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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 $$\boxed{xg(g(x)) -(4x+1)g(x)+4x=0}$$
ə̷̶̸͇̘̜́̍͗̂̄︣͟
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nonuser
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  • 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. – AryanSonwatikar Jul 22 '20 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 $$f(x)>\frac{20x+4}{x(4x+1)},.$$ – Batominovski Jul 22 '20 at 12:32
  • Look at this comparison plot. I think the supremum of the values after the repeated iterations suggested above will be the value of $f$. – Batominovski Jul 22 '20 at 12:33
  • @Batominovski Yeah, nice! Anyway, the continuation of my comment, which Batominovski has already done/shown: Then I obtain$$4+\frac{1}{xf(x)-4}=f((4x+1)f(x)-20-\frac 4x)$$ Then use $f(x)>4/x$ on RHS. Rearrange to obtain above inequality, the right hand side of which is always greater than $\frac 4x$. – AryanSonwatikar Jul 22 '20 at 12:37
  • @Batominovski: Did you get the second iteration stuff by hand? If yes, could you sketch the method? – AryanSonwatikar Jul 22 '20 at 13:03
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    @AryanSonwatikar Not by hand. By my best friend. It looks hopeless for the third iteration, though. – Batominovski Jul 22 '20 at 13:04
  • Asymptotically, I think $f(x)\approx \dfrac{16}{3x}$. See how good this approximation is here. – Batominovski Jul 22 '20 at 16:10
  • @Batominovski But if you plug that in, it nearly works out: $$4x-1=4x$$ An impossible condition :( – AryanSonwatikar Jul 22 '20 at 16:51
  • @AryanSonwatikar I know. I assumed $f(x)=k/x$ to see what would be the best $k$. It is a bit painful that it almost worked out. – Batominovski Jul 22 '20 at 16:56
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    $f(x)=4+\frac 4x$ works if I'm not mistaken. – timon92 Jul 22 '20 at 17:05
  • I'm a little bit confuse but see here https://math.stackexchange.com/questions/2699486/using-steffensens-method-trying-to-prove-phix-0e – Miss and Mister cassoulet char Jul 27 '20 at 16:36
  • @timon92 How you found it? – nonuser Jan 03 '21 at 21:50
  • @Aqua I don't remember, this was like 6 months ago... – timon92 Jan 03 '21 at 22:52
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    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 $,y=4+4/x,$. – Han de Bruijn Jul 26 '21 at 13:49

2 Answers2

6

Partial answer:

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 $0<j\le n$. Thus if $f$ is a finite Laurent polynomial then the only solution to the functional equation is $$f(x)=a+\frac ax.$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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  • 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 $b=0$. – Batominovski Jul 23 '20 at 12:58
5

Partial answer

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 $(0,2)$.

Szeto
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