Just a note: generalization of this problem appeared in IMC 2000 as problem 5. http://www.mathlinks.ro/Forum/viewtopic.php?p=357658#p357658

orl wrote: Find all functions $f$ defined on the non-negative reals and taking non-negative real values such that: $f(2)=0,f(x)\ne0$ for $0\le x<2$, and $f(xf(y))f(y)=f(x+y)$ for all $x,y$. Can someone confirm this? The Engel book says no solution...

More or less the same problem has been given to us in the 5th German TST 2004, only made a bit harder by stating that $f\left(x\right)\neq 0$ holds for all 0 < x < 2 (instead of $0\leq x<2$), so that we had to prove that $f\left(0\right)\neq 0$ by an additional argument. Towersfreak2006's solution is correct, and is more or less identic with the solution I gave on the exam. Engel is sometimes really strange about giving solutions in his book, so "no solution" should neither mean that the problem is too hard, nor that it is too easy. But there is always MathLinks, Kalva, ... Darij

Consider $ 0\leq y < 2$. \[ x: =\frac{2}{f(y)} \Rightarrow f(\frac{2}{f(y)}+y)=f(2)f(y)=0 \Rightarrow \frac{2}{f(y)}+y \geq 2 \Rightarrow f(y) \leq \frac{2}{2-y} \\ x: =(2-y) \Rightarrow f((2-y)f(y))f(y)=0 \Rightarrow (2-y)f(y)\geq 2 \Rightarrow f(y) \geq \frac{2}{2-y}\]

Let's find all $ f: [0,\infty)\to [0,\infty)$ with $ (*)$ $ f(xf(y))f(y) = f(x + y)$ for all $ x,y\ge 0$ First assume $ f(x) = 0$ for some $ x\ge 0$ and take $ c: = \inf\{x\ge 0: f(x) = 0\}\ge 0$. Then for all $ z > c$ we find $ y < z$ with $ f(y) = 0$, and for $ x = z - y$ in $ (*)$ we get $ f(z) = f(x + y) = f(xf(y))f(y) = 0$. This gives $ f(x) = 0$ for all $ x > c$ and $ f(x)\not = 0$ for all $ 0\le x < c$. Then for all $ 0\le y < c$ we have $ f(y)\not = 0$ and $ x > c - y\Rightarrow f(x + y) = 0 \Rightarrow f(xf(y)) = 0 \Rightarrow f(y)\ge {c\over x}$ and $ 0 < x < c - y\Rightarrow f(x + y)\not = 0 \Rightarrow f(xf(y))\not = 0 \Rightarrow f(y)\le {c\over x}$ for all $ x$. So $ f(y)\ge \sup_{x > c - y}{c\over x} = {c\over c - y}$ and $ f(y)\le \inf_{0 < x < c - y}{c\over x} = {c\over c - y}$ for all $ 0\le y < c$. This gives $ f(x) = {c\over c - x}$ for all $ 0\le x < c$, which leaves to find $ f(c)$. Assume $ f(c)\not = 0$. Then for all $ x > 0$ and $ y = c$ in $ (*)$ we get $ f(xf(c))f(c) = f(x + c) = 0\Rightarrow f(xf(c)) = 0\Rightarrow xf(c)\ge c$. So $ c\le \inf_{x > 0}xf(c) = 0$, and therefore $ c = 0$ and $ f(0)\not = 0$. But $ x = y = 0$ in $ (*)$ gives $ f(0)^2 = f(0)\Rightarrow f(0) = 1$. So $ f(x) = \{ {1\text{ for }x = 0\atop 0\text{ for }x\not = 0}$. Otherwise $ f(c) = 0$ and $ f(x) = \{ {{c\over c - x}\text{ for }0\le x < c\atop 0\text{ for }x\ge c}$ for some $ c\ge 0$. (This is $ f\equiv 0$ for $ c = 0$) Finally assume $ f(x) > 0$ for all $ x\ge 0$. If $ f(y) > 1$ for some $ y\ge 0$, then for $ x = {y\over f(y) - 1}\ge 0$ we have $ xf(y) = x + y$, and $ (*)$ gives $ 1 < f(y) = {f(x + y)\over f(xf(y))} = 1$. Contradiction! So $ f(x)\le 1$ for all $ x\ge 0$, and $ (*)$ gives $ f(y)\ge f(xf(y))f(y) = f(x + y)$ for all $ x,y\ge 0$. So $ f$ is monotonically decreasing. Assume $ f(a) = f(b)$ for some $ 0\le a < b$. Then by $ (*)$ we get $ f(x + a) = f(xf(a))f(a) = f(xf(b))f(b) = f(x + b)$ for all $ x\ge 0$. For $ x = n(b - a)$ we get $ f(n(b - a) + a) = f(n(b - a) + b) = f((n + 1)(b - a) + a)$ for all $ n\in\mathbb{N}_0$, and so by induction $ f(a) = f(n(b - a) + a)$ for all $ n\in\mathbb{N}_0$. But then for all $ x\ge a$ we find $ n\in\mathbb{N}_0$ with $ x < n(b - a) + a$, and then $ f(a)\ge f(x)\ge f(n(b - a) + a) = f(a)$. So $ f(x) = f(a)$ for all $ x\ge a$, and for all $ y\ge 0$ we can find $ x > 0$ with $ xf(y)\ge a$ and $ x + y\ge a$, and using $ (*)$ we get $ f(y) = {f(x + y)\over f(xf(y))} = {f(a)\over f(a)} = 1$. This gives $ f\equiv 1$, or otherwise $ f$ is strictly monotonically decreasing. Assume the latter. Then replacing $ x$ by $ {x\over f(y)}$ in $ (*)$ we get $ f(x)f(y) = f({x\over f(y)} + y)$ for all $ x,y\ge 0$, and using the symmetry of the left side we get $ f({x\over f(y)} + y) = f({y\over f(x)} + x)$ and $ {x\over f(y)} + y = {y\over f(x)} + x$ for all $ x,y\ge 0$. Then for $ y = 1$ and $ c: = {1\over f(1)} - 1\ge 0$ we get $ f(x) = {1\over 1 + cx}$ for all $ x\ge 0$. (This is $ f\equiv 1$ for $ c = 0$) We therfore get $ 3$ types of solutions of $ (*)$: $ f(x) = \{ {1\text{ for }x = 0\atop 0\text{ for }x\not = 0}$ or $ f(x) = \{ {{c\over c - x}\text{ for }0\le x < c\atop 0\text{ for }x\ge c}$ or $ f(x) = {1\over 1 + cx},x\ge 0$ for some $ c\ge 0$. Easy to see, that these are legit.

orl wrote: Find all functions $ f$ defined on the non-negative reals and taking non-negative real values such that: $ f(2) = 0,f(x)\ne0$ for $ 0\le x < 2$, and $ f(xf(y))f(y) = f(x + y)$ for all $ x,y$. Looking at this old IMO problem I get a strange result. If I put $ y=2$ in the property $ f(xf(y))f(y) = f(x + y)$ this implies $ \implies f(x+2)=0$ for all $ x\ge0$. This should imply that $ f(x)=0$ for all $ x\ge0$. Where do I get wrong? The problem statement is correct?

prester wrote: orl wrote: Find all functions $ f$ defined on the non-negative reals and taking non-negative real values such that: $ f(2) = 0,f(x)\ne0$ for $ 0\le x < 2$, and $ f(xf(y))f(y) = f(x + y)$ for all $ x,y$. Looking at this old IMO problem I get a strange result. If I put $ y = 2$ in the property $ f(xf(y))f(y) = f(x + y)$ this implies $ \implies f(x + 2) = 0$ for all $ x\ge0$. This should imply that $ f(x) = 0$ for all $ x\ge0$. Where do I get wrong? The problem statement is correct? No : $ f(x + 2) = 0$ for all $ x\ge 0$ implies $ f(x)=0$ $ \forall x\ge 2$ and not $ f(x) = 0$ for all $ x\ge 0$ So the required function is non zero on $ [0,2)$ and zero on $ [2,+\infty)$

pco wrote: prester wrote: orl wrote: Find all functions $ f$ defined on the non-negative reals and taking non-negative real values such that: $ f(2) = 0,f(x)\ne0$ for $ 0\le x < 2$, and $ f(xf(y))f(y) = f(x + y)$ for all $ x,y$. Looking at this old IMO problem I get a strange result. If I put $ y = 2$ in the property $ f(xf(y))f(y) = f(x + y)$ this implies $ \implies f(x + 2) = 0$ for all $ x\ge0$. This should imply that $ f(x) = 0$ for all $ x\ge0$. Where do I get wrong? The problem statement is correct? No : $ f(x + 2) = 0$ for all $ x\ge 0$ implies $ f(x) = 0$ $ \forall x\ge 2$ and not $ f(x) = 0$ for all $ x\ge 0$ So the required function is non zero on $ [0,2)$ and zero on $ [2, + \infty)$ Ok pco, I got a stupid result. $ x=y-2 \ge 0$ $ \implies f(y)=0 \;\;\;\forall y\ge 2$ Thanks for your quick remark

Taking $y=2$ gives $f(x+2)=0 \forall x\ge 0$, or equivalently $f(x)=0 \forall x\ge 2$. Taking $y=2-x$ for $0\le x<2$ yields $$f(xf(2-x))f(2-x)=f(2)=0$$Since $0<2-x\le 2$, we must have $f(xf(2-x))=0$, which forces $$xf(2-x)\ge 2\implies f(x)\ge \frac{2}{2-x}\forall 0\le x<2.$$Now taking $x=2/f(y)$, for $0\le y<2$ gives, $$f(y+2/f(y))=f(2)f(y)=0\implies y+2/f(y)\ge 2.$$Thus, $$f(y)\le \frac{2}{2-y}\forall 0\le y<2$$Hence, this forces $f(x)=2/(2-x)$ for $0\le x<2$, giving a final answer of \[ f(x)=\begin{cases} 0 & \text{ if }x \ge 2 \\ \frac{2}{2-x} & \text{ if } 0 \le x < 2 . \end{cases} \]

Solved with L567 - posting merely for storage and hoping that we get a P5 as easy as this at this year's IMO

The only function is $$f(x) = \begin{cases} \frac 2{2-x} &0\leq x<2 \\ 0 & x\geq 2.\end{cases}$$First the following claim: Claim. $f(x) = 0$ for $x \geq 2$. Let $y=2$, from where we obtain $f(x+2) = 0$ for all $x$ in the domain of $f$, or all $x \geq 0$. Hence $f(x) = 0$ for $x\geq 2$. $\blacksquare$ Now set $x = 2-y$ for $x, y > 0$, such that $$f((2-y)f(y))f(y) = f(2) = 0.$$By the third condition, $f(y) \neq 0$, so $$f((2-y)f(y)) = 0 \iff f(y)(2-y) \geq 2.$$This means that $f(y) \geq \frac 2{2-y}$ for all $0 \leq y < 2$. On the other hand, set $x = 2-\varepsilon - y$. Because $f(2-\varepsilon) \neq 0$, we know that $$f((2-\varepsilon-y)f(y)) \neq 0 \iff f(y)(2-\varepsilon-y) < 2.$$This means that $f$ is bounded $$\frac 2{2-\varepsilon-x} < f(x) \leq \frac 2{2-x}$$for all $0\leq x < 2$. It follows that $f(x) = \frac 2{2-x}$ for all $0\leq x<2$, as required. It suffices to verify this function for $x, y < 2$ and $x+y<2$. Indeed, the LHS equals $$f\left(\frac {2x}{2-y}\right) \cdot \frac 2{2-y} = \frac 2{2-\frac{2x}{2-y}} \cdot \frac 2{2-y} = \frac 2{2-(x+y)}.$$

Let $P(x,y)$ denote the assertion. Claim: $f(x)=0$ for $x\ge2$. $P(x,2): f(x+2)=0$, which proves our claim. Claim: $f(x)\ge\frac{2}{2-x}$ for all $0\ge x<2$. Proof: For $x<2$, $P(2-x,x): f((2-x)f(x))f(x)=0$. Since $f(x)>0$, \[f((2-x)f(x))=0\implies (2-x)f(x)\ge 2\implies f(x)\ge \frac{2}{2-x}.\] $P(0,x): f(0)f(x)=f(x)$. If $x<2$, we divide both sides by $f(x)$ and get $f(0)=1$. $P(\frac{2}{f(x)},x<2):0=f(\frac{2}{f(x)}+x)$. Then \[\frac{2}{f(x)}+x\ge 2\implies \frac{2}{f(x)}\ge 2-x\implies \frac{2}{2-x}\ge f(x)\] So $f(x)=\frac{2}{2-x}\forall x<2$. We claim that the only solution, $$\boxed{f(x) = \begin{cases} \frac 2{2-x} &0\leq x<2 \\ 0 & x\geq 2.\end{cases}}$$works. Proof: We prove this with casework. Case 1: $x,y\ge 2$. Then $0=0$, which is true. Case 2: $x\ge 2, y<2$. We have $f(y)=\frac{2}{2-y}\ge1$. So $xf(y)\ge 2$. Thus, we get $0=0$, which is true. Case 3: $x<2, y\ge2$. Then we get $0=0$. Case 4: $x,y<2$. Then we have $\frac{2f(\frac{2x}{2-y})}{2-y}=f(x+y)$. Subcase 4.1: $x+y\ge 2$. Then $x\ge 2-y\implies \frac{2x}{2-y}\ge 2$. So $0=0$. Subcase 4.2: $x+y< 2$. Then $x< 2-y\implies \frac{2x}{2-y}<2$. So we have \[\frac{\frac{4}{2-\frac{2x}{2-y}}}{2-y}=\frac{4}{(2-y)(2-\frac{2x}{2-y})}=\frac{4}{4-2y-2x}=\frac{2}{2-x-y},\]which is true.

Label the conditions as follows: AoPS wrote: (i) $f(xf(y)) f(y) = f(x + y)$ for all $x$, $y \ge 0$, (ii) $f(2) = 0$, (iii) $f(x) \neq 0$ for $0 \le x < 2$. Let $P(x,y)$ denote assertion $(\text i)$. $P(x,2)\Rightarrow f(x)=0\forall x\ge2$ by $(\text{ii})$, so $f(x)=0\Leftrightarrow x\ge2$ by $(\text{iii})$. Let $x\in[0,2)$, then $f(x)\ne0\ne f(2-x)$ and: $P(2-x,x)\Rightarrow f((2-x)f(x))=0\Rightarrow (2-x)f(x)\ge2\Rightarrow f(x)\ge\frac2{2-x}$ $P\left(\frac2{f(x)},x\right)\Rightarrow f\left(\frac2{f(x)}+x\right)=0\Rightarrow\frac2{f(x)}+x\ge2\Rightarrow f(x)\le\frac2{2-x}$ Thus $\boxed{f(x)=\begin{cases}\frac2{2-x}&\text{if }x\in[0,2)\\0&\text{if }x\ge2\end{cases}}$, which works by simple casework.

Let $P(x,y)$ denote the assertion $f(xf(y))f(y)=f(x+y)~\forall x,y \geq 0.$ $P(x,2)\implies f(x)=0 ~\forall x\geq 2,$ which works. $P(0,0)\implies f(0)=1$ or $f(0)=0,$ the latter is false. For $x\in (0,2),$ we have $f(x)\geq 2/(2-x),$ by $P(2-x,x).$ Assume $f(x)>2/(2-x)$. $\exists u,v\in (0,2): f(u)>2/(2-u)$ and $f(v)=2/(2-u).$ Case 1. $v<u\implies f(v)=2/(2-v)\implies u=v,$ contradiction. Case 2. $v>u.$ But then, $P(2-v,u)\implies u\geq v,$ contradiction. Thus the solution for $x\in [0,2)$ is $f(x)=2/(2-x),$ which works.

Alright, I love this problem! Usual tricks worked! phew! Solution: Denote $P(x,y)$ as the assertion to the functional equation. The solution will be divided into two parts. Answer and Verification The answer is \[f(x) = \begin{cases} \frac{2}{2-x} & \text{for all $x \in [0,2)$} \\ 0 & \text{for all $x \ge 2$.} \end{cases}\]We will now show that it works. Consider two cases based on size of $x+y$. Say $x+y < 2$, then clearly $\frac{2x}{2-y}<2$. Plugging everything back, we get that it works. If $x+y >2$, then note that if $y \ge 2$ then we're already done. So, assume $y \le 2$. Since $x+y > 2 \implies \frac{2x}{2-y}>2$, we find a tautology which verifies the solution. Uniqueness of Solution Start with $P(x,2)$ to get that $f(x) = 0$ for $x \ge 2$. $P(k,2)$ for $k \in [0,2]$ gives $f\left(2f(k) \right) = 0$. Due to problem condition, we must have $2f(k) \ge 2 \iff f(k) \ge 1$. From here, apply the cancellation trick and substitute $P\left(x,\frac{kf(k)}{f(k)-1} \right)$. This yields \[f\left( \frac{kf(k)}{f(k)-1} \right) = 0 \iff \frac{kf(k)}{f(k)-1} \ge 2 \iff f(k) \le \frac{2}{2-k}. \tag{1} \label{1} \]Here's the final blow! Consider $P(k,2-k)$. This would yield \[f\left( kf(2-k) \right) = 0 \iff kf(2-k) \ge 2 \iff f(k) \ge \frac{2}{2-k}. \tag{2} \label{2}\]Finally, on combining $(1)$ and $(2)$, we get a strict equality which concludes the solution. $\blacksquare$

We claim the only such function is \[\boxed{f(x)=\begin{cases} \dfrac{2}{2-x} &\text{if }0\leq x<2 \\ 0&\text{if }2\leq x \\ \end{cases}}.\]We first show this function satisfies the conditions. We can see that $f(2)=0$ and $f(x)\neq 0$ for all $0\leq x< 2.$ If $y\geq 2,$ then since $x+y\geq 2,$ $f(x(f(y))f(y)=0=f(x+y).$ If $0\leq y <2,$ then $f(xf(y))f(y)=f\left(\dfrac{2x}{2-y}\right)f(y).$ If we also have $\dfrac{2x}{2-y}\geq 2,$ then $\dfrac{x}{2-y}\geq 1,$ so $x+y\geq 2,$ so $f(xf(y))f(y)=f\left(\dfrac{2x}{2-y}\right)f(y)=0=f(x+y),$ and if $x+y\geq 2,$ then we have $x\geq 2-y,$ so $\dfrac{2x}{2-y}\geq 2,$ so $f(x+y)=0=f\left(\dfrac{2x}{2-y}\right)f(y)=f(xf(y))f(y).$ As such, $f$ satisfies all of the conditions of the problem. We now prove that this is the only solution to the functional equation. Denote $P(x,y)$ as the condition that $f(xf(y))f(y)=f(x+y).$ From $P(x,2)$ for $x\geq 0,$ we have $f(xf(2))f(2)=0=f(x+2),$ so $f(y)=0$ for all $y\geq 2.$ From $P(2-x,x)$ for $0\leq x<2,$ we have that $f((2-x)f(x))f(x)=f(2)=0,$ so since $f(x)\neq 0,$ we have $f((2-x)f(x))=0,$ so $(2-x)f(x)\geq 2,$ and $f(x)\geq \dfrac{2}{2-x}.$ From $P\left(\dfrac{2}{f(x)},x\right)$ for $0\leq x<2$ (which is valid as $f(x)\neq 0$), we get $f\left(\dfrac{2}{f(x)}\cdot f(x)\right)f(x)=f(2)f(x)=0=f\left(\dfrac{2}{f(x)}+x\right),$ so $\dfrac{2}{f(x)}+x\geq 2,$ so $\dfrac{2}{f(x)}\geq 2-x,$ so $\dfrac{2}{2-x}\geq f(x)\geq \dfrac{2}{2-x},$ so $f(x)=\dfrac{2}{2-x}$ for all $0\leq x <2.$ As such, the only such function is $f(x)=\begin{cases} \dfrac{2}{2-x} &\text{if }0\leq x<2 \\ 0&\text{if }2\leq x \\ \end{cases},$ as desired. $\Box$

Note that $f(x) = 0 \forall x \ge 2$. Now the idea is to take $x = \frac{2}{f(y)} - \epsilon$ for epsilon small to get one bound (in particular, consider the limit as epsilon tends to zero), and $x = \frac{2}{f(y)}$ for the other side bound. This should give $f(x) = \frac{2}{2-x} \forall x \in [0, 2)$. Oh, and this function works. $\square$