From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: C-1 Date: 28 Aug 2000 23:08:26 -0400 Newsgroups: sci.math Summary: [missing] In article <8of2sr$23pk$1@mortar.ucr.edu>, John Baez wrote: :In article <8oep2c$atl$1@nnrp1.deja.com>, wrote: :>I was meaning to find a proof that: :> :>"If f is a C-1 function, then f is differentiable". :> :>Apparently it requires the MVT (Mean Value Theorem). : :What's your definition of a C^1 function? Mine is: :a differentiable function whose derivative is continuous. :With this definition, it's obvious that C^1 implies :differentiability. In more variables, some textbooks make this distinction: C^1 consists of functions whose all first partial derivatives are continuous on an open set, differentiable functions are those which have a (Frechet) derivative at every point of that open set. Although Rudin's definition (3rd Ed. of Principles, Def. 9.20) calls C^1 the functions whose derivative exists and depends continuously, as a linear-mapping-valued function, on the point (as it should), Rudin does formulate and prove (Theorem 9.21) the implication "continuous partial derivatives => C^1". And mean value theorem, one variable at a time, is indeed used there. Hope it helps, ZVK(Slavek).