Find a possible f(x) and g(x)

Question

Do you think that it is possible to find a $f$ and a $g$ such that $f(x) \neq x$ and $\forall x \gt 1, y \gt 1$ then

$$f\left({1\over 1-{1\over x}-{1\over y}}\right) = {1\over 1-{1\over g(x)}-{1\over g(y)}}$$

EDIT 1:

and

$$f(2x) = 2f(x)$$ $$g(2x) = 2g(x)$$

EDIT 2:

and

$$f(x) \gt 0$$ $$g(x) \gt 0$$

Answer 1

(began answering before the $f(2x)=2f(x),g(2x)=2g(x)$ was added)

$$f\left({1\over 1-{1\over x}-{1\over y}}\right) = {1\over 1-{1\over g(x)}-{1\over g(y)}}$$

Let $A(x)=\frac{1}{f\left(\frac{1}{x}\right)}, B(x)=\frac{1}{g\left(\frac{1}{x}\right)},u=\frac{1}{x},v=\frac{1}{y}$. Then:

$$\frac{1}{A(1-u-v)}=\frac{1}{1-B(u)-B(v)}\implies A(1-u-v)=1-B(u)-B(v)$$

Now, let $C(x)=1-A(1-x)$, then:

$$C(u+v)=B(u)+B(v)$$

Letting $u+v=x$, this implies that $\forall \alpha, B(u)+B(x-u)$ must be constant in $u \implies \frac{d}{du}(B(u)+B(x-u))=0$

Thus $B'(u)-B'(x-u)=0$ for all $u$ and all $x$, meaning that $B'$ is constant and thus that $B(x)=mx+b$ for some $m,b$. This would lead to $C(x)=mx+2b$, and, working backwards:

$$\implies A(x)=1-m(1-x)-2b=mx+(1-2b-m)$$

$$\implies f(x)=\frac{x}{m+(1-2b-m)x}, g(x)=\frac{x}{m+bx}$$

[I realise I've made assumptions about differentiability - currently thinking about how to remove them]

Answer 2

From $f(2x)=2f(x)$ and $g(2x)=2g(x)$, we can deduce that $\forall n\in\mathbb Z : f(2^nx)=2^nf(x)$ and $g(2^nx)=2^ng(x)$.

So from

$$f\left({1\over 1-{1\over x}-{1\over y}}\right) = {1\over 1-{1\over g(x)}-{1\over g(y)}}$$

when $x=y$, and using $f(2x)=2f(x)$ and $g(2x)=2g(x)$, we can get to

$$f\left(\frac{1}{x-1}+1\right)=\frac{1}{g(x)-1}+1\quad (1)$$

In fact, using our first deduction, we obtain a more general form

$$f\left(\frac{1}{x-{1\over2^n}}+2^n\right)=\frac{1}{g(x)-{1\over2^n}}+2^n,\quad \forall n\in\mathbb Z\quad (2)$$

Since both $f$ and $g$ are continuous, we can work with limits. In $(2)$, when $n\rightarrow\infty$, the result is also infinite. More generally, this means that $f(x)\rightarrow\infty$ as $x$ approaches infinity.

Knowing this, let's try $x\rightarrow1^+$ in $(1)$ and solve for $g(1)$. We get $g(1)=1$. Thus $g(2^n)=2^n\quad \forall n\in\mathbb Z$.

My input doesn't really answer your question, but I hope it gives some insight for what to expect.