How to prove that if a continuous function satisfies $f(a b)=f(a) + f(b)$, this function must be a log function?

Question

How to prove that if a continuous function satisfies $f(ab)=f(a)+f(b)$ and both $a$ and $b$ are positive real numbers, this function must be a log function? i.e., proof of uniqueness. Thanks

If you don't assume continuity, this is false. To build a counterexample, start with $F(x)$, additive but not linear (so $F(a+b)=F(a)+F(b)$ but $F(\lambda a)$ not always equal to $\lambda F(a)$. Then let $f(x)=F(\log x)$. — lulu, Dec 13 '16 at 01:45

score 3 · Answer 1 · edited Apr 13 '17 at 12:20

If $f(xy) = f(x) + f(y)$ then taking $g(x) = f(e^x)$ we get $g(x+y) = g(x) + g(y)$ which is Cauchy's functional equation. This equation have been discussed in many questions on this site, see Overview of basic facts about Cauchy functional equation for a very good overview with many links. If $f$ is assumed to be continuous (at a single point) then the only solutions $f:\mathbb{R_{>0}}\to \mathbb{R}$ are given by $f(x) = C \log(x)$ for some constant $C$. If continuity is not assumed there there does not have to be a unique solution (see the link above for how to construct explicit examples).

See also the more directly related questions:

user383778 · Answer 2 · 2016-12-14T00:24:15.487

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Here's a quick proof which neatly sidesteps all the $(1+1/n)^n$ stuff in Euler's original:

Let $$ {df\over dx} = g(x) $$ Applying first principles $$ {df\over dx} = \lim_{h\to 0} {f(x+h) - f(x) \over h} $$ $$ = \lim_{h\to 0} {f(1+{h\over x})\over h} $$ by virtue of the functional equaiton.

Now let $t=h/x$ and rewrite as a limit in $t$: $$ {df\over dx} = \lim_{t\to 0} {f(1+t) \over tx} $$ $$ = {g(1)\over x} \quad \text{by de l'Hopital's rule} $$ This shows that only solutions to the functional equation have derivatives of the form $g(1)/x$.

We have a free choice of $g(1)$, which is equivalent to the choice of base for the log. Setting it to the 'natural choice' of $1$ gives us natural logs.

It's not hard to show that the differentiable solutions span all continuous solutions. Suppose there is a continuous solution $f'$. Select a value $x$ and consider the differentiable $f$ such that $f(x) = f'(x)$. Then $f$ and $f'$ must be equal for all rational powers of $x$. As these are dense, we can apply a limit to any real value and the two functions are therefore equal.

edited Dec 14 '16 at 00:24

answered Dec 13 '16 at 03:11

user383778

881

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How do you get differentiability? – dejongbrent Dec 13 '16 at 03:42
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I didn't :-) This shows that the only $C_1$ solution is log. – user383778 Dec 13 '16 at 19:16
This is awesome, thank you, Scott! The other solution seems good too, but I don't really know about Cauchy's functional equation. haha – gjbyu Dec 13 '16 at 22:31
However, I actually didn't really say that f has to be differentiable. – gjbyu Dec 13 '16 at 22:33

score 0 · Answer 3 · answered Dec 13 '16 at 01:52

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We know that: $f(ab) = f(a) + f(b)$, that give us :

$f(1\cdot a) = f(1) + f(a)$ which give us :

$f(1) = 0$

Now : $f(\frac{1}{a} \cdot a) = f(1) = 0 = f(\frac{1}{a}) + f(a)$

Also : $f(a^{n}) = nf(a)$ , and using some kind of induction we could get: $f(a_{1}^{\alpha_{1}} \dots) = \sum \alpha_{i}f(a_{i})$.

So this function return us zero in $1$. Return us sum of power product with their multiples. And equals negate function in inverse point.

So this is $log_{a}(x)$.

answered Dec 13 '16 at 01:52

openspace

6,470

Sorry for my English! – openspace Dec 13 '16 at 01:53
Under what assumptions does this show uniqueness? – Farewell Dec 13 '16 at 01:57
@Farewell I use the definition of log function. – openspace Dec 13 '16 at 01:58
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What is $a$? In the original question, $a$ is just a variable argument for the function. Here, it has become the base of the logarithm. What are you assuming (or proving) about $a$? – lulu Dec 13 '16 at 01:59
@lulu actually this moment should be upgrade.. – openspace Dec 13 '16 at 02:00
Ah, now your English might be an issue. What did you mean by that? – lulu Dec 13 '16 at 02:01
Regardless of the question about $a$, you really do need to say something about continuity. Your claim that $f(a^n)=nf(a)$ is fine so long as $n\in \mathbb Q$, but for general $n\in \mathbb R$ there is no reason to assume this. Of course, continuity lets you deduce it, but the OP does not assume continuity. – lulu Dec 13 '16 at 02:03
@lulu But it seems that we need more than continuity, because zero function $z(x)=0$ for every positive $x$ also satisfies functional equation, right? – Farewell Dec 13 '16 at 02:04
@Farewell All you need is continuity plus some value for which $f(c)\neq 0$. If I have such a $c$ then we can solve for $f(d)=1$ and that becomes the base of the log (obviously, since we can write any $x$ as $d^{\log_d x}$ so $f(x)=f(d^{\log_d x})=\log_d x$ – lulu Dec 13 '16 at 02:07
@lulu Do you know does the functional equation implies continuity anywhere? Or we can have everywhere discontinuous solutions? – Farewell Dec 13 '16 at 02:10
@Farewell oh, it can be fantastically discontinuous. Take a basis for $\mathbb R$ as a vector space over $\mathbb Q$, then take $f(x)$ to be the projection along one of the basis factors. – lulu Dec 13 '16 at 02:16
Sorry I guys, I should have said that f() is continuous. – gjbyu Dec 13 '16 at 02:20
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The truly beautiful way to prove this is to go through Euler's derivation of $\int {1\over x} dx$. This shows that the integral must satisfy the functional equation, and there's a constant to resolve, which gives us $e$. – user383778 Dec 13 '16 at 02:21

How to prove that if a continuous function satisfies $f(a b)=f(a) + f(b)$, this function must be a log function?

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