I've been trying to add math rigor to a solution of the functional equation in [1], eq. (22). It is: $$ f\left(\frac{x}{f(x)}\right) = \frac{1}{f(x)}\,, $$ where you know that $f(0)=1$ and $f(-x) = f(x)$.

I've been trying to fill in the missing steps. Let's use a substitution $g(x) = \frac{x}{f(x)}$. So we rewrite the functional equation as $$ x = \frac{\frac{x}{f(x)}}{f\left(\frac{x}{f(x)}\right)} $$ and get $$ x = g(g(x))\,. $$ Then we calculate $g'(0)$ as follows: $$ g'(0) = \left.\left(\frac{x}{f(x)} \right)'\right|_{x=0} = \left.\frac{f(x)-xf'(x)}{f^2(x)}\right|_{x=0} = \frac{f(0)}{f^2(0)} = 1\,. $$ Also we can use that $g(x)$ is odd $$ g(-x) = \frac{-x}{f(-x)} = -\frac{x}{f(x)} = -g(x)\,. $$ Whenever $g(g(x)) = x$, I have found that this is called involution. That can have many solutions, but using $g(-x)=-g(x)$ and $g'(0)=1$, there should be a way to prove that $g(x) = x$, from which $f(x)=1$ as the only solution.

The article [1] just says, that since $g(x) = g^{-1}(x)$, the graph of $g(x)$ and its inverse must be symmetric along the $y=x$ axis, and since the tangent at $x=0$ is equal to this same line, then $g(x)=x$ must be the only solution. They do mention that this is not completely rigorous proof.

You can assume that $f(x)$ is differentiable. It'd be nice if it can be proven only for continuous functions $f(x)$, but any proof at first would be a good start, even with stronger assumptions.

Update: another idea is to use the fact, that if $a$ and $b$ are two points in the domain of the $g$ function for which $g(a)=g(b)$, then it follows that $g(g(a)) = g(g(b))$ and $a = b$, which proves that the function is one-to-one. Given that $g$ is continuous, it means that it is strictly increasing or decreasing [2]. From that, let's say it's increasing, then we can use $y=g(x)$, if $y < x$, then $g(y) < g(x)$ from which $g(g(x)) < y$ and $x < y$ which is a contradiction. Similarly for $y > x$. So we must have $y=x$, from which $g(x)=x$. If $g$ is decreasing, then we set $h(x)=-g(x)$, obtain $h(x)=x$ and $g(x)=-x$. Unless I made a mistake we got $g(x)=\pm x$. And using $g'(0)=1$, we get $g(x)=x$ as the only solution. But I don't feel very good about this proof yet.

[1] Levy-Leblond, J.-M. (1976). One more derivation of the Lorentz transformation. American Journal of Physics, 44(3), 271–277.

[2] A continuous, injective function $f: \mathbb{R} \to \mathbb{R}$ is either strictly increasing or strictly decreasing.

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    $\begingroup$ Because $g(x)=1/x$ is also a solution of $g(g(x))$ and somehow it got eliminated. So something is not right. (Of course, it would get eliminated later anyway due to $g'(0)=1$, but that's not the point.) $\endgroup$ Jan 16, 2013 at 7:58
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    $\begingroup$ Because $g(x)=1/x$ is not defined on $\mathbf R$, only on $\mathbf R^*$. Notice that it's also neither increasing nor decreasing. The answer you linked to uses the intermediate value theorem, which you can only apply on an interval, and $\mathbf R^*$ is not an interval. $\endgroup$
    – jathd
    Jan 16, 2013 at 8:01
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    $\begingroup$ Your proof in your update is almost certainly exactly what the authors had in mind, but didn't type out. Briefly: if $g$ is not $x\mapsto x$, then there is some $h>0$ such that $g(h)$ is close to $h>0$, yet $g(h)\neq h$. Then the reflection of the point $(h,g(h))$ will be on the other side of the line $y=x$, contradicting the monotonicity of the function.$\Box$ $\endgroup$
    – Samuel
    Jan 16, 2013 at 8:13
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    $\begingroup$ Two things. For the increasing case, you assume (without saying it) that $g$ is defined on an interval $I$, and that $g$ sends $I$ into $I$ (because you use $y<x\Rightarrow g(y)<g(x)$), which explain $g(x)=-1/x$ on $(0,\infty)$. The other thing is that for the decreasing case, you set $h(x)=-g(x)$, and if you write that case in detail you'll see that you use $g(-x)=-g(x)$, which you can't say about $g(x)=1/x$ on $(0,\infty)$, because if $x\in(0,\infty)$ then $-x\notin(0,\infty)$. $\endgroup$
    – jathd
    Jan 16, 2013 at 8:13
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    $\begingroup$ Given that $f(x)$ is symmetric, you also know that all odd derivatives of $1/f(x/f(x))$ are zero. If you manage to express th even ones, $(1/f(x/f(x)))^{(2n)}$, analytically, you might get some information out, maybe even conclude that all are zero such that $f(x)=1$. $\endgroup$
    – Nikolaj-K
    Jan 16, 2013 at 13:07

3 Answers 3


Some results described above.
(i) $g$ is monotone.
(ii) $g$ = $g^{-1}$
(iii) $g$ is odd.
(iv) $g$ differentiable at $0$
Suppose $g$ is not the identity. Then for $\forall r$, we know $g$ has a symmetrical counterpart to $g(r)$ w.r.t. $y=x$ from (ii). Now consider, $(r,g(r))$ for $r>0$ and his symmetric counterpart. Later consider $(-r,g(-r))$ and his symmetric counterpart. There is no way to ensure monotonicity between these coordinates for, in any case, the points will form a rectangle with two sides parallel and two perpendicular to $y=x$. So in adhering to the symmetry, $g$ must either fail to be a function or $g$ must violate monotonicity somewhere by crossing $y=x$. So $g$ must be the identity. Thus $f=1$.


Hmm, it seems to me that the only solution is the constant function $f(x)=1$
Let's assume that $f(x)$ has a power series, then

  • the constant term must be 1, because $f(0)=1$
  • we have only nonzero coefficients at the powers of x where the exponents are even because $f(x)=f(-x)$

so we assume $$f(x) = 1 + ax^2 + bx^4 + cx^6 + O(x^8) $$ Then for the rhs of your defining equation we have $$ {1 \over f(x)} = 1 - ax^2 + (a^2-b) x^4 + (-c + (-a^3 + 2ba)) x^6 + O(x^8) $$ For the lhs of your defining equation we have $$ f(x {1 \over f(x)}) = 1 + ax^2 + (b -2 a^2)x^4 + (c + 3 a^3 - 6ab) x^6 + O(x^8) $$ and equating coefficients requires

  • $a=0 \to (f"(0)=0) $ and
    then $ \qquad \qquad {1 \over f(x)} = 1 -b x^4 -c x^6 + O(x^8) $
    must equal $ f(x {1 \over f(x)}) = 1 + b x^4 + c x^6 + O(x^8) $
  • thus is required that $b=0 \to f^{(4)}(0)=0 $ and
    then $ \qquad \qquad {1 \over f(x)} = 1 -c x^6 + O(x^8) $
    must equal $ f(x {1 \over f(x)}) = 1 + c x^6 + O(x^8) $
  • thus is required that $c=0 \to f^{(6)}(0)=0 $ ...

... and I assume, that all following coefficients must equal zero. Thus: the only solution should be $$f(x)=1$$ and your $$g(x)={x \over f(x)}=x$$

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    $\begingroup$ Very nice! So if the function $f(x)$ is analytic, then you prove that $f(x)=1$. Assuming only that $f(x)$ is continuous, then the only possible other solutions are not analytic. That helps a lot. $\endgroup$ Jan 16, 2013 at 16:15
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    $\begingroup$ The constant function $-1$ works too. $\endgroup$
    – 2'5 9'2
    Jan 15, 2014 at 15:19

Given what you have already established, it will suffice to show that if $g$ is a continuous function with $g(g(x)) = x$, and $g(0) = 0,\ g'(0) = 1$, then $g(x) = x$ for all $x$. (Note that $g'(0) = 1$ because $g(x) = x + x(\frac{1}{f(x)} - 1) = x + o(x)$, using only continuity of $f$ at $0$).

Clearly, $g$ is injective: If $g(x_0) = g(x_1)$ then $x_0 = g(g(x_0)) = g(g(x_1)) = x_1$.

Thus, $g$ is monotonous: If it were not, it would have a local extreme at some $x_0$, and would not be injective in the neighbourhood of $x_0$. Because $g$ is increasing at $0$, it is therefore increasing on the whole domain.

We will now show that $g(x) = x$ for all $x$. Clearly, this holds for $x=0$. For a proof by contradiction, suppose that we have $x_0$ with $g(x_0) \neq x_0$. Let $x_1 := g(x_0)$. Then $x_0,x_1$ are distinct, and $g$ ''swaps'' these two points: $g(x_0) = x_1,\ g(x_1) = x_0$. Suppose without loss of generality that $x_0 < x_1$. We have $g(x_0) = x_1 > x_0 = g(x_1)$. But this contradicts the assumption that $g$ is increasing. Thus, we indeed have $g(x) = x$ for all $x$. $\square$


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