Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f$ be strictly increasing and such that $f(x)+f^{-1}(x)+1=e^x$. Is it true that $f$ has at most one fixed point?

I am told the answer is yes, but I am having trouble proving it. It's obvious that it must have at most two, but why can it not have two?

share|cite|improve this question
Is $f$ a smooth function? – Dario Jun 23 '14 at 20:45
up vote 14 down vote accepted

If $x$ is a fixed point, then

$$x + x + 1 = e^x,$$

and that equation has only two solutions (in $\mathbb{R}$), one of them is $0$.

Note that since $f^{-1}$ is everywhere defined, $f$ must be surjective, and hence continuous.

Suppose $0$ were a fixed point. Then for small $x > 0$, you have either $0 < f(x) < x$ or $x < f(x)$. Switching the roles of $f$ and $f^{-1}$ if necessary, let's assume that $x < f(x)$ for $0 < x < \varepsilon$.

Then the functional equation implies

$$x < f(x) < e^x - 1$$

for $0 < x < \varepsilon$. Squeezing shows that $f$ has a right derivative in $0$, and

$$D_+ f(0) = \lim_{x\downarrow 0} \frac{f(x)-f(0)}{x-0} = 1.$$

But then $f^{-1}$ also has a right derivative in $0$, and

$$D_+ f^{-1}(0) = \frac{1}{D_+ f(0)} = 1$$

too. But that means the right derivative of $g\colon x\mapsto f(x) + f^{-1}(x)$ in $0$ is $D_+ f(0) + D_+ f^{-1}(0) = 2$, which contradicts $g(x) = e^x - 1$, from which we obtain

$$D_+ g(0) = g'(0) = e^0 = 1.$$

So $0$ cannot be a fixed point of $f$.

share|cite|improve this answer
$f$ can only have a fixed point in two places. Everywhere else, you must have $f(x) > x$ or $f(x) < x$. Since $f^{-1}$ is everywhere defined, $f$ must be surjective, and hence it is continuous, so $f(x) - x$ can change the sign only in at most two points. – Daniel Fischer Jun 23 '14 at 23:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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