Functional equation

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#1 h Oct 18, 2013, 11:11 AM • 2 Y

Y by Adventure10, Mango247

Determine all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that $f (x + f (x + y)) = f (2x) + y,$ for all $x,y \in \mathbb{R}^+$

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#2 h Oct 18, 2013, 7:54 PM • 2 Y

Y by socrates, Adventure10

Take $z=x+y$ . Clearly $z$ can be any number $>x$ .
Now we have $f(x+f(z))=f(2x)+z-x$ where $z$ can be any number $>x$ .
Let $inf(f(x))=l$ Now let's take $x=a_i$ such that $lim_{i\rightarrow \infty} (f(2a_i))=l \Longrightarrow$ if we set $y=$ any positive number $\Longrightarrow f(x+f(z))$ can take all positive values $>l \Longrightarrow f(x)$ is $surjective$ over the interval $(l,+\infty)$ (1).
Now let's take any $x>l$ and from (1) $\Longrightarrow$ for some number $r$ holds $f(r)=x$ . If we suppose that for that $r$ holds also $r>x \Longrightarrow$ when we put $z=r \Longrightarrow f(x+x)=f(2x)+r-x \Longrightarrow r=x$ , contradiction
$\Longrightarrow$ for any $x>l$ if we take any $r$ such that $f(r)=x$ (there clearly exists such) $\Longrightarrow r\le x \Longrightarrow f(r)\ge r$ (2). But when $x$ goes through the interval $(l,+\infty) \Longrightarrow r$ can take any positive value except those $r$ for which $f(r)=l$ (3)(if there even exist such $r$ ). And form (2) and (3) $\Longrightarrow$ for every $x$ holds $f(x)\ge x$ or $f(x)=l$ (4).
$f(x+f(z))=f(2x)+z-x\ge l+z-x>l$ and from (4) $\Longrightarrow f(x+f(z))\ge x+f(z) \Longrightarrow f(2x)+z-x \ge x+f(z)$ for any $z>x \Longrightarrow f(x)+y \ge x+f(y)$ for any $y>x/2$ . Now for any $x,y: y>x/2$ and $x>y/2$ hold $\Longrightarrow f(x)+y=x+f(y)$ . Now let $x, y$ be random numbers $\in (1;2) \Longrightarrow$ the conditions above hold $\Longrightarrow f(x)+y=x+f(y) \Longrightarrow f(x)=x+f(1,6)-1,6=x+c,$ for any $x\in (1;2)$ (where $c=f(1,6)-1,6$ ) $\Longrightarrow f(x)-x=c$ for $x\in (1;2)$ . Now we do the same with the interval $(1,5;3)$ and we get $f(x)-x=$ the same $c$ . We can analogically pick such intervals covering the interval $(\varepsilon;+\infty)$ , where $\varepsilon$ is any positive number $\Longrightarrow f(x)-x=c$ for all $x>0 \Longrightarrow f(x)\equiv x+c$ . Plugging this into the equation gives only the solution $f(x)\equiv x$ .

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#3 h Oct 19, 2013, 10:22 PM • 2 Y

Y by Adventure10, Mango247

Here is my solution:

The given equation rewrites $f(x+f(y))=f(2x)+y-x, \ \forall y>x>0 \ (*).$

f(x+f(x))=2f(x)
f(x)\geq x
f(x)-x is decreasing and eventually constant
f(x)=x

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#4 h Feb 26, 2016, 4:38 PM • 5 Y

Y by socrates, john111111, BobaFett101, Adventure10, Mango247

Hello.

A different solution.

Let $P(x,y)$ be the given property of the function.

I"ll first prove that $f$ is injective.Consider $x_1,x_2>0$ such that $f(x_1)=f(x_2)$ .

Obviously there exist $x_0,y_1,y_2>0$ such that $x_0+y_1=x_1$ and $x_0+y_2=x_2$ .

$P(x_0,y_1)$ and $P(x_0,y_2)$ give the desired result.

Now, $P(x,x)$ and $P(x,f(2x))$ imply $f(2x+f(2x))=2f(2x) , \ \forall x>0$ hence $f(x+f(x))=2f(x) , \ \forall x>0 \ (1)$ .

We combine $P(x,f(4x))$ and $P(2x,f(2x))$ and injectivity to obtain $f(x+f(4x))=x+f(2x+f(2x)) , \ \forall x>0$ .

We use $P(x,3x)$ and $(1)$ and the latter relation is written as $f(2x)+3x=x+2f(2x) , \ \forall x>0$

whence we easily obtain $f(x)=x , \ \forall x>0$ .

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#5 h Mar 5, 2016, 1:29 AM • 1 Y

Y by Adventure10

Here's a solution which is again slightly different.

Let $P(x,y)$ be the assertion $f(x+f(x+y))=f(2x)+y$

First I will show that $f$ is injective
Let $f(a)=f(b)$
We can always chose an $x \in \mathbb{R}^+$ such that $x \le a,b$
$P(x,a-x) \implies f(x+f(a))=f(2x)+a-x$ $P(x,b-x) \implies f(x+f(b))=f(2x)+b-x$
That is $a=b$

Hence we have $f$ is surjective.

$P(x,f(x+y)) \implies f(x+f(x+f(x+y)))=f(2x)+f(x+y)$ $\implies f(x+y+f(2x))=f(2x)+f(x+y)$ Setting $2x=s, x+y=t$
$\implies f(t+f(s))=f(s)+f(t)$ for all $t,s \in \mathbb{R}^+$ (since we can find such $x,y$
But this also implies $f(s)+f(t)=f(s+f(t))$

Thus we have
$f(s+f(t))=f(t+f(s))$ Using injectivity and setting $s=1$
$f(t)=t+f(1)-1$
That is $f$ is linear.

Putting $f \equiv t+c$ , we get $c=0$

Hence $\boxed{f(t)=t}$ is the only solution.

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#6 h Mar 6, 2016, 3:29 PM • 2 Y

Y by Adventure10, Mango247

utkarshgupta wrote:

$P(x,f(x+y)) \implies f(x+f(x+f(x+y)))=f(2x)+f(x+y)$ $\implies f(x+y+f(2x))=f(2x)+f(x+y)$ Setting $2x=s, x+y=t$
$\implies f(t+f(s))=f(s)+f(t)$ for all $t,s \in \mathbb{R}^+$ (since we can find such $x,y$
.

This is not true.

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