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DNCT1

#1 h Oct 10, 2020, 6:43 AM • 1 Y

Y by tiendung2006

Find all functions $f:\mathbb{R^+}\rightarrow \mathbb{R^+}$ such that
$f(2f(x)+2y)=f(2x+y)+y\quad\forall x,y,\in\mathbb{R^+}$

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#2 h Oct 10, 2020, 9:30 AM

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Any ideas?

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Mr.C

#3 h Oct 10, 2020, 12:46 PM • 2 Y

Y by tiendung2006, Blackbeam999

not bad i guess ... i had another solution which is not really nice , i think this is better if anyone intrested i can send the other aswell .
.
here is the easy one, let
$p(x,y)=f(2f(x)+2y)=f(2x+y)+y$
now
$p(x,y+f(z)-f(x))$ gives by letting $f(z)-f(x)=t$
$f(2x+t+y)+t=f(2z+y)$
so if $2z-2x-t=k$ we have
$f(y+k)=t+f(y)$ for big enough $y$
so let $x$ be big now from
$p(x+k,y)$ we have
$f(2f(x)+2t+2y)-f(2f(x)+2y)=2t$ so
$f(2f(x)+2t+2y)=f(2f(x)+2y+2k)$
we now prove $f$ is injective let
$f(a)=f(b)$ so from
$p(a,y)-p(b,y)$ we get
$f(2a+y)=f(2b+y)$ so $f$ is periodic let the period be $s$ we have from
$p(x,y+s)-p(x,y)$ that $s=0$ so
$a=b$
now going back at
$f(2f(x)+2t+2y)=f(2f(x)+2y+2k)$
we get $t=k$ but
$f(z)-f(x)=t=k=2z-2x-f(z)+f(x)$ so
$f(z)-z=f(x)-x$
which gives $f(x)=x+c$
putting in the relation we get $c=0$
.
.
@below
you can't put $y=0$ in the equation

This post has been edited 2 times. Last edited by Mr.C, Oct 10, 2020, 3:31 PM

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#6 h Apr 29, 2025, 7:58 AM

Y by

Mr.C wrote:

not bad i guess ... i had another solution which is not really nice , i think this is better if anyone intrested i can send the other aswell .
.
here is the easy one, let
$p(x,y)=f(2f(x)+2y)=f(2x+y)+y$
now
$p(x,y+f(z)-f(x))$ gives by letting $f(z)-f(x)=t$
$f(2x+t+y)+t=f(2z+y)$
so if $2z-2x-t=k$ we have
$f(y+k)=t+f(y)$ for big enough $y$
so let $x$ be big now from
$p(x+k,y)$ we have
$f(2f(x)+2t+2y)-f(2f(x)+2y)=2t$ so
$f(2f(x)+2t+2y)=f(2f(x)+2y+2k)$
we now prove $f$ is injective let
$f(a)=f(b)$ so from
$p(a,y)-p(b,y)$ we get
$f(2a+y)=f(2b+y)$ so $f$ is periodic let the period be $s$ we have from
$p(x,y+s)-p(x,y)$ that $s=0$ so
$a=b$
now going back at
$f(2f(x)+2t+2y)=f(2f(x)+2y+2k)$
we get $t=k$ but
$f(z)-f(x)=t=k=2z-2x-f(z)+f(x)$ so
$f(z)-z=f(x)-x$
which gives $f(x)=x+c$
putting in the relation we get $c=0$
.
.
@below
you can't put $y=0$ in the equation

How can you know that y+f(z)-f(x)>0?

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11352 posts

#7 h Apr 29, 2025, 5:27 PM • 1 Y

Y by Blackbeam999

Blackbeam999 wrote:

Mr.C wrote:

not bad i guess ... i had another solution which is not really nice , i think this is better if anyone intrested i can send the other aswell .
.
here is the easy one, let
$p(x,y)=f(2f(x)+2y)=f(2x+y)+y$
now
$p(x,y+f(z)-f(x))$ gives by letting $f(z)-f(x)=t$
$f(2x+t+y)+t=f(2z+y)$
so if $2z-2x-t=k$ we have
$f(y+k)=t+f(y)$ for big enough $y$
so let $x$ be big now from
$p(x+k,y)$ we have
$f(2f(x)+2t+2y)-f(2f(x)+2y)=2t$ so
$f(2f(x)+2t+2y)=f(2f(x)+2y+2k)$
we now prove $f$ is injective let
$f(a)=f(b)$ so from
$p(a,y)-p(b,y)$ we get
$f(2a+y)=f(2b+y)$ so $f$ is periodic let the period be $s$ we have from
$p(x,y+s)-p(x,y)$ that $s=0$ so
$a=b$
now going back at
$f(2f(x)+2t+2y)=f(2f(x)+2y+2k)$
we get $t=k$ but
$f(z)-f(x)=t=k=2z-2x-f(z)+f(x)$ so
$f(z)-z=f(x)-x$
which gives $f(x)=x+c$
putting in the relation we get $c=0$
.
.
@below
you can't put $y=0$ in the equation

How can you know that y+f(z)-f(x)>0?

since $f(z)-f(x)=t>0$

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