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CarlFriedrichGauss-1777

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CarlFriedrichGauss-1777

#1 h Jun 4, 2021, 1:40 PM

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Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that:
$f(2021+xf(y))=yf(x+y+2021)$

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#2 h Jun 5, 2021, 2:29 AM • 1 Y

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$x,y$ are real numbers or positive real?????

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#3 h Jun 5, 2021, 2:39 AM

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Assuming it holds for all $x,y\in\mathbb R^+$ .
If $\exists u:f(u)>1$ then
$P\left(\frac u{f(u)-1},u\right)\Rightarrow\exists k:f(k)=0$ , contradiction.
So $f(x)\le1$ for all $x$ . Then $f(2021+xf(y))\le y$ , and taking $x=\frac1{f\left(\frac12f(2022)\right)}$ and $y=\frac12f(2022)$ is a contradiction.

This post has been edited 1 time. Last edited by jasperE3, Jun 5, 2021, 2:40 AM

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#4 h Apr 28, 2025, 7:35 AM

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jasperE3 wrote:

Assuming it holds for all $x,y\in\mathbb R^+$ .
If $\exists u:f(u)>1$ then
$P\left(\frac u{f(u)-1},u\right)\Rightarrow\exists k:f(k)=0$ , contradiction.
So $f(x)\le1$ for all $x$ . Then $f(2021+xf(y))\le y$ , and taking $x=\frac1{f\left(\frac12f(2022)\right)}$ and $y=\frac12f(2022)$ is a contradiction.

I don’t get it why f(k)=0 for some k?

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11287 posts

#5 h Apr 28, 2025, 5:03 PM

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Blackbeam999 wrote:

I don’t get it why f(k)=0 for some k?

what a terrible solution who wrote that

lemme rewrite it and see if it's more clear:

CarlFriedrichGauss-1777 wrote:

Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that:
$f(2021+xf(y))=yf(x+y+2021)$

Claim: $f(x)\le1$ for all $x>2021$
Suppose otherwise, then for some $u>2021$ we have $f(u)>1$ .
Plugging in $x=\frac u{f(u)-1}$ and $y=u$ we get:
$(u-1)f\left(\frac{uf(u)}{f(u)-1}\right)=0$ and since $u>2021$ we have $u-1\ne0$ and so $f\left(\frac{uf(u)}{f(u)-1}\right)=0$ , which contradicts that the codomain of $f$ is $\mathbb R^+$ .
So the claim is proven.

Then since $x+y+2021>2021$ , $P(x,y)$ implies:
$f(2021+xf(y))\le y,$ so $f(x+2021)\le y$ for all $x,y>0$ , but taking $y=\frac12f(x+2021)$ gives an immediate contradiction.
Hence no solutions.

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lksb

#6 h Apr 28, 2025, 9:59 PM

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$x\to \frac{y}{f(y)-1}\implies f\left(2021+\frac{yf(y)}{f(y)-1}\right)=yf\left(2021+\frac{yf(y)}{f(y)-1}\right)\implies y=1 \text{, Abs!}$
Therefore, $f(x)=1$ (obviously doesn't work) or $f(y)-1\leq0\implies f(y)\leq1\ \forall y\in\mathbb{R}$
$\implies f(2021+xf(y))\leq y\implies\ x\to\frac{1}{f(\frac{1}{2}f(2022))}, y\to \frac{1}{2}f(2022)\implies f(2022)\leq \frac{1}{2}f(2022)<f(2022),\ \text{Abs!}$
Therefore, no such function exists

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