From: vaillant Subject: Re: absolutely continuous funcs (urgent) Date: Sun, 07 May 2000 06:39:24 -0700 Newsgroups: sci.math >Let int(f,[a,b]) mean the integral of f in [a,b] and var(f,[a,b]) mean >the total variation of f on [a,b]. > >Can someone give a hint as to how to show that if f is an absolutely >continuous function, then int(|f'|,[a,b])=var(f,[a,b]) i.e., the >integral of the absolute value of the derivative of f is equal to the >total variation of f? f being assumed to be absolutely continuous, and since you mention f', my understanding is that you assume granted the theorem stating that if f:R->R is absolutely continuous, there exists h:R->R measurable and locally integrable such that: f(t)=int(h,[0,t]) for all t. Moreover, f is almost everywhere differentiable with f'=h almost surely. See http://www.probability.net [Tutorial 16, th 9, p 11] for a precise statement [for functions defined on R+] Having accepted this result, your question amounts in showing that the total variation function |f| [not the modulus !] of f defined by f(t)=int(h,[0,t]) for some locally integrable h, is given by |f|(t)=int(|h|,[0,t]). A more general statement of that result [at least when defined on R+] can be found in Tutorial 15, th 3,p 4. To handle the particular case at hand, I would try the following: for simplicity let us assume f and h are defined on R+. and recall that |f|(t) is defined as the supremum of all sums Sum(|f(ti)-f(ti-1)|) taken over all finite sequences t0<=t1<=...<=tn in [0,t]. Consider (ti), i=0,...,n to be one such sequence. Each term |f(ti)-f(ti-1)| is bounded by int (|h|,[ti-1,ti]). It follows that int(|h|,[0,t]) is an upper bound of all sums involved in the definition of |f|(t), and we see that |f|(t)<=int(|h|,[0,t]). Err... I think I was over ambitious, as I cannot prove the reverse inequality in any simple way. I shall outline elements of the [only] proof I know: 1. Consider the case when h is integrable on R+ [localisation] 2. From f(t)=int(h,[0,t]) conclude that the [complex or signed] stieltjes measure defined by f [denoted df] has density h with respect to the lebesgue measure on R+ 3. The total variation of the measure df, denoted |df| has therefore a density |h| with respect to the lebesgue measure on R+ [see Tutorial 12, th 6, page 8] 4. The Stieltjes measure associated with the total variation |f| of f,denoted d|f|, is in fact the total variation of the [signed] stieltjes measure associated with f, i.e. d|f|=|df|. [See Tutorial 14, th 6, page 11] 5. It follows that d|f| has density |h| with respect to the lebesgue measure on R+, and finally |f|(t)=d|f|([0,t])=int (|h|,[0,t]). I d be happy to learn a more direct route. :-) Regards. Noel. -- http://www.probability.net * Sent from RemarQ http://www.remarq.com The Internet's Discussion Network * The fastest and easiest way to search and participate in Usenet - Free! ============================================================================== From: AIommi@aol.com (A. Iommi) Subject: Re; absolutely continuous funcs (urgent) Date: 7 May 2000 17:06:56 -0400 Newsgroups: sci.math On Sun, 07 May 2000 02:09:26 GMT chuleta2099@my-deja.com wrote: >Let int(f,[a,b]) mean the integral of f in [a,b] and var(f,[a,b]) >mean the total variation of f on [a,b]. > >Can someone give a hint as to how to show that if f is an absolutely >continuous function, then int(|f'|,[a,b])=var(f,[a,b]) i.e., the >integral of the absolute value of the derivative of f is equal to >the total variation of f? > For a proof of this result see for example pages 47-48 of the book F. Riesz, B. Sz.-Nagy, Functional Analysis, Dover Publ., Inc., New York (1990). Greetings A. Iommi