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
http://www.math.niu.edu/~rusin/known-math/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
.