7
$\begingroup$

I thought of this question somewhat randomly on a walk, and have discussed it with another friend of mine (we both have pure mathematics degrees). We have made some headway, and we think we have generated a proof, but we would appreciate any additional insight and proof verification.

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $f(ax)=f(x)+f(a)$. How much can we say about the behavior of $f(x)$? What additional restrictions, if any, allow us to show $f(x)=\log_b(x)$?

We conjecture that, with the restriction that $f$ is not $0$ everywhere, $f$ must be the logarithm. We have, by the log rules,

$$\log_b(|ax|)=\log_b(|x|)+\log_b(|a|)$$

So the logarithm is indeed one such $f$. I will give what we have been able to show about $f$ below, followed by our general proof. If you can offer any insight into this problem, we would greatly appreciate it.

Edit: Turns out this can be shown fairly easily by considering the Cauchy Functional Equation and showing $g(x)=f(e^x)=cx$ for some $c$. The proof given below does not use this fact.

From here on, we assume $f$ is not trivial.


Result 1: $f(1)=0$

We note that

$$f(a)=f(a\cdot1)=f(1)+f(a)$$

which shows $f(1)=0$.


Result 2: $f(0)$ is not defined

We see that

$$f(0)=f(a\cdot0)=f(0)+f(a)$$

which implies $f(a)=0$ for all $a$. However, we have assumed $f(x)\neq0$ for some $x$, giving a contradiction. Thus $f(0)$ must not be defined. So, we redefine $f$ as $f:\mathbb{R}\setminus\{0\}\to\mathbb{R}$.


Result 3: $f(-x)=f(x)$

If we allow $x<0$, we then have

$$0=f(1)=f(-1\cdot-1)=2f(-1)$$

Showing that $f(-1)=0$ as well. Using this result, we have

$$f(-a)=f(-1)+f(a)=f(a)$$


Our Proof:

From here on, we use results from elementary abstract algebra and analysis. We realized that the condition on $f$ is that of a group homomorphism $\varphi:(\mathbb{R}\setminus\{0\},*)\to(\mathbb{R},+)$ where

$$\varphi(xy)=\varphi(x)+\varphi(y)$$

Similarly, you can show the above results for $\varphi$. We noted that if we also define the continuous group homomorphism $\psi:(\mathbb{R},+)\to(\mathbb{R}\setminus\{0\},*)$ where

$$\psi(x+y)=\psi(x)\psi(y)$$

$\psi$ and $\varphi$ seem like they could possibly be inverses with some additional restrictions. We see that

$$\psi(x)=\psi(0+x)=\psi(0)\psi(x)=1$$

So that $\psi(0)=1$. We can then say for any integer $n\geq0$,

$$\psi(n)=\psi(1+1+1+...+1)=\psi(1)\cdot\psi(1)\cdot\cdot\cdot\psi(1)=\psi(1)^n$$

Note, $\psi(1)<0$ may give us complex results, so we restrict $\psi(1)>0$. For any integer $n<0$,

$$\psi(n)=\psi(-1-1+...-1)=\psi(-1)^{|n|}=(\psi(1)^{-1})^{|n|}=\psi(1)^{-|n|}=\psi(1)^n$$

Then, for any rational number $\frac{p}{q}$,

$$\psi(\frac{p}{q})=\psi(\frac{1}{q})^p=\psi(q^{-1})^p=(\psi(q)^{-1})^p=\psi(1)^{\frac{p}{q}}$$

Let $x\in\mathbb{R}\setminus\mathbb{Q}$. Since the rationals are dense in the reals, we can find rational $\frac{p}{q}<x<\frac{r}{s}$ arbitrarily close to $x$. Note that $\psi$ is strictly increasing over the rationals, and thus

$$\psi(1)^{\frac{p}{q}}<\psi(x)<\psi(1)^{\frac{r}{s}}$$

Since $\psi$ is continuous, this implies $\psi(x)=\psi(1)^x$ for all $x\in\mathbb{R}$. Therefore,

$$\psi(x)=\psi(1)^x$$

This shows that $\psi$ must be the exponential function. Note that it is completely characterized by its value at $1$. Noting that $\varphi$ seems like it should be related to the inverse of $\psi$, we would expect $\varphi(x)=\log_b(x)$. We note that for integer $n\geq0$ and real $a>0$,

$$\varphi(a^n)=\varphi(a)+\varphi(a)+...+\varphi(a)=n\varphi(a)$$

For $n<0$, we have

$$\varphi(a^n)=\varphi(\frac{1}{a})+...+\varphi(\frac{1}{a})=|n|\varphi(a^{-1})=-|n|\varphi{a}=n\varphi(a)$$

The rational case is slightly trickier. We note that $a^{\frac{n}{n}}=a$, and thus

$$\varphi(a)=\varphi(a^{\frac{n}{n}})=n\varphi(a^{\frac{1}{n}})$$

And thus $\varphi(a^{\frac{1}{n}})=\frac{1}{n}\varphi(a)$. Therefore, for any rational $\frac{p}{q}$

$$\varphi(a^{\frac{p}{q}})=p\varphi(a^{\frac{1}{q}})=\frac{p}{q}\varphi(a)$$

Using the same argument as before, we can extend this to all $x\in\mathbb{R}$. Therefore,

$$\varphi(a^x)=x\varphi(a)$$

Combining these, we have

$$\varphi(\psi(x))=\varphi(\psi(1)^x)=x\varphi(\psi(1))$$

Thus, $\varphi$ and $\psi$ are inverses up to multiplication by a constant, which confirms that $\psi$ must be the logarithm. If we restrict $\varphi(\psi(1))=1$, these are exactly inverses. If $\psi(1)>0$, we must have $\varphi_{\psi(1)}(x)=\log_{\psi(1)}(x)$. This proof requires that $x>0$. However, we showed earlier that $\varphi(-x)=\varphi(x)$, and for positive $x$, $\varphi_b(x)=\log_b(x)$. To make this function even, we modify it as

$$\varphi_b(x)=\log_b|x|$$

and this is the only possible continuous solution for $\varphi$ defined on all of $\mathbb{R}\setminus\{0\}$. In other words, $f$ must be the logarithm. We have shown the logarithm rules are unique to logarithms!


One thing that concerns us is the restriction that $f$ is continuous. Are there discontinuous $f$ that satisfy this property? Please let us know if we made any invalid assumptions somewhere, or if our statements about the uniqueness of these functions is incorrect. We would also appreciate any alternate proofs you may have.

$\endgroup$
  • 1
    $\begingroup$ Check out the Cauchy functional equation: math.stackexchange.com/questions/423492/… and in particular: math.stackexchange.com/questions/98673/… $\endgroup$ – Chris Culter Nov 17 '17 at 7:44
  • $\begingroup$ @ChrisCulter I'm a bit unsure of how this is Cauchy's functional equation. I see in one of the links that any $f(e^x)$ must be an additive function, but how does that show $f(xy)=f(x)+f(y)$ is also additive? $\endgroup$ – superckl Nov 17 '17 at 7:49
  • 1
    $\begingroup$ Think of the function $f\circ\exp$. It satisfies Cauchy's functional equation. The functions that take addition to addition, multiplication to addition, addition to multiplication, or multiplication to itself are all about the same (modulo some exponentials and logarithms). $\endgroup$ – Robert Wolfe Nov 17 '17 at 7:51
  • $\begingroup$ @Robert So we have $f(e^x)=cx$ for some $c$ which tells us $f$ must be the logarithm. I see, thanks! $\endgroup$ – superckl Nov 17 '17 at 7:54
  • 1
    $\begingroup$ @superckl Well, no, $f$ isn't additive. The two functional equations are different -- but only superficially! Once you develop a theory of solutions to the standard Cauchy functional equation, there is no need to start from scratch when considering a slight variant. You can apply what is already known about the standard form; then you can focus on cleaning up any gaps in the translation, like zero and negative numbers. $\endgroup$ – Chris Culter Nov 17 '17 at 7:55
1
$\begingroup$

This function is equivalent to the known as the cauchy functional equation. "Most" of the solutions are in fact discontinuous.

However all noncontinuous functions have a graph that is dense. So it suffices to show, that the graph is not dense (eg. f is monotone).

The original cauchy functional equation is $g(x+y) = g(x) +g(y)$. Let $g_0$ be a solution of this functional equation, then $h(x) = g_0(\log(x))$ is a solution to your equation. This is why (sometimes) your equation is also called the cauchy functional equation, because it's equivalent.

Futhermore the equations $f(x+y) = f(x)\cdot f(y)$ and $f(x\cdot y) = f(x)\cdot f(y)$ are equivalent aswell.

$\endgroup$
  • $\begingroup$ Could you explain why this is Cauchy's functional equation? The definition I find for it is $f(x+y)=f(x)+f(y)$, but here we have $f(xy)=f(x)+f(y)$. $\endgroup$ – superckl Nov 17 '17 at 7:51
1
$\begingroup$

Now, you see

$f^{'}(x)= \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$

$ \implies f^{'}(x)= \lim \limits_{h \to 0} {\frac{f(x(1+\frac{h}{x}))-f(x)}{h}}$

Now since, $ f(ax)=f(x)+f(a)$

$ \implies f^{'}(x)= \lim \limits_{h \to 0} {\frac{f(x)+f(1+\frac{h}{x})-f(x)}{h}} $

$ \implies f^{'}(x)= \lim \limits_{h \to 0} {\frac{f(1+\frac{h}{x})}{h}} $

Now, since we've considered $ f$ to be a derivable function, the limit is of the $ \frac{0}{0}$ form.

Using L'Hospital's rule and solving,

$ f^{'}(x)=f^{'}(1)\frac{1}{x}$

So, taking the anti-derivative and since $ f^{'}(1)$ is constant,

$ f(x)=c\ln |x|+\ln k$

Or, $ f(x)=c\ln |kx|$

Now, since $ f(1)=0$, implies either c=0 or k=1.

And, since for c=0, doesn't satisfy the original condition, k=1;

So, $ f(x)=c\ln |x|$

$\endgroup$
  • $\begingroup$ Interesting approach! This makes a much stronger assumption that $f$ is differentiable though, which turns out to not matter with some outside knowledge. Our first attempts were actually to show $f'(x)=c/x$, but we didn't manage it. $\endgroup$ – superckl Nov 17 '17 at 8:32
  • $\begingroup$ We first noticed the continuity of f and we then examine the differentiability by checking in the neighbourhoods of x i.e., x+h and x-h. So, it is completely reasonable and there pretty much isn't any assumption considered. And, we did show that f'(x)=c/x and then only took the anti-derivative. I don't understand what are you referring to? $\endgroup$ – Your IDE Nov 17 '17 at 8:53
  • $\begingroup$ I'm saying the first thing we tried to do was a proof of this form, but we couldn't manage it. I'm glad you did so I could see how it worked out. $\endgroup$ – superckl Nov 17 '17 at 8:58
  • $\begingroup$ Oh...OK. I didn't understand that "we" was referring to you and your friend and I thought you were referring to my solution when you stated "Our first attempts were actually to show f′(x)=c/x, but we didn't manage it". Sorry about that..I had only seen the heading and solved it. But, I went through your solution now and realised what you meant in your first comment $\endgroup$ – Your IDE Nov 17 '17 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.