R+ FE with arbitrary constant

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R+ FE with arbitrary constant

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Source: APMO 2023/4

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#1 h Jul 5, 2023, 7:22 PM • 5 Y

Y by Johnson100, Rounak_iitr, Maths-I25, GeoKing, Funcshun840

Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that $f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.$

This post has been edited 2 times. Last edited by CyclicISLscelesTrapezoid, Jul 5, 2023, 7:24 PM

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#2 h Jul 5, 2023, 9:03 PM • 4 Y

Y by centslordm, Maths-I25, sttsmet, MS_asdfgzxcvb

The answer is $f(x)=2x$ only, which clearly works. Now we prove that this is the only solution.

Let $P(x,y)$ denote the assertion. If $f(y)<2y$ for any $y$ , then from $P(x,\tfrac{2y-f(y)}{c})$ we find that $2cx=0$ : contradiction. Therefore we may let $f(x)=g(x)+2x$ , where $g: \mathbb{R}^+ \to \mathbb{R}_{\geq 0}$ , and the functional equation becomes
$g((c+1)x+g(y)+2y)+2g(y)=g(x+2y).$ Let $k=2+f(1)$ . Picking an arbitrary $t>0$ , by $P(t,k)$ , we find that $g(t+k)\geq g((c+1)t+k+f(k))$ , which really just means that we can always find some $t'\geq(c+1)t$ such that $g(t'+k)\leq g(t+k)$ .

By repeatedly applying this, we find that there exists some $t''\geq 2t+k$ such that $g(t''+k)\leq g(t+k)$ . But then from $P(t''-2t-k,t+k)$ , we find that $g((c+1)(t''-2t-k)+g(t+k)+2(t+k))\leq 0$ , with equality holding only if $g(t''+k)=g(t+k)=0$ . Therefore $g(y)=0$ when $y>k$ . But then from $P(k,x)$ where $x>0$ is arbitrary, we find that $g(x)=0$ for all $x$ , hence $g \equiv 0$ and we extract the desired solution. $\blacksquare$

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kred9

#3 h Jul 5, 2023, 9:11 PM • 2 Y

Y by StarLex1, solasky

The answer is $f(x) = 2x$ only. This works.

Notice that equation $(c+1)x+f(y) = x+2y$ can never have solutions in the positive reals, since otherwise $x$ or $c$ would have to be $0$ . This implies that $f(y) \ge 2y \phantom{jj} (*)$ for all $y > 0$ . We have two cases.

Case 1. $f(t) = 2t$ for some $t > 0$ .
By $*$ , we have $f(x+2y) = f((c+1)x + f(y)) - 2cx \ge 2(c+1)x + 2f(y) -2cx\ge 2x + 4y.$ Pick some $(x,y)$ with $x+2y = t$ . Then the above inequality is actually an equality, so we have $f(y) = 2y$ and $f((c+1)x + 2y) = 2(c+1)x + 4y$ for all $x+2y = t$ . The second equality simplifies to $f(t + cx) = 2t + 2cx$ for all $x < t$ . Since $c+1 > 1$ , repeating this process implies that $f(x) = 2x$ for all $x \ge t$ , since the function $(c+1)^n$ grows without bound.

In particular, $f(2t) = 4t$ , so we have $f(y) = 2y$ for all $y < t$ as well, showing that $f(x) = 2x$ for all $x \in \mathbb{R}_{>0}$ .

Case 2. $f(t) > 2t$ for all $t>0$ . Then let $g(x) = f(x) - 2x$ . Clearly the codomain of $g$ is also $\mathbb{R}_{>0}$ .

The given equation simplifies to $g(x+2y) = 2g(y) + g((c+1)x+g(y) + 2y)$ . This implies $g(x+2y) > 2g(y)$ , so for all pairs $(x,y)$ with $x>2y$ , we have $g(x) > 2g(y)$ .

Let $g(1) = M>0$ . Then for all $x > 2$ , $g(x) > 2M$ . Taking $y = 1$ into the given equation,
$\begin{align*} g(x+2) &= 2M + g(\textcolor{blue}{(c+1)x+M}+2) \\ &= 2M + 2M + g((c+1)(\textcolor{blue}{(c+1)x+M}) +M + 2) \\ &= 2M + 2M + 2M + g((c+1)((c+1)(\textcolor{blue}{(c+1)x+M})+M) +M + 2) \\ &= \cdots, \end{align*}$ which is a contradiction as $M > 0$ . We are done.

This post has been edited 3 times. Last edited by kred9, Jul 5, 2023, 11:46 PM

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#4 h Jul 5, 2023, 9:31 PM • 2 Y

Y by Rounak_iitr, jolynefag

Kind of easy P4, anyway, as it's FE here's my solution : $~$ :play_ball:

:play_ball:

Let $P(x,y)$ be the assertion :
$f((c+1)x+f(y))=f(x+2y)+2cx$ Claim : $f(x)\ge 2x ~\forall x \in \mathbb{R_+}$

Proof : if $\exists w \in \mathbb{R_+}~ ; f(w)<2w :$
$P(\frac{2w-f(w)}{c},w): f(2w+\frac{2w-f(w)}{c})=f(2w+\frac{2w-f(w)}{c})+2(2w-f(w))$ $\implies f(w)=2w$ contradiction. $\blacksquare$

Define : $g:\mathbb{R_+} \to \mathbb{R}_{\geq 0} ~,~ g(x)=f(x)-2x\implies f(x)=g(x)+2x$ $\bullet ~P(x,y)$ becomes $:$
$g((c+1)x+2y+g(y))+2g(y)=g(x+2y)$ $g((c+1)x+2y+g(y))\ge 0\implies g(x+2y)\ge 2g(y)$ and so $x>2y\implies g(x)\ge 2g(y)$

$\bullet$ Now assume that $: \exists\alpha\in\mathbb{R_+}~ ; g(\alpha)>0$
$P(x-\frac{2\alpha+g(\alpha)}{c+1},\alpha) ~\text{with}~ x>\frac{2\alpha+g(\alpha)}{c+1} :$ $g((c+1)x)+2g(\alpha)=g(x+\frac{2c\alpha-g(\alpha)}{c+1})$
$a:=c+1>1,r:=\frac{1}{a}<1,t:=2g(\alpha)>0,d:=\frac{2c\alpha-g(\alpha)}{c+1}\in\mathbb{R}$ $\implies g(ax)+t=g(x+d)\implies g(ax)=g(x+d)-t$ § Now change $x\to ax:$
$g(a^2x)=g(ax+d)-t=g(a(x+rd))-t=g(x+rd+d)-2t$ § again change $x\to ax:$
$g(a^3x)=g(ax+rd+d)-2t=g(a(x+r^2d+rd))-2t=g(x+r^2d+rd+d)-3t$ § By induction we get $:$
${g(a^nx)+nt=g(x+d\frac{1-r^n}{1-r}) ~ \forall n \in\mathbb{N}}$ $\bullet$ Now put $x=x_0$ where $x_0$ is a fixed positive real number satisfying $:$

$x_0+d\frac{1-r^n}{1-r}>0, \forall n\in\mathbb{N}$ which is exist

because $\frac{1-r^n}{1-r}<\frac{1}{1-r}~$ (bounded)

So $x_0+d\frac{1-r^n}{1-r}$ is bounded $\forall n\in\mathbb{N}$
$\implies \exists l \in\mathbb{R_+}~ ; l>x_0+d\frac{1-r^n}{1-r} ~\forall n\in\mathbb{N}$ $\implies 2l>2(x_0+d\frac{1-r^n}{1-r})\implies g(2l)\ge 2g(x_0+d\frac{1-r^n}{1-r})~\forall n\in\mathbb{N}$ So $g(x_0+d\frac{1-r^n}{1-r})$ is bounded $\forall n\in\mathbb{N}$

$\bullet$ Now take $n\to +\infty$ in the equation $:$
$g(a^nx_0)+nt=g(x_0+d\frac{1-r^n}{1-r})$ we see that $RHS$ is bounded, but $LHS$ goes to infinity, so we get a contradiction and so no such $\alpha$ exist.
$\implies g(x)=0 ~ \forall x\in\mathbb{R_+}\implies \boxed{ f(x)=2x ~ \forall x \in\mathbb{R_+} }$ remark

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yofro

#5 h Jul 5, 2023, 9:48 PM • 1 Y

Y by LLL2019

See this thread from 2019. The solution should adapt.

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#6 h Jul 6, 2023, 12:51 AM

Y by

The only solution is $f(x)=2x$ , which evidently works. From now on, assume that $f$ is a solution which is different from this function. We prove a series of Claims:

Claim 1: $f(x) \geq 2x$ for all $x>0$ .
Proof: Indeed, if $f(u)<2u$ for some $u>0$ , then we may take $x=\dfrac{2u-f(u)}{c}$ and $y=u$ in the given equation to obtain that $2c \cdot \dfrac{2u-f(u)}{c}=0,$ a contradiction $\blacksquare$

Claim 2: If $x>2y$ , then $f(x)-2x \geq 2(f(y)-2y)$ .
Proof: Indeed, note that

$f(x+2y)-2(x+2y) =f((c+1)x+f(y))-2cx-2x-4y \geq 2((c+1)x+f(y))-2cx-2x-4y=2(f(y)-2y),$

and so $f(x)-2x \geq 2(f(y)-2y)$ for all $x,y$ such that $x>2y$ $\blacksquare$

To the problem, note that

$(c+1)x+f(y)>f(y)>2y,$

and so

$f((c+1)x+f(y))-2((c+1)x+f(y)) \geq 2(f(y)-2y),$

implying that

$(f(x+2y)+2cx)-2((c+1)x+f(y)) \geq 2(f(y)-2y)$ . Therefore,

$f(x+2y)-2x-2f(y) \geq 2f(y)-4y,$

or equivalently

$f(x+2y) \geq 2x+4f(y)-4y$ .

This readily implies that $f(x+2y)-2(x+2y) \geq 4(f(y)-2y)$ , and so

$f(x)-2x \geq 4(f(y)-2y)$ for all $x,y>0$ such that $x>2y$ . We may generalize this property:

Claim 3: If $x,y>0$ such that $x>2^ny$ for some positive integer $n$ , then $f(x)-2x \geq 4^n(f(y)-2y)$ .
Proof: We proceed by induction on $n$ . The base case is already established. Now, suppose that $f(x)-2x \geq 4^n(f(y)-2y)$ for all $x,y$ such that $x>2^ny$ .

Then, if $x>2^{n+1}y$ we may pick an $s$ such that $s \in (2y, \dfrac{x}{2^n})$ . Therefore, $x>2^ns$ and $s>2y$ . Thus, using the inductive hypothesis and the base case,

$f(x)-2x \geq 4^n(f(s)-2s) \geq 4^{n+1}(f(y)-2y),$

as desired $\blacksquare$

To the problem, let $x,y>0$ such that $x>2y$ . Let $n=\lfloor \log_2{\dfrac{x}{y}} \rfloor \geq 1$ . Then, $x>2^n y$ , and so by Claim 3 we infer that

$f(x)-2x \geq 4^{n}(f(y)-2y) \geq 4^{\log_2{\dfrac{x}{y}}-1}(f(y)-2y),$

or equivalently that

$f(x)-2x \geq \dfrac{x^2(f(y)-2y)}{4y^2}$ .

Now, take a $y_0$ such that $f(y_0) \neq 2y_0,$ and let $\dfrac{f(y_0)-2y_0}{4y_0^2}=A>0$ . Then,

$f(x) \geq Ax^2+2x$

for all large $x$ . Since $A>0$ , we obtain that

$f(x)> 4x+1-c$

for all large $x$ .

Therefore,

$(f(y)+c+1)-2(2y+1)=f(y)-4y+c-1 >0,$

if $y$ is large, and so by Claim 2 we obtain that

$f(2y+1)+2c=f(c+1+f(y)) \geq 2(c+1+f(y))+2(f(2y+1)-2(2y+1)),$

or equivalently that

$f(2y+1)+2f(y) \leq 8y+2,$

which in light of Claim 1 implies that equality must hold, and so $f(x)=2x$ for all large $x$ , say $x>N$ . Now, for any $y$ we may take a positive integer $n$ such that $2^ny>N$ , and so for all $x>2^ny$ we obtain that $f(x)=2x$ . By Claim 3, this implies that

$0=f(x)-2x \geq 4^n(f(y)-2y) \geq 0,$

and so equality must hold everywhere, thus $f(x)=2x$ for all $x$ , which is a contradiction to our initial assumption.

To sum up, the only solution is $f(x)=2x$ for all $x>0$ .

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vsamc

#7 h Jul 6, 2023, 1:59 AM

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Solution

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TLP.39

#8 h Jul 6, 2023, 2:27 AM • 2 Y

Y by StarLex1, Zaro23

CyclicISLscelesTrapezoid wrote:

Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that $f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.$

$P(x+2z,y)-P(x+2y,z)$ implies that $\frac{(2z(c+1)+f(y))-(2y(c+1)+f(z))}{4c(z-y)}$ is constant (for $y\neq z$ ). The rest is easy.

(The linked lemma can't be applied directly, but the same proof can be used here.)

This post has been edited 1 time. Last edited by TLP.39, Jul 6, 2023, 3:53 AM
Reason: clarification

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#9 h Jul 6, 2023, 3:50 AM

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A standard R+ functional equation lol.

Let $P(x, y)$ be the assertion. First, if there exists $y > 0$ with $f(y) < 2y$ , there exists $x > 0$ such that $(c+1)x+f(y)=x+2y$ . With the original equation, we deduce that $x = 0$ which is a contradiction. Therefore, $\boxed{f(y) \geq 2y, \, \forall y > 0}$ . Then we have
$f(x + 2y) + 2cx = f((c + 1)x + f(y)) \geq 2((c + 1)x + f(y)) \implies f(x + 2y) \geq 2(x + f(y)).$ In particular, $\boxed{f(x + 2y) > f(y)}$ for all $x, y > 0$ .

Now we will prove the injectivity. Suppose that $f(a) = f(b)$ for some $a > b > 0$ . Compare with $P(x, a)$ and $P(x, b)$ we know that $f(2a + x) =f(2b + x)$ . It is a periodic function with $f(2b) = f(2b + (2a - 2b) \cdot k)$ for all $k \in \mathbb{N}$ . We can choose $k$ sufficiently large so that $2b + (2a - 2b) \cdot k > 2 \cdot 2b$ . Combined with the previous inequality, it is a contradiction. Hence $f$ is injective.

Next, compared with $P(\frac{z}{c+1} + 2x, y)$ and $P(\frac{z}{c + 1} + 2y, x)$ we have
$f(2(c + 1)x + f(y) + z) - f(2(c + 1)y + f(x) + z) = 4(x - y).$ Also, compared with $P(2x, y)$ and $P(2y, x)$ , the above equation also holds for $z = 0$ . For given $x, y > 0$ , we select $t > x, y$ , then
$f(2(c + 1)(x + t) + f(y + t)) - f(2(c + 1)(y + t) + f(x + t)) = 4(x - y).$ Since $f(x + t) > f(x)$ and $2(c + 1)(y + t) > 2(c + 1)y$ , there exists $z > 0$ with
$2(c + 1)y + f(x) + z = 2(c + 1)(y + t) + f(x + t).$ Compared with the previous two equations and by the injectivity of $f$ we get
$2(c + 1)x + f(y) + z = 2(c + 1)(x + t) + f(y + t).$ Thus we have
$f(x) - f(y) = f(x + t) - f(y + t).$ The equation also holds for all $t > 0$ since for given $x, y, z > 0$ we can choose a $A > \max\{x + 2z, y + 2z\}$ , then
$f(x) - f(y) = f(x + A) - f(y + A) = f((x + z) + (A - z)) - f((y + z) + (A - z)) = f(x + z) - f(y + z)$
With this strong equation, we can substitute the original equation into
$Q(x, y): f(cx + f(y)) = f(2y) + 2cx \implies f(x + f(y)) = f(2y) + 2x$ Compared with $Q(f(x), y)$ and $Q(f(y), x)$ we have $f(2y) = 2f(y)$ , so
$Q(x, y) \implies f(x + f(y)) = 2(x + f(y)),$ which also implies there exists a constant $M$ such that $f(t) = 2t$ for all $t > M$ . Finally, for all $x > 0$ , there exists $k \in \mathbb{N}$ such that $2^k x > M$ , then we have $f(x) = \frac{f(2^kx)}{2^k} = \frac{2 \cdot 2^kx}{2^k} = 2x$ . Hence the unique solution is $\boxed{f(x) \equiv 2x}$ .

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Prod55

#10 h Jul 6, 2023, 11:59 AM • 2 Y

Y by Batapan, PRMOisTheHardestExam

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

$(c+1)x+f(y)\neq x+2y\Rightarrow 2y-f(y)\leq 0\Leftrightarrow f(x)\geq 2x,x>0$

Define $g:\mathbb{R}_{>0}\to \mathbb{R}_{\geq 0}$ , $g(x)=f(x)-2x$

Now $P(x,y)$ becomes $Q(x,y): 2g(y)+g((x+1)c+f(y))=g(x+2y)$ .

$Q(x,y)\Rightarrow g(x)\geq 2g(y)$ whenever $x> 2y$

Now $(x+1)c+f(y)>2y$ so $g((x+1)c+f(y))\geq 2g(y)$ and thus $Q(x,y)\Rightarrow g(x)\geq 4g(y),x> 2y$ .

Continuing in the same way we find that $g(x)>kg(y)$ , when $x>2y$ for arbitrary large $k$ 's and thus $g(y)=0$ etc

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DNCT1

#11 h Jul 6, 2023, 7:51 PM

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Let $P(x,y)$ denote the assertion. If $f(y)<2y$ for some $y$ then $P\left(x,\frac{2y-f(y)}{c}\right)$ lead the contracdition. So $f(x)\ge 2x\quad\forall x\in\mathbb{R}^+ \ (1)$ .
$P(x,y)\implies f(x+2y)+2cx\ge 2(c+1)x+2f(y)\iff f(x+2y)\ge 2f(y)+2x\quad\forall x,y\in\mathbb{R}^+.$ Let denote this assertion is $Q(x,y)$ . There are two cases, Let $V= \{x\in\mathbb{R^+}| \ f(x)=2x\}$ .
$\textbf{Case 1:}$ $V \ \text{is empty}$ .
We obviously have $f(x)>2x$ for $x>0$ . Pick any $u$ such that $g(u)=f(u)-2u=A>0$ .
$P(x-2u,u)\implies f(cx+x-2uc+A)=f(x)+2cx-4cu,\quad\forall x>2u.$ $Q\left(2cx-4u,x\right)\implies f(cx+x-2uc+A)=f(x)+2cx-4cu\le \frac{f\left(2x+2cx-4cu\right)}{2},\quad\forall x>2u.$ Let $t=cx+x-2uc$ implies that $f(t+A)\le \frac{f(2t)}{2} \quad\forall t>2u. \ (2)$
If $A=f(u)-2u>2u$ then let $t=A$ leads the contracdition, so $f(u)\le 4u$ . We claim that if $f(x)>2x$ then $f(x)\le 4x$ .
And so in this case $2x<f(x)\le 4x \ \forall x\in\mathbb{R}^+.$ And so $\lim_{x\to 0^+} f(x)=0$ .
We use $(2)$ to have $f(2t)\ge 2f(t+A)> 4t+4A \implies f(t)\ge 2t+4A\quad\forall t>4u.$
Let $y>2u$ , since $f(y)>2y>4u$ , then by $P(x,y)\implies f(x+2y)+2cx> 2(c+1)x+2f(y)+4A\implies f(x+2y)>2x+2f(y)+4A \ \quad\forall x>0, y>2u. \ (99)$
Obvious, we also have $4x+8y\ge f(x+2y)>2x+2f(y)+4A\quad\forall x>0, y>2u.$ Let $x\to 0$ , then $f(y)\le 4y-2A\quad\forall y>2u.$
By $Q(x,y)$ , we have $4x+8y-2A\ge f(x+2y)>2x+2f(y)+4A\quad\forall x>0, y>2u. \ (100)$ Let $x\to 0$ again, we get that $f(y)\le 4y-3A\quad\forall y>2u$ , we can do similarly $(100)$ to obtain that $f(y)\le 4y-4A, \ \forall y>2u$ . And so $2x+2f(y) \le f(x+2y)\le 4x+8y-4(f(y)-2y)=4x+16y-4f(y)\quad\forall x>0, y>0. \ (101)$
Let $x\to 0$ , we get that $f(y)\le \frac{8}{3}y, \ y>0$ .
By $(99)$ we have $2x+2f(y)+4A<f(x+2y)\le \frac{8}{3}(x+2y)\quad\forall x>0, y>2u.\ (102)$ Let $x\to 0$ to see that $f(y)\le \frac{8}{3}y-2A\quad\forall y>2u$ , do similarly $(100)$ we have $f(y)\le\frac{8}{3}y-4A$ forall $y>2u$ . From this clause, similarly $(102)$ , which have $2x+2f(y)\le f(x+2y)\le \frac{8}{3}(x+2y)-4(f(y)-2y)\quad\forall x,y>0.$ We can do similarly each step in $(100),(101),(102)$ to see that if $f(x)\le v_nx \ \forall x>0$ then $f(x)\le v_{n+1}x$ which sequence $(v_n)$ is determined by $\begin{cases} \displaystyle \ v_1=4 \\ \ \displaystyle v_{n+1}=\frac{v_n+4}{3}. \end{cases}$ It's easy to see that $(v_n)$ is decreasing and $\lim v_n=2$ , and so $2x< f \le 2x\quad\forall x>0$ , which is asburd.
$\textbf{Case 2:} \ \text{there exists} \ a\in V$
$Q\left(x,\frac{a-x}{2}\right)\implies 2a=f(a)\ge 2f\left(\frac{a-x}{2}\right)+2x\quad\forall 0<x<a.\implies f\left(\frac{a-x}{2}\right)\le a-x\implies f\left(\frac{a-x}{2}\right)=a-x.\ \forall 0<x<a.$ Which means that $f(x)=2x\quad\forall 0<x<\frac{a}{2}.$
$P\left(x,\frac{a}{8}\right)\implies f\left((c+1)x+\frac{a}{4}\right)=2cx+f\left(x+\frac{a}{4}\right),\quad\forall x>0.$ Let $x\in\left(0,\frac{a}{4}\right)$ implies that $f\left((c+1)x+\frac{a}{4}\right)=2cx+2x+\frac{a}{2}\quad\forall \ 0<x<\frac{a}{4}$ .
Let $y=(c+1)x+\frac{a}{4}$ then $f(y)=2y\quad\forall y\in\left(\frac{a}{4},\frac{ca}{4}+\frac{a}{2}\right) \ (3)$ .
Denote the sequence $(x_n)$ $\begin{cases} \displaystyle \ x_1=\frac{a}{2} \\ \ \displaystyle x_{n+1}=\left(x_n-\frac{a}{4}\right)(c+1)+\frac{a}{4} \ \end{cases} \quad\forall n\in\mathbb{Z}^+.$ It's easy to see that $x_n=\frac{a}{4}(c+1)^{n-1}+\frac{a}{4}$ , hence $\lim x_n=+\infty$ , and use induction to see that $f(x)=2x\quad\forall x\in \left(\frac{a}{4},x_n\right)$ (continue the process similarly $(3)$ is not hard), but since $f=2x$ in $\left(0,\frac{a}{2}\right)$ so $f(x)=2x\quad\forall x\in (0,x_n)$ .
Thus, the only sol is $\boxed{f(x)=2x\quad\forall x\in\mathbb{R}^+}.$

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DS68

#12 h Jul 8, 2023, 1:38 PM • 1 Y

Y by PRMOisTheHardestExam

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

$\textbf{Claim: } f(x) \geq 2x$
$\textit{Proof}$ If we assume contrary, we can have $P\left(\frac{2y-f(y)}{c}, y\right)$ to reach a contradiction. $\square$

Now define $g: \mathbb{R^+} \rightarrow \mathbb{R}_{\geq 0}, g(x) := f(x) - 2x$ . We get that $g((c+1)x + g(y)+2y)+2g(y) = g(x+2y)$ and redefine $P(x,y)$ for this equation. Notice immediately we get $2g(y) \le g(x+2y)$ . As $g$ is bounded on intervals, by Bolzano-Weirstrass theorem, considering a converging sequence $x_n \rightarrow 0$ such that $y = x_i, x+2y = x_j$ gives us that $g(x_n) \rightarrow 0$ . As $2g(y) \le g(x+2y)$ , we can easily see that every sequence limiting to $0$ gives $g \rightarrow 0$ , i.e $\lim_{x \rightarrow 0} g(x) = 0$ . Now taking only $y \rightarrow 0$ gives us that $g(Nx) \leq g(x)$ where $N$ can be values arbitrarily close to $c+1$ . This implies $g(Mx) \leq g(x)$ for values $M$ arbitrarily close to $(c+1)^n, \forall n \in \mathbb{N}$ . Now then having $\frac{g(z+2x)}{2} \geq g(x) \geq g(Mx)$ , letting $z = Mx-2x$ for a sufficiently large $n$ gives $g \equiv 0 \rightarrow f \equiv 2x$ .

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#13 h Jul 14, 2023, 9:16 AM

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Let $P(x,y)$ be the original equation. If $f(y)<2y$ for some $y\in\mathbb{R}^+$ , then
$P\left(\dfrac{2y-f(y)}{c},y\right)\implies 2(2y-f(y))=0\ (\text{contradict}).$ So we have $f(y)\geq 2y$ . If $f(a)=f(b)$ for some $a-b>0$ . Comparing $P(x,a)$ and $P(x,b)$ ,
$f(x+d)=f(x)\quad\forall x\gg 0,$ where $d=2(a-b)$ . Take $n$ sufficiently large such that exists $x_0>0$ s.t.
$(c+1)x_0+f(1)=x_0+2,$ then $P(x_0,1)\implies 2cx_0=0$ (contradict). Hence $f$ is injective. Consider
$f((c+1)x+f(y))+2cz=f(x+2y)+2c(x+z),$ and apply the original equation on both sides, we have
$f\left((c+1)z+f\left(\dfrac{(c+1)x+f(y)-z}{2}\right)\right)=f\left((c+1)(x+z)+f\left(y-\dfrac{z}{2}\right)\right),$ for $2y>z$ (this implies $f(y)\geq 2y>z$ ). By injective, we have
$f\left(\dfrac{(c+1)x+f(y)-z}{2}\right)=(c+1)x+f\left(y-\dfrac{z}{2}\right).$ Compare above equation with $P\left(\dfrac{(c+1)x}{2c},\dfrac{1}{2}\left(y-\dfrac{z}{2}-\dfrac{(c+1)x}{2}\right)\right)$ and by injective, we have
$\dfrac{(c+1)x+f(y)-z}{2}=\dfrac{(c+1)^2x}{2c}+f\left(\dfrac{1}{2}\left(y-\dfrac{z}{2}-\dfrac{(c+1)x}{2}\right)\right).$ Let $y=2w+\dfrac{z+(c+1)x}{2}$ , we have
$f\left(2w+\dfrac{z+(c+1)x}{2}\right)=2f(w)+(c+1)x+z \quad \forall x,z>0,$ that is $f(x+2w)=2f(w)+2x$ $\forall w,x\in\mathbb{R}^+$ , denoted this equation by $Q(x,w)$ . Compare $Q(2x,w)$ and $Q(2w,x)$ , we get $f(x)=2x$ for all $x\in\mathbb{R}^+$ .

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Strudan_Borisov

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Strudan_Borisov

#14 h Jul 30, 2023, 5:56 PM • 2 Y

Y by PRMOisTheHardestExam, Assassino9931

Let $P(x,y)$ be the given functional equation. Obviously $f(x)=2x$ is a solution. Let's prove that it is the unique one.

Claim 1. $f(x)\geq 2x$ .

Proof. Assume there is $x_{0}$ such that $f(x_{0})<2x_{0}$ .

$P(\frac{2x_{0}-f(x_{0})}{c}, x_{0})$ : $2(2x_{0}-f(x_{0}))=0$ - absurd. The claim follows.

Now, let $g(x)=f(x)-2x$ and rewrite $P(x,y)$ in terms of $g$ . We get the following $g((c+1)x+2y+g(y))+2g(y)=g(x+2y).$ We'll prove that $g \equiv 0$ which will prove the desired result.
Assume there is $y_{0} \in \mathbb{R^{+}}$ such that $g(y_{0})>0$ (the fact that $f(x)\geq 2x$ means that $g(x) \geq 0$ ). In the last assertion we take $x=p-2y_{0}, y=y_{0}$ to get that: $g((c+1)p-2y_{0}c+g(y_{0}))+2g(y_{0})=g(p), \forall p>2y_{0}; \longrightarrow Q(p)$ Now, lets fix $p>2y_{0}$ and define the following sequence $\textbf{a} =\{a_{i}\}_{i=0}^{\infty}: a_{0}=p, a_{n}=(c+1)a_{n-1}+g(y_{0})-2cy_{0}$ . We see that $a_{n}-a_{n-1}=c(a_{n-1}-2y_{0})+g(y_{0})$ . Because of $a_{0}>2y_{0}$ by induction we get that $\textbf{a}$ is strictly increasing, so $a_{n}>2{y_0}$ . It means that $g(a_{i+1})+2g(y_{0})=g(a_{i})$ from $Q(a_{i})$ . Summing the last one for $i=0,1,...,n-1$ we get that $g(a_{n})+2ng(y_{0})=g(a_{0})$ , so $g(a_{0})\geq2ng(y_{0})$ which contradicts the fact $n$ can tend to infinity. So, $g\equiv 0$ and we are done.

This post has been edited 2 times. Last edited by Strudan_Borisov, Jul 30, 2023, 5:57 PM

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#15 h Jul 30, 2023, 7:33 PM • 2 Y

Y by PRMOisTheHardestExam, egxa

Let $P(x,y)$ be the denotation of the given equation.
Lemma: The function $f$ is injective.
Proof: Assume not. Then there exists $a<b$ such that $f(a)=f(b)$ . From comparing $P(x,a)$ and $P(x,b)$ we obtain
$f(x+2a)=f(x+2b)$ In other words $f(x)=f(x+2(b-a))$ for all $x>2a$ . Thus $f$ is periodic on $(2a,\infty)$ . Let the smallest period be $T$ . From comparing $P(x,y)$ and $P(x+\frac{T}{c+1},y)$ we obtain $f(x+2y)=f(x+\frac{T}{c+1}+2y)+\frac{2cT}{c+1}$ and with induction if $z=x+2y$ then
$f(z)=f(z+\frac{nT}{c+1})+\frac{2cnT}{c+1}$ for all positive integer $n$ . But if $n>\frac{(c+1)f(z)}{2cT}$ we'll get a contradiction since the right-hand side is obviously bigger than the left-hand side.
Thus the conclusion follows.
From compairing $P((c+1)x,(c+1)^2x+f(y))$ and $P((3c+1)x,(c+1)x+2y)$ we get the equation
$f((5c+3)x+4y)+4c^2x=f((2c^2+5c+3)x+2f(y))..........(1)$ From equation $(1)$ , $P(2cx,2y+\frac{3(c+1)}{2}x)$ and injectivity we get $f(2y+\frac{3(c+1)}{2}x)=3(c+1)x+2f(y)$ . And with rewriting $\frac{3(c+1)}{2}x$ as $x$ we obtain
$f(x+2y)=2x+2f(y)$ After writing this equation on the original equation and rewriting $(c+1)x$ as $x$ we get
$f(x+f(y))=2x+2f(y)=f(x+2y)$ With injectivity, we obtain $f(x)=2x$ for all positive reals $x$ and obviously, this function works on the original equation. Thus the only function that satisfies this equation is $f(x)=2x$ for all positive reals $x$

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#16 h Nov 19, 2023, 1:04 PM

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We uploaded our solution https://calimath.org/pdf/APMO2023-4.pdf on youtube https://youtu.be/yRc3q7caBvk.

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#18 h Jan 23, 2024, 6:45 PM

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Stupid Problem,
The only solution is $f(x)=2x$
Claim 1: $f(x) \geq 2x$
Proof
Now the idea is to define $g(x)=f(x)-2x$ , the equation now becomes $g((c+1)x+g(y)+2y)+2g(y)=g(x+2y).$
Claim 2 $g \equiv 0$
Proof
Hence we're done

This post has been edited 1 time. Last edited by math_comb01, Jun 17, 2024, 6:11 PM
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#19 h Mar 27, 2024, 3:36 AM

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kred9 wrote:

In particular, $f(2t) = 4t$ , so we have $f(y) = 2y$ for all $y < t$ as well, showing that $f(x) = 2x$ for all $x \in \mathbb{R}_{>0}$ .

I don’t see why $f(2t) = 4t$ implies that $f(y) = 2y$ for all $y < t$ . Could you explain this to me?

This post has been edited 1 time. Last edited by solasky, Mar 27, 2024, 3:37 AM

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Marius_Avion_De_Vanatoare

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Marius_Avion_De_Vanatoare

#20 h Jun 9, 2024, 7:13 PM

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Looks like all the solutions have very alike ideas, however everybody explains them differently, I will take a try.
First, denote the assertion by $P(x;y)$ , assuming there exists $y$ such that $f(y)<2y$ , from $P(\frac{2y-f(y)}{c};y)$ we get a contradiction. Assume there exists some $y_1$ such that $f(y_1) \neq 2y_1$ , I will achieve a contradiction. First, apply $f(a) \ge a$ to the $LHS$ to obtain after reducing that $2x+2f(y) \le f(x+2y)$ . Now define $g:\mathbb{R_+} \rightarrow \mathbb{R}_{\ge0}$ , $g(x)=f(x)-2x$ . The given property reduces to $g(x+2y) \ge 2g(y)$ , call it $Q(x;y)$ .
First, I will prove $g$ is unbounded. It is not hard, as $Q(x;x) \Rightarrow g(3x) \ge 2g(x)$ , and iterating for $y_1$ , we achieve the desired result.
Now I will prove that for any given constant $k$ there exists another constant $N_k$ such that $g(x) \ge kx, \forall x \ge N_k$ .
For this just consider $Q(\epsilon y;y)$ and iterate to get $g((2+\epsilon)^ny_0) \ge 2^ng(y_0)$ . Take a sufficiently large $g(y_0)$ . Considering the largest $n$ such that $2^ny_0 <x$ , and taking $\epsilon$ such that $(2+\epsilon)^ny_0=x$ , I get of course together with $2^ny_0 \ge \frac{x}{2}$ , that for $x \ge 2y_0$ , $g(x) \ge \frac{x}{2}g(y_0)$ , which concludes the claim.
From here, just pick a large enough $k$ and $x \ge N_k$ , and applying the inequality $f(x) \ge (2+k)x$ to the $LHS$ of $P(x;y)$ , together with small iterations shows that for all big enough $x$ , $f(x)$ is unbounded, which obviously can't be true.
This is a very good example of FE, where you use the most standart $\mathbb{R}_+ \rightarrow \mathbb{R}_+$ techniques.

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lnmfx

#21 h Jun 17, 2024, 5:41 PM

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Solution

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#22 h Sep 4, 2024, 12:51 PM • 1 Y

Y by iStud

I hope there is nothing wrong :maybe:

:maybe:

.

The only function is $f(x)=2x \ \forall \ x \in \mathbb{R^+}$ . Easy to check that this satisfies, now we'll prove that this is the only one. Let $P(x,y)$ be the assertion. All variables mentioned should be in $\mathbb{R^+}$ unless specified.

Claim 1. $f(x)\ge 2x \ \forall \ x \in \mathbb{R^+}$
Proof. Suppose there exist $k \in \mathbb{R^+}$ such that $f(k)<2k$ . Then $P(\frac{2k-f(k)}{c},k)$ gives
$f\left( \frac{2k-f(k)}{c}+f(k) \right)=f\left( \frac{2k-f(k)}{c}+f(k) \right) +2c\left( \frac{2k-f(k)}{c} \right)$ Clearly this isn't possible since $2c\left( \frac{2k-f(k)}{c} \right) >0$ . As desired.

Claim 2. $f$ is injective
Proof. Suppose there exist distinct $a,b \in \mathbb{R^+}$ such that $f(a)=f(b)$ , we may assume $a>b$ .
Comparing $P(x,a)$ and $P(x,b)$ yields
$f(x+2a)=f(x+2b) \Leftrightarrow f(x)=f(x+(2a-2b)) \ \forall \ x > 2a$ . By Induction, we have
$f(x)=f(x+n(2a-2b)) \ \forall \ x > 2a$ for all $n \in \mathbb{N}$ . Note that $x+n(2a-2b)$ goes to infinity as $n$ goes to infinity.
Since $f(x)\ge 2x \ \forall \ x \in \mathbb{R^+}$ , $f(x)$ is infinity for all $x>2a$ . Clearly this is absurd because technically infinity is not a real number. So this is a contradiction. As desired.

Claim 3. $f(x)-2x$ is constant
Proof. From $P\left( x+y,\frac{z}{2} \right)$ , we have
$f\left( (c+1)(x+y)+f\left( \frac{z}{2} \right) \right)=f(x+y+z)+2cx+2cy \ (1)$ . From $P\left( x,\frac{y+z}{2} \right)$ , we have
$f\left( (c+1)x+f\left( \frac{y+z}{2} \right) \right)=f(x+y+z)+2cx \ (2)$ . Plugging $(2)$ to $(1)$ yields
$f\left( (c+1)(x+y)+f\left( \frac{z}{2} \right) \right)=f\left( (c+1)x+f\left( \frac{y+z}{2} \right) \right)+2cy \$ . Note that $\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2}>0$ . From $P\left( y, \frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2}\right)$ , we have
$f\left( (c+1)y + f\left(\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2} \right)\right)=f\left( (c+1)x+f\left( \frac{y+z}{2} \right) \right) + 2cy$ . Injectivity gives
$(c+1)(x+y)+f\left( \frac{z}{2} \right)= (c+1)y + f\left(\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2} \right)$ $\Leftrightarrow (c+1)x+f\left( \frac{z}{2} \right)=f\left(\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2} \right)$ . Change $x$ to $\frac{y}{c+1}$ to get
$y+f\left( \frac{z}{2} \right)=f\left(\frac{f\left( \frac{y+z}{2} \right)}{2} \right)$ . The RHS is symmetric in $y$ and $z$ , so swapping them yields
$y+f\left( \frac{z}{2} \right)=z+f\left( \frac{y}{2} \right) \Leftrightarrow f\left( \frac{y}{2} \right)-y \ is \ constant$ . Just change $y$ to $2y$ and we are done.

Now, plug $f(x)=2x + c \ \forall \ x \in \mathbb{R^+}$ for a constant $c \in \mathbb{R}$ to $P(x,y)$ . Easy to check that $c=0$ by pairing the coefficients. As desired.

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AblonJ

#23 h Feb 23, 2025, 12:09 PM

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Why This seems too easy.
First note that $f(x) \geq 2x$ , otherwise $P(\frac{2y-f(y)}{c},y)$ gives a contradiction.
Introduce $g(x)=f(x)-2x$ . We see that $g(x) \geq 0$ .
Now, from the original equation,
$g((c+1)x+g(y)+2y)=g(x+2y)-2g(y)$ As $g((c+1)x+g(y)+2y) \geq 0$ , $g(x+2y) \geq 2g(y)$ .
$g(x+2y)-2g(y)=g((c+1)x+g(y)+2y) \geq 2g(y)$ Now, $g(x+2y) \geq 4g(y)$ . Note that we can continue this process and have that
$g(x+2y) \geq Ng(y)$ for arbitrarily large $N$ , which force $g(y)=0$ .
Hence, $f(x)=2x$ .

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bo18

#24 h Mar 8, 2025, 8:50 PM

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I don’t want to write why in injective, but I will just say that follows from periodicity

Here is the solution of me:

Given the function:

$f((c+1)x+f(y)) = f(x+2y) + 2cx$
Let us denote this statement as $P(x, y)$ .

$P(x, (c+1)x+f(y))$
$f((c+1)x+f(x+2y)+2cx) = f(x+2((c+1)x+f(y)))+2cx$
The left-hand side can be rewritten as:

$f((c+1)x+f(x+2y)+2cx) = f\left( x \cdot \frac{3c+1}{c+1} \cdot (c+1) + f(x+2y) \right)$
$= f\left( x \cdot \frac{3c+1}{c+1} + 2(x+2y) \right) + 2c x \cdot \frac{3c+1}{c+1}$
Thus:

$f\left( x \cdot \frac{3c+1}{c+1} + 2(x+2y) \right) + 2c x \cdot \frac{2c}{c+1} = f(x+2((c+1)x+f(y)))$
Applying $P\left( x \cdot \frac{2c}{c+1}, \frac{4y+3x}{2} \right)$ :

$f(2cx+f((4y+3x)/2)) = f\left( x \cdot \frac{2c}{c+1} + 4y+3x \right) + 2c x \cdot \frac{2c}{c+1}$
$= f\left( x \cdot \frac{3c+1}{c+1} + 2(x+2y) \right) + 2c x \cdot \frac{2c}{c+1}$
Since this matches the previous expression, we conclude:

$f(2cx+f((4y+3x)/2)) = f(x+2((c+1)x+f(y)))$
By injectivity:

$2cx + f((4y+3x)/2) = x + 2cx + 2x + 2f(y)$
Thus:

$f((4y+3x)/2) = 3x + 2f(y)$
Denoting this as $Q(x, y)$ .

Applying $Q(x/3, y)$ :

$f((4y+x)/2) = x + 2f(y)$
Denoting this as $T(x, y)$ .

Let $f(x) = 2x + g(x)$ , where $f(x) > 2x$ .

If $2x > f(x)$ , then applying $P\left( \frac{2y - f(y)}{c}, y \right)$ :

$2c \cdot \frac{2y - f(y)}{c} = 0$
$\Rightarrow 2y = f(y)$
which leads to a contradiction.

Thus, from the properties of $g(x)$ , we obtain:

$g\left( \frac{4y+x}{2} \right) + \frac{4y+x}{2}\cdot{2}= x + 2g(y) + 4y$
$\Rightarrow g\left( \frac{4y+x}{2} \right) = 2g(y)$
The right-hand side depends on $x$ , but the left-hand side does not.

Substituting:

$x \to 4y \Rightarrow g(4y) = 2g(y)$
$x \to 2y \Rightarrow g(3y) = 2g(y)$
$x \to 8y \Rightarrow g(6y) = 2g(y)$
$g(3y) = g(6y)$
If we substitute $y \to y/3$ , we obtain:

$g(y) = g(2y)$
$g(y) = g(2y) = g(4y) = 2g(y)$
Thus:

$g(y) = 0$
Therefore:

$f(x) = 2x + g(x) = 2x$

This post has been edited 4 times. Last edited by bo18, Mar 8, 2025, 9:00 PM
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SomeonesPenguin

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SomeonesPenguin

#25 h Mar 10, 2025, 6:50 PM

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I think this is new.

Claim: $f$ is injective.

sketch

Fix a positive real $a>0$ and let $b=f(a)$ .

$\bullet$ $y\mapsto a\implies f((c+1)x+b)=f(x+2a)+2cx$

$\bullet$ $y\mapsto (c+1)y+b\implies f((c+1)x+f(y+2a)+2cy)=f(x+2(c+1)y+2b)+2cx$

$\begin{align*} f(x+2(c+1)y+2b)+2cx&=f((c+1)x+f(y+2a)+2cy)\\&=f\left(\left(c+1\right)\left(x+\frac{2cy} {c+1}\right)+f(y+2a)\right)\\ &=f\left(x+\frac{2cy}{c+1}+2y+4a\right)+2cx+\frac{4c^2y}{c+1}\\ \end{align*}$
If we let $\frac{2c^2}{c+1}=\alpha$ and $4a-2b=\beta$ (which are fixed) we get $f(x+\alpha y+\beta)=f(x)+2\alpha y$ Henceforth $y\mapsto \frac{cx}{\alpha}$ and $x\mapsto x+2y$ yield $f(x+2y+cx+\beta)=f(x+2y)+2cx=f((c+1)x+f(y))$ So injectivity further yields $f(y)=2y+\beta$ , but this only holds for sufficiently large $y>>\beta$ . Anyway, for $x,y>>\beta$ in the original equation we get $\beta=0$ which implies that $f(a)=2a$ for all positive reals $a$ , which is indeed a solution.

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#26 h Mar 10, 2025, 9:43 PM

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Let $P(x,y)$ be the assertion $f((c+1)x+f(y))=f(x+2y)+2cx$ .

Claim 1: $f(x)\ge2x$
Suppose $f(u)<2u$ for some $u$ , then:
$P\left(\frac{2u-f(u)}c,u\right)\Rightarrow2(2u-f(u))=0$
which is a contradiction.

Claim 2: $f$ is injective
Suppose $f(a)=f(b)$ for some $a>b$ and let $p=2a-2b>0$ . Then:
$P(x,a)\Rightarrow f((c+1)x+f(a))=f(x+2a)+2cx$
$P(x,b)\Rightarrow f((c+1)x+f(a))=f(x+2b)+2cx$
So $f(x+2a)=f(x+2b)$ . Then $f(x+p)=f(x)$ for all sufficiently large $x>2b$ , and by induction we get that $f(x+np)=f(x)$ for all $x>2b$ and $n\in\mathbb N$ . Then fixing $x>2b$ and taking $n\gg\frac{|f(x)-2x|}{2p}$ sufficiently large we get:
$f(x)=f(x+np)\ge2x+2np>2x+|f(x)-2x|\ge f(x),$ contradiction. Hence $f$ is injective.

Claim 3: if $f(x+a)=f(x)+b$ for all $x>c$ for some $a,b,c\in\mathbb R^+$ then $b=2a$
Note that $f(x+2a)=f(x+a)+b=f(x)+2b$ (again considering only $x>c$ ). Then:
$P(x,y+a)\Rightarrow f((c+1)x+f(y)+b)=f(x+2y)+2cx+2b$
So we have:
$f((c+1)x+f(y)+b)=f((c+1)x+f(y))+2b=f((c+1)x+f(y)+2a)$ and so $b=2a$ by injectivity.

Claim 4: $\boxed{f(x)=2x}$ for all $x\in\mathbb R^+$
$P(x,(c+1)y+f(z))\Rightarrow f((c+1)x+f(y+2z)+2cy)=f(x+2(c+1)y+2f(z))+2cx$
$P\left(x+\frac{2cy}{c+1},y+2z\right)\Rightarrow f((c+1)x+f(y+2z)+2cy)=f\left(x+\frac{2cy}{c+1}+2y+4z\right)+2cx+\frac{4c^2y}{c+1}$
Comparing these, we get:
$f(x+2(c+1)y+2f(z))=f\left(x+\frac{2cy}{c+1}+2y+4z\right)+\frac{4c^2y}{c+1},$ and taking $x\mapsto x-\frac{2cy}{c+1}-2y-4z$ we have:
$f\left(x+\frac{2c^2y}{c+1}+2f(z)-4z\right)=f(x)+\frac{4c^2y}{c+1}$ for all $x>\frac{2cy}{c+1}+2y+4z$ . By Claim 3 (fixing $y,z$ ) we get $4f(z)-8z=0$ , so $f(x)=2x$ for all $x$ which indeed fits.

This post has been edited 4 times. Last edited by jasperE3, Apr 23, 2025, 6:19 PM

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#27 h Apr 22, 2025, 10:08 AM

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The answer is $f(x) = 2x$ , the check is trivial.

$P(\frac{2y - f(y)}{c}, y) : 0 = 2cx$ which is clearly a contradiction thus $f(x) \geq 2x$ for all $x \in \mathbb{R}^+$ .

Now let us manipulate the original equation:
$f((c+1)x + f(y)) = f(x + 2y) + 2cx$ $2cx + 2x + 2f(y) \geq f(x + 2y) + 2cx$ $2x + 2f(y) \geq f(x + 2y)$ Let $g(x) = f(x) - 2x$
$2x + 4y + 2g(y) \geq 2x + 4y + g(x + 2y)$ $2g(y) \geq g(x + 2y)$ Now assume that if there exist $a$ s.t. $g(a) = 0$ . Notice that for all $x > 2a$ , we have $g(x) = 0$ .
Then, if you pick $x, y$ in the original equation such that $g(y) > 0$ and $x >2a$ , then you have the LFS $= 2x + 4y + 2cx$ and the RHS is larger than $2x + 4y + 2cx$ which is a contradiction, thus if such $a$ exists, we have $f(x) = 2x$ for all $x \in \mathbb{R}^+$ .

If no such $g(a)$ exists, then $f(x) > 2x$ for all $x \in \mathbb{R}^+$ . Let $[0, b]$ be some fixed interval, and pick any number $k \in [0, b]$ . If $\min(g(k)) = 0$ (i.e. there exists $k$ s.t. $g(k)$ is arbritarily close to 0), then notice that for any $x > 2b$ , we have to have $g(x) = 0$ which is a contradiction. Thus for any interval $[0, b]$ , we have that the lower bound of $g(x)$ in that range is higher than 0.

If the lower bound of $g(x)$ is $d > 0$ (for all $x \in \mathbb{R}^+$ ), then let $x + 2y$ be a value such that $g(x + 2y)$ is arbitarily close to d. Then the RHS is arbitarily close to $2x + 4y + 2cx + d$ while the LHS must be bigger than $2x + 4y + 2cx + 2d$ which is a contradiction. Thus the lower bound of $g(x)$ is 0.

Now from the interval claim we proved, we can let $y = 1$ and $x + 2y$ is a value such that $g(x + 2y)$ is arbritarily close to 0. In a similar way as before, this creates a contradiction.

Thus $g(x) = 0$ and $f(x) = 2x$ .

This post has been edited 1 time. Last edited by DeathIsAwe, Apr 22, 2025, 10:09 AM

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