TITLE: If $f(x)-f^{-1}(x)=e^{x}-1$, what is $f(x)$?
QUESTION [28 upvotes]: $f(x)$ is an increasing, differentiable function satisfying $f(x)-f^{-1}(x)=e^{x}-1$ for every real number $x$
I couldn't figure it out whether such function $f(x)$ exists or not.
And if it exists, I want to know the method to find what $f(x)$ is.
Thank you.

REPLY [14 votes]: This is not an answer but only collects some properties which $f$ must neccesarily have.
Let us assume $f : \mathbb R \to \mathbb  R$. Then

$f$ is a strictly increasing differentiable bijection.
$f'(x) > 0$ for all $x \in \mathbb R$. [Note that this does not follow from 1. as the example $f(x) = x^3$ shows.]
$f^{-1}$ is a strictly increasing differentiable bijection. We have
$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} > 0 .$$
Note that here it is essential to know that $f'$ does not have zeros.
One has the equation
$$f'(x) - \frac{1}{f'(f^{-1}(x))} = e^x .$$
$f(0) = 0$.
$f'(0) = \frac{1+\sqrt{5}}{2} = \phi$, $(f^{-1})'(0) = \frac{1}{\phi} = \frac{-1+\sqrt{5}}{2}$.
$\lim_{x \to \infty}f'(x) = \infty$, $\lim_{x \to \infty}(f^{-1})'(x) = 0$.
If $\lim_{x \to -\infty}f'(x)$ exists, then it has the value $1$.
If $f$ has higher derivatives, we get additional functional equations. They can easily be computed recursively. Let us write $y = f^{-1}(x)$. Then the basic functional equation is
$$f(x) - y = e^x -1 .$$
For $n > 0$ we get
$$f^{(n)}(x) - y^{(n)} = e^x .$$
The $y^{(n)}$ can be computed recursively. We have
$$y' = \frac{1}{f'(y)} ,$$
$$y'' = -\frac{f''(y) \cdot y'}{(f'(y))^2} = -\frac{f''(y)}{(f'(y))^3} ,$$
$$y''' = -\frac{f'''(y) \cdot y' \cdot (f'(y))^3 - f''(y)\cdot 3(f'(y))^2 \cdot f''(y) \cdot y'}{(f'(y))^6} \\ = - \frac{f'''(y) \cdot f'(y) - 3(f''(y))^2 }{(f'(y))^5}$$
etc. In this context see http://vixra.org/pdf/1703.0295v1.pdf which shows that there seems to exist no simple formula for $y^{(n)}$. Anyway, this allows to compute $f^{(n)}(0)$ and $(f^{-1})^{(n)}(0)$ recursively by noting that for $x= 0$ we have $y = 0$. This gives a linear equation for $f^{(n)}(0)$ whose solution can be expressed by the collection of $f^{(i)}(0)$ with $i < n$. Once we have determined the $f^{(i)}(0)$ with $i \le n$, we get $(f^{-1})^{(n)}(0)$ by inserting these values into the above equation for $y^{(n)}$. For example we get
$$f''(0) = \frac{\phi^3}{1+\phi^3} =  \frac{\phi}{2} , (f^{-1})''(0) = -\frac{\frac{\phi}{2}}{\phi^3} = -\frac{1}{2\phi^2} .$$

Let us give proofs.

In order to have an inverse $f^{-1}$ the function $f$ must be injective (in which case $f^{-1}$ is defined on $f(\mathbb R)$) and in order that $f(x) + f^{-1}(x)$ is defined for all $x \in \mathbb R$ the function $f$ must be surjective.
Since $f$ is increasing, we have $f'(x) \ge 0$ for all $x \in \mathbb R$. Replacing $x$ by $f(x)$ the functional equation yields
$$f(f(x)) - x = f(f(x)) - f^{-1}(f(x)) = e^{f(x)} -1 .$$
Differentiating gives
$$f'(x) \cdot f'(f(x)) - 1 = f'(x)\cdot e^{f(x)} .$$
This shows that $f'(x) \ne 0$ for all $x \in \mathbb R$.
$f^{-1}$ is trivially a bijection. Since $f'$ does not have zeros, it is differentiable with $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} > 0$. Hence $f^{-1}$ is increasing. Note that this also follows easily from the fact that $f$ is increasing (without using differentiabilty).
This is obtained by differentiating the functional equation.
We have $f(0) = f^{-1}(0)$. Assume $f(0) > 0$. Then $0 = f^{-1}(f(0)) > f^{-1}(0)$ which is impossible. Similarly $f(0) < 0$ is impossible.
By 4.
$$f'(0) - \frac{1}{f'(0)} = f'(0) - \frac{1}{f'(f^{-1}(0))} = e^0 = 1 .$$
This implies (note $f'(0) > 0$) $f'(0) = \frac{1+\sqrt{5}}{2} = \phi$ and $(f^{-1})'(0) = \frac{1}{f'(0))} = \frac{1}{\phi}$.
This follows from $f'(x) = e^x + \frac{1}{f'(f^{-1}(x))}$ and 3.
We have $\lim_{x \to -\infty}(f'(x) - \frac{1}{f'(f^{-1}(x))}) = 0$. Assume that $a = \lim_{x \to -\infty}f'(x)$ exists, $0 \le a  \le \infty$. Then also $\lim_{x \to -\infty}f'(f^{-1}(x)) = a$ since $f^{-1}$ is an increasing bijection. $a = 0$ is impossible since then $\lim_{x \to -\infty}(f'(x) - \frac{1}{f'(f^{-1}(x))}) = -\infty$. $a = \infty$ is impossible since then $\lim_{x \to -\infty}(f'(x) - \frac{1}{f'(f^{-1}(x))}) = \infty$. Hence $0 < a  < \infty$ and $a - \frac{1}{a}= 0$ which yields $a= 1$.