Let's say I have two functions $h,k:\Bbb R\to \Bbb R$. I want to find $f,g:\Bbb R\to \Bbb R$ such that $g\circ f=h$ and $f\circ g=k$. I know that $f,g$ might not exist (for example, Functional equation involving composition and exponents). Do we know at least a condition for $h,k$ such that $f,g$ exist?
Which condition guarantees the uniqueness of $f,g$ (provided that they exist)? Note that there are $h,k$ such that $f,g$ are not unique. For example, $h=k=0$, where $f=0$ works and $g$ is any function s.t. $g(0)=0$. Or when $h=k$ is the identity function, and we take $f$ to be any bijection and $g=f^{-1}$.
At the very least, what do we know about this problem when $h,k$ are polynomial functions? Is there a simple test that tells us there are polynomials $f,g$ that satisfy the conditions for a given pair of polynomials $h,k$? Again, what about the uniqueness of polynomial solutions?
If the general problem is too hard, I am most interested in this specific problem. I want to find $f,g:\Bbb R\to\Bbb R$ such that
$$g\circ f(x)=x^3+1$$
and
$$f\circ g(x)=x^3+3x^2+3x+2.$$
Clearly $f,g$ are bijective functions if they exist. So, can we determine the value of $g\circ f^{-1}(-7)$?
I found $f,g$ that almost work. When $f(x)=x^3$ and $g(x)=x+1$, we have $g\circ f(x)=x^3+1$ but $f\circ g(x)=x^3+3x^2+3x+1$. Unfortunately they don't quite work. I know also that there are no polynomial functions $f,g$ that work.
Note that
$$f(x^3+1)=f(x)^3+3f(x)^2+3f(x)+2$$
and
$$g(x^3+3x^2+3x+2)=g(x)^3+1.$$
$\therefore$ if $a,b$ are the unique real numbers such that $a^3+1=a$ and $b^3+3b^2+3b+2=b$, we see that $f(a)=b$ and $g(b)=a$. These are the only values of $f$ and $g$ that I know. But I can also see that
$$ f^{-1}(-7)=g(-3)$$
if that helps.
Let $h(x)=x^3+1$ and $k(x)=x^3+3x^2+3x+2$. Due to $f\circ g(x)$ and $g\circ f(x)$ are given; find $f$ and $g$, if $f=f_0$ and $g=g_0$ satisfy the conditions, then $f=f_0\circ \phi$ and $g=\phi^{-1}\circ g_0$ form a solution for any bijection $\phi:\Bbb R\to\Bbb R$ such that $h\circ \phi=\phi\circ h$. Because any iteration of $h$ commutes with $h$, we can see that there are infinitely many $f$ and $g$, if $f_0,g_0$ exist. How do I see whether $f_0,g_0$ exist?
Best Answer
This is an addendum to the very brilliant analysis already given by orangeskid. In light of their analysis, I'll provide some easy facts about topological conjugation over the reals.
Claim 1: If $f:\mathbf{R}\to\mathbf{R}$ is strictly increasing, continuous, unbounded above and below, and such that $f(0)>0$, then there is a strictly increasing and continuous $\varphi:\mathbf{R}\to\mathbf{R}$ such that $\varphi(0)=0$ and $f\circ\varphi(x)=\varphi(x+1)$. Moreover if $f(x)>x$ for all $x\in\mathbf{R}$, then $\varphi$ is also unbounded above and below.
Proof: Since we know $f(0)>0$, let $\varphi(a)=af(0)$ for all $a\in[0,1)$. We will define the rest of $\varphi$ by extending in the obvious fasion: $\varphi(x)=f^{(\lfloor x\rfloor)}\circ\varphi\left(x-\lfloor x\rfloor\right)$, where $f^{(-)}$ denotes functional iteration, as $f$ is bijective. Clearly the next thing to do is to check that this fits the requirements:
We forced $f\circ\varphi(x)=\varphi(x+1)$ by contsruction, so that is done.
To check continuity, note that $f^{(\lfloor x\rfloor)}$ is always continuous, so by functional composition $\varphi$ is continuous over $\mathbf{R}\smallsetminus\mathbf{Z}$. To check continuity on $\mathbf{Z}$, it suffices to check continuity as $x\to 1^-$. For this note that $$\varphi(1)=f\circ\varphi(0)=f(0)=\lim_{x\to 1^-}\varphi(x)$$
To see $\varphi$ is strictly increasing, note that $f^{(\lfloor x\rfloor)}$ is strictly increasing by assumption and that $\varphi$ is strictly increasing over $[0,1)$, so we get $\varphi$ is strictly increasing over all intervals $[z,z+1)$ where $z\in\mathbf{Z}$. However $\varphi$ is continuous, and so it is strictly increasing over $\mathbf{R}$.
Now to check the "moreover" part.
Claim 2: If $f:[0,\infty)\to[0,\infty)$ is strictly increasing and continuous, such that $f(0)=0$ and $f(x)>x$ for all $x>0$, then there is a strictly increasing, continuous, and unbounded $\varphi:[0,\infty)\to[0,\infty)$ such that $\varphi(0)=0$ and $f\circ\varphi(x)=\varphi(2x)$.
Proof: Let $g:\mathbf{R}\to\mathbf{R}$ be given by $g(x)=\log_2 f(2^x)$. By Claim 1, there is some $\psi:\mathbf{R}\to\mathbf{R}$ that is strictly increasing, continuous, unbounded above and below, and such that $g\circ\psi(x)=\psi(x+1)$. Then let $\varphi(x)=2^{\psi(\log_2 x)}$, so we see that $$\varphi(2x)=2^{\psi(1+\log_2 x)}=2^{g\circ\psi(\log_2 x)}=f(2^{\psi(\log_2 x)})=f\circ\varphi(x)$$
Claim 3: If $f:\mathbf{R}\to\mathbf{R}$ is strictly increasing, continuous, and has exactly one unstable fixed point $c$, that is, $f(x)>x$ for all $x>c$ and $f(x)<x$ for all $x<c$, then there is an increasing homeomorphism $\varphi:\mathbf{R}\to\mathbf{R}$ such that $\varphi^{-1}\circ f\circ \varphi(x)=2x$.
Proof: Let $g:\mathbf{R}\to\mathbf{R}$ be given by $g(x)=f(x+c)-c$, thus $g$ shares all properties with $f$ except $0$ is the fixed point of $g$. By Claim 2, there are increasing homeomorphisms $\varphi_{\pm}:[0,\infty)\to[0,\infty)$ such that $\varphi_{\pm}(0)=0$, and moreover both $\varphi_+^{-1}\circ g\circ\varphi_+(x)=2x$ and $\varphi_-^{-1}(-g(-\varphi_-(x)))=2x$. Let $\psi:\mathbf{R}\to\mathbf{R}$ be given by $$\psi(x)=\begin{cases} \varphi_+(x)&\text{if }x\ge 0\\ -\varphi_-(-x)&\text{if }x<0 \end{cases}$$ Then it is not hard to see that $\psi$ is an increasing homeomorphism such that $\psi^{-1}\circ g\circ\psi(x)=2x$. Finally let $\varphi:\mathbf{R}\to\mathbf{R}$ be given by $\varphi(x)=\psi(x)+c$, so then $$2x=\varphi^{-1}(\psi(2x)+c)=\varphi^{-1}(g\circ\psi(x)+c)=\varphi^{-1}\circ f\circ\varphi(x)$$
As a corollary, note that both $x^3+1$ and $x^3+2$ satisfies Claim 3, so both are conjugate to $2x$.
Also note that it is completely possible to modify the proof such that both $x^3+1$ and $x^3+2$ are conjugate to $2x$ via a homeomorphism that is smooth on all of $\mathbf{R}$ except at the fixed point.
This is unavoidable:
Added Claim 4: Consider the two linear functions $f(x)=2x$ and $g(x)=4x$. Let $\varphi:\mathbf{R}\to\mathbf{R}$ be any homeomorphism such that $\varphi\circ f=g\circ\varphi$. Then $\varphi$ cannot be continuously twice differentiable at $0$.
Proof: Assuming not, then by Taylor's theorem we have $$\varphi(x)=ax+bx^2+h(x)\cdot x^2$$ where $h$ is continuous at $h(0)=0$. Then by expanding on $\varphi\circ f=g\circ\varphi$, we eventually get $$h(2x)-h(x)=\frac{a}{2x}$$ Taking the limit $x\to 0$ on both sides, we see that $a=0$, and $h(2x)=h(x)$. However the continuity of $h$ at $0$ implies that $h$ is identically $0$, meaning that $\varphi(x)=bx^2$, and $\varphi$ cannot be a homeomorphism.