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Assume that there exists a function $f:\mathbb{R}\to\mathbb{R}$ that is bijective and satisfies $$ f(x) + f^{-1}(x)=x $$ for all $x$. Here $f^{-1}$ is the inverse function. Show that $f$ is odd.

This was a brain-teaser given to me by a friend. Two other related questions are:

  1. Show that $f$ is discontinuous
  2. Give an example of such a function (if indeed one exists).

Edit: As an initial idea, maybe approaching the problem graphically would help? A function and its inverse are reflections of each other about $y=x$ on the $x$-$y$ plane. Does this lead to anywhere?

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  • $\begingroup$ I assume the domain must be $\mathbb R$? $\endgroup$ – 5xum Aug 17 '15 at 12:53
  • $\begingroup$ Yes the function is from the Reals to the Reals.\ $\endgroup$ – Iconoclast Aug 17 '15 at 13:08
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    $\begingroup$ One can show that $f(0)=0$ as follows: Let $f(0)=a$, then $f(0)+f^{-1}(0)=0$ or that $f^{-1}(0)=-a$. Therefore, $f(-a)=0$ and $f(0)=a$. Now, $f(-a)+f^{-1}(-a)=-a$, but since $f(-a)=0$, $f^{-1}(-a)=-a$ or that $f(-a)=-a$ so $-a=0$. $\endgroup$ – Michael Burr Aug 17 '15 at 13:10
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Here is finally a constructive example of a solution, continuous except on a countable set.

Let $\phi = \frac{1+\sqrt5}{2}$, and let $$f(x) = \begin{cases} 0 & \text{if }x = 0 \\ -\phi x & \text{if }|x| \in [\phi^{3k}, \phi^{3k+1}), k \in \mathbb Z \\ \phi x & \text{if } |x| \in [\phi^{3k+1}, \phi^{3k+2}), k \in \mathbb Z \\ \phi^{-2} x & \text{if } |x| \in [\phi^{3k+2}, \phi^{3k+3}), k \in \mathbb Z \\ \end{cases} $$

Then $f$ is a bijection, whose inverse is $$f^{-1}(x) = \begin{cases} 0 & \text{if }x = 0 \\ \phi^2 x & \text{if }|x| \in [\phi^{3k}, \phi^{3k+1}), k \in \mathbb Z \\ -\phi^{-1} x & \text{if } |x| \in [\phi^{3k+1}, \phi^{3k+2}), k \in \mathbb Z \\ \phi^{-1} x & \text{if } |x| \in [\phi^{3k+2}, \phi^{3k+3}), k \in \mathbb Z \\ \end{cases} $$ Checking each case shows that $f(x)+f^{-1}(x) = x$, as required.

Plot of f

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    $\begingroup$ What intuition did you follow to arrive at this result? $\endgroup$ – templatetypedef Aug 18 '15 at 2:26
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    $\begingroup$ @templatetypedef I tried to build the function on $[1, \lambda)$ by setting $f(x)=-\alpha x$. See my other answer to see why this leads to a cycle $\pm[1, \lambda) \to \pm[\alpha, \alpha\lambda) \to \pm[\alpha+1, (\alpha+1)\lambda)$. Letting $\alpha=\lambda$ and $\alpha^2=\alpha+1$ enables the three intervals to be contiguous. At this point we have built the bijection on $[1,\phi^3)$. We can restart from $\phi^3$ to build the bijection on $[\phi^3, \phi^6)$, and so on. $\endgroup$ – yoann Aug 18 '15 at 10:40
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    $\begingroup$ A "golden" idea.... $\endgroup$ – Piquito Aug 30 '15 at 11:56
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By plugging in $x=f(y)$ we obtain: $$ f(f(y))=f(y)-y $$ Call this assertion $P(y)$ and let $f^{(n)}(x)=\underbrace{f(f(…f(x)...))}_{n \space times}$. Now we have: $$ P(f(x)): f^{(3)}(x)=f(f(x))-f(x)=f(x)-x-f(x)=-x \iff f^{(4)}(x)=f(-x) $$ But: $$ P\left(f(f(x))\right): f^{(4)}(x)=f^{(3)}(x)-f^{(2)}(x)=-x-(f(x)-x)=-f(x) $$ By combining these equations, we obtain $f(-x)=-f(x)$ and therefore, $f$ is odd.

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    $\begingroup$ Nicely done. but what about continuity and existence? $\endgroup$ – Iconoclast Aug 17 '15 at 13:30
  • $\begingroup$ Thank you. I'm working on it :) $\endgroup$ – Redundant Aunt Aug 17 '15 at 13:33
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Proof that the function is odd (if it exists):

Suppose that $f(b)=a$. Now, consider $$f(b)+f^{-1}(b)=b.$$ Then, $f^{-1}(b)=b-a$ or that $f(b-a)=b$. Now, consider $$f(b-a)+f^{-1}(b-a)=b-a.$$ By substitution, we have $b+f^{-1}(b-a)=b-a$ or that $f^{-1}(b-a)=-a$. Therefore, $f(-a)=b-a$. Now, consider $$ f(-a)+f^{-1}(-a)=-a. $$ By substitution, $b-a+f^{-1}(-a)=-a$ or that $f^{-1}(-a)=-b$. In other words, $f(-b)=-a$, so the function is odd.

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  • $\begingroup$ Better than accepted answer. Did you use bijective anywhere? $\endgroup$ – BCLC Aug 18 '15 at 12:56
  • $\begingroup$ Btw what exactly are b and a? I suspect that that is where bijection is used $\endgroup$ – BCLC Aug 18 '15 at 12:58
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    $\begingroup$ I only used the bijection to get that $f^{-1}$ exists. In this answer, $a$ and $b$ are arbitrary (satisfying the condition that $f(b)=a$). (The accepted answer is quite clean and provides insight on how to solve other similar questions.) $\endgroup$ – Michael Burr Aug 18 '15 at 13:03
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    $\begingroup$ googles more on bijection Strange. It seems that the fact of declaring the existence of $f^{-1}$ should be sufficient to mean that the function is bijective. Anyway, you mean repeated compositioning can solve other similar questions? $\endgroup$ – BCLC Aug 18 '15 at 13:08
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First, as others have noted, we must have $f(0) = 0$. Let us now "construct" such a function on $\mathbb R^{*}$ using the Axiom of Choice.

Using $f(x) = x - f^{-1}(x)$, it is easy to see that, if $A \in \mathbb R^{*}$ is such that $f(A) = B$, then the following cycle must occur: $$A \mapsto B \mapsto B - A \mapsto -A \mapsto -B\mapsto A-B \mapsto A$$ (Note that this proves that the function is odd.)

Therefore, an idea to construct such a function is for instance to set $B = -2A$, so that we have the following cycle: $$A \mapsto -2A \mapsto -3A \mapsto -A \mapsto 2A \mapsto 3A \mapsto A$$

If we could partition $\mathbb R^{*}$ into sets of the form $\{-3A,-2A,-A,A,2A,3A\}$, we could easily define the function on each of these sets using the cycle above.

To do so, we can use the Axiom of Choice and take the quotient of $\mathbb R^{*}$ by the equivalence relation $$A \equiv B \Leftrightarrow \exists (p,q) \in \mathbb Z^2, A = \pm 2^p 3^q B$$

The Axiom of Choice gives us a set $X$ such that $\mathbb R^{*} = \coprod_{A \in X} \bigcup_{(p,q)\in\mathbb Z^2}\{\pm2^p3^qA\}$.

Then, for each $A\in X$, we can construct $f$ on $E_A = \bigcup_{(p,q)\in\mathbb Z^2}\{\pm2^p3^qA\}$. Indeed, since $\mathbb Z^2$ is countable, we can order it and use induction to build our cycles. When considering the $n$-th element of $\mathbb Z^2$, either it does not appear in a cycle formed by one of the previous elements of $\mathbb Z^2$, in which case we build $f$ on the cycle formed by the current element; or it does, in which case we skip the element.

It is easy to check that the function built this way has the required properties.

Note that I have no idea if we can build such a function without using the Axiom of Choice, but I doubt so.

Edit: I finally found a constructive solution, see my other answer. (If it is inappropriate to double-post, I will merge both answers.)

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  • $\begingroup$ could you kindly tell me what is $\mathbb R^{*}$? Sorry for the rather basic question. $\endgroup$ – Iconoclast Aug 17 '15 at 17:10
  • $\begingroup$ @Iconoclast Just a notation: $\mathbb R^{*} = \mathbb R \setminus \{0\}$ $\endgroup$ – yoann Aug 17 '15 at 17:12
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    $\begingroup$ In my opinion it's nice to have your two answers separate. $\endgroup$ – user21820 Aug 18 '15 at 17:16
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If $f$ were continuous, then it would have to be strictly monotonic, since it's bijective. Suppose $f$ is decreasing, $x>0$. Then, since $f(0)=0$ we get $f(x)<0$ and $f^{-1}(x)<0$, but $x=f(x)+f^{-1}(x)$. So $f$ is increasing and $f(x)>0$ when $x>0$. But then, $f^{-1}(x)$ is also positive for $x>0$, so $f(x)<x$. Since $f$ is increasing, $f^{-1}(x)>x$. But that's not possible.

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  • $\begingroup$ So does this mean that $f(x)$ is discontinuous everywhere? i.e., continuous on no internal? $\endgroup$ – Iconoclast Aug 17 '15 at 13:49
  • $\begingroup$ This only shows that it has a discontinuity. I suspect such a function ( if one exists ), is very discontinuous, but I'm still thinking about it. $\endgroup$ – Callus Aug 17 '15 at 13:56
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Let's call $(E)$ the equation $f(y) + f^{-1} (y) = y$

Let $a \in \mathbb{R}$. Since $f$ is bijective, there exists $x \in \mathbb{R}$ such that $f(x) = a$. Now, since $f^{-1}(a) = x$, we have (using $(E)$ with $y = a$) $f(a) = a - x$. Now, using $(E)$ with $y = x$:

$$f^{-1}(x) = x - a$$ Applying $f$ on both side: $$f(x-a) = x $$ Using $(E)$ with $y = x-a$: $$f^{-1}(x-a) = x - a - f(x -a) = x -a -x = a$$ Applying $f$ on both side: $$f (-a ) = x -a $$.

Thus $f(a) = a - x = -f(-a)$. This is true for all $a \in \mathbb{R}$: $f$ is therefore odd.

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