FE from R_>=0 to R_>=0

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FE from R_>=0 to R_>=0

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Source: Ukrainian mathematical olympiad 2018 10.4 and 11.3

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#1 h Jul 22, 2018, 6:07 PM • 3 Y

Y by Amir Hossein, qwertyboyfromalotoftime, Adventure10

Find all functions $f:[0,+\infty) \mapsto [0,+\infty)$ , which for all nonnegative $x,y$ satisfy $f(f(x)+f(y))=xyf(x+y)$ Proposed by Igor Voronovich

This post has been edited 1 time. Last edited by Mindstormer, Jul 24, 2018, 8:22 AM

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#2 h Jul 23, 2018, 8:26 AM • 2 Y

Y by Amir Hossein, Adventure10

Let $P(x, y)$ be the proposition that $f(f(x)+f(y))=xyf(x+y)$ .
Let $f(0) = a$ .
$\begin{align*} P(0, y) :\ & f(a+f(y))=0\\ P(0, a+f(y)) :\ & f(a+f(a+f(y))) = 0\\ &\ f(a)= 0\\ P(a, f(y)) :\ & f^3(y) = af(y)f(a+f(y)) = 0\\ \end{align*}$ Substituting $y=0$ into the last statement gives $f^3(0) = 0$ , but $f(a) = f^2(0) = 0$ , hence $f(0)=0$ .
$\begin{align*} P(0,y) :\ & f(f(y))=0\\ P(f(x),y) :\ & yf(x)f(f(x)+y) = f(f(f(x))+f(y)) = f(f(y)) = 0\\ \end{align*}$ In the last statement, if $y > 0$ ,
$0 = f(x)f(f(x)+y)$ In other words, if $f(x) > 0$ , $f(f(x)+y) = 0$ for all nonnegative $y$ . Now consider some $k$ such that $f(k) > 0$ .
$\begin{align*} P(k, y): f(f(k)+f(y)) &= kyf(k+y)\\ 0 &= kyf(k+y)\\ \end{align*}$ Which means for any $j > k$ , $f(j) = 0$ . Hence there exists at most $1$ such $k$ .
$P(k, k): f(2f(k)) = k^2f(2k) = 0$ Hence $f(k) \ne \frac k 2$ . Also, since $f(f(k)) = 0$ , $f(k) \ne k$ .

Hence either $f(x) = 0$ , or there exists some $k \ne 0$ where $f(k) = c \ne 0, k, \frac k 2$ , and $f(x) = 0$ otherwise.
(Edit Oopsie, see #6 below) It can be checked that these functions work.

This post has been edited 5 times. Last edited by joyce_tan, Jul 23, 2018, 2:47 PM

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#3 h Jul 23, 2018, 9:04 AM • 2 Y

Y by Adventure10, Mango247

joyce_tan wrote:

Substituting $y=0$ into the last statement gives $f^3(0) = 0$ , but $f(a) = f^2(0) = 0$ , hence $f(0)=0$ .

srr if its a dumb question but how do we get that $f(0) = 0$ shouldn`t we prove function is injective first?

This post has been edited 1 time. Last edited by deimis1231, Jul 23, 2018, 9:04 AM
Reason: ss

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TLP.39

#4 h Jul 23, 2018, 9:06 AM • 2 Y

Y by Adventure10, Mango247

@Above
$0=f^3(0)=f(f^2(0))=f(0)$

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#5 h Jul 23, 2018, 11:43 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Mindstormer wrote:

Find all functions $f:[0,+\infty) \mapsto [0,+\infty)$ , which for all nonnegative $x,y$ satisfy $f(f(x)+f(y))=xyf(x+y)$ Proposed by Igor Voronovich

I think, the following problem will be more complicated:
find all functions $f\colon (0,+\infty)\to(0,+\infty)$ such that $f(f(x)+f(y))=xyf(x+y).$ Here we get non-trivial solution $f(x)=\frac 1 x$ . Is it true that there are no more solutions?

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Fever

#6 h Jul 23, 2018, 2:39 PM • 2 Y

Y by Adventure10, Mango247

joyce_tan wrote:

Let $P(x, y)$ be the proposition that $f(f(x)+f(y))=xyf(x+y)$ .
Let $f(0) = a$ .
$\begin{align*} P(0, y) :\ & f(a+f(y))=0\\ P(0, a+f(y)) :\ & f(a+f(a+f(y))) = 0\\ &\ f(a)= 0\\ P(a, f(y)) :\ & f^3(y) = af(y)f(a+f(y)) = 0\\ \end{align*}$ Substituting $y=0$ into the last statement gives $f^3(0) = 0$ , but $f(a) = f^2(0) = 0$ , hence $f(0)=0$ .
$\begin{align*} P(0,y) :\ & f(f(y))=0\\ P(f(x),y) :\ & yf(x)f(f(x)+y) = f(f(f(x))+f(y)) = f(f(y)) = 0\\ \end{align*}$ In the last statement, if $y > 0$ ,
$0 = f(x)f(f(x)+y)$ In other words, if $f(x) > 0$ , $f(f(x)+y) = 0$ for all nonnegative $y$ . Now consider some $k$ such that $f(k) > 0$ .
$\begin{align*} P(k, y): f(f(k)+f(y)) &= kyf(k+y)\\ 0 &= kyf(k+y)\\ \end{align*}$ Which means for any $j > k$ , $f(j) = 0$ . Hence there exists at most $1$ such $k$ .
$P(k, k): f(2f(k)) = k^2f(2k) = 0$ Hence $f(k) \ne \frac k 2$ . Also, since $f(f(k)) = 0$ , $f(k) \ne k$ . Hence either $f(x) = 0$ , or there exists some $k \ne 0$ where $f(k) = c \ne 0, k, \frac k 2$ , and $f(x) = 0$ otherwise. It can be checked that these functions work.

By $P(\dfrac{k}{2}, \dfrac{k}{2})$ we will get that $0 = \dfrac{k^2}{4}f(k)$ so $f(k) = 0$ and this is contradiction with assumption that exists a positive value of $f$ , so its constantly zero.

This post has been edited 1 time. Last edited by Fever, Jul 23, 2018, 2:39 PM
Reason: forgot $

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Fever

#7 h Jul 23, 2018, 2:54 PM • 2 Y

Y by Adventure10, Mango247

@joice_tan, after your edit its still incorrect: there is no non-zero solution, check my comment properly

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#8 h Jul 23, 2018, 6:38 PM • 2 Y

Y by Adventure10, Mango247

Yeah, I wanted to leave my original wording there, only in red, just so it's not that confusing when you quote it. I realize that $f(x) = 0$ is the only solution.

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Anar24

#9 h Jul 24, 2018, 7:41 AM • 3 Y

Y by MNJ2357, Adventure10, Mango247

I don't think that problem must be such easy.Igor Voronovich is master in FE and Mindstormer titled this problem as R+ so maybe this question was proposed as R+ in contest.Am i true about it?

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#10 h Jul 24, 2018, 8:23 AM • 3 Y

Y by Anar24, Adventure10, Mango247

@above the problem is exactly what I wrote; my mistake though, changed the title from R+ to R_>=0

This post has been edited 1 time. Last edited by Mindstormer, Jul 24, 2018, 8:24 AM

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