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I've been stumped by this problem:

Find three non-constant, pairwise unequal functions $f,g,h:\mathbb R\to \mathbb R$ such that $$f\circ g=h$$ $$g\circ h=f$$ $$h\circ f=g$$ or prove that no three such functions exist.

I highly suspect, by now, that no non-trivial triplet of functions satisfying the stated property exists... but I don't know how to prove it.

How do I prove this, or how do I find these functions if they do exist?

All help is appreciated!

The functions should also be continuous.

amWhy
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Franklin Pezzuti Dyer
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    easy thing to notice is that all three functions when applied twice look exactly the same $f \circ g \circ h = h^2 = f^2 = g \circ h \circ f = g^2$. – mdave16 Jul 10 '17 at 21:21
  • Did you mean pairwise unequal? If you just mean that all three can't be equal then $f(x)=-x=g(x), h(x)=x$ works. – lulu Jul 10 '17 at 21:21
  • @lulu Yes. I will clarify in the question. – Franklin Pezzuti Dyer Jul 10 '17 at 21:22
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    @mdave16 Yes, but the problem is, functions have lots of functional square roots. Just as $$x^2=y^2$$ does not imply $x=y$, $$f(f(x))=g(g(x))$$ does not imply $f=g$. – Franklin Pezzuti Dyer Jul 10 '17 at 21:29
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    i know, i thought it might help in some way – mdave16 Jul 10 '17 at 21:31
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    This is relevant, I'm sure. – Clive Newstead Jul 10 '17 at 21:34
  • @CliveNewstead Hey, that was pretty educational! :) – Evgeny Jul 10 '17 at 22:32
  • @CliveNewstead Yeah, I think that answers my question! :D – Franklin Pezzuti Dyer Jul 10 '17 at 22:33
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    @mdave16 Your observation is $$f\circ f = g\circ g = h\circ h = f\circ g\circ h = g\circ h\circ f = h\circ f\circ g$$ Using stuff like $f\circ h\circ g = (g\circ h)\circ h\circ g = g\circ (h\circ h)\circ g = g\circ (g\circ g)\circ g = (g\circ g)\circ (g\circ g)$ we can see a bit more. In the answers given, I think $f\circ f$ etc. is the identity function. I wonder if a solution exists where for example $f\circ g \ne g\circ f$. – Jeppe Stig Nielsen Jul 10 '17 at 22:51
  • I did think that $f^2$ was the identity function, but didn't have a proof at the ready, nor paper which is why i just left it. So now the question reduces to finding three continuous square roots of the identity function which still hold the love triangle property. EDIT: i misread and thought you had proved that $f^2$ need be the identity. :/ but i still have a hunch that it's true – mdave16 Jul 10 '17 at 23:10
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    @mdave16: Sage tells me that the group presented by these relations is the quaternion group $Q_8$. It's easy to check that $(f,g,h) \mapsto (i,j,k)$ gives a group homomorphism, so $f \circ f$ is only required to be an involution. –  Jul 11 '17 at 02:07
  • $f(x) = 0+x\mod3$, $g(x) = 1+x\mod3$, $h(x) = 2+x\mod3$ – Bergi Jul 11 '17 at 02:59
  • @Hurkyl It is well-known that $Q_8$ is non-abelian. However, all the functions $f$, $g$, and $h$ that solve the question above, seem to commute, $f\circ g = g\circ f$ and so on. Is there any reason for that? Or is it even true? – Jeppe Stig Nielsen Jul 11 '17 at 14:04
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    @Jeppe: Take any existing solution. Choose any collection of 8 points closed under the action of $f,g,h$. There exists a faithful action of $Q_8$ on that set of 8 points, and you can modify $f,g,h$ to act as $i,j,k$ do on those points under that action. –  Jul 11 '17 at 16:32
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    @Nilknarf Please stop rolling back edits to the title so you can keep your absolutely irrelevant click-bait as a title. I'll continue to roll back to my edit, until your post becomes community wiki, or until a mod intervenes to decisively over rule your click bait. – amWhy Jul 15 '17 at 23:25
  • @amWhy Sorry, I didn't mean for it to be click-bait. It just seems that the other title is too wordy. – Franklin Pezzuti Dyer Jul 15 '17 at 23:27
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    Then say something relevant: Composition of three functions from R$\to$ R – amWhy Jul 15 '17 at 23:29

6 Answers6

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Here is a proof that no such continuous functions exist. I'll use juxtaposition to denote function composition. As noted in the comments, $$ h^2=fgh=f^2=ghf=g^2. $$ Let $e=f^4=g^4=h^4$. Note that $fgf=hf=g$ and $gfg=gh=f$, so $$ ef=g^4f=gf^2gf=gfg=f, $$ $$ fe=fg^4=fgf^2g=gfg=f. $$ Similarly $eg=g=ge$ and $eh=h=he$. In particular $e^2=ef^4=f^4=e$, so $e$ is idempotent. Consider its image $X=\mathrm{im}\,e\subseteq\mathbb R$, so $e|_X$ is the identity. We have $\mathrm{im}\,f\subseteq X$ since $ef=f$. Moreover $f|_X^{4}=e|_X=\mathrm{id}_X$, so $f|_X$ is a permutation of $X$, and similarly for $g$ and $h$. (aside: it now follows that $f|_X$, $g|_X$ and $h|_X$ generate a quotient of the quaternion group).

Suppose $|X|>1$. Since $e$ is continuous, $X$ is a (possibly infinite) interval. Any continuous permutation of such a set is strictly monotone (either increasing or decreasing). Moreover the only strictly increasing involution of $X$ is the identity. Indeed suppose $\sigma$ is such a function. If $\sigma(x)>x$ then $$ x=\sigma^2(x)>\sigma(x)>x, $$ a contradiction. Similarly we cannot have $\sigma(x)<x$, so $\sigma$ is the identity.

In particular $f|_X$, $g|_X$ and $h|_X$ are strictly monotone. Since $f|_X=g|_Xh|_X$, they can't all be decreasing. wlog suppose $f|_X$ is increasing. Then $f|_X^2$ is increasing. Since $f|_X^4=\mathrm{id}_X$, applying the above result twice gives $f|_X^2=\mathrm{id}_X$ and then $f|_X=\mathrm{id}_X$.

If $|X|=1$, then we also have $f|_X=\mathrm{id}_X$. In either case, for any $x\in\mathbb R$ we have $g(x)\in X$, so $$ h(x)=f(g(x))=g(x), $$ whence $h=g$.

stewbasic
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  • Could you clarify for a nonexpert (me) how $f^4\rvert_X = \operatorname{id}X$ implies that $f\rvert_X$ is a permutation of $X$? By analogy, something like $q(z) = iz$ for $x\in\mathbb{R}$ satisfies a similar criterion that $q^4\rvert{\mathbb{R}} = \operatorname{id}{\mathbb{R}}$, but that case we cannot conclude that $q\rvert{\mathbb{R}}$ is a permutation of $\mathbb{R}$. I'm missing what makes the logic different in your answer. – David Z Jul 11 '17 at 00:19
  • @DavidZ, A standard exercise in functions states that if $v \circ u$ is injective, then $u$ is injective, and if $v \circ u$ is surjective, then $v$ is surjective. Thus, $(f|X \circ f|X) \circ (f|X \circ f|X)$ is bijective implies that $(f|X \circ f|X)$ is bijective, implies that $f|X $is bijective. A bijection $X$ to $X$ is a permutation. – Charles Baker Jul 11 '17 at 00:24
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    @DavidZ Your function $q$ is not analogous because it doesn't map $\mathbb R$ to itself, whereas in this case we know $f(X)\subseteq X$. – stewbasic Jul 11 '17 at 00:27
  • I have two issues in the second paragraph. 1) (Quibble-sized) Continuity alone does not force $X$ to be closed; e.g., $\arctan x$ maps $\mathbb{R}$ to the open interval $(- \pi/2, \pi/2)$. We have that $X$ is an increasing union of closed intervals $[\min_{[-n, n]} e(x), \max_{[-n, n]} e(x)]$, but that can be half-open or open. 2) (More important) How do you show that $X$ is not a singleton, that $e$ is not the project-to-$0$ map? (This presumably is where $f$, $g$, $h$ nonidentical nonconstant is used) – Charles Baker Jul 11 '17 at 00:27
  • @stewbasic Ah, yep, that's what I missed. Thanks. – David Z Jul 11 '17 at 00:29
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    @user52733 1) Yes, the closed part also uses that $e$ is idempotent (though I think I don't need that $X$ is closed anyway). 2) The statement still holds if we allow constant functions. The argument goes through when $X$ is a singleton (except you need to replace "can't all be decreasing" by "can't all be not increasing"). – stewbasic Jul 11 '17 at 00:30
  • @stewbasic, agreed on 1. For 2, I would phrase it differently, but I agree that case works. [If $X = \lbrace x_0 \rbrace$, just skip the increasing/decreasing and say that $g(x) \in X$, $h(x) \in X$, $f(x) \in X$ forces $f(x) \equiv g(x) \equiv h(x) = x_0$ for all real $x$.] – Charles Baker Jul 11 '17 at 00:44
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    @user52733 I split that case out for clarity. – stewbasic Jul 11 '17 at 00:48
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(This is by no means a comprehensive and strict answer, just few observations I made).

If you removed $0$ from the domain, the functions might be:

$$f_0 (x) \equiv -x$$ $$g_0(x) \equiv \frac 1 x$$ $$h(x) \equiv -\frac 1 x$$

Or better don't remove $0$ but allow $\infty$ into the domain and the image, making each of them the projectively extended real line. This is interesting because if you take a broader look at complex numbers and their representation as the Riemann sphere, you will notice that:

  • $f_0$ corresponds to rotating the sphere half turn, while the axis goes through $0$ and $\infty$;
  • $g_0$ is a similar rotation, the axis goes through $-1$ and $1$;
  • $h$ is also a similar rotation, the axis goes through $-i$ and $i$.

Rotating comes easy to my imagination so I will stick to this interpretation for a while, but we should remember that every half turn rotation is equivalent to some axial reflection.

So in this case the three functions (and their compositions) correspond to certain operations (and their compositions) in 3D space where we imagine the Riemann sphere is.

In general you could choose any three half turns with respect to mutually perpendicular axes. In this case however each of them must map the extended real line (which is a great circle on the Riemann sphere) onto itself. This means one of the axes must go through $-i$ and $i$, i.e. one of the functions must be our $h(x)$.

If I did my calculations right, more general forms are ($\alpha \in \mathbb R$):

$$f_\alpha (x) \equiv \frac {-x+\alpha} {\alpha x + 1},$$ $$g_\alpha (x) \equiv \frac {\alpha x + 1} {x-\alpha} $$ $$h(x) \equiv -\frac 1 x$$

The following picture shows the projectively extended real line as a cross section of the Riemann sphere. Few possible axes are drawn.

projectively extended real line

I said our functions correspond to certain reflections (or rotations) in 3D. Now in 2D the interpretation is:

  • $f_\alpha$ and $g_\alpha$ correspond to reflections about certain axes;
  • $h$ corresponds to the reflection through the point $S$ (or a half turn, if you wish).

To me the most surprising conclusion is that: $$f_0 (x) \equiv -x$$ and $$g_0(x) \equiv \frac 1 x$$ are more similar than I ever thought.

Kamil Maciorowski
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An example:

$$f(x) = \begin{cases} x, & \text{if $x\in \mathbb Q$} \\ -x, & \text{if $x\notin \mathbb Q$} \end{cases}$$

$$g(x) = \begin{cases} -x, & \text{if $x\in \mathbb Q$} \\ x, & \text{if $x\notin \mathbb Q$} \end{cases}$$

$$h(x)=-x$$

lulu
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A simple answer that almost works is $f(x)=g(x)=-x,h(x)=x$ with the only problem being $f=g$ but we can patch that up by making each one the identity on different parts of the real line. Split the reals into $|x| \gt 2,1 \le |x| \le 2, |x| \lt 1$. Make each the identity on one part and $-x$ on the other two.

Explicitly $$f(x)=\begin {cases} x& x \lt -2\\-x & -2 \le x \le 2 \\x & x \gt 2 \end {cases}\\g(x)=\begin {cases} -x& x \lt -2\\x & -2 \le x \le -1 \\-x & -1 \lt x \lt 1\\x&1 \le x \le 2\\-x&2 \lt x \end {cases}\\h(x)=\begin {cases} -x& x \le -1\\x & -1 \lt x \lt 1 \\-x & x \gt 1 \end {cases}$$

Ross Millikan
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    Piecewise continuous, but still not continuous. – Jeppe Stig Nielsen Jul 10 '17 at 22:02
  • Fix a bounded interval $I=[-a,a]$ with zero as its midpoint (either open or closed, not half-open). Comparing this answer and that of lulu, we can also see a solution with $f(x)=x$ for $x\in I$ and $f(x)=-x$ for $x\notin I$; and $g(x)=-x$ for $x\in I$ and $g(x)=x$ for $x\notin I$; and $h(x)=-x$ everywhere. – Jeppe Stig Nielsen Jul 10 '17 at 22:11
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A simple, isometric solution exists in $\mathbb{R}^n$ for $n \ge 2$: let $f$ be a reflection in the first coordinate, and $g$ be a reflection in the second. A bijection between $\mathbb{R}^n$ and $\mathbb{R}$ then yields a solution to the stated problem, though continuity is lost.

Jim Ferry
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For the non-continuous version, you can just consider the three imaginary quaternion units, acting by left multiplication on the space of quaternions (or just the unit quaternions) -- it has the same cardinality as the reals, so it resolves the problem in the non-continuous case.

tomasz
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