f_n(c) converges, then {f_n} converges uniformly to a differentiable function f, and f'=g. Variants with relaxed conditions also exist. Hope it helps, ZVK(Slavek).