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Hard FE R^+

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DNCT1

#1 h Dec 30, 2020, 12:45 PM • 2 Y

Y by toanhocmuonmau123, Mango247

Find all functions $f:\mathbb{R^+}\to\mathbb{R^+}$ such that
$f(3x+f(x)+y)=f(4x)+f(y)\quad\forall x,y\in\mathbb{R^+}$

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DNCT1

#2 h Dec 31, 2020, 1:29 AM • 2 Y

Y by Mango247, Mango247

I just can prove that $f(x)\geq x\quad\forall x\in\mathbb{R^+}.$ $\quad$ Any idea ?

This post has been edited 1 time. Last edited by DNCT1, Dec 31, 2020, 1:29 AM

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TuZo

#3 h Dec 31, 2020, 3:32 PM

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DNCT1 wrote:

I just can prove that $f(x)\geq x\quad\forall x\in\mathbb{R^+}.$ $\quad$ Any idea ?

How you proved this?

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Mgh

#4 h Dec 31, 2020, 4:22 PM

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TuZo wrote:

DNCT1 wrote:

I just can prove that $f(x)\geq x\quad\forall x\in\mathbb{R^+}.$ $\quad$ Any idea ?

How you proved this?

Consider there is a positive integer x fod which $x-f (x)$ is greter than zero, now $p (x, x-f (x)$ gives contradiction. I thinks it's good to write it: $f (x)=x+g (x)$

This post has been edited 1 time. Last edited by Mgh, Dec 31, 2020, 4:23 PM

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DNCT1

#5 h Jan 1, 2021, 4:56 PM

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Well, here's a solution which I have
Let $P(x,y)$ be the assertion of $f(3x+f(x)+y)=f(4x)+f(y)\quad\forall x,y\in\mathbb{R^+}$

Assume there exist $x_0\in\mathbb{R^+}$ such that $f(x_0)<x_0$ so
$P(x_0,x_0-f(x_0))\implies f(x_0-f(x_0))=0\quad(\text{impossible})$ Hence we have $f(x)\geq x\quad\forall x\in\mathbb{R^+}$ .

By $P(x,y)$ there exxists $a,b\in\mathbb{R^+}$ such that
$f(x+a)=f(x)+b\quad\forall x\in\mathbb{R^+}.$ So we also have $f(x+na)=f(x)+nb\quad\forall x\in\mathbb{R^+},n\in\mathbb{N^+}$
$P(x+a,y)\implies f(3x+3a+f(x)+b+y)=f(4x)+f(y)+4b\quad\forall x,y\in\mathbb{R^+}$ $P(x,y+4a)\implies f(3x+f(x)+y+4a)=f(4x)+f(y)+4b\quad\forall x,y\in\mathbb{R^+}$ And by two equations we have
$f(3x+3a+f(x)+b+y)=f(3x+f(x)+y+4a)\quad\forall x,y\in\mathbb{R^+}$ If $a\neq b$ we assume that $a<b$ and $y$ such that $y>4a+3+f(1)$
$P(1,y-4a-3-f(1))\implies f(y)=f(y+b-a)\quad\forall y\in\mathbb{R^+},y>4a+3+f(1)$ Hence $f$ must be period $a-b$ with $x>4a+3+f(1)$ but we have $f(x)\geq x\quad\forall x\in\mathbb{R^+}$ that implies $f$ must be period $0$ (absurd). $\quad$ Hence $a=b$ .
Otherwise we have $f(h)=f(h+n(b-a))\quad\forall n\in\mathbb{N^+}$ for some $h>4a+3+f(1)$ so $d=f(h)\geq h+n(b-a)$ so choose $n\in\mathbb{N^+}$ such that $n>\frac{d}{b-a}$ we get the contracdition.
And so we have
$f(4x)=f(x)+3x\quad\forall x\in\mathbb{R^+}$ $P(x,y)$ becomes
$f(f(x)+y)=f(x)+f(y)\quad\forall x,y\in\mathbb{R^+}$ $P(x,y-3x)\implies f(f(x)+y)=f(4x)+f(y-3x)=f(x)+3x+f(y-3x)\quad\forall x,y\in\mathbb{R^+}, y>3x$ And so $3x+f(y-3x)=f(y)\quad\forall x,y\in\mathbb{R^+}, y>3x$ $\implies f(y)-f(y-x)=x\quad\forall x,y\in\mathbb{R^+}, y>x (1)$ Let $y\to x+1$ in $(1)$ we have $f$ is linear $\forall x>1$ or $f(x)=x+f(1)-1\quad\forall x>1$
By $(1)$ we let $(x,y)\to (1-x,1)$ we have $f(x)=x+f(1)-1\quad\forall 0<x<1$ and it's also true with $x=1$ .
And so
$\boxed{\text{Solution:}\quad f(x)=x+f(1)-1\quad\forall x\in\mathbb{R^+}}$

This post has been edited 2 times. Last edited by DNCT1, Jan 1, 2021, 5:02 PM

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#6 h Apr 21, 2025, 12:59 AM • 1 Y

Y by truongphatt2668

Let $P(x,y)$ be the assertion $f(3x+f(x)+y)=f(4x)+f(y)$ .

Claim 1: $f(x)\ge x$ for all $x$
Suppose $f(u)<u$ for some $u$ , then:
$P(u,u-f(u))\Rightarrow f(u-f(u))=0$ which is impossible.

Claim 2: if $f(x+a)=f(x)+b$ for all $x>0$ , given some $a,b>0$ , then $f(x+b)=f(x)+b$
Note that $f(x+na)=f(x)+nb$ for $n\in\mathbb N$ , by simple induction. Then:
$P(x+a,y)\Rightarrow f(3x+f(x)+y+b)=f(4x)+f(y)+b$
$P(x,y+b)\Rightarrow f(3x+f(x)+y+b)=f(4x)+f(y+b)$
and the claim is proven.

Claim 3: $f$ is injective.
Suppose $f(a)=f(b)$ for some $a>b>0$ .
$P(x,a)\Rightarrow f(3x+f(x)+a)=f(4x)+f(a)$
$P(x,b)\Rightarrow f(3x+f(x)+b)=f(4x)+f(a)$
Fix $x$ and apply claim $2$ to $P(x,y)$ , we get $f(f(4x)+y)=f(4x)+f(y)$ , which implies $f(f(x)+y)=f(x)+f(y)$ . Then:
$f(x)+f(3x+a)=f(3x+f(x)+a)=f(3x+f(x)+b)=f(x)+f(3x+b)$ so $f(3x+a)=f(3x+b)$ . This transforms to $f(x+a-b)=f(x)$ for all $x>b$ . Let $x>b$ , then:
$P(x,y+a-b)\Rightarrow f(3x+f(x)+y)=f(3x+f(x)+y+a-b)=f(4x)+f(y+a-b)\Rightarrow f(y)=f(y+a-b)$ so $f$ is periodic in general
By induction, $f(x)=f(x+n(a-b))$ for $n\in\mathbb N$ , but by Claim 1 we have:
$f(x)\ge x+n(a-b),$ taking $n\to\infty$ gives a contradiction.

Finish:
Recall that $f(f(x)+y)=f(x)+f(y)$ , so swapping $x,y$ and using injectivity gives $\boxed{f(x)=x+c}$ which works for any constant $c>0$ .

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