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Previously I have discussed about odd number of intersect points (See : If the graphs of $f(x)$ and $f^{-1}(x)$ intersect at an odd number of points, is at least one point on the line $y=x$?) Now , I want to know the even condition . For example $f(x) = \sqrt{x}$ and $f^{-1}(x) = x^2 , x\ge 0$ intersects each other in $(0,0)$ and $(1,1)$ points and these points located on $y=x$ line

Edit : Consider $f$ is continuous function.

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    $\begingroup$ How would any intersection point not lie on $y=x$? The graphs of the function and its inverse (if it exists) is symmetric about $y=x$, so any intersection must be on $y=x$ $\endgroup$
    – user160738
    Feb 23 '17 at 15:58
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    $\begingroup$ @user160738 No , for example consider $f(x) = \frac{1}{x}$ and $f^{-1}(x) = \frac{1}{x}$ $\endgroup$
    – S.H.W
    Feb 23 '17 at 16:00
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    $\begingroup$ @user160738 another example is $f(x) = -x^3$ and $f^{-1}(x) = -\sqrt[3]{x}$ . They intersect each other in $(1 , -1)$ that is not on $y=x$ line. See : math.stackexchange.com/questions/1452098/… $\endgroup$
    – S.H.W
    Feb 23 '17 at 16:08
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The continuous case:

If the domain is not an interval, we can still find a counterexample. Define $$ f(x)=\cases{2x+2 & if $x\in(0,2)$,\cr x-3 & if $x\in(3,5)$,} $$ then $$ f^{-1}(x)=\cases{x/2-1 & if $x\in(2,6)$,\cr x+3 & if $x\in(0,2)$,} $$ and intersections are $(1,4)$, $(4,1)$.

However, if the domain is connected, it turns out to be true.

Proof of the theorem in the case of a connected domain (interval)

For contradiction, let us suppose that there is an even number of intersections and an intersection is out of diagonal.

  1. We assume that $f^{-1}$ exists which means that $f$ is injective. Moreover, $f$ is continuous and its domain is interval, so $f$ is increasing or decreasing.
  2. $f$ and $f^{-1}$ are symmetric by diagonal, therefore the set of intersections is symmetric by the diagonal.
  3. Let denote $(x_1, x_2)$ an intersection out of diagonal. By symmetry, there is an intersection $(x_2, x_1)$. We can therefore WLOG assume that $x_1<x_2$.
  4. $f(x_1) > x_1$ and $f(x_2) < x_2$, so by continuity, $f$ intersects the diagonal.
  5. There is an even number of intersections and by symmetry, there is an even number of intersections out of diagonal. So there is even number of intersections on the diagonal. So there are at least two of them: $f(x_3) = x_3$, $f(x_4) = x_4$
  6. $x_1<x_2$ and $f(x_1) > f(x_2)$, so $f$ is decreasing (on the whole, by 1).
  7. $x_3<x_4$ and $f(x_3) < f(x_4)$, so $f$ is increasing.

Contradiction :-)

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  • $\begingroup$ Okay but it's true in many cases in connected domain . Can you provide an example that it's wrong in connected domain ? $\endgroup$
    – S.H.W
    Mar 3 '17 at 10:45
  • $\begingroup$ Okay so why it's true for many examples ? $\endgroup$
    – S.H.W
    Mar 3 '17 at 15:14
  • $\begingroup$ You proved the theorem is wrong .okay , now provide some examples that shows theorem is false . $\endgroup$
    – S.H.W
    Mar 3 '17 at 15:38
  • $\begingroup$ Oh , I'm really sorry let me to read again . $\endgroup$
    – S.H.W
    Mar 3 '17 at 15:40
  • $\begingroup$ Sorry, I don't understand step 4 in your purported proof. $\endgroup$ Mar 4 '17 at 10:45
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Assume that $f$ is a continuous and invertible real function with a connected domain. Then $f$ is either strictly decreasing or strictly increasing over its domain (or it would assume some value twice, and would not be one-to-one). Consider these two cases:

  1. $f$ is decreasing. Then, since $f(x)-x$ is also decreasing, $f$ can't cross $y=x$ more than once. If it never crosses $y=x$, then it lies entirely above or below that line, and its inverse lies entirely on the other side; they never meet, and the theorem holds vacuously. If, on the other hand, it crosses $y=x$ exactly once, then the total number of intersections between $f$ and $f^{-1}$ is odd, since off-diagonal intersections come in pairs. The theorem holds in this case too.

  2. $f$ is increasing. Then it cannot intersect $f^{-1}$ at any point off the line $y=x$. Suppose it did, at a pair of points $(x,y)$ and $(y,x)$ with $x<y$: then we would have $f(x)=y > x=f(y)$, contradicting the fact that $f$ is increasing. In this case, then, all intersections are on $y=x$, and the theorem holds.

Since the theorem holds whether $f$ is increasing or decreasing, it is true in general.

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  • $\begingroup$ For all decreasing functions we can say $f(x) - x$ is also decreasing ? Is it right ? $\endgroup$
    – S.H.W
    Mar 9 '17 at 6:03
  • $\begingroup$ Sure. A decreasing function minus an increasing function decreases even faster. $\endgroup$
    – mjqxxxx
    Mar 9 '17 at 20:48
  • $\begingroup$ Okay and how you can conclude that it intersects at most one time ? $\endgroup$
    – S.H.W
    Mar 9 '17 at 20:55
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Your question, as it currently stands definitely does not hold, as you do not specify the domain, continuity or conditions on inverse. This answer does define an $f$ on the whole real line that has a proper inverse.

Consider the function: $$ \begin{align} f(x) = \begin{cases} 1 &\text{if $x = 0$}\\ 0 &\text{if $x = 1$}\\ x^2 &\text{if $x > 0$ and $x\neq 1$}\\ -2x &\text{if $x < 0$} \end{cases} \end{align} $$ which has inverse $$ \begin{align} f^{-1}(y) = \begin{cases} 1 &\text{if $y = 0$}\\ 0 &\text{if $y = 1$}\\ \sqrt{x} &\text{if $x > 0$ and $x\neq 1$}\\ -\tfrac{x}{2} &\text{if $x < 0$} \end{cases} \end{align} $$ and so $f(x) = f^{-1}(y)$ only at $x = 0$ and $x = 1$, but at these points $f(x) \neq x$.

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  • $\begingroup$ Consider $f$ is continuous $\endgroup$
    – S.H.W
    Feb 23 '17 at 17:24
  • $\begingroup$ I will have a think @S.H.W, I imagine the answer might be that such a function cannot exist: if they intersect at two points then they are equal on an interval (that is my hunch anyway, YMMV). $\endgroup$
    – Cryvate
    Feb 23 '17 at 21:17
  • $\begingroup$ I think this statement is true but I can't prove it. $\endgroup$
    – S.H.W
    Feb 23 '17 at 21:32
  • $\begingroup$ Actually, I don't think that is quite true: consider $f(x) = x^2$ for $x \geqslant 0$ and $f(x) = -2x$ for $x < 0$, then $f^{-1}(x) = \sqrt{x}$ for $x \geqslant 0$ and $f^{-1}(x) = -\tfrac{x}{2}$ for $x < 0$ and so they only intersect at $x = 0$ and $1$. $\endgroup$
    – Cryvate
    Feb 23 '17 at 21:59
  • $\begingroup$ if $f(x) = x^2$ then $f^{-1}(x) = \sqrt{x} , x \ge 0$ and the intersection points are $(0,0)$ and $(1,1)$ which they are in $y=x$ line $\endgroup$
    – S.H.W
    Feb 23 '17 at 22:03
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For $f$ and $f^{-1}$ to intersect there must be a point(s) $(x,f(x)) = (x,f^{-1}(x))$. Assume $x \neq f(x)$. Because $f$ and $f^{-1}$ are symmetric around y=x then there must be a corresponding point(s) $(f(x),x) = (f^{-1}(x),x)$.

The line defined by the points $(x_0,f(x_0)$ and $(f(x_),x_0)$ is $$ (y-f(x_0) = \frac{f(x_0)-x_0}{x_0-f(x_0)}(x-x_0) $$ and simplified becomes $$ y=-x+x_0+f(x_0) $$ This line, with a slope of -1, is not parallel with $y=x$, therefore must cross $y=x$. Because the two points are symmetric around $y=x$, they must lie on different sides of $y=x$. By the intermediate value theorem both $f$ and $f^{-1}$ must cross $y=x$.

Hence there is at least one point of $f$ on $y=x$

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