Functional equation

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160 posts

#1 h Apr 24, 2006, 10:51 AM • 15 Y

Y by tenplusten, Illuzion, Adventure10, jhu08, megarnie, HWenslawski, ImSh95, horia36, Mango247, NicoN9, cubres, aidan0626, PikaPika999, and 2 other users

We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
$f(x)f(y)=2f(x+yf(x))$
for all positive real numbers $x$ and $y$ .

Proposed by Nikolai Nikolov, Bulgaria

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cefer

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cefer

#2 h Apr 26, 2006, 5:39 PM • 8 Y

Y by Adventure10, jhu08, megarnie, ImSh95, UpvoteFarm, cubres, farhad.fritl, PikaPika999

It is from the short-list 2005 (i think A2) and it's offisial solution is very beautiful

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cefer

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cefer

#3 h Apr 26, 2006, 5:56 PM • 8 Y

Y by Adventure10, jhu08, ImSh95, UpvoteFarm, Mango247, cubres, farhad.fritl, PikaPika999

You can also see this problem similar it http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=f%28x%29f%28y%29%3Df%28x%2Byf%28x%29%29&t=14110[/list]

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grobber

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grobber

#4 h Apr 26, 2006, 6:55 PM • 25 Y

Y by anhchangtoanhoc97, jt314, anantmudgal09, A-Thought-Of-God, Jerry122805, Rustem-E303, Adventure10, jhu08, chystudent1-_-, guptaamitu1, ImSh95, KhongMinh, Mathlover_1, UpvoteFarm, tiendung2006, Tellocan, imabc, Sedro, cubres, ashboy_14, elfi, and 4 other users

I think the only solution is the constant function $f\equiv 2$ .

Suppose we can find $x$ such that $f(x)<1$ . Then $y=\frac x{1-f(x)}>0$ , and $x+yf(x)=y$ . Plug these values of $x,y$ into the equation to get $f(x)f(y)=2f(y)\Rightarrow f(x)=2$ , contradicting $f(x)<1$ . We thus find $f(x)\ge 1,\ \forall x>0\ (*)$ . Now assume we can find $y$ such that $f(y)<2$ . Choose $x_0>0$ arbitrarily, and put $x_{n+1}=x_n+yf(x_n),\ \forall n\ge 0$ . We have $f(x_n)=\left(\frac{f(y)}2\right)^nf(x)\to 0$ as $n\to\infty$ , but this contradicts $(*)$ applied to $x_n$ . This means that $f(x)\ge 2,\ \forall x>0\ (**)$ . $(**)$ applied to $y$ instead of $x$ , together with the initial equation, show that $f(x)\le f(x+yf(x)),\ \forall x,y>0$ , i.e. $f$ is non-decreasing.

If $x_1<x_2$ and $f(x_1)=f(x_2)$ (i.e. if $f$ is not one-to-one), then it follows from the equation that $\forall t>0,\ f(x_1+t)=f(x_2+t)$ , and together with the fact that $f$ is non-decreasing, this implies that $f$ is eventually constant. This constant can only be $2$ , so $f\equiv 2$ .

Now assume that $f$ is not constant. From the paragraph above we see that $f$ must be one-to-one. Since $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$ , we get $x+yf(x)=y+xf(y),\ \forall x,y>0$ . This means that for some $a>0,\ f(x)=ax+1,\ \forall x>0$ , but this contradicts $(**)$ .

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#20002 pwns

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#20002 pwns

#5 h Jun 16, 2006, 2:59 AM • 8 Y

Y by Adventure10, Adventure10, jhu08, ImSh95, UpvoteFarm, Mango247, NicoN9, cubres

hmmm...

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Nima Ahmadi Pour

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Nima Ahmadi Pour

#6 h Jun 16, 2006, 11:50 AM • 7 Y

Y by Adventure10, jhu08, ImSh95, UpvoteFarm, Mango247, NicoN9, cubres

#20002 pwns wrote:

hmmm...

Nope, $0$ is not in the domain ( $\mathbb R^+=\{x\in\mathbb R|x>0\}$ )

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abdurashidjon

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abdurashidjon

#7 h Jun 21, 2006, 6:17 PM • 6 Y

Y by Adventure10, jhu08, ImSh95, UpvoteFarm, Mango247, cubres

grobber wrote:

Now assume that $f$ is not constant. From the paragraph above we see that $f$ must be one-to-one. Since $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$ , we get $x+yf(x)=y+xf(y),\ \forall x,y>0$ . This means that for some $a>0,\ f(x)=ax+1,\ \forall x>0$ , but this contradicts $(**)$ .

Can you explain this part

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cefer

294 posts

cefer

#8 h Jun 23, 2006, 1:43 PM • 9 Y

Y by anhchangtoanhoc97, A-Thought-Of-God, jhu08, ImSh95, UpvoteFarm, Adventure10, Mango247, cubres, farhad.fritl

abdurashidjon wrote:

grobber wrote:

Now assume that $f$ is not constant. From the paragraph above we see that $f$ must be one-to-one. Since $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$ , we get $x+yf(x)=y+xf(y),\ \forall x,y>0$ . This means that for some $a>0,\ f(x)=ax+1,\ \forall x>0$ , but this contradicts $(**)$ .

Can you explain this part

@abdurashidjon
I think he replaced $x$ by $y$ and $y$ by $x$ and got $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$ , and from property one to one he got
$x+yf(x)=y+xf(y)$ $\Longrightarrow \frac{f(x)-1}{x}=\frac{f(y)-1}{y}=constant=a$
So $f(x)=ax+1$ ,contradiction.

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abdurashidjon

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abdurashidjon

#9 h Jun 30, 2006, 8:18 PM • 5 Y

Y by Adventure10, jhu08, ImSh95, UpvoteFarm, cubres

@Cefer thanks for answer

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Zhero

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Zhero

#10 h Oct 24, 2010, 3:33 AM • 12 Y

Y by DanDumitrescu, Illuzion, shon804, karitoshi, jhu08, ImSh95, UpvoteFarm, Adventure10, Mango247, cubres, and 2 other users

Let $P(x,y)$ be the assertion that $f(x)f(y) = 2f(x+yf(x))$ . $P(x+zf(x), y)$ yields $f(x+zf(x))f(y) = 2f(x+zf(x) + yf(x+zf(x))$ . $f(x+zf(x)) = \frac{f(x)f(z)}{2}$ , so $f(x)f(y)f(z) = 4f(x+zf(x) + \frac{yf(x)f(z)}{2})$ . Similarly, $P(x, z+yf(z))$ yields $f(x)f(y)f(z) = 4f(x+(z+yf(z))f(x)) = 4f(x + zf(x) + yf(x)f(z))$ . Hence, $f(x+zf(x) + \frac{yf(x)f(z)}{2}) = f(x+zf(x) + yf(x)f(z))$ for all positive $x,y,z$ .

Let $a$ and $b$ be any positive reals such that $2a > b > a$ . Let $x = a - \frac{b}{2}$ , $z = \frac{2a-b - x}{f(x)}$ , and $y = \frac{2(b-a)}{f(x)f(z)}$ . Then $x + zf(x) + \frac{yf(x)f(z)}{2} = 2a-b + (b-a) = a$ , and $x + zf(x) + yf(x)f(z) = 2a-b + 2(b-a) = b$ , so $f(a) = f(b)$ . It follows that $f$ is constant over all intervals of the form $(c, 2c)$ .

For all integers $k$ , let $S_k = ((3/2)^k, 2 (3/2)^k)$ . From the above, $f$ is constant over each $S_k$ . Since $2(3/2)^k > (3/2)^{k+1}$ , $S_{k+1} \cap S_k \neq \emptyset$ , so the constant value of $f$ over any $S_k$ is equal to that of $S_{k+1}$ . Since $\bigcup_{k=-\infty}^{\infty} S_k = \mathbb{R}^+$ , $f$ must be constant over $\mathbb{R}^+$ . It follows that $f(x) = 2$ for all $x$ .

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dnkywin

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dnkywin

#11 h Dec 25, 2010, 8:54 PM • 19 Y

Y by jt314, tenplusten, dangerousliri, Cindy.tw, parola, A-Thought-Of-God, Rustem-E303, jhu08, ImSh95, UpvoteFarm, Adventure10, Ikbal_gk, imabc, Jermainelamarrcole, Davud29_09, cubres, and 3 other users

Here is a solution I came up with a long time ago:
Note that from the given, we have:
$f\left(x\right)=\frac2{f\left(y\right)}f\left(x+yf\left(x\right)\right)$
Thus we have:
$\begin{eqnarray*}2f\left(x\right)f\left(y+2z\right)&=&f\left(x\right)f\left(y\right)f\left(\frac{2z}{f\left(y\right)}\right)\\ &=&2f\left(x+yf\left(x\right)\right)f\left(\frac{2z}{f\left(y\right)}\right)\\ &=&4f\left(x+yf\left(x\right)+\frac{2z}{f\left(y\right)}f\left(x+yf\left(x\right)\right)\right)\\ &=&4f\left(x+yf\left(x\right)+zf\left(x\right)\right)\\ &=&2f\left(x\right)f\left(y+z\right) \end{eqnarray*}$
Thus dividing both sides by $f\left(x\right)$ gives $f\left(y+z\right)=f\left(y+2z\right)$ for all positive reals $y,z$ . Now repeatedly replacing $y$ with $y+z$ gives $f\left(y+z\right)=f\left(y+nz\right)$ for all reals $y,z$ and positive integers $n$ . Now suppose $a,b$ are arbitrary positive real numbers with $a>b$ . Let $n$ be an integer greater than $a/b$ , and let
$y=\frac{nb-a}{n-1}, z=\frac{a-b}{n-1}$
Thus $y+nz=a$ and $y+z=b$ , so we have $f\left(a\right)=f\left(b\right)$ . However, since $a,b$ were arbitrary, $f\left(x\right)\equiv c$ for some $c$ , so plugging this back into the given equation gives $f\left(x\right)\equiv 2$ as the only solution. $\blacksquare$

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Mewto55555

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Mewto55555

#12 h Apr 25, 2011, 3:09 AM • 5 Y

Y by jhu08, ImSh95, Adventure10, Mango247, cubres

Not particularly nice

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math154

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math154

#13 h Apr 25, 2011, 3:18 AM • 6 Y

Y by IndoMathXdZ, jhu08, ImSh95, Adventure10, Mango247, cubres

Mewto55555 wrote:

Thus, $f$ is periodic. Let it have period $p \ge 0$ (if $p=0$ , $f$ is constant)
[...]
$\implies 2f(\frac{p}{2})=2f(p) \implies f(\frac{p}{2})=f(p) \implies p|\frac{p}{2}$ , which is impossible.

This is only a contradiction if there exists a minimal period $p$ , which is not necessarily true. (You don't rule out the case in which $f$ takes one value over the rationals and another over the irrationals, in which case $p$ could be any rational number.)

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Mewto55555

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Mewto55555

#14 h Apr 25, 2011, 7:58 PM • 5 Y

Y by jhu08, ImSh95, Adventure10, Mango247, cubres

Whoops, but that's easily fixed if we replace that last $\frac{p}{2}$ with an $x$ , meaning $p$ divides all real $x$ , impossible.

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math154

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math154

#15 h Apr 25, 2011, 8:02 PM • 5 Y

Y by jhu08, ImSh95, Adventure10, Mango247, cubres

I don't see how you're getting that $p$ divides all real $x$ . Your solution doesn't address the case I mentioned in my last post.

Edit: Actually, I missed an even more glaring mistake:

Mewto55555 wrote:

Thus, $f(x+yf(x))=f(x+yf(x)+pf(x)) \implies p|pf(x)$ , so $f(x)$ is a positive integer for all $x$ .

You're going to need some more algebraic manipulation with the original equation to get more information.

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cursed_tangent1434

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cursed_tangent1434

#101 h Mar 6, 2024, 3:02 AM

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The answer is $f(x)=2$ for all $x\in \mathbb{R}_{>0}$ . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We first prove the following.

Claim : $f(x)\geq 2$ for all $x\in \mathbb{R}_{>0}$ .
Proof : First, assume there exists $\alpha \in \mathbb{R}_{>0}$ such that $f(\alpha) <1$ . Then, plugging in $x=\alpha$ and $y= \frac{\alpha}{1-f(\alpha)}$ gives
$\begin{align*} f(\alpha)f\left(\frac{\alpha}{1-f(\alpha)}\right) &= 2f\left(\alpha + \frac{\alpha}{1-f(\alpha)}\cdot f(\alpha)\right) \\ f(\alpha)f\left(\frac{\alpha}{1-f(\alpha)}\right) &= 2f\left(\frac{\alpha}{1-f(\alpha)}\right) \end{align*}$ from which it follows that $f(\alpha)=2$ which is a very clear contradiction. Thus, it follows that $f(x) \geq 1$ for all $x\in \mathbb{R}_{>0}$ . Now, note that we have
$f(x)^2 = 2f(x+xf(x))\geq 2$ from which it follows that $f(x) \geq \sqrt{2}$ for all $x\in \mathbb{R}_{>0}$ . Similarly, a straightforward induction gives,
$f(x) \geq 2^{\sum{\frac{1}{2^i}}}=2$ for all $x\in \mathbb{R}_{>0}$ , which proves the claim.

Now, we show that there exists $r\in \mathbb{R}_{>0}$ such that $f(r)=2$ . First note that swapping $x$ and $y$ gives us that
$f(x+yf(x))=f(y+xf(y))$ for all positive real numbers $x$ and $y$ . Then, say $x+yf(x)=y+xf(y)$ for all $x,y \in\mathbb{R}_{>0}$ . Then, plugging in $y=1$ gives $f(x)=cx+1$ for some constant $c$ for all positive reals $x$ which is clearly false for any $c\in \mathbb{R}_{>0}$ upon substitution. This means, there exists $x_0,y_0 \in \mathbb{R}_{>0}$ such that $x_0+y_0f(x_0) \neq y_0+x_0f(y_0)$ . WLOG, assume that $x_0+y_0f(x_0)>y_0+x_0f(y_0)$ . Then, let $b= y_0+x_0f(y_0)$ and $a=\frac{x_0+y_0f(x_0) - (y_0+x_0f(y_0))}{f(y_0+x_0f(y_0))}>0$ .
Then, plugging in $x=b$ and $y=a$ yields,
$f(b)f(a)=2(f(b+af(b)))=2f(x_0+y_0f(x_0)) = 2f(b)$ from which it follows that $f(a)=2$ . Thus, there indeed exists $r\in \mathbb{R}_{>0}$ such that $f(r)=2$ .
We now have our final key claim.

Claim : $f$ is non-decreasing.
Proof : Simply note that
$2f(x+yf(x))=f(x)f(y)\geq 2f(x)$ since $f\geq 2$ . Thus, $f(x) \leq f(x+yf(x))$ for all $x,y \in\mathbb{R}_{>0}$ . Thus, it follows that $f(x)\leq f(x+\epsilon)$ for any $x, \epsilon \in \mathbb{R}_{>0}$ (by varying $y$ ) which implies that $f$ is non-decreasing as claimed.

Now, consider $\alpha \in \mathbb{R}_{>0}$ such that $f(\alpha)=2$ . Plugging in $x=y=\alpha$ gives
$4=f(\alpha)^2=2f(\alpha+\alpha f(\alpha))=2f(3\alpha)$ which implies that $f(3\alpha)=2$ . Thus, it follows by a straightforward induction that $f(3^ra)=2$ for all $r\in \mathbb{N}$ , from which it follows that there exists arbitrarily large values of $x\in\mathbb{R}_{>0}$ such that $f(x)=2$ . Putting this together with the non-decreasing nature of $f$ it follows that $f$ is constant 2, as desired.

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asdf334

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asdf334

#102 h Apr 8, 2024, 12:38 AM

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Let $P(x,y)$ be the assertion.

Claim: $f(x)\ge 1$ .
Proof: Suppose otherwise. If there exists $a$ where $f(a)<1$ then we can find
$y=a+yf(a)$ hence $P(a,y)$ yields $f(a)=2$ , a contradiction.

Claim: $f(x)\ge 2$ .
Proof: Suppose otherwise. Let $S$ be the range of $f$ . From $P(x,x)$ with $v=f(x)<2$ we find
$v\in S\implies \frac{v^2}{2}\in S$ hence we can generate a sequence $v_0,\dots,v_n\dots$ and note that
$v_{i+1}\le \frac{v_0}{2}\cdot v_i$ thus values become arbitrarily small. Hence at some point values are less than $1$ , violating the first claim.

Claim: $f$ is nondecreasing.
Proof: Take any $a$ and $b$ with $a<b$ . Notice we can always choose $x$ and $y$ such that
$a=x$ $b=x+yf(x).$ Then since $f(y)\ge 2$ we find $f(a)\le f(b)$ .

Claim: Either $f$ is increasing or $f\equiv 2$ .
Proof: Suppose there exist $a$ and $b$ such that $f(a)=f(b)$ . Then $f$ is constant in the interval $[a,b]$ . At this point take $P(a,y)$ such that
$a+yf(a)\in (a,b].$ For any such $y$ we have $f(y)=2$ . Hence there exists an interval where $f\equiv 2$ . Take $x$ in this interval and $P(x,x)$ implies $f(3x)=2$ . Hence we can obtain arbitrarily large values outputting $2$ , and so $f\equiv 2$ everywhere.

Claim: $f$ increasing is not possible.
Proof: Swap $x$ and $y$ . Then
$f(x+yf(x))=f(y+xf(y))\implies x+yf(x)=y+xf(y)\implies \frac{f(x)-1}{x}=\frac{f(y)-1}{y}$ hence $f(x)=cx+1$ for some $c$ . However taking $x\to 0$ is a contradiction as we would have $f(x)\to 1$ which is less than $2$ .

To conclude, we have one solution (which clearly works): $\boxed{f\equiv 2}$ .

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WLOGQED1729

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WLOGQED1729

#103 h May 29, 2024, 3:50 PM

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Very instructive problem! This reviews many useful techniques in FE.

The answer is $f(x)=2~~\forall x \in \mathbb{R}^+$ which is clearly satisfy the problem condition.
We’ll show that this is the only solution.
Denote $P(x,y)$ by the condition $f(x)f(y)=2f(x+yf(x))~~ \forall x,y \in \mathbb{R}^+$
Claim 1: $f(x)\geq 1~~ \forall x \in \mathbb{R}^+$
Proof: FTSOC, suppose there exist $t \in \mathbb{R}^+$ s.t. $f(t)<1$
Consider $P(t,\frac{t}{1-f(t)})$ , it follows that $f(t)=2$ which leads to contradiction. $\square$

Claim 2: $f(x)\geq 2~~ \forall x \in \mathbb{R}^{+}$
Proof: Define a sequence ${a_n}$ s.t. $a_1=1$ and $a_{n+1}=\sqrt{2a_{n}}~~ \forall n \in \mathbb{N}$
It is easy to see that $lim_{n\to \infty}a_n=2$ and $a_n<2~~ \forall n \in \mathbb{N}~~(\clubsuit)$
We prove that $f(x)\geq a_n ~~\forall x \in \mathbb{R}^+ , n \in \mathbb{N}$ by induction on $n$ .
The base case is proven in Claim 1
Suppose that $f(x)\geq a_m~~ \forall x \in \mathbb{R}^+$
Consider $P(x,x)$ , $f(x)^2=2f(x+xf(x))\geq 2a_m~~ \forall x \in \mathbb{R}^+$
$\implies f(x)\geq \sqrt{2a_m}=a_{m+1}~~\forall x \in \mathbb{R}^+$ So, $f(x)\geq a_n ~~\forall x \in \mathbb{R}^+ , n \in \mathbb{N}$
Combine this with $(\clubsuit)$ , $f(x)\geq 2~~ \forall x \in \mathbb{R}^{+}$ , as desired. $\square$

Claim 3: $f$ is non decreasing
Proof: Let $a<b$ be any two distinct positive reals
Consider $P(a,\frac{b-a}{f(a)})$ , $f(a)f(\frac{b-a}{f(a)})=2f(b)$
Since $f(x)\geq 2$ for all $x$ , it follows that $f(a)\leq f(b)~~\square$

Now we divide into 2 cases
Case 1) There exist $c \in \mathbb{R}^+$ such that $f(c)=2$
Note that if we have $f(x)=2$ , it implies that $f(x+xf(x))=2$ (This followed from $P(x,x)$ )
We can easily prove that $f(3^nc)=2 \forall n \in \mathbb{N}$ by induction on $n$
Combine this result with Claim 2 and Claim 3, we conclude that $f(x)=2~~\forall x \in \mathbb{R}^+$

Case 2) $f(x)>2~~ \forall x \in \mathbb{R}^+$
Claim: $f$ is strictly increasing
Proof: We can easily copied the proof in Claim 3. $\square$
This implies that $f$ is injective.

The trick is to swap the variable in the problem assertion.
Consider $P(x,y)$ and $P(y,x)$ , $f(x+yf(x))=f(y+xf(y)) \forall x,y \in \mathbb{R}^+$
By injectivity, $x+yf(x)=y+xf(y) \forall x,y \in \mathbb{R}^+$
$\implies \frac{f(x)-1}{x}$ is a constant $\forall x \in \mathbb{R}^+$
So, $f(x)=cx+1 \forall x \in \mathbb{R}^+$ for some constant $c \in \mathbb{R}^+$ which can be easily checked that it fails. $\blacksquare$

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joshualiu315

2534 posts

joshualiu315

#104 h Aug 31, 2024, 5:44 AM

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The answer is $f(x) = \boxed{2}$ , which works. Denote the given assertion as $P(x,y)$ . We will solve this problem through a series of claims:

Claim 1: $f(x) \ge 1$ for all $x \in \mathbb{R}^+$

Proof: FTSOC suppose that $x$ satisfies $f(x)<1$ . Plugging in $P(x, \tfrac{x}{1-f(x)})$ yields

$f(x)f\left(\frac{x}{1-f(x)}\right) = 2f \left(\frac{x}{1-f(x)}\right) \implies f(x) = 2,$
a contradiction. $\square$

Claim 2: $f(x) \ge 2$ for all $x \in \mathbb{R}^+$

Proof: $P(x,x)$ yields

$f(x)^2 = 2f(x+f(x)) \ge 2.$
Hence, $f(x) \ge \sqrt{2}$ . In fact, we can show that

$f(x) \ge \lim_{k \to \infty} 2^{\tfrac{2^k-1}{2^k}} = 2$
using an inductive process on $P(x,x)$ , as desired. $\square$

Claim 3: $f$ is nondecreasing

Proof: Note that

$2f(x+yf(x)) = f(x)f(y) \ge 2f(x),$
which means $f(x) \le f(x+yf(x))$ . Take $y = \varepsilon$ to be an arbitrarily small number, so that $f(x) \le f(x+\varepsilon)$ and $x < x+ \varepsilon$ . $\square$

Claim 4: If there exists any value $x \in \mathbb{R}^+$ such that $f(x) = 2$ , then $f \equiv 2$ .

Proof: If $f(x) = 2$ , then

$f(3x) = f(x+xf(x)) = \frac{f(x)^2}{2} = 2.$
Thus, the set of $x$ such that $f(x)=2$ is unbounded, and Claim 3 finishes. $\square$

Claim 5: $f$ is not injective.

Proof: Assume FTSOC that $f$ is injective. Then, comparing $P(x,y)$ and $P(y,x)$ gives

$f(x+yf(x)) = f(y+xf(y)) \implies x+yf(x) = y+xf(y)$ $\implies x(1-f(y)) = y(1-f(x)) = k.$
Rearraging yields $f(y) = 1-\tfrac{k}{x}$ , which implies $f(k) = 0$ , contradiction. $\square$

Claim 6: $f>2$ cannot always hold.

Proof: Suppose that $x$ and $y$ satisfy $x+yf(x) < y+xf(y)$ . Such values must exist due to Claim 5 preventing $x+yf(x) = y+xf(y)$ . Note that all values in the interval $[x+yf(x), y+xf(y)]$ are constant.

The trick is to raise $y \to y'$ such that $x+y'f(x) = y+xf(y) \le y'+xf(y')$ . Hence, we can expand the constant interval to $[x+yf(x), y'+xf(y')]$ . It suffices to show that increasing $y$ infinitely many times causes the interval to be unbounded.

Let $\ell$ be the size of the current interval. Each time, we increase $y$ by $\tfrac{\ell}{f(x)}$ , which increases the size of the next interval by at least $\tfrac{\ell}{f(x)}$ . Hence, the new interval at least has size

$\ell\left(1+\frac{1}{f(x)} \right),$
which is evidently unbounded. $\square$

Claims $4$ and $6$ imply that $f$ is constant, which gives our solution set.

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ezpotd

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ezpotd

#105 h Oct 16, 2024, 11:10 PM

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The answer is the constant $2$ only, it is easy to check that this works. Let $P(x,y)$ denote the assertion.

First, observe is $f(x)$ is below $1$ at any point, we can find $y$ such that $x + yf(x) = y$ , plugging in such $y$ forces $f(x) = 2$ , thus $f$ can never be below $1$ (since we already said $f(x)$ was below $1$ ). If $f(x)$ is below $2$ at any point, we can always see there exists a value of $k$ such that $f(k) = \frac 12 f(x)^2$ , which eventually goes below $1$ (it is bounded by a geometric series with common ratio less than $1$ ), so this returns to the case where $f(x)$ is below $1$ at some point. So we know $f \ge 2$ .

Claim: $f$ is $2$ at some point.
If $f$ is always above $2$ , we see $f(x) \frac{f(y)}{2} =f(x +yf(x))$ , so for all $a > x$ , setting $y = \frac{a - x}{f(x)}$ forces $f(a) > f(x)$ . Now take $2f(x + yf(x)) = f(x)f(y) = f(y)f(x) = 2f(y +xf(y))$ , since $f$ is increasing we know $f$ is injective, this gives $x + yf(x) = xf(y) + y$ , for sufficiently small $y$ the left side is below $2x$ and the right side is greater, which is a contradiction, so this is impossible.

Claim: $f$ is periodic.
Proof: Take $c$ with $f(c) = 2$ , then we get $P(x,c)$ forces $f(x) = f(x + 2c)$ .

Claim: $f(x) = 2$ for all $x$ .
Proof: We first show $f(x) = 1$ or $f(x) = 2$ for all $x$ . Assume $f(x) \neq 1$ , we show $f(x) = 2$ . By varying $y$ , we can find some value of $y$ such that there exists an integer $k$ with $x + yf(x) - y = 2kc$ , this forces $f(y) = f(x + f(y))$ , so $P(x,y)$ gives $f(x) = 2$ . Now take some value with $f(x) = 1$ , then $P(x,x)$ gives $f(x + xf(x)) = \frac 12$ , which is impossible, so such a value cannot exist, giving the desired conclusion.

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cosdealfa

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cosdealfa

#106 h Nov 11, 2024, 9:46 AM • 2 Y

Y by pb_ana, Mathandski

I hope this is right :blush:

Denote by $P(x, y)$ the given assertion.
Claim 1: $f$ is monotonically increasing.
Proof: take two numbers $a < b$ and assume FTSOC $f(a) > f(b)$ .
Let $r$ be the solution of the equation $a + rf(a) = b+ rf(b)$ (it exists because of the way we ordered these two numbers).

$P(a, r): f(a)f(r)=2f(a + rf(a))$
$P(b, r): f(b)f(r)=2f(b + rf(b))$
So $f(a)=f(b)$ , contradiction. $\square$

Now we will prove that $f$ is not injective( i.e strictly increasing)
Suppose FTSOC that $f$ is injective.
Swapping $x$ and $y$ in the original equation gives:
$x + yf(x) = y + xf(y), \forall x > 0$
$\Rightarrow \frac{f(x)-1}{x} = k, \forall x > 0$
Plugging this in the original equation, we get a contradiction. $\square$

Since $f$ is not injective, we can find $a < b$ such that $f(a) = f(b) = c$ . By monotonicity, we get that $\forall x \in [a, b], f(x) = c$ .
$P(a, \frac{b-a}{c}): f(a)f(\frac{b-a}{c})=2f(b)$ so we can find $z \in \mathbb{R}$ such that $f(z)=2$
$P(z, x): f(x) = f(2x + 2), \forall x \in \mathbb{R}$
Set $x=b$ $\Rightarrow f(b) = f(2b+2) =c$ . Using monotonicity again, we get that $\forall x \in [b,2b+2], f(x) = c$ Similarily, $\forall x > b, f(x) =c$
Finally, $P(a, x): cf(x) = 2(a + xc)$ . Since the second argument $>a$ we get that $f(x) = 2 \forall x \in \mathbb{R}$ which is clearly the only solution $\blacksquare$

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SomeonesPenguin

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SomeonesPenguin

#107 h Nov 12, 2024, 12:26 PM

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storage

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OronSH

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OronSH

#108 h Nov 17, 2024, 3:41 AM

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We claim $f\equiv 2$ is the only solution, which clearly works.

First, if $f(x)<1$ then $P(x,\tfrac x{1-f(x)})$ gives $f(x)=2$ . Thus $f(x)\ge 1$ .

Next, we prove by induction that $f(x)\ge 2^{1-2^{-n}}$ . From the above, the base case $n=0$ works, and for the inductive step $P(x,x)$ gives $f(x)^2=2f(x+xf(x))\ge 2^{2-2^{-n}}$ implying $f(x)\ge 2^{1-2^{-(n+1)}}$ , completing the induction. Taking $n$ arbitrarily large implies $f(x)\ge 2$ for all $x$ .

Next we show $f$ is constant or strictly increasing. From $f(y)\ge 2$ we get $f(x)\le f(x+yf(x))$ . If $f$ is not strictly increasing we can choose some $x$ and some $\varepsilon$ for which $f(x)=f(x+kf(x))$ for $0<k\le\varepsilon$ , and by equality case $f(k)=2$ for $0<k\le\varepsilon$ . Now $P(x+\varepsilon f(x),k)$ implies $f(x)=f(x+(k+\varepsilon)f(x))$ , so by equality case again $f(k)=2$ for $0<k\le 2\varepsilon$ . Repeating this, $f(k)=2$ for all $k>0$ .

Now if $f$ is strictly increasing, it is injective. Taking $P(x,1)$ and $P(1,x)$ implies $x+f(x)=1+xf(1)$ , which fails for $x$ small enough since $f(x)\ge 2$ .

Thus $f\equiv 2$ is the only possible solution.

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Mathandski

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Mathandski

#109 h Mar 4, 2025, 1:31 AM • 1 Y

Y by OronSH

Subjective Rating (MOHs)
Please contact westskigamer@gmail.com if there is an error with any of my solution for cash bounties by 3/18/2025.

Solution

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cj13609517288

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cj13609517288

#110 h Mar 4, 2025, 4:02 PM • 1 Y

Y by OronSH

The answer is $\boxed{f(x)=2}$ only, which clearly works.

Note that
$\begin{align*} f(x)f(y)f(z) &= f(y)f(z)f(x) \\ 2f(x+yf(x))f(z) &= 2f(y+zf(y))f(x) \\ 4f(x+yf(x)+zf(x+yf(x))) &= 4f(x+(y+zf(y))f(x)) \\ 4f(x+yf(x)+\frac12\cdot zf(x)f(y)) &= 4f(x+yf(x)+zf(x)f(y)). \end{align*}$ We want to prove that $f(a+\frac12 b)=f(a+b)$ . To do so, choose some $x<a$ , then choose $y$ such that $x+yf(x)=a$ , then choose $z$ such that $zf(x)f(y)=b$ .

So now we claim that $f$ is constant. But note that
$f(x)=f(1.5x)=f(1.5^2 x)=f(1.5^3 x)=\dots=f(\alpha\cdot 1.5^n x)$ where $1\le\alpha<1.5$ . This is enough to show that $f(x)=f(y)$ when $x<y$ , so we are done. $\blacksquare$

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N3bula

296 posts

N3bula

#112 h Mar 8, 2025, 2:50 AM

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Let $P(x, y)$ denote the assertion.
Suppose that $f$ , is injective, I will prove $f$ is not injective.
$P(x, y)-P(y, x)$ $f(x+yf(x))=f(y+xf(y))$ Thus from injective:
$f(x)=\frac{y-(f(y)-1)x}{y}$ If $f(y)>1$ we can choose $x$ such that $y-(f(y)-1)x$ is negative which is a contradiction. Now suppose $f(y)< 1$ . If $f(y)<1$ we can choose $x$ such that:
$x=y+xf(y)$ which means that $f(x)f(y)=2f(x)$ and thus $f(y)=2$ , however we supposed that $f(y)<1$ , which is a contradiction thus $f(y)\geq 1$ . If $f(y)=1$ we
get that $f$ is constant and so not injective thus $f$ cannot be injective.
Thus $f(x)=\frac{2-x}{c}$ , however for no $c$ is this a solution. Thus
$f$ is not injective. Now suppose that $f(a)=f(b)$ and $a\neq b$ .
$P(a, y)$ $f(a)f(y)=f(a+yf(a))$ $P(b, y)$ $f(a)f(y)=f(b+yf(a))$ $\therefore f(a+yf(a))=f(b+yf(a))$ Thus $f$ is periodic with period $a-b$ . Let $\vert a-b \vert=k$ . Let $u$ be a value such that for some fixed $x$ and $i$ and suppose that $f(x+i)>f(x)$ , $uf(x+i)-uf(x)=nk-i$ let $n$ be a value such that $nk>i$ . Note if
$f(x)>f(x+i)$ we can find a similar positive $u$ .
$P(x, u)$ $f(x)f(u)=f(x+uf(x))$ $P(x+i, u)$ $f(x+i)f(u)=f(x+i+uf(x+i))$ As: $uf(x)-uf(x+i)-i=nk$ , we get $f(x+uf(x))=f(x+i+uf(x+i))$ , thus this implies $f(x)=f(x+i)$ , which implies $f$ is constant which means $f(x)=2$ for all $x$ .

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EpicBird08

1755 posts

EpicBird08

#114 h Apr 6, 2025, 6:04 PM • 2 Y

Y by PikaPika999, OronSH

This is the strangest FE I've ever solved.

The only solution is $f(x) = 2,$ which clearly works. Now we will show that this is the only solution. Let $P(x,y)$ denote the assertion.

Claim: $f(x) \ge 1$ for all $x.$
Proof: Otherwise $P\left(x, \frac{x}{1-f(x)}\right)$ gives $f(x) = 2$ , contradiction.

Claim: $f(x) \ge 2^{1 - \frac{1}{2^n}}$ for all nonnegative integers $n.$
Proof: We prove this inductively, with the base case $n=0$ being the claim above. Then $P(x,x)$ gives $f(x)^2 = 2 f(x + x f(x)) \ge 2 \cdot 2^{1 - \frac{1}{2^n}} = 2^{2 - \frac{1}{2^n}}$ by the inductive hypothesis. Thus $f(x) \ge 2^{1 - \frac{1}{2^{n+1}}},$ so the induction is complete.

By taking $n \to \infty,$ we get that $\boxed{f(x) \ge 2}$ for all $x.$

Claim: $f$ is NOT injective.
Proof: Otherwise, by looking at $P(x,1)$ and $P(1,x),$ we get $f(x+f(x)) = f(1 + x f(1)),$ so $x + f(x) = 1 + x f(1),$ meaning $f(x) = 1 + x(f(1) - 1).$ But taking $x \to 0$ implies $f(x) < 2,$ contradiction.

Thus we have $f(r) = f(s)$ for some $r < s.$ Then $P\left(r, \frac{s-r}{f(r)}\right)$ gives $f(r) f\left(\frac{s-r}{f(r)}\right) = 2 f(s) \implies f\left(\frac{s-r}{f(r)}\right) = 2.$ For simplicity, let $a = \frac{s-r}{f(r)}.$ Then $P(a,a)$ gives $4 = 2 f(a + a f(a)) = 2 f(3a),$ so $f(3a) = 2.$ Similarly, $P(a, 3a)$ gives $4 = 2 f(a + 3a f(a)) = 2 f(7a),$ so $f(7a) = 2.$ In general, this logic can be repeated to get $f((2^n - 1)a) = 2$ for all $n \in \mathbb{N}.$

Finally, if $b$ is such that $f(b) = 2$ and $x < b,$ then $P\left(x, \frac{b-x}{f(x)}\right)$ gives $f(x) f\left(\frac{b-x}{f(x)}\right) = 2 f(b) = 4.$ But $f(x), f\left(\frac{b-x}{f(x)}\right) \ge 2,$ so this forces $f(x) = 2$ for all $x < b.$ Since $b$ can be arbitrarily large, this finishes.

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ezpotd

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ezpotd

#115 h May 8, 2025, 6:17 PM

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We claim the answer is only the constant function $f(x) = 2$ .

Assume $f(x) < 1$ for some $x$ , then setting $y = \frac{x}{1 - f(x)}$ gives $f(x) = 2$ , contradiction. So $f(x) \ge 1$ for all $x$ . Then assume $f(a) < 2$ , repeatedly taking $f(x + af(x)) = f(x) \cdot \frac a2$ and replacing $x$ with $x + af(x)$ allows us to get $f$ with a value below one in finitely many applications of the assertion, so $f(a) \ge 2$ for all $a$ . By varying $y$ over all reals, we get $f(x) \le \frac 12 f(x)f(y) \le f(x + c)$ for any $c$ , so $f$ is nondecreasing.

Now assume $f$ is never equal to $2$ , so $f$ is strictly increasing as the previous inequality becomes strict. Thus $f$ is injective, so consider swapping $x,y$ so we have $x + yf(x) = y + xf(y)$ , taking extremely large $y$ and sufficiently small $x$ to ensure $xf(y) < y$ forces $x + yf(x) > 2y > y + xf(y)$ , contradiction, so our assumption that $f$ can be strictly increasing is false.

Thus $f$ is $2$ at some input $a$ , and by nondecreasing it is equal to $2$ for all of $(0,a]$ . Now $f(a)f(a) = 2f(3a)$ gives $f(3a) = 2$ , so $f$ is equal to $2$ over all of $(0,3a]$ , repeating this infinitely allows us to acquire $f(x) = 2$ for all $x$ .

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youochange

180 posts

youochange

#116 h May 9, 2025, 6:27 AM

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Nima Ahmadi Pour wrote:

We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
$f(x)f(y)=2f(x+yf(x))$ for all positive real numbers $x$ and $y$ .

Proposed by Nikolai Nikolov, Bulgaria

Can someone pls verify my sol??
fakesolve

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jasperE3

11386 posts

jasperE3

#117 h May 9, 2025, 5:11 PM • 1 Y

Y by youochange

youochange wrote:

Nima Ahmadi Pour wrote:

We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
$f(x)f(y)=2f(x+yf(x))$ for all positive real numbers $x$ and $y$ .

Proposed by Nikolai Nikolov, Bulgaria

Can someone pls verify my sol??

Let $P(x,y)$ be the assertion.

$P(0,0): f(0)^2=2f(0)$

Problem says that $P(x,y)$ is true for all positive real number $x$ and $y$ , but $0$ isn't a positive real number.

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