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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?

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Is $f$ a smooth function? –  Dario Jun 23 at 20:45

1 Answer 1

up vote 13 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 $f$.

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$f$ 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 at 23:10

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