From: Dave Rusin Date: Mon, 29 Jun 1998 10:49:20 -0500 (CDT) To: sb0264@uni-essen.de Subject: Re: f(x+y)=f(x)*f(y) >are there functions f other than exp that suffice f(x+y)=f(x)*f(y) ? Yes, although other than f(x) = exp(a*x) for fixed but arbitrary a, one must be willing to explore functions which are not continuous, or which have a limited domain. For example, one may define a function F : R -> Q which projects the set of real numbers to the set of rational numbers (so that F(r) = r if r is already rational) using a basis for the rational vector space R; then one may let f(x) = exp(F(x)). Without assumptions of continuity or anything, you're simply asking for the homomorphisms between the groups (R, +) and (R-{0}, *). These are both Abelian groups. You could consult index/20-XX.html to get started. Kaplansky has a nice book on Abelian group theory. dave ============================================================================== From: s_starr@ix.netcom.com Newsgroups: sci.math Subject: Re: f(x+y)=f(x)*f(y) Date: Mon, 29 Jun 1998 14:15:27 GMT On 29 Jun 1998 14:36:44 +0100, mtx014@coventry.ac.uk (Robert Low) wrote: >Robin Chapman wrote: >OK, I'll bite. I can see how to assume no more than >differentiability at one point, but I don't see how >you can get away with just continuity. >-- >Rob. http://www.mis.coventry.ac.uk/~mtx014/ You can see that f(n*x) = f(x)^n for any positive integer n. Then you can see that f((a/b)x) = f(x)^(a/b) because f(b*(a/b)x) = f(a*x)=f(x)^a and also equals f((a/b)x)^b. If we take x=1 then we see that f(q) = f(1)^q = exp(c q) for all rational q, where c=log(f(1)). Thus if f is continuous then, since it equals the exponential map on a dense set Q, it equals the exponential map on the whole real line. And also, continuity at any point y and finiteness of f, guarantees continuity at every point x by translation: |f(x+h) - f(x)| = |f(x-y)|*|f(y+h) - f(y)| -> 0 as h->0. So if f(x) is finite and continuous at any point, then it must be an exponential map. ============================================================================== From: robjohn9@idt.net (Rob Johnson) Newsgroups: sci.math Subject: Re: f(x+y)=f(x)*f(y) Date: 29 Jun 1998 14:49:08 GMT In article <6n859c$s5i@leofric.coventry.ac.uk>, mtx014@coventry.ac.uk (Robert Low) wrote: >Robin Chapman wrote: >(On when f(x+y)=f(x)f(y) implies that f(x)=exp(x)) >>With more work we can reduce the assumption to continuity and then >>to continuity at one point. > >OK, I'll bite. I can see how to assume no more than >differentiability at one point, but I don't see how >you can get away with just continuity. Letting y = 0, f(x+y) = f(x)f(y) implies that f(x)(f(0) - 1) = 0 for all x. So either f(0) = 1 or f(x) = 0 for all x. Let's take the case where f(0) = 1. It is easy to prove by induction that for any x and any integer k, f(kx) = f(x)^k [1] This means that f(1/q)^q = f(1), that is to say, f(1/q) = f(1)^{1/q}. Therefore, f(p/q) = f(1)^{p/q}. This means that f(x) = f(1)^x for all rational x. If f is continuous at any point, then f(x+y) = f(x)f(y) allows us to prove that f is continuous everywhere. Since we know that f(x) = f(1)^x on a dense subset of the reals and f is continuous, f(x) must be equal to f(1)^x for all real x. f(1) must be positive since f(1) = f(1/2)^2, so we can write f(1) = e^c and so f(x) = e^{cx}. Rob Johnson robjohn9@idt.net ============================================================================== From: orjanjo@math.ntnu.no (Orjan Johansen) Newsgroups: sci.math Subject: Re: f(x+y)=f(x)*f(y) Date: 1 Jul 1998 15:22:59 GMT In article <6n7s8r$mnn$1@nnrp1.dejanews.com>, Robin Chapman wrote: >In article <6n7i7i$7if$1@nnrp1.dejanews.com>, > Urs.Schreiber@uni-essen.de wrote: >> >> Are there (analytic) functions f other than f=exp that suffice >> f(x+y)=f(x)*f(y)? Which? > [snip proof] >Therefore f(x) = exp(kx). > >What have we asssumed? Just differentiability. We can reduce this >to differentiability at one point by the equation f(x+y) = f(x) f(y). >With more work we can reduce the assumption to continuity and then >to continuity at one point. It is also enough to assume (for real functions) that the function is Lebesgue measurable. Use the nice theorem (Exercise 1.5.31 of Folland's Real Analysis): If E is a set with positive measure, then E-E contains a neighborhood of 0. Because then you can let E be the set of points where |ln f| and .