Slovenia 2019 TST2 P2

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Source: 2019 Slovenia 2nd TST Problem 2

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pj294

#1 h Feb 14, 2019, 8:01 PM • 2 Y

Y by RAMUGAUSS, Adventure10

Determine all non-negative real numbers $a$ , for which $f(a)=0$ for all functions $f: \mathbb{R}_{\ge 0}\to \mathbb{R}_{\ge 0}$ , who satisfy the equation $f(f(x) + f(y)) = yf(1 + yf(x))$ for all non-negative real numbers $x$ and $y$ .

This post has been edited 3 times. Last edited by pj294, Feb 18, 2019, 11:12 PM

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#2 h Feb 15, 2019, 5:42 AM • 1 Y

Y by Adventure10

Partial Solution: Since $0$ lies in the co-domain of $f$ , we have $f\left(f\left(y\right)\right) = yf\left(1\right) \xrightarrow[\text{}]{\text{iterate}} f\left(yf\left(1\right)\right) = f\left(y\right)f\left(1\right)$ In particular, $f\left(0\right) = f\left(0\right)f\left(1\right)$ , which means $f\left(0\right) = 0$ or $f\left(1\right) = 1$ . If the latter holds, we get $f\left(f\left(y\right)\right) = y$ and thus $f\left(f\left(0\right)\right) = 0 = f\left(2f\left(0\right)\right)$ . It follows that $f\left(0\right) = 0$ in this case. We conclude that $f\left(0\right)$ is always zero. Finally, $0 = f\left(f\left(x\right)\right) = xf\left(1\right)$ , $f\left(1\right)$ equals to zero as well.

This post has been edited 3 times. Last edited by FEcreater, Feb 15, 2019, 1:37 PM
Reason: So many mistakes ...

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XbenX

#3 h Feb 15, 2019, 12:41 PM • 3 Y

Y by FEcreater, Adventure10, Mango247

Let $P(x,y)$ denote the assertion.
$P(x,0)$ and $P(0,0)$ gives that $f(f(x)+f(0))=0$ and $f(2f(0))=0$
$P(2f(0),y)$ and $P(2f(0),0)$ gives $f(f(y))=yf(1)$ and $f(f(0))=0$ .

If $f(1) \neq 0$ we have that $f$ is bijective , and $f(2f(0))=f(f(0))$ implies $f(0)=0$ and from $f(f(x)+f(0))=0$ we have $f(f(x))=0$ , but this contradicts surjectivity.
So we must have $f(1)=0$ . $P(1,1)$ gives $f(1)=f(0)=0$

See below for complete solution.

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L3435

#4 h Feb 15, 2019, 1:01 PM • 3 Y

Y by Morskow, Adventure10, Mango247

It's not hard to get $f(0)=f(1)=0$ . Suppose $f(p)\ne 0$ , $p>1$ . Let $q=f(p)+f\left(\frac{p-1}{f(p)}\right)$ . $P\left(p, \frac{p-1}{f(p)}\right)$ gives us $f(q)=p-1.$ Comparing $P(q, 0)$ with $P(q, 1)$ gives $0=f(f(q))=f(p),$ a contradiction. All such numbers are $0$ and all numbers greater than or equal to $1$ , because the function $f(x)=\begin{cases}1\textrm{, if }0<x<1\\0\textrm{ otherwise}\end{cases}$ solves the equation.

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256 posts

#5 h Oct 14, 2021, 2:00 AM • 1 Y

Y by jasperE3

Anyone can solve this functional equation which $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ ?

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11102 posts

#6 h Mar 18, 2025, 5:59 AM

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you should try this problem

kent2207 wrote:

Anyone can solve this functional equation which $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ ?

pj294 (modified) wrote:

Find all functions $f:\mathbb R^+\to\mathbb R^+$ for which:
$f(f(x)+f(y))=yf(1+yf(x))$ for all $x,y\in\mathbb R^+$ .

Claim 1: $f$ is surjective
Use $\left(1+x,\frac x{f(1+x)}\right)$ instead of $(x,y)$ in the equation, we get:
$f\left(f(1+x)+f\left(\frac x{f(1+x)}\right)\right)=x.$
Now setting $f(x)\mapsto x$ in the original equation we get the assertion $P(x,y)$ :
$f(x+f(y))=yf(1+xy).$ Claim 2: $f(x+1)\le1$ for all $x\in\mathbb R^+$
Suppose FTSOC that $f(u+1)>1$ for some $u$ :
$P\left(\frac{f(u+1)-1}u,u+1\right)\Rightarrow uf\left(\frac{(u+1)f(u+1)-1}u\right)=0$ ,
contradiction.

Claim 3: $f$ is injective
Suppose $f(a)=f(b)$ for some $a<b$ , and let $p=\frac ab<1$ :
$P\left(\frac xb,a\right)\Rightarrow\frac1bf\left(\frac xb+f(a)\right)=pf(1+px)$
$P\left(\frac xb,b\right)\Rightarrow\frac1bf\left(\frac xb+f(a)\right)=f(1+x)$
Comparing, we have $f(1+x)=pf(1+px)$ , and by simple induction $f(1+x)=p^nf(1+p^nx)$ for all $n\in\mathbb N$ . But then:
$f(1+x)=p^nf(1+p^nx)\le p^n$ so fixing $x$ and taking sufficiently large $n\gg\left\lvert\frac{\ln(f(1+x))}{\ln p}\right\rvert$ we reach a contradiction.

Claim 4: $f$ is nonincreasing
Suppose FTSOC that there are some $c>b$ with $f(c)>f(b)$ , then let $a=\frac cb-1$ , we have $a>0$ and $f(ab+b)>f(b)$ . Then:
$P(x+f(ab+b)-f(b),b)\Rightarrow f(x+f(ab+b))=bf(1+xb+bf(ab+b)-bf(b))$
$P(x,ab+b)\Rightarrow f(x+f(ab+b))=(ab+b)f(1+xb+xab)$
Comparing these, we get $f(1+xb+bf(ab+b)-bf(b))=(a+1)f(1+xb+xab)$ .
$P(xb,a+1)\Rightarrow f(xb+f(a+1))=(a+1)f(1+xb+xab)$
Comparing these, we get $f(1+xb+bf(ab+b)-bf(b))=f(xb+f(a+1))$ , so by injectivity (the $x$ terms cancel):
$f(a+1)=1+b(f(ab+b)-f(b)).$ However, $f(a+1)\le1$ and $1+b(f(ab+b)-f(b))>1$ , contradiction.

Claim 5: $f(f(x))=x$ for all $x\in\mathbb R^+$
Note that since $f$ is nonincreasing and bijective, it must be continuous.
$P(1,1)\Rightarrow f(1+f(1))=f(2)\Rightarrow f(1)=1$
Then take the limit as $x\to0^+$ :
$f(f(y))=\lim_{x\to0^+}f(x+f(y))=\lim_{x\to0^+}yf(1+xy)=yf(1)=y.$
Claim 6: $\boxed{f(x)=\frac1x}$ for all $x\in\mathbb R^+$
$P(1,x)\Rightarrow f(1+f(x))=xf(1+x)$
$P(1,f(x))\Rightarrow f(1+x)=f(x)f(1+f(x))$
Comparing these, we get:
$f(1+x)=xf(x)f(1+x),$ or $xf(x)=1$ . We verify that $f(x)=\frac1x$ fits $P(x,y)$ , and we're done.

This post has been edited 3 times. Last edited by jasperE3, Mar 18, 2025, 4:33 PM

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