Scary Functional Equation

by utkarshgupta, Nov 8, 2015, 4:49 PM

Problem : (ISL 2007)
Find all functions $f: \mathbb{R}^{ + }\to\mathbb{R}^{ + }$ satisfying $f\left(x + f\left(y\right)\right) = f\left(x + y\right) + f\left(y\right)$ for all pairs of positive reals $x$ and $y$ . Here, $\mathbb{R}^{ + }$ denotes the set of all positive reals.

Solution 1 :
Let $f(x)=g(x)+x$
the problem statement can be rewritten as
$g(x+y+g(y)) = g(x+y)+y$ Also,

$f(y) \ge y$
Suppose $f(y)<y$ for some $y$
$P(y-f(y),f(y)) \implies f(2y-f(y))=0$ a contradiction.

$\implies g$ is a well defined function from $\mathbb{R}^+ \to \mathbb{R^+}$

Let $P(x,y)$ be the assertion $g(x+y+g(y)) = g(x+y)+y$
Using $P(x,y)$ and $P(y,x)$ we have,
$g(x+y+g(y))+x = g(x+y+g(x))+y$ Let $Q(x,y)$ be the assertion $g(x+y+g(y))+x = g(x+y+g(x))+y$

$P(x,g(y)) \implies g(x+g(y)+g(g(y))) = g(x+g(y))+g(y)$
Let $z = 2x+2g(y)+g(g(y))$

$Q(x+g(y)+g(g(y)),x+g(y)) \implies$
$g(z+g(x+g(y)))+x+g(y)+g(g(y)) = g(z+g(x+g(y)+g(g(y)))+x+g(y)$ $\implies g(z+g(x+g(y)+g(g(y)))) - g(z+g(x+g(y))) = g(g(y))$ Using $P(x,g(y))$
$\implies g(z+g(x+g(y))+g(y)) - g(z+g(x+g(y))) = g(g(y))$
Choosing $x >y$ , we have $z > 2x >y$
$P(z+g(x+g(y))-y, y)$
$\implies g(z+g(x+g(y))+g(y)) = g(z+g(x+g(y))) + y$ $\implies g(g(y))=y$
Now the problem should be easy (I will add soon)

Solution 2 :
Let $P(x,y )$ be the assertion $f\left(x + f\left(y\right)\right) = f\left(x + y\right) + f\left(y\right)$

Lemma 1 $f(y) \ge y$
Suppose $f(y)<y$ for some $y$
$P(y-f(y),f(y)) \implies f(2y-f(y))=0$ a contradiction.
Thus the lemma

Lemma 2 : $f$ is injective.
Let $f(a)=f(b)$ and without loss of generality, $a>b$
$P(f(x),y) \implies$
$f(f(x)+f(y)) = f(x+y) + f(x) + f(y)$ $\implies f(a+y) = f(b+y)$
That is $f$ is periodic in the interval $(a, \infty)$ .
But this can't be true because of lemma 1.
Hence $f$ is injective.

Main Proof :
$P(x,y) \implies f(x+f(y))=f(x+y)+f(y)$ $P(y,x) \implies f(y+f(x))=f(x+y)+f(x)$ $\implies f(x+f(y))-f(y+f(x))=f(y)-f(x)$
Let $y>x$
$P(x,y-x) \implies f(x+f(y-x))=f(y)+f(y-x)$
$\implies f(x+f(y))+f(x)-f(y+f(x))=f(y)=f(x+f(y-x))-f(y-x)$ $\implies f(x+f(y))+f(x)+f(y-x)=f(x+f(y-x))+f(y+f(x))$
$P(f(y),f(x)) \implies f(f(y)+f(x))=f(x+f(y)+f(x)$
Using this,
$f(f(x)+f(y))+f(y-x) = f(x+f(y-x))+f(y+f(x))$
Obviously, by Lemma 2, $f(x)+f(y)>y-x$
$P(f(x)+f(y)+x-y,y-x) \implies f(f(x)+f(y)+x-y+f(y-x))=f(f(x)+f(y))+f(y-x)$
Thus we have,
$f(f(x)+f(y)+x-y+f(y-x))= f(x+f(y-x))+f(y+f(x))$

Case 1 $x+f(y-x)>y+f(x)$
$P(x+f(y-x)-y-f(x),y+f(x)) \implies f(x+f(y-x))+f(y+f(x))=f(x+f(y-x)-y-f(x)+f(y+f(x)))$ $\implies f(x+f(y-x)-y-f(x)+f(y+f(x)))=f(f(x)+f(y)+x-y+f(y-x))$ Using injectivity,
$x+f(y-x)-y-f(x)+f(y+f(x))=f(x)+f(y)+x-y+f(y-x)$ $\implies f(y+f(x))=2f(x)+f(y)$ $\implies f(x+y)=f(x)+f(y)$ $\implies f$ is additive.
Replacing $y$ by $x+a$ , we get
If $f(a)>a+f(x)$
$\implies f(x)+f(x+a)=f(x+2a)$
$\implies f(x+f(a))=f(x+2a)$
$\implies f(a)=2a$

Case 2 $x+f(y-x)<y+f(x)$
$P(y+f(x)-x-f(y-x),x+f(y-x)) \implies f(x+f(y-x))+f(y+f(x))=f(y+f(x)-x-f(y-x)+f(x+f(y-x)))$ $\implies f(y+f(x)-x-f(y-x)+f(x+f(y-x)))=f(f(x)+f(y)+x-y+f(y-x))$ Injectivity implies,
$y+f(x)-xf-f(y-x)+f(y)+f(y-x)=f(x)+f(y)+x-y+f(y-x)$ $\implies f(y-x)=2(y-x)$ Replacing $y$ by $x+a$ ,
$\implies f(a)=2a$

For each $a$ we can find such an $x$ such that either of the two conditions hold.
Suppose we cannot find such an $x$
$\implies x+f(a)=x+a+f(x)$
$\implies f(x)=f(a)-a$
$\implies$ $f$ is a constant function.
A contradiction.

The only solution is
$\boxed{f(x)=2x}$

Solution 3 :

We can establish injectivity as in the above solution.

Let $P(x,y)$ be the assertion $f(x+f(y))=f(x+y)+f(y)$
$P(x,x+f(z)) \implies f(x+f(x+f(z)))=f(x+x+f(z))+f(x+f(z))$ $\implies f(x+f(x+z)+f(z))=f(2x+f(z))+f(x+f(z))$ $\implies f(x+f(x+z)+f(z))=f(2x+z)+f(z)+f(x+z)+f(z)$
Now observe $P(x+f(z),x+z)$
$\implies f(x+f(z)+f(x+z))=f(2x+z+f(z))+f(x+z)$ $\implies f(x+f(z)+f(x+z))=f(2x+2z)+f(z)+f(x+z)$
That is we have
$f(2x+z)+f(z)+f(x+z)+f(z)=f(2x+2z)+f(z)+f(x+z)$ $\implies f(2x+2z)=f(2x+z)+f(z)$ $\boxed{f(x+y)=f(x)+f(y)}$
That is $f$ is additive.

$P(x,x) \implies f(x+f(x))=f(2x)+f(x)$ $\implies f(x)+f(f(x))=f(2x)+f(x)$ $\implies f(f(x))=f(2x)$ $\implies \boxed{f(x)=2x}$

This post has been edited 4 times. Last edited by utkarshgupta, Dec 17, 2017, 12:53 PM

functional equation

ISL

2007

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3 Comments

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Wow. The length :O

by TheOneYouWant, Nov 18, 2015, 11:27 AM

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This must be incorrect.

Put x=0,
$f(f(y)) = 2 f(y)$
Let $f(y) = t$
$f(t) = 2t$

by kgdpsdurg, Dec 31, 2015, 7:25 PM

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Of course @above

The domain is positive reals and surjectivity has not yet been established.

by anantmudgal09, Jan 3, 2016, 1:41 PM

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Here goes first post of 2025! Great blog.
by math_holmes15, Jan 14, 2025, 8:53 AM
First post of 2024
by Yiyj1, Feb 8, 2024, 5:40 AM
First post of 2023
by HoRI_DA_GRe8, Jul 22, 2023, 7:45 AM
Nice blog ! Your isogonality lemma is really powerful !
by 554183, Oct 14, 2021, 8:55 AM
Post plss....
by samrocksnature, Apr 11, 2021, 10:12 PM
alas,this is ded
by Hamroldt, Mar 18, 2021, 4:13 PM
Thanks for the nice blog.
by Feridimo, Mar 6, 2020, 4:17 PM
I think this might be silly but ... when should we expect to have another post ?? I am very keen to see it
by gamerrk1004, Nov 4, 2019, 4:54 PM
Let's all echo what's written in the blog description - Stay Insane / 'Cause it's your labor, will and pain/ That takes you to the top of soda fountain
by Kayak, Oct 2, 2017, 7:18 PM
hey utkarsh jee is over now ... continue your elementary blog pleaseeeeeee!
by kk108, Jun 17, 2017, 11:19 AM
Congrats on becoming a contest moderator!
by Ankoganit, Mar 9, 2017, 5:22 AM
INTERSTING BLOG
by kk108, Feb 19, 2017, 2:04 PM
I have no plans for this blog right now....
No time here people !
But lets see....
I may try some combinatorics
by utkarshgupta, Feb 15, 2017, 12:47 PM
Thanks for the nice blog!
by Orkhan-Ashraf_2002, Feb 13, 2017, 6:34 PM
Revive it!!!
Best blog out there, for sure!
by rmtf1111, Jan 12, 2017, 6:02 PM
Oh you certainly will..
Bu I don't think I will
by achilles04, Feb 11, 2015, 6:24 PM
Thanks
Hope you are there too ...

And hope I get selected
by utkarshgupta, Feb 11, 2015, 11:14 AM
Nice blog ..
keep it up and you might one day become a mathematician lIke me
( just joking )
Btw don't forget us after going to imotc.
by achilles04, Feb 10, 2015, 4:49 PM
nice blog!!!!!
by pradhit, Feb 7, 2015, 11:51 AM
No ideas about the cut off but should be less than 60 according to me...
So you never know
by utkarshgupta, Feb 6, 2015, 4:06 PM
Hi utkarsh,
what do u expect the cutoff to be this year??

by kgdpsdurg, Feb 6, 2015, 12:41 PM
Hi utkarsh
How many did you clear in INMO 2015?
by moldevort, Feb 1, 2015, 3:59 PM
@ Anonymous Bunny
If we both (that should have included me) are lucky enough !
by utkarshgupta, Sep 17, 2014, 2:44 AM
Nice blog. Great job!

If I am lucky enough, hopefully we shall meet at IMOTC.
by AnonymousBunny, Sep 16, 2014, 8:08 PM
Thanx all !!

If anyone wish to contribute anything just pm !

I will add u to the contributors' list !
by utkarshgupta, Sep 16, 2014, 6:35 AM
doing good progress!! btw nice blog.
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Nice blog.
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Hi Utkarsh! Wonderful blog. Keep it up.
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I need some shouts and characters

by utkarshgupta, Jul 20, 2014, 4:22 AM

48 shouts

Elementary

Scary Functional Equation

by utkarshgupta, Nov 8, 2015, 4:49 PM

3 Comments