Let $\mathbb{R}_+$ be the set of positive real numbers. Find all functions $f\colon \mathbb{R}_+ \to \mathbb{R}_+$ such that \[f(xy + x + y) + f \left( \frac1x \right) f\left( \frac1y \right) = 1\]for every $x$, $y\in \mathbb{R}_+$. Proposed by Li4 and Untro368.
Problem
Source: 2024 Taiwan TST Round 2 Independent Study 2-A
Tags: functional equation, algebra
KI_HG
28.03.2024 11:11
ah, $\frac{\sqrt{5}-1}{2}$ is a fixed point of $f(x)$.
Tintarn
28.03.2024 18:32
This was fun.
Let $\varphi$ be the unique positive solution of $\phi^2+\phi=1$.
Plugging in $x=y=\frac{1}{\varphi}$ we see that $f\left(\frac{1}{\varphi}\right)=\varphi$.
Next, note that clearly $f(x)<1$ for all $x$.
Putting $y=\frac{1}{x+1}$ we get that $f(x+1)=\frac{1}{1+f\left(\frac{1}{x}\right)}$.
From here it is easy to get that $f(x)>\frac{1}{2}$ for all $x>1$, then $f(x)<\frac{2}{3}$ for all $x<2$, then $f(x)>\frac{3}{5}$ for all $x>\frac{2}{3}$ etc.
Iterating and taking the limit, we get that $f(x) \ge \varphi$ for all $x \ge \frac{1}{\varphi}$ and $f(x) \le \varphi$ for all $x \le \frac{1}{\varphi}$. Moreover, $f$ is continuous at $x=\varphi$.
Fixing $y$ and considering $x_n \to \frac{1}{\varphi}$ we get that $f$ is continuous at $\varphi y+y+\varphi$ and hence on $(\varphi,\infty)$ and by one more application we get that $f$ is continuous everywhere.
Next we want to bootstrap the inequalities in a similar way to get monotonicity.
Indeed, whenever $x \le \varphi$, we have $f(xy+x+y) \le f(\varphi y+\varphi +y)$.
From here, it's easy to get that $f$ is monotonic on $(\varphi,\varphi+1]$, then similarly on $[\varphi+1,\varphi+2]$ etc. and hence on $x \ge \varphi$ and by one more application of the equation we get that $f$ is monotonic everywhere.
In particular, we get that $f$ has a limit at $x \to 0$ and $x \to \infty$ and so can be continuously extended to $x \ge 0$ (and the functional equation continues to hold).
Letting $x \to 0$ we then get that $f(y)+Mf\left(\frac{1}{y}\right)=1$ where $M=\lim_{x \to \infty} f(x) \le 1$.
If $M<1$, this immediately shows that $f$ is constant and hence $\boxed{f(x)=\varphi}$ for all $x$ which is indeed a solution.
On the other hand, if $M=1$, we get that $f(x)+f\left(\frac{1}{x}\right)=1$ and in particular $f(1)=\frac{1}{2}$.
From this equation and the previous relation with $f(x+1)$ it is now easy to get $f(x)=\frac{x}{x+1}$ for all rational $x$ and then by continuity we also get $\boxed{f(x)=\frac{x}{x+1}}$ for all real $x>0$ which is indeed another solution.
math_comb01
29.06.2024 22:32
Cool Problem! We claim that the only solutions are $f(x) \equiv \frac{x}{x+1}$ and $f(x) \equiv \varphi$ both of which clearly work. Claim 1: $f\left(\frac{1}{\varphi}\right)=\varphi$
$P(\varphi,\varphi)$ yields $f(\frac{1}{\varphi})^2+f(\frac{1}{\varphi})=1$ which implies $f(\frac{1}{\varphi})=\varphi$
Claim 2: $\forall x \geq \varphi-1$ we have $f(x) \geq \varphi$, and $\forall x \leq \varphi -1$ we have $f(x) \leq \varphi$
This follows from $P(x,1)$, noting $f(x)<1 \forall x$ and then iteration.
Claim 3: $f$ is continuous.
We start by observing that $f$ is continuous at $\varphi$, now define a sequence $a_i$ such that $a_i \mapsto \frac{1}{\varphi}$, which yields $f$ continuous for $\geq \varphi$ now similarly we get $f$ is continuous everywhere
Claim 4: $f$ is monotonic.
By $P(x,\varphi)$ notice that we get $f(xy+x+y) \leq f(x\varphi+x+\varphi)$ which implies monotonicity on $[\varphi,\varphi+1]$ and similarly we extend the monotonicity to everywhere.
By Claim 3 and Claim 4, we conclude there exists a limit of $f$ at $x \mapsto 0, \infty$, so by $P(\epsilon,y)$ we get $f(y)+L_{\infty}f\left(\frac{1}{y}\right)=1$ if $L_{\infty} < 1$ then we conclude that $f$ is constant (and equal to $\varphi$) and if $L_{\infty}=1$ we get $f(1)=1/2$ which implies $f(x)=x/(x+1)$ over rationals which finishes by continuity.
sami1618
17.08.2024 06:27
Let $\phi$ and $\Phi$ denote the little and big golden ratios. The functions that satisfy the equation are $f\equiv \phi$ and $f(x)=\frac{x}{x+1}$ which can both be checked to work. Let the assertion be $P(x,y)$. Claim: $f(x)$ is continuous at $x=\Phi$ First notice $f(\Phi)=\phi$ by $P(\phi,\phi)$. The assertion $P(\frac{z}{z+1},\frac{1}{z})$ gives that $$f\left(\frac{z+1}{z}\right)=\frac{1}{1+f(z)}$$Then define the sequences $a_0=z$ with $a_{n+1}=\frac{a_n+1}{a_n}$ and $b_0=f(z)$ and $b_{n+1}=\frac{1}{1+b_n}$. By induction we have that $f(a_n)=b_n$. Define the Fibonacci sequence by $F_0=0$, $F_1=1$, and $F_{n+2}=F_{n+1}+F_{n}$. Then we have that by easy induction $$a_n=\frac{F_{n+1}z+F_n}{F_n z+F_{n-1}}\;\;\;\;\;\text{and}\;\;\;\;\; b_n=\frac{F_{n-1} f(z)+F_{n}}{F_{n} f(z)+F_{n+1}}$$Thus by varying $z$ we have that $f$ maps the interval $(F_{n}/F_{n-1}, F_{n+1}/F_{n+1})$ to the interval $(F_{n-1}/F_{n}, F_{n+1}/F_{n+1})$. As it is well known that $\lim _{n\rightarrow \infty}{F_n/F_{n-1}}=\Phi$ and that $\Phi$ lies on $(F_{n}/F_{n-1}, F_{n+1}/F_{n+1})$ we have that $\lim_{x\rightarrow \Phi} f(x)=\phi$ by the epsilon-delta definition of the limit. Claim: $f$ is continuous As the function of $x$, $1-f(1/y)f(1/x)$, is continuous at $x=\phi$, the function $f(xy+x+y)$ is continuous at $x=\phi$, so the function $f(x)$ is continuous at $x=\phi y+\phi +y$ and hence on $x\in (\phi,\infty)$. Repeating this trick at $x=B$ gives that $f$ is continuous on $(1/B,\infty)$, implying the result by taking $B\rightarrow \infty$. Claim: Either $f\equiv \phi$ or $f$ achieves values arbitrarily close to $0$ and $1$ Notice that clearly we have $0<f<1$ so let $M$ and $m$ be the supremum and infimum of $f$. Then we must have that $m+M^2\leq 1\leq M+m^2$. Graphing the parabolas $m+M^2\leq 1$ and $1\leq M+m^2$ as well as the line $m\leq M$ one can easily see that the only solutions are $(m,M)=(\phi,\phi)$ or $(m,M)=(0,1)$. Claim: $f(x)+f(1/x)=1$ By Bolzano–Weierstrass theorem applied to the sequence $f(1), f(2), \dots$, there exists an infinite convergent sub-sequence with limit $m$. Then taking $x$ to be the reciprocal of the terms of this sequence we get that $f(y)+mf(1/y)=1$. If $m<1$ then taking $f(y)$ close to $0$ gives a contradiction. Now we have $f(1)=\frac{1}{2}$. Let $S$ denote the set of all $x$ such that $f(x)=\frac{x}{x+1}$. Then if $x\in S$ then $\frac{1}{x}\in S$ and by the $P(\frac{1}{z+1},z)$ assertion if $1/x\in S$ then $x+1 \in S$. Thus we have that $a+\frac{1}{b}\in S$ for all natural $a$ and $b$. Applying again $c+\frac{b}{ab+1}\in S$ for all natural $a$, $b$, and $c$. But then clearly $S$ is dense in $\mathbb{R^+}$ so as $f$ is continuous, the result follows.