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Source: Iran TST 2025 Test 2 Problem 3

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M11100111001Y1R

130 posts

M11100111001Y1R

#1 h Jun 5, 2025, 5:07 PM • 1 Y

Y by sami1618

Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for all $x, y >0$ we have:
$f(f(f(xy))+x^2)=f(y)(f(x)-f(x+y))$

This post has been edited 2 times. Last edited by M11100111001Y1R, Jun 5, 2025, 5:09 PM

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CatinoBarbaraCombinatoric

117 posts

CatinoBarbaraCombinatoric

#2 h Jun 5, 2025, 6:25 PM • 1 Y

Y by Lorenzo_degli_atti

$f$ is decresing because $RHS>0$ .
Suppose exist $k$ s.t. $f(x)>k>0$ for every $x$ .
Let $k=\inf(f)$
Fix $y$ and for large $x$
$k<f(y)(f(x)-f(x+y))<f(y)(f(x)-k)$
But $RHS$ can be arbitrarily small so we have a contradiction.
So
$\lim_{x\to \infty} f(x)=0$
Suppose that $f$ is upper bounded.
Given that $f$ is decresing exist $x_0$ such that
$\lim_{x\to x_0^+} f(x)=f(x_0)$
because there is at most $|\mathbb{N}|$ of points where there is a jump.
Fix $x=x_0$ and $y$ that approach $0$ .
$LHS$ is bounded from below by $f(x^2+M)$ where $M$ is the sup of $f$ , but $RHS\leq M(f(x_0)-f(x_0+y))$ that became arbitrarily small so we have a contradiction.
Now fix any $x$ and take $y$ that approach $0$ .
$LHS$ is bounded from above by $f(x^2)$ but if we look at $RHS$ , $f(y)$ became arbitrarily big so $f(x)-f(x+y)$ became arbitrarily small so
$\lim_{x\to x_0^+} f(x)=f(x_0)$
Now usa the sostitution $x\to x-y$ and with a similar argoument we find
$\lim_{x\to x_0^-} f(x)=f(x_0)$
So $f$ is continues.
Set $y=k/x$ for a positive real $k$ and consider the limit for $x\to 0^+$ .
$f(f(f(k)))=\lim_{x\to 0^+} f(k/x)f(x)$
And this gives also
$f(f(f(k)))=\lim_{x\to \infty} f(k/x)f(x)$

$\dfrac{f(f(f(a)))}{f(f(f(b)))}=\lim_{x\to \infty} \dfrac{f(a/x)}{f(b/x)}=\lim_{x\to 0^+} \dfrac{f(ax)}{f(bx)}$
So
$\lim_{x\to 0^+}\dfrac{f(g(x))}{f(h(x))}=f(f(f(\lim_{x\to 0^+} \dfrac{g(x)}{h(x)})))$

Rewrite the original equation as
$\dfrac{f(f(f(xy))+x^2)}{f(x)}=f(y)(1-\dfrac{f(x+y)}{f(x)})$
Take limit for $x\to 0$
$f(f(f(\lim_{x\to 0^+} f(f(xy))/x)))=f(y)$
$f(f(\lim_{x\to 0^+} f(f(xy))/x))=y$
But
$\lim_{x\to 0^+} f(f(xy))/x=\lim_{x\to 0^+} f(f(yx/y))/(x/y)=Cy$ for same costant $C$ .
This costant exist because it is $f^{-1}(f^{-1}(1))$ and inverse function is well defined because $f$ continues and decresing (+unbounded) gives $f$ is bigective.
So we find $f(f(Cy))=y$
$C=\lim_{y\to 0^+} f(f(Cy))/Cy=1/C$
So $f(f(y))=y$ .

But before we proved
$\dfrac{f(f(f(a)))}{f(f(f(b)))}=\lim_{x\to 0^+} \dfrac{f(ax)}{f(bx)}$
Which dipends only from ratio $a/b$
Now we have that $f$ is involution so we can replaced $f^3$ with $f$ and
$f(a)/f(b)=f(a/b)/f(1)$ which is similar to moltiplicative and from here is easy to conclude.
The original equation became
$f(x+y)f(x)=f(1)f(y)f(x)-f(1)f(y)f(x+y)$
Change $x$ with $y$ and subtracting and then dividing by $f(x+y)$
$f(1)f(x)+f(y)=f(1)f(y)+f(x)$
So $f(1)=1$ or $f$ is constant but constant function didn't work.
$f(x+y)(f(x)+f(y))=f(x)f(y)$
$x=y$ gives
$2f(2x)=f(x)$
We can prove by induction $nf(nx)=f(x)$
Sostitution $y=nx$
$f((n+1)x)(f(nx)+f(x))=f(x)f(nx)$
$f((n+1)x)(f(x)/n+f(x))=f(x)^2/n$
$(n+1)f((n+1)x)=f(x)$
From this we have also
$f(x/m)=mf(x)$
So $f(qx)=f(x)/q$ for every rational $q$ .
$f(q)=1/q$ and so
$f(x)=1/x$ because of continuity.
It's easy to see that this satisfies the equation.

This post has been edited 4 times. Last edited by CatinoBarbaraCombinatoric, Jun 5, 2025, 6:54 PM

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11417 posts

#3 h Jun 19, 2025, 4:50 AM

Y by

already posted elsewhere in the official thread

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