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I have a question regarding graphical intersection between a continuous function and its inverse (if exists).

Suppose $f$ is a real continuous function and $f^{-1}$ exists.

Can anyone assist in proving the following problem:

If $f$ and $f^{-1}$ intersect graphically at some points, then the points must lie on the line $y=x$.(*)

For here, I have some trouble regarding the meaning of "some" points.

If it is uncountable many points of intersection, $y=\frac{1}{x}$ is a counterexample.

What if it is countable (including finite many or countably infinite)?

Intuitively, (*) is true when we draw the functions pictorially, but can anyone provide hints or steps to the proof?

If the statement is wrong at the first place, how should we amend?

Thank you in advance.

Novice
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    What do you mean by the intersection of two functions? Do you mean the intersection of their graphs? If so, please edit that into the body of the question. – Gerry Myerson Sep 04 '14 at 13:00
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    Surely there must be some condition on f, otherwise let f be a function that swaps two given reals and fixes all the others – Belgi Sep 04 '14 at 13:01
  • Thank you for the answer. Perhaps I change the question to continuous functions. – Novice Sep 04 '14 at 13:03
  • @Belgi I'm missing how that is a counter example. As far as I can tell, all points $a$ such that $f(a)=f^{-1}(a)$ are in ${(x,x)\colon x\in \mathbb R}$. – Git Gud Sep 04 '14 at 13:09
  • Also, $f(a)=f^{-1}(a)$ doesn't mean the point of intersections are on line $y=x$, a counterexample will be $y=\frac{1}{x}$ – Novice Sep 04 '14 at 13:13
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    @GitGud the point is (a, f(a)), say f swaps 0 and 1, then f agrees with its inverse for all reals, but does not coincide with y=x(excuse me for the lack of tex, I'm using a mobile phone) – Belgi Sep 04 '14 at 13:18
  • @Belgi You're right, I made a geometric mistake in my mind. – Git Gud Sep 04 '14 at 13:20

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The graphical proof can be algebra-fied: Let's look at a certain $x=x_1$ where $f(x_1)=x_1$ i.e. $f(x)$ intersects the $y=x$ at that $x_1$. Now since they intersect we have $x_1-f(x_1)=0 \Rightarrow x_1=f(x_1)$. Now we take the inverse $f^{-1}$ of both sides: $f^{-1}(x_1)=x_1 \Rightarrow f^{-1}(x_1)-x_1=0$ which means $f^{-1}(x)$ also intersects the $y=x$ line at that $x_1$ which means $f(x)$ intersects $f^{-1}(x)$ on the line $y=x$.

An alternative solution would be to say that in order for an intersection to take place we want $f^{-1}(x)=f(x)$ and the $x$ which satisfies this is where the intersection happens. Taking the $f$ of both sides gives us $x=f(f(x))$ which is a functional equation. Now $f(x)=x$ is a solution to that functional equation which indicates that the intersections happen at $y=x$. The only issue is that I'm not sure how to prove that there are no other solutions to that functional equation.

I hope that helps somewhat.

Sheheryar Zaidi
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  • By your first paragraph, it seems that you are proving the converse of the statement. – Novice Sep 04 '14 at 14:03
  • True, I tried to show that if $f(x)$ intersects $y=x$ then at that point we also have $f(x)$ intersecting $f^{-1}(x)$. – Sheheryar Zaidi Sep 04 '14 at 14:06
  • Your second question is the crux. Pictorially it is evident, but $\frac{1}{x}$ is the spoiler. – Novice Sep 04 '14 at 14:09
  • Such a function is an involution (I found on Wikipedia). Apparently, $y=-x$ is an involution. The function coincides with its inverse, yet the intersections is the line itself, not $y=x$. – Novice Sep 04 '14 at 14:14
  • Which is why we probably cannot show that all intersections take place on the line $y=x$. Interesting problem I must say. – Sheheryar Zaidi Sep 04 '14 at 14:17
  • But if suppose the continuous function and its inverse (if exists) intersect countable many times (be it finite or countably many), perhaps y=x is their intersections. For finitely many, pictorially it helps, but algebraically I cannot prove it. Not to say countably infinite. – Novice Sep 04 '14 at 14:20
  • But, what I was trying to get at was that only for the $x$'s in which the inverse and the $f$ intersect we have the functional equation $x=f(f(x))$. I guess that should be specified clearly. – Sheheryar Zaidi Sep 04 '14 at 14:20
  • The original problem is something like what you say: if there exists $a$ such that $f(a)=f^{-1}(a)$, then $f(a)=a$. – Novice Sep 04 '14 at 14:21
  • Your functional equation has a name. I got it from Wiki: Babbage's functional equation – Novice Sep 04 '14 at 14:24
  • Using $a$ is certainly more clear. Thanks for the functional equation's name. Googled it myself earlier, couldn't find anything on it! – Sheheryar Zaidi Sep 04 '14 at 14:26
  • Haha. Perhaps I am afraid that I might misinterpret it as the intersection only occurs at one point. – Novice Sep 04 '14 at 14:37
  • What do you say about the function $f(x)=1-x^3$ and its inverse$f^{-1}(x)=\left(1-x\right)^{\frac{1}{3}}$.There is only one point of intersection on $y=x$ whereas the points $(1,0)$ and $(0,1)$ also are points of intersection which do not lie on $y=x$.There are two more points of intersection. Also,consider the function $f(x)=\sin x-x$ and its inverse. – Maverick Apr 29 '20 at 06:20