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Find functions $f(x) : \mathbb{R}_+ \rightarrow \mathbb{R}_+$ such that $f(b)/f(a)-1 = g(b/a - 1)$

$a, b \in \mathbb{R}_+$

$g$ doesn't have many limitations on it but it can not have any dangling $a$ or $b$ terms.

Here is an example that I found $f(x) = x^2$ works. In that case $g(x) = x(x+2)$

$b^2/a^2 - 1 = (b/a-1)(b/a-1+2)$

Are there any others? This works due to difference of squares.

This is not homework, just a puzzle that I bumped into. It identifies a specific kind of invariant that I am after.

7zf
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  • You're asking for those functions $f$ for which $f(a)/f(b)$ only depends on the ratio $a/b$ (and not $a$ or $b$) itself. (Do you see why?) So, any $f$ of the form $f(x) := a x^{b}$ works. – Aryaman Maithani Jul 15 '21 at 15:44
  • $f$ cannot depend on $a$ and $b$ – 7zf Jul 15 '21 at 15:48
  • Aah, my bad. The use of notation was bad. I meant that any function of the form $x \mapsto c_1 x^{c_2}$. For example, $5x$, $x^2$, $32 \sqrt{x}$, et cetera. – Aryaman Maithani Jul 15 '21 at 15:50
  • I am trying to solve for $g$ in your $f(x)=c_1 x^{c_2}$ but not seeing anything jumping out at me when $c_2$ is not 2 – 7zf Jul 15 '21 at 15:59
  • For $5x$, you have $f(a)/f(b) = a/b$. So you can choose $g(x) = x$. For $\sqrt{x}$, you have $f(a)/f(b) = \sqrt{a/b}$. So you can choose $g(x) = \sqrt{x + 1} - 1$. In general, for $f(x) = c_1 x^{c_2}$, you can choose $g(x) = (x + 1)^{c_2} - 1$. – Aryaman Maithani Jul 15 '21 at 16:02
  • WOW that is great! Thank you. I see the clever minus 1 plus 1 stuff going on for sure now. So, we are close to $g(x) = f(x + 1) - 1$ whenever $f(a/b) = f(a) / f(b)$ but not quite because then we would have the dangling $c_1$ whenever $c_1 \neq 1$ .. is there a more general principle here? – 7zf Jul 15 '21 at 16:15
  • As I mentioned in my first comment, the question is equivalent to just asking for $f(a)/f(b) = g(a/b)$. And I've given an "obvious" family of examples that do work. However, I'm not sure how to find more/conclude that these are the only ones. From my experience with similar questions, I feel like some assumption of continuity/differentiability somewhere would help and that we should have counterexamples otherwise. But again, I'm not able to formulate it. – Aryaman Maithani Jul 15 '21 at 16:33

1 Answers1

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Note that you want a function $f : \Bbb R_+ \to \Bbb R_+$ such that there exists a function $g$ satisfying $$\frac{f(a)}{f(b)} = 1 + g\left(\frac{a}{b} - 1\right).$$ This is equivalent to asking for the existence of a function $h$ such that $$\frac{f(a)}{f(b)} = h\left(\frac{a}{b}\right) \tag{1}$$ for all $a, b \in \Bbb R_+$.

Note that as $a, b$ vary over $\Bbb R_+$, the ratio also varies precisely over $\Bbb R_+$. Thus, we take the domain of $h$ to be $\Bbb R_+$. $(1)$ also tells us that $h$ maps into $\Bbb R_+$. Thus, we have $h : \Bbb R_+ \to \Bbb R_+$.

Claim 1. If $f$ is of the form $f(x) = c_1 x^{c_2}$ for $c_1 > 0$ and $c_2 \in \Bbb R$, then $f$ satisfies the condition.
Proof. It is easy to see that $f$ does define a function from $\Bbb R_+$ to itself. Check that $h(x) = x^{c_2}$ does the job. $\qquad \Box$

Now, the next question to ask is if there are any more functions. We show that under reasonably nice conditions, there are not. First, we make observations.

Claim 2. $f(xy) = f(x)h(y)$ for $x, y \in \Bbb R_+$.
Proof. Follows from $(1)$ by putting $a = xy$ and $b = x$. $\qquad \Box$

Corollary 3. $f(y) = f(1) h(y)$ for all $y > 0$. $\qquad \Box$

Corollary 4. $h : \Bbb R_+ \to \Bbb R_+$ satisfies the functional equation $$h(xy) = h(x)h(y). \tag{2}$$ Proof. Follows from the last two results by cancelling $f(1)$. (It is nonzero.) $\qquad \Box$

Now, we do something which may not seem very motivated. Consider the function $\varphi : \Bbb R \to \Bbb R_+$ defined as $$\varphi = h \circ \exp. \tag{3}$$ (Note that the function composition is well-defined.)
The reason for doing this is that we get the following result.

Claim 5. For all $x, y \in \Bbb R$, we have $$\varphi(x + y) = \varphi(x) \varphi(y). \tag{4}$$ Proof. Use $(2)$ and the fact that $\exp(x + y) = \exp(x) \exp(y)$. $\qquad \Box$

The upshot of the last result is that we do know something about $\varphi$.

Theorem 6. Let $a := \varphi(1)$. Then, $\varphi(x) = a^x$ holds for all $x \in \Bbb R$, under any one of the following assumptions:

  1. $\varphi$ is monotonic.
  2. $\varphi$ is continuous.
  3. $\varphi$ is differentiable.
  4. $\varphi$ is continuous at $0$.
  5. $\varphi$ is differentiable at $0$.

Proof. See the proof here. $\qquad \Box$

Note that the conditions on $\varphi$ translate directly to $h$. Since $\exp$ is monotonic and differentiable (with a differentiable inverse), we see that $\varphi$ is monotonic iff $h$ is. Moreover, $\varphi$ is continuous (differentiable) at $0$ iff $h$ is continuous (differentiable) at $1$.
Moreover, in view of Corollary 3., we can replace $h$ by $f$ in the previous paragraph. So, under the modest assumption that $f$ is continuous at $1$, we see that $\varphi(x) = a^x$. In turn, this gives us that $$h(x) = \varphi(\ln(x)) = a^{\ln x} = x^{\ln a}.$$ Again, by Corollary 3., we get that $$f(x) = f(1) x^{\ln a},$$ which is the same form as in Claim 1. (We can recover $a$ as well. It is $\varphi(1) = h(e) = f(e)/f(1)$.)

Conclusion. If $f$ is continuous at $1$, then $f$ is of the form $$f(x) = c_1 x^{c_2}$$ for some $c_1 > 0$ and $c_2 \in \Bbb R$. Conversely, every such function does give a valid choice of $f$.

Aryaman Maithani
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