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Source: Bulgaria NMO 2024, Problem 3

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1390 posts

#1 h Apr 15, 2024, 11:01 AM • 5 Y

Y by PRMOisTheHardestExam, MathLuis, rightways, topologicalsort, sami1618

Find all functions $f:\mathbb {R}^{+} \rightarrow \mathbb{R}^{+}$ , such that $f(af(b)+a)(f(bf(a))+a)=1$ for any positive reals $a, b$ .

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topologicalsort

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topologicalsort

#2 h Apr 15, 2024, 11:30 AM • 3 Y

Y by sami1618, BorisAngelov1, kkloveMinecraft

What a fun and easy functional equation for a 3rd problem

Let $P(a,b)$ be the given equation.
$P(a,b)$ for a large enough $a$ $\implies$ $f(x)$ has an infinitely small value.
Let $f(c)$ be small enough. Then $P(a,{c}/{f(a)})$ means that $f$ is surjective.
Let $f(bf(a))+a=k$ for some parameters $a$ and $k$ , and variable $b$ . This equation has solution $\iff$ $a<k$ , since $f$ is surjective. This means that $Q(a,k) = f(af(b)+a) = 1/k$ also has a solution (for the same values of $a$ and $k$ ) $\iff$ $a<k$ . We also have that $af(b)+a = a(f(b)+1)$ , takes on all values larger than $a$ , again due to surjectivity.
Let $f(z) = 1/k$ . $Q(k,k)$ doesn't have a solution $\implies z \leq k$ . But $Q(k-\epsilon,k)$ does have a solution $\implies z > k-\epsilon \implies z \geq k \implies z=k \implies \boxed{f(k) = 1/k}$ which clearly works.

Remark

This post has been edited 4 times. Last edited by topologicalsort, Apr 15, 2024, 11:47 AM

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#3 h Apr 15, 2024, 12:21 PM • 1 Y

Y by sami1618

Let $P(a,b)$ the assertion of the F.E.
Claim 1: $\lim_{x \to \infty} f(x)=0$
Proof: By $P \left(\frac{a}{f(b)+1}, b \right)$ we have:
$f(a)=\dfrac{1}{f \left(b f \left(\frac{a}{f(b)+1} \right) \right)+\frac{a}{f(b)+1}}<\frac{f(b)+1}{a}$ This holds for any $a,b \in \mathbb R^+$ so fix $b$ and take $a$ really huge and by definition of the limit at infinity our claim is proven.
Claim 2: $xf(x) \le 1$ for all $x \in \mathbb R^+$
Proof: In the inequality from Claim 1 just set $b \to \infty$ and done.
Claim 3: $f$ is surjective.
Proof: By $P \left(a-f(b), \frac{b}{f(a)} \right)$ for $a>f(b)$ we have:
$f \left((a-f(b)) \left(f \left(\frac{b}{f(a)} \right)+1 \right) \right)=\frac{1}{a}$ As we can make $f(b)$ arbitrarily small we have proven this claim.
Claim 4: $f$ is injective.
Proof: Suppose $f(m)=f(n)$ for $m>n$ then by $P(a,m)-P(a,n)$ and Claim 3 we can easly get $f(mx)=f(nx)$ for all $x \in \mathbb R^+$ , now this means:
$f(x)=f \left(\left(\frac{m}{n} \right)^N \cdot x \right) \le \dfrac{1}{\left(\frac{m}{n} \right)^N \cdot x} \overset{N \to \infty}{\implies} f(x) \le 0 \; \text{contradiction!}$ Finishing: Clearly for $c>a$ there exists a $b$ such that for such $a$ and fixed $c$ we have $f(af(b)+a)=\frac{1}{c}=f(z)$ (because by $P(a,b)$ we get $f(bf(a))=c-a$ so we choose a suitable $b$ for any $a<c$ ) , note also $c \ge z>a$ (because $z=af(b)+a>a$ and $\frac{1}{c} \le \frac{1}{z}$ ) then , but a suitable $b$ lets us choose $a=c-\epsilon$ for any $\epsilon$ really really small so $c \ge z>c-\epsilon$ which in fact would mean by trivial analysis stuff that $c \ge z \ge c$ therefore $c=z$ and $f(c)=\frac{1}{c}$ , since in this process $c$ was not bounded or anything...
The only solution to this F.E. is $\boxed{f(x)=\frac{1}{x} \; \forall x \in \mathbb R^+}$ thus we are done :cool:

:cool:

This post has been edited 1 time. Last edited by MathLuis, Apr 15, 2024, 12:21 PM

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#4 h Apr 22, 2024, 7:40 AM • 1 Y

Y by topologicalsort

Easy functional equation by surjectivity

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910 posts

#5 h Jun 6, 2024, 3:32 PM

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The answer is $\boxed{f\equiv 1/x}$ , which can be verified to work:

Claim: $\lim_{x\rightarrow \infty}f(x)=0$
Taking $a\rightarrow \infty$ this follows by the sandwich theorem.
Claim: $f$ is surjective
Notice that $1/(f(bf(a))+a)$ is in the domain of $f$ , by varying $a$ and choosing $bf(a)$ to be a fixed large number finishes.
Claim: $f(x)\leq 1/x$
Taking $a=x/(1+f(b))$ have that $f(x)\leq (1+f(b))/x$ so simply take $b\rightarrow \infty$ .
Claim: $f$ is injective
If $f(b)=f(c)$ with $c<b$ then $f(bf(a))=f(cf(a))$ or $f(\frac{b}{c}x)=f(x)$ or that $f(\frac{b^n}{c^n}x)=f(x)$ , contradicting the first claim.

Fix $a$ . Now notice that $f(af(b)+a)$ achieves all values smaller than $\frac{1}{a}$ or that $f:(a,\infty)\rightarrow (0,\frac{1}{a})$ is a bijection. However $f:(a+\epsilon,\infty)\rightarrow (0,\frac{1}{a+\epsilon})$ is also a bijection so $f:(a,a+\epsilon)\rightarrow(\frac{1}{a+\epsilon},\frac{1}{a})$ is also a bijection. Taking $\epsilon\rightarrow 0$ and varying $a$ finishes the problem.

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#6 h Sep 19, 2024, 7:50 AM • 4 Y

Y by sami1618, internationalnick123456, seoneo, Ali_Vafa

Let $P(x,y)$ be the assertion $f\left( xf\left( y \right)+x \right)=\frac{1}{f\left( yf\left( x \right) \right)+x}$

1) $\lim_{x\rightarrow +\infty}f(x)=0$
Proof

2) $f\left( x \right)\le \frac{1}{x},\forall x>0$
Proof

3) $f$ is surjective
Proof
4) $f$ is injective
Proof
5) $f\left( a \right)<1\Rightarrow a>1$
Proof

5) $y>z\Rightarrow f(y)<f(z)$
Proof

6) $\boxed{f\left( x \right)=\frac{1}{x},\forall x>0}$
Proof

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seoneo

#7 h Oct 4, 2024, 10:40 AM

Y by

@topologicalsort wrote

Quote:

Let $P(a,b)$ be the given equation.
$P(a,b)$ for a large enough $a$ $\implies$ $f(x)$ has an infinitely small value.
Let $f(c)$ be small enough. Then $P(a,{c}/{f(a)})$ means that $f$ is surjective.
Let $f(bf(a))+a=k$ for some parameters $a$ and $k$ , and variable $b$ . This equation has solution $\iff$ $a<k$ , since $f$ is surjective. This means that $Q(a,k) = f(af(b)+a) = 1/k$ also has a solution (for the same values of $a$ and $k$ ) $\iff$ $a<k$ . We also have that $af(b)+a = a(f(b)+1)$ , takes on all values larger than $a$ , again due to surjectivity.
Let $f(z) = 1/k$ . $Q(k,k)$ doesn't have a solution $\implies z \leq k$ . But $Q(k-\epsilon,k)$ does have a solution $\implies z > k-\epsilon \implies z \geq k \implies z=k \implies \boxed{f(k) = 1/k}$ which clearly works.

Except for the last two lines of the solution, it shows that $f^{-1}(1/k) \cap (a,\infty) \neq \emptyset \iff a<k$ . However, we also need to exclude the possibility that $f^{-1}(1/k)$ has $k$ as a limit point without $k$ being a member.

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#8 h Apr 8, 2025, 10:31 AM

Y by

$f(af(b)+a)(f(bf(a))+a)=1$ Answer is $f(x)=\frac{1}{x}$ . Let $P(a,b)$ be the assertion. First, $P(a,b/f(a))$ gives $f(af(\frac{b}{f(a)})+a)=\frac{1}{a+f(b)}$ hence $f$ is not bounded below by a positive value and this implies $f(b)$ can take sufficiently small values thus, $f$ is surjective. Let $f(t)=1$ . $P(t,b)$ gives $f(tf(b)+t)(f(b)+t)=1$ . Replace $a$ with $f(b)$ to get $f(ta+t)=\frac{1}{t+a}$ so for $a>t$ we have $f(a)=\frac{t}{a+t^2-t}$ . Also $P(\frac{t}{f(b)+1},b)$ yields $f(bf(\frac{1}{f(b)+1}))+\frac{t}{f(b)+1}=1$ hence $\frac{t}{f(b)+1}<1$ or $t<f(b)+1$ and since $f$ is surjective, we have $t\leq 1$ . If $t<1$ , then $2t-t^2>t$ so $f(2t-t^2)=\frac{t}{2t-t^2+t^2-t}=1$ . Comparing $P(a,t)$ and $P(a,2t-t^2)$ implies $f(tf(a))=f((2t-t^2)f(a))$ so $f(a)=f((2-t)a)$ . However comparing $P(a,b)$ with $P(a(2-t),b)$ gives a contradiction. Thus, $t=1$ and for $a>1$ we see that $f(a)=\frac{1}{a}$ . For $a>b>1$ we have $\frac{1}{\frac{a}{b}+a}(f(\frac{b}{a})+a)=1$ so $f(\frac{b}{a})=\frac{a}{b}$ where $b/a<1$ thus, $f(x)=1/x$ for all positive reals as desired. $\blacksquare$

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