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While thinking of perfect numbers, I came across the functional equation $g\circ f-f\circ g=g\circ f\circ g$ where the unknowns $f$ and $g$ are functions from $\mathbb{R}$ to itself. I only know one non trivial solution: $(f,g)=(x\mapsto 2^x,x\mapsto x-1)$. Can someone tell me whether it is the only one or not? Thanks in advance.

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  • $\begingroup$ $g = \Bbb 0$, $f(0)=0$ $\endgroup$ – xavierm02 Feb 17 '14 at 15:45
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    $\begingroup$ @xavierm02 this is considered a trivial solution $\endgroup$ – Omnomnomnom Feb 17 '14 at 15:46
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    $\begingroup$ If $g$ is invertible, the condition is equivalent to $f\circ h-h\circ f=f$ (where $h=g^{-1}$). And then if $f=k\circ k$ for some invertible $k$ it is equivalent to $k\circ h-h\circ k = \text{id}$. $\endgroup$ – Hagen von Eitzen Feb 17 '14 at 15:50
  • $\begingroup$ I don't know if this helps, but $g\circ f=(I+g)\circ f\circ g$. $\endgroup$ – user122283 Feb 17 '14 at 15:52
  • $\begingroup$ @Hagen von Eitzen: I know nothing about Lie algebras, but the equation $k\circ h-h\circ k=id$ might be related to this theory. $\endgroup$ – Sylvain Julien Feb 17 '14 at 16:19
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Let $g\colon \mathbb R\to\mathbb R$ be any fixpoint-free homeomorphism with inverse $h$ and $g(0)<0$ (hence $h(0)>0$). Then $k\colon x\mapsto x+h(x)$ is also invertible. Let $x_0=0$ and recursively $x_{n+1}=h(x_n)$, $x_{n-1}=g(x_n)$. Note that $\lim_{n\to\pm\infty}x_n=\pm\infty$ as otherwise we'd have convergence to a fixed point of $g$. Let $\phi\colon[x_0,x_1]\to\mathbb R$ be any continuous function with $\phi(x_1)=k(\phi(x_0))$.

Proposition. For $n\in\mathbb N$ there exists a unique continuous function $f_n\colon[x_{1-n},x_n]\to\mathbb R$ with $f_n|_{[x_0,x_1]}=\phi$ and $f_n(h(x))=k(f_n(x))$ whenever $\{x,h(x)\}\subseteq [x_{1-n},x_n]$.

Proof: For $n=1$ this follows with $f_1=\phi$. Assume we have already shown existence and uniqueness for $f_{n-1}\colon[x_{2-n},x_{n-1}]\to\mathbb R$. Then we must define $$f_n(x)=\begin{cases}f_{n-1}(x)&\text{if }x_{2-n}\le x\le x_{n-1}\\ k(f_{n-1}(g(x))&\text{if }x_{n-1}\le x\le x_n\\ k^{-1}(f_{n-1}(h(x)))&\text{if }x_{1-n}\le x\le x_{2-n}\\\end{cases} $$ by uniqueness on the smaller interval and the required functional equation. Fortunately, we can define this way and obtain a continuous function (because $f_{n-1}(x_{2-n})=k^{-1}(f_{n-1}(h(x_{2-n}))$ and $ f_{n-1}(x_{n-1})=k(f_{n-1}(g(x_{n-2}))$ by induction hypothesis) that satisfies the condtion. $_\square$

Then it is well-defined to let $f(x)=f_n(x)$ for any $n$ with $x_{1-n}\le x<x_n$. Then $f\colon \mathbb R\to\mathbb R $ is continuous with $f|_{[x_0,x_1]}=\phi$ and $f\circ h=k\circ f$, i.e. $f\circ h-h\circ f=f$. We conclude that $$\tag1g\circ f-f\circ g = g\circ(f\circ h-h\circ f)\circ g = g\circ f\circ g.$$

Conclusion. For any $g$ and $\phi$ as above there exists a unique continuous $f$ such that $(1)$ holds.

Remark. We could do the same with $g(0)>0$. If we drop the fixed-point-freeness condition, we can argue as above for all intervals bounded by consecutive fixed points. One has to take care a bit to check the continuity of $f$ at the fixed-points in this case: If $x$ is a fixed point of $g$, we need to have $f(x)=0$ and this restricts the choice of $\phi$ (possibly too much). Or we recall that continuity wasn't even among the requirements from the problem statement.

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  • $\begingroup$ Thank you very much for this beautiful answer. Conversely, do we know $\phi$ and $g$ if we know $f$ (supposed to be continuous)? $\endgroup$ – Sylvain Julien Feb 17 '14 at 18:50
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There are quite a lot of solutions to this. To simplify the setting a bit, set $f := id$. Then the equation becomes $g \circ g = 0$, which has a number of non-trivial solutions, e.g.: $g(x) = -x$ for $x \geq 0$ and $g(x) = 0$ for $x \leq 0$.

Edit: Starting from the choice of $g$ and revisiting the choice of $f$, note that $g$ also works just fine together with $x \mapsto 2x$ in place of $f$, or any other positive real in place of $2$.

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  • $\begingroup$ Yes, but to me it sounds like a trivial solution to set $f=id$. $\endgroup$ – Sylvain Julien Feb 17 '14 at 16:36
  • $\begingroup$ This is a very subjective matter, of course, but I would only call a solution trivial if either both parameters are set to "obvious" examples, or if setting one parameter to an obvious example removes all constraints on the second one. However, I've modified my answer to provide a solution with two non-trivial components $\endgroup$ – Arno Feb 17 '14 at 17:08

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