Are logarithms the only continuous function on (0,∞) such that f(xy)=f(x)+f(y)?

Question

ramirezhereva · Accepted Answer

Other than f(x)=0 for all x, logarithms are the only continuous functions with that property. We show this in steps.  f(1)=f(1⋅1)=f(1)+f(1), so f(x0)=f(1)=0=0⋅f(x). Obviously f(x1)=1⋅f(x).  f(x2)=f(x⋅x)=f(x)+f(x)=2f(x), and by induction we can show that f(xn)=nf(x) for all natural numbers n (including zero).  For rational number y=ab,  f(xy)=f(xab)=1b⋅bf(xab)=1bf((xab)b)=1b(xa)=abf(x)  =yf(x)  So f(xy)=yf(x) for all rational numbers y. By continuity we can extend this to all real values y.  If f(x) is ever non-zero, we can find b&amp;gt;0 and c≠0 such that f(b)=c. Then  0≠c=f(b)=f(eln⁡b)=ln⁡b⋅f(e)  so f(e)≠0. Let a=e1f(e), so f(e)=1ln⁡a. Then for all x&amp;gt;0,  f(x)=f(eln⁡x)=ln⁡x⋅f(e)=ln⁡xln⁡a=logax  Therefore, f(x) truly is a logarithm function.  If we give up the continuity restriction, I believe that the axiom of choice will show other such functions are possible. I think this is done by setting a well-order on the real numbers and using transfinite induction. But the details are probably beyond me.

Debbie Moore · Accepted Answer

some hints for an approach to the other part of your question.
you can show
$f (1) = 0$
and
$f (x) = - f (\frac{1}{x})$
also, for a set of distinct 's $p_{k}$ , and integers $a_{k}$
$f (\prod_{k = 1}^{n} p_{k}^{a_{k}}) = \sum_{k = 1}^{n} a_{k} f (p_{k})$
so that f is defined on $Q$ and can be extended to $R$ by continuity.
for integers $a, b$ we have
$f (n^{\frac{b}{a}}) = \frac{b}{a} f (n)$
using continuity and rational approximation $f (a^{s}) = a$ for $a$ and $s \in (0, \infty)$
if $f$ is not identically zero then there is a $b$ for which $f (b) = 1$

RizerMix · Accepted Answer

Well this is a partial answer: It is the only such function differentiableat x=1, since we have f(x+h)−f(x)h=1xf(1+h/x)h/x→f′(1)xas h→0, and the fundamental theorem of calculus (combined with f(xy)=f(x)+f(y)) gives that f(x)=f′(1)ln⁡x, so they are all of the form clogx. So now it would be enough to show that any such function is differentiable at x=1.Edit: Actually one can derive f(1+h)h=yf(y+hy)−f(y)hy , so it is enough to show that there exists one point at which f is differentiable.

Are logarithms the only continuous function on (0,\infty) such that

Answered question

Answer & Explanation

New Questions in Algebra II