# Find all functions $f:\mathbb{R}^+\to \mathbb{R}$ such that $xf(xf(x)-4)-1=4x$

Find all functions $$f:\mathbb{R}^+\to \mathbb{R}$$ such that for all $$x\in\mathbb{R}^+$$ the following is valid: $$xf\big(xf(x)-4\big)-1=4x$$

All I could do is:

• $$f(x)> {4\over x}$$ for all $$x$$ so $$f(x)>0$$ for all $$x$$.
• $$(4,\infty )\subseteq {\rm Range}(f)$$, since $$f(xf(x)-4)={4x+1\over x} >4$$
• Function $$g(x)=xf(x)-4$$ is injective: \begin{align}g(x_1)=g(x_2) &\implies f(g(x_1))=f(g(x_2))\\&\implies {4x_1+1\over x_1}={4x_2+1\over x_2} \\&\implies x_1=x_2\end{align}
• Function $$g$$ satisfies $$\boxed{xg(g(x)) -(4x+1)g(x)+4x=0}$$
• This is probably unnecessary but I managed to tighten the lower bound: $$f(x)>\frac{20x+4}{4x^2+x}$$ The process is way to long to be actually typed in a comment, so a sketch-ish something, words only: I tried to "reuse" the given condition by saying $xf(x)-4$ is my "new" $x$. Hope that makes sense. Commented Jul 22, 2020 at 12:30
• I think the idea is to substitute $x$ by $g(x)=x\,f(x)-4$ repeatedly. You will then see that $$g(x)\,f\big(g(x)\big)>4\tag{*}\,.$$ However, you know $f\big(g(x)\big)=4+\dfrac1x$, so (*) becomes $$f(x)>\frac{20x+4}{x(4x+1)}\,.$$ Commented Jul 22, 2020 at 12:32
• @AryanSonwatikar Not by hand. By my best friend. It looks hopeless for the third iteration, though. Commented Jul 22, 2020 at 13:04
• $f(x)=4+\frac 4x$ works if I'm not mistaken. Commented Jul 22, 2020 at 17:05
• The following recurrence is easily derived. $$x_{k+1} = x_ky_k-4 \\ y_{k+1} = 4+1/x_k$$ When $\,y_k = 4+4/x_k\,$ we have $\,x_{k+1} = 4x_k\,$ giving $\,y_{k+1} = 4+4/x_{k+1}\,$ as expected. For $\,x_k=1\,$ and $\,y_k=5\,$ we find a single stationary point: $\,x_{k+1}=1\,$ and $\,y_{k+1}=5\,$. Numerical experiments indicate that all other solutions for $\,x\to\infty\,$ are approximated best by $\,y=4+4/x\,$. Commented Jul 26, 2021 at 13:49

Consider the equation $$xf(xf(x)-a)-1=ax$$ for $$a>0$$ so that $$f(xf(x)-a)=a+\frac1x.$$ This means that $$\lim\limits_{x\to+\infty}f(xf(x)-a)=a$$ so that $$\lim\limits_{x\to+\infty}f(x)=a$$. Further, we have $$\lim_{x\to0^+}f(xf(x)-a)=+\infty$$ and since $$f(x)>a/x\implies\lim\limits_{x\to0^+}f(x)=+\infty$$, it follows that $$\lim\limits_{x\to0^+}xf(x)=a$$.

Let $$m,n$$ be integers such that $$m<-1$$ and $$n>0$$. Notice that $$f(x)=\sum\limits_{k=m}^na_kx^k$$ implies $$\lim\limits_{x\to0^+}xf(x)=a$$ so $$a_{-1}=a$$ and $$a_i=0$$ for all $$m\le i<-1$$. Likewise we have $$\lim\limits_{x\to+\infty}f(x)=a$$ so $$a_0=a$$ and $$a_j=0$$ for all $$0. Thus if $$f$$ is a finite Laurent polynomial then the only solution to the functional equation is $$f(x)=a+\frac ax.$$

• I think the last paragraph of your answer is incorrect (unless you have some constraint on $b$, for example, $b>0$). But the Laurent series analysis is great. When $a>1$, $f(x)=\dfrac{a^2}{(a-1)x}$ is a solution to $x\,f\big(x\,f(x)-a\big)-b=ax$ when $b=0$. Commented Jul 23, 2020 at 12:58

Let $$g(x)=xf(x)-4$$.

Our functional equation becomes

$$\frac{g\circ g(x)+4}{g(x)}=4+\frac1x$$

If there exists an invertible function $$\phi(x):\mathbb R^+\to\mathbb R^+$$ such that $$g(x)=\phi(\phi^{-1}(x)+1)$$, direct substitution gives $$\frac{\phi(z+2)+4}{\phi(z+1)}=4+\frac1{\phi(z)}$$ or $$\phi(z+2)=\left[4+\frac1{\phi(z)}\right]\phi(z+1)-4$$ by substituting $$z=\phi^{-1}(x)$$.

Clearly, this recurrence relation extends a $$\phi(x)$$ defined arbitrarily on $$(0,2)$$ to the whole $$\mathbb R^+$$. By computing $$\phi^{-1}(x)$$ this method generates a large class of solution to the functional equation.

(Of course there are certain restrictions on $$\phi(x)$$ on $$(0,2)$$ so that $$\phi(x)$$ is invertible.)

The solution $$f(x)=4+\frac4x$$ corresponds to $$g(x)=4x$$ and $$\phi(x)=k\cdot 4^x$$ where $$k$$ is a positive constant.

A feature of this special solution is that $$\phi(x)$$ is continuous. In contrast, if rather arbitrary values are assigned to $$\phi(x)$$ on $$(0,2)$$, a discontinuous solution may yield. Below is the graph of $$\phi(x)$$ where $$\phi(x) = x+1$$ on $$(0,2)$$.