# What function satisfy: $f(x)+f^{-1}(x)=2x$?

What function satisfy: $f(x)+f^{-1}(x)=2x$?

I have tried to substitute $x=f(x)$ to get $f^{(2)}(x)+1=2f(x)$ and subsequently plug in values to try to find $f(x)$ but to no avail. Please help thank you in advance!

• Merely observing the function y = x satisfies the condition – Umashankar Sasikumar Jun 23 '15 at 16:50
• Let $f(x)=f^{-1}(x)=x$. – TomGrubb Jun 23 '15 at 16:50
• Maybe from a geometric point of view ? – Vim Jun 23 '15 at 17:15
• Let $f(x)=x+c$, where $c$ is any constant. Then $f^{-1}(x)=x-c$. – vadim123 Jun 23 '15 at 17:16
• Two points are unclear: 1. What is the expected mapping of $f(x)$ ? I guess OP wants $\mathbb{R} \rightarrow \mathbb{R}$ but can't be absolutely certain. 2. Does OP wants a complete solution, or just a single example? – Abel Cheung Jun 23 '15 at 17:29

## 1 Answer

I think that I can prove that (under some conditions), linear functions $f(x) = x + c$ are the only solutions.

The equation (implicitly) assumes that $f$ has an inverse function $f^{-1}$ with the same domain as $f$. I am only considering bijective functions $f : \mathbb R \to \mathbb R$ here. Note that such functions are strictly monotonic if and only if they are continuous.

My claim is:

Let $f : \mathbb R \to \mathbb R$ be continuous and bijective such that $$\tag 1 f(x) + f^{-1}(x) = 2x \text{ for all } x \in \mathbb R \, .$$

Then $f(x) = x+f(0)$ for all $x \in \mathbb R$.

If $f(x) \equiv x$ then we are done, so let's assume that $f(a) \ne a$ for some $a \in \mathbb R$. Without loss of generality we can assume that $f(a) > a$, otherwise consider $g := f^{-1}$ instead of $f$. So $$d := f(a) - a > 0 \, .$$

It follows from $(1)$ that $$f^{(2)}(x) = 2 f(x) - x \tag 2$$ and then via induction for all $n \in \mathbb N$ $$f^{(n)}(x) = n f(x) - (n-1)x \, . \tag 3$$ In particular, $$f^{(n)}(a) = n\,f(a) - (n-1) a = n (a+d) - (n-1) a = a + nd \, .$$ Applying the same calculation to $f^{-1}$ gives $$f^{(-n)}(a) = a - nd \, .$$ Together it follows that $$f(a + kd) = a + (k+1)d \tag 4$$ for all $k \in \mathbb Z$.

$f$ is strictly monotonic, so $(4)$ implies that each interval from $a + kd$ to $a + (k+1)d$ is mapped onto the "next" interval: $$f \bigl([a + kd, a + (k+1)d]\bigr) = [a + (k+1)d, a + (k+2)d] \tag {5}$$

Now let $x \in \mathbb R$ and choose $k \in \mathbb Z$ such that $$a + kd \le x < a + (k+1)d \, . \tag 6$$ For all $n \in \mathbb N$ it follows from $(5)$ and $(6)$ that $$x + (n-1)d \le a + (k+n)d \le f^{(n)}(x) \le a + (k+n+1)d \le x + (n+1) d \, .$$ Substituting this in $(3)$ gives $$x + (n-1)d \le n f(x) - (n-1)x \le x + (n+1) d$$ or $$\frac{nx + (n-1)d}{n} \le f(x) \le \frac{n x + (n+1)d}{n} \, .$$ Finally, $n \to \infty$ gives $$x + d \le f(x) \le x + d \, .$$

• How does (2) follow from (1)? – Sean Clark Oct 30 '15 at 21:50
• Never mind, I see. I was thinking about it totally wrong. – Sean Clark Oct 30 '15 at 21:53