$f:\mathbb{R_{\geq 0}} \to \mathbb{R_{\geq 0}}$ such that for all $x$ we have $xf(1+xf(y))=f(f(x)+f(y))$

Question

Find all nonnegative real number $a$, such that $f(a)=0$ for any function $f$ satisfying: $xf(1+xf(y))=f(f(x)+f(y))$ with all $x,y$ are nonnegative real number.

I don't know why this problem only ask the number $a$ for $f(a)=0$ instead of the function $f(x)$. Is there any special thing here? Is it possible to show the bijection? I will show what I get from trying to solve this: $f(f(x)+f(0))=f(1+xf(0))=0$. How to find $t=f(0)$, show me your solution or any idea please, thank all.

I don't understand your calculation. If we let $y=0$ we get $xf(1+xf(0))=f(f(x)+f(0))$ which is not what you wrote. Better, I think, to let $x=0$, which yields $0=f(f(0)+f(y))$. — lulu, Jun 23 '19 at 10:56
One might think that $f(x)\equiv 0$ is the only solution. As a counterexample, consider the function $$f(x)=\begin{cases} 1 \text{ if $0<x<1$} \ 0 \text{ else }\end{cases}$$ which works — user574848, Jun 24 '19 at 01:00

score 2 · Answer 1 · answered Jun 23 '19 at 11:53

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I have started writing couse I thought I will solve it. But no, sorry. Perhaps you will get some idea from here. If you don't like it I will delete it.

We have: $$xf(1+xf(y))=f(f(x)+f(y))$$

Suppose there exists $a$ such that $f(a)=0$.

If $y=a$ we get: $bx = f(f(x))$ where $b=f(1)$. So if $\boxed{b\ne 0}$ then $f$ is injective (so $f$ has only one zero), so for $x=1$ we get $f(1+f(y)) = f(b+f(y))\implies b=1$. Now for $x=0$ we get $f(f(0)+f(y))=0$, so $f(0)+f(y)=a $ for all $y$, thus $f$ is constant. A contradiction. So $\boxed {b=0}$ and $f(f(x))=0$ for all $x$. So in special case (if we put $x=a$) we get $f(0)=0$.
If $x=a$ we get: $af(1+af(y)) = f(f(y))=0$

answered Jun 23 '19 at 11:53

nonuser

90,026

1

Not sure if it helps but the right side is symmetric in x, y so we have $xf(1+xf(y))=yf(1+yf(x))$ – kingW3 Jun 23 '19 at 11:56
@Aqua Do you know the solution for function $f$, It does not seem to be enough..., can you tell me the solution for $a$? – user628755 Jun 24 '19 at 02:03
You asked your sister? – nonuser Jun 24 '19 at 07:27
@Aqua. It is a problem in $belarus$ training for imo 2018 – user628755 Jun 26 '19 at 02:56
Help me please everyone... – user628755 Jun 26 '19 at 03:20
@Aqua help me pls... – user628755 Jun 27 '19 at 16:35
I would if I could... Sorry :( – nonuser Jun 27 '19 at 16:46

$f:\mathbb{R_{\geq 0}} \to \mathbb{R_{\geq 0}}$ such that for all $x$ we have $xf(1+xf(y))=f(f(x)+f(y))$

1 Answers1