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Does a function $f:\mathbb{R}\to\mathbb{R}$ such that $f^{-1}(x)=1-f(x)$ exist?


Edit: I tried with different “simple” forms of function ($ax+b$, polynomes, $ae^{bx}+c$) without any inside as where to go. So I don't even know how to solve equations with a function and its inverse in it.

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    $\begingroup$ Are you able to provide any context or motivation for the problem? Any thoughts on what you could do (or have done, but it ultimately failed)? $\endgroup$ – Clayton Aug 8 '18 at 13:10
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    $\begingroup$ Isn't that the inverse of $f$ on the LHS? Your function should be bijective @TheSimpliFire $\endgroup$ – asdf Aug 8 '18 at 13:19
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    $\begingroup$ A function with this property will certainly not be monotonic because the inverse has the same monotonicity as $f$ but here they are different. The function is one-one though so it is not continuous. $\endgroup$ – Μάρκος Καραμέρης Aug 8 '18 at 13:24
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    $\begingroup$ Any function of the form $f(x)=ax+b$ wouldn't work, since the inverse would be $f^{-1}(x)=\frac{x-b}{a}$, so comparing the coefficient of $x$ in the equation $f^{-1}(x)=1-f(x)$, we would have $a^2=-1$. We may not choose $f$ to be simply affine. $\endgroup$ – Suzet Aug 8 '18 at 13:25
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    $\begingroup$ Why „on hold as off-topic“? There are a multitude of questions of the same form (for example: math.stackexchange.com/questions/897113/…, math.stackexchange.com/questions/1906059/…, math.stackexchange.com/questions/1184899/…). Shouldn't that ones also be put on-hold then? $\endgroup$ – mr_georg Aug 9 '18 at 7:17
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Partition $\mathbb{R}$ according to the equivalence relation $x\sim y$ if and only if $x=y$, or $x=1-y$, or $1-x=y$.

There is one class with one element, $\{1/2\}$ and the rest have two elements.

Partition the classes with two elements in two disjoint sets of the same cardinality $A$, $B$. Let $H$ be a bijection $H:A\to B$. Fix an order within each element of $A$ and within each element of $B$, such that for an element $\{a_1,a_2\}\in A$ we can name the first element $a_1$, and the second $a_2$. Likewise for elements of $B$.

Consider each element $\{a_1,a_2\}\in A$ and its corresponding pair in $B$ defined by the bijection $H$, say $\{b_1,b_2\}=H(\{a_1,a_2\})\in B$, define $g(a_1)=b_1$, $g(a_2)=b_2$, $g(b_1)=a_2$, and $g(b_2)=a_1$. Define also $g(1/2)=1/2$.

Then $g:\mathbb{R}\to\mathbb{R}$ is a bijection of $\mathbb{R}$ such that $1-g(1-g(x))=1-x$.

Define $f(x)=1-g(x)$. Then $f:\mathbb{R}\to\mathbb{R}$ is bijective, since it is a composition of the bijections $g$ and $1-x$, is a function that satisfies $f(f(x))=1-x$. Therefore, composing with $f^{-1}$, we get $$f^{-1}(x)=1-f(x)$$

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    $\begingroup$ Composing with $f$ yields $f(1-f(x)) = x$ right? $\endgroup$ – Pjotr5 Aug 8 '18 at 13:27
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    $\begingroup$ @Pjotr5 It depends on which side in which you compose. Since that is not the result I wrote, then you understood the wrong side. $\endgroup$ – user582578 Aug 8 '18 at 13:32
  • $\begingroup$ Right. I see what you are saying now. $\endgroup$ – Pjotr5 Aug 8 '18 at 13:33
  • $\begingroup$ @floodbaharak So what is the result ? $\endgroup$ – S.H.W Aug 8 '18 at 13:35
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    $\begingroup$ @mr_georg Again, one can compose from the left and from the right. You composed from the left and didn't get what I meant. Compose instead from the right: $f(f(f^{-1}(x)))=1-f^{-1}(x)$ gives $f(x)=1-f^{-1}(x)$, or what is the same $f^{-1}(x)=1-f(x)$. $\endgroup$ – user582578 Aug 9 '18 at 15:03
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Here is a geometric approach which is equivalent to the solution from user582578 above:

Consider the graph of $f$ i.e. $\{(x,f(x))| x \in \mathbb{R} \}$. The transformation $f \to f^{-1}$ reflects this graph in the line $x=f(x)$. The transformation $f \to 1-f$ reflects this graph in the line $f(x)=\frac{1}{2}$.

So $f^{-1} = 1-f$ means that $f$ has a graph such that its reflection in $x=f(x)$ and its reflection in $f(x)=\frac{1}{2}$ are the same.

Given any $a,b \in \mathbb{R} \backslash \{ \frac{1}{2} \}$ with $a \ne b$ consider the (irregular) octagon with vertices

$\{P_0=(a,b),P_1=(b,a),P_2=(b,1-a),P_3=(1-a,b),$ $P_4=(1-a,1-b),P_5=(1-b,1-a),P_6=(1-b,a),P_7=(a,1-b)\}$

$f \to f^{-1}$ transforms this octagon as follows:

$P_0 \leftrightarrow P_1, P_2 \leftrightarrow P_3, P_4 \leftrightarrow P_5, P_6 \leftrightarrow P_7$

whereas $f \to 1-f$ transforms the octagon as follows:

$P_0 \leftrightarrow P_7, P_1 \leftrightarrow P_2, P_3 \leftrightarrow P_4, P_5 \leftrightarrow P_6$

so if the graph of $f$ contains the points $\{P_0, P_2, P_4, P_6\}$ then both transformations will map these points to $\{P_1, P_3, P_5, P_7\}$ and we have $f^{-1}=1-f$ for the restriction of $f$ to the domain $\{a,b,1-a,1-b\}$. Reading off the co-ordinates of $\{P_0, P_2, P_4, P_6\}$ we have

$f(a) = b\\f(b) = 1-a\\f(1-a) = 1-b\\f(1-b) = a$

and so

$f^{-1}(a) = 1-b = 1-f(a)\\f^{-1}(b) = a = 1-f(b)\\f^{-1}(1-a) = b = 1-f(1-a)\\f^{-1}(1-b) = 1-a = 1-f(1-b)$

As user582578 points out, we can partition $\mathbb{R} \backslash \{ \frac{1}{2} \}$ into pairs $\{a,1-a\}$, combine these pairs into pairs of pairs $\{ \{a, 1-a\}, \{b, 1-b\}\}$ and then use this construction to define $f$ on the four values in each pair of pairs. We then just add the special case $f(\frac{1}{2}) = \frac{1}{2}$ to complete the definition of a $f$ for which $f^{-1}=1-f$.

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I offer a proof that $f$ cannot be a real polynomial...

Putting $f(x)$ back into $f^{-1}(x)=1-f(x)$ gets:

$$f^{-1}(f(x))=1-f(f(x)) \\ \implies x=1-f(f(x)) \\ \implies f(f(x))=1-x$$

Suppose that $f(x)$ was a polynomial of degree $n$. Then $f(f(x))$ has degree $n^2$.

$f(f(x))=1-x$ then gives us that $n^2=1$ and so $n=1$.

So $f(x)$ must be linear and we can write $f(x)=ax+b$ for some $a,b \in \mathbb{R}$ and:

$$ f(f(x))=a(ax+b)+b=a^2x+ab+b$$

Equating this to $1-x$ gives:

$$ \left\{ \begin{array} \\ a^2=-1 \\ ab+b=1 \end{array} \right. $$

$$ \implies \left\{ \begin{array} \\ a=i \\ b=\frac{1}{2}(1-i) \end{array} \right. $$

... which is obviously not real.

It does nicely give $f(x)=ix+\frac{1}{2}(1-i)$ as a function that works on $\mathbb{C}$ though!

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  • $\begingroup$ It is easy to prove that $f$ can't be continuous. Since $1-x$ has only one fixed point $1/2$, then $f(x)$ can only have $1/2$ as fixed point. Since $f$ is invertible, if it were continuous, it must be strictly monotonic. But then $f(f(x))$ would be strictly increasing near $1/2$. But $1-x$ is decreasing. Therefore, $f$ cannot be continuous. $\endgroup$ – user582578 Aug 8 '18 at 14:36

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