A routine Functional Equation

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A routine Functional Equation

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Source: Iran TST 2007, Day 1

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#1 h May 5, 2007, 5:38 PM • 7 Y

Y by tenplusten, Adventure10, Mango247, and 4 other users

Find all solutions of the following functional equation: $f(x^{2}+y+f(y))=2y+f(x)^{2}.$

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#2 h May 5, 2007, 6:36 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

See here http://www.mathlinks.ro/Forum/viewtopic.php?p=18289#18289

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#3 h Sep 29, 2009, 9:58 AM • 2 Y

Y by Adventure10, Mango247

Here's my Ugly Proof... Let given functional eq as T-eq

We know $f$ is surjective cuz for given $x$ , $2y$ moves all in Real number.

Let $f(a) = 0$ for some $a$ , take $y = 0$ in T-eq then $f(x^2 + f(0)) = f(x)^2$ ,

Thus $f(x)^2 = f(x^2 )$ , so $f( - a) = 0$

Take $x = 0, y = a, - a$ , we get $f(a) = 2a + f(0)^2$ , $f( - a) = - 2a + f(0)^2$ so $a = 0$

Here, We get $f(x) = 0 \Longleftrightarrow x = 0$

Now put $y = - \frac { f(x)^2}{2}$ , we get $f(x^2 - \frac { f(x)^2}{2} + f( - \frac { f(x)^2}{2})) = 0$

which means $x^2 - \frac { f(x)^2}{2} + f( - \frac { f(x)^2}{2}) = 0$ Thus $f(p) = f(q) \Longrightarrow p^2 = q^2$

If $f(p) = f( - p) = r$ , where $p > 0$

put $y = 0$ in T-eq, $f(x^2 ) = f(x)^2$ Thus $f(x) \ge 0$ when $x \ge 0$ , Thus $r \ge 0$

put $y = p, - p$ in T-eq, we get $f(x^2 + p + r ) = 2p + f(x^2)$ , $f(x^2 - p + r) = - 2p + f(x^2 )$

Put $x = 0$ in first one, $f(p + r) = 2p$ , put $x = \sqrt {p + r}$ in Second one, we get $f(2r) = - 2p + f(p + r) = 0$ so $r = 0$ means $p = 0$

Therefore, $f(p) = - f( - p)$

Thus $f(p) = f(q)$ means $p^2 = q^2$ so $p = q$ which means $f$ is injective

Now, take $x = 0$ in T-eq, we get $f(y + f(y)) = 2y$ . We know $f$ is bijective.

Thus $y + f(y)$ moves all of real numbers. So we can set $y + f(y) = t$ with $\forall t \in R$

Let's reconsider about T-eq, we get $f(x + t) = f(x) + f(t)$ with $x\ge 0$

Then $f(x + y) = f(|x| + x + y - |x|) = f(|x| - x) + f(y - |x|) = f(|x|) + f(x) - f(|x| - y)$

$= f(x) + f(y)$ Thus $f(x + y) = f(x) + f(y)$ Thus $f(q) = kq$ for $q \in Q$

Then $\forall \alpha \ge 0$ , $f(x + \alpha) = f(x) + f(\sqrt {\alpha})^2 \ge f(x)$ which means $f$ in increasing function.

$\forall x \in R$ , let two rational sequence $(a_n), (b_n)$ which are converge to $x$ with $a_n < x$ , $b_n > x$ for all $n in N$

Then $f(a_n ) = k a_n \le f(x) \le f(b_n ) = k b_n$ , so $f(x) = kx$ for all $x \in R$

cuz $f(x^2 ) = f(x)^2$ , $f(1) = 1$ . so we get $f(x) = x$ for all $x \in R$

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#4 h Apr 29, 2014, 3:15 PM • 2 Y

Y by Adventure10, Mango247

we have $f$ is surjective and $f(0)=0$ , $f(y+f(y))=2y$ and $f(x^2)=f(x)^2$ . We can $f(a)=0$ $\Rightarrow$ $a=0$ . So $g(x)=f(x)+x$ is surjective in $g:(-\infty ,0]\rightarrow (-\infty, 0]$ . For all $z\le 0$ , $x\in R$ , we have $f(x+z)=f(x)+f(z)$ and $f(x)=-f(-x)$ - odd function. So $g$ is surjective in all $R$ . For all $x,y\ge 0$ we get that $f(x+y)=f(x)+f(y)$ and $f$ is monotone increasing.

Answer: $f(x)=x$ .

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#5 h Jul 7, 2014, 4:48 AM • 2 Y

Y by Tawan, Adventure10

$P(x,y)$ and $P(-x,y)$ give us $f(x)^{2}=f(-x)^{2}$
So $f(x)=f(-x)$ or $f(x)=-f(-x)$

Suppose that $f(0)\neq 0$ ,
If there exist a $t$ satisfy $f(t)=-f(-t)$
then $P(0,t)$ and $P(0,-t)$ give us $f(0)=0$ , a condradiction.
Since $f(x)^{2}=f(-x)^{2}$ , $f$ is even.
If there exist $p$ s.t. $f(p)\leq 0$ and $p\neq 0$
We can get that $p=0$ from $P(\sqrt{-f(p)},p)$ and $P(\sqrt{-f(p)},-p)$
So $f(x)>0$ for all non-zero real number $x$
a contracdition with $P(x,-y)$ for a sufficient large $y$

So $f(0)=0$ , and it's easy to get $f(x^{2})=f(x)^{2}$ and $f(x+y)=f(x)+f(y)$
Hence $f(x)=x$

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#6 h Nov 19, 2017, 6:29 PM • 4 Y

Y by Anar24, Ferid.---., Adventure10, Mango247

Here is a solution by my friend and teammate RAMİL JABİYEV (İ like his fe solutions that is why I post here)

Claim 1: $f (a)=0$ $\Longrightarrow$ $a=0$
Proof:Just see $P (0,a)$ and $P (0,-a)$ (Note:Here we used that $f (x)^2=f (-x)^2$ ) Then done...

Claim 2: $f (f (x))+f (x)=2x$
Proof:Firstly lets look at $P (x,-\frac {f (x)^2}{2})$
which gives us $x^2-\frac {f (x)^2}{2}-f (-\frac {f (x)^2}{2})=0$ .Then $P (0,-\frac {f (x)^2}{2})$ yields that $f$ is odd.Now $P (x,-x^2)$ gives the desired relation .

Claim 3: $f(2x)=2f (x)$
Proof:Using the above claim it is nothing

Claim 3: $f (x)=x$ is a solution
Proof: $P(x,f (y))$ and using the fact that $f$ is odd we conclude that $f$ is additive and since $f$ is bounded in some interval we get $f (x)=x$ for any $x\in R$ .

This post has been edited 2 times. Last edited by tenplusten, Nov 19, 2017, 6:51 PM

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#7 h Nov 19, 2017, 6:35 PM • 3 Y

Y by tenplusten, Tawan, Adventure10

tenplusten wrote:

... and since $f$ is bounded we get $f (x)=x$ for any $x\in R$ .

Quite funny

"since $f(x)$ is bounded ...", we get ... an unbounded function ....
Do you think this sentence is quite consistant ?

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#8 h Nov 19, 2017, 6:47 PM • 3 Y

Y by Tawan, Adventure10, Mango247

$f (x^2)=f (x)^2\ge 0$ .So f is bounded in some interval

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#9 h Nov 20, 2017, 8:46 AM • 4 Y

Y by tenplusten, Tawan, Adventure10, Mango247

tenplusten wrote:

$f (x^2)=f (x)^2\ge 0$ .So f is bounded in some interval

According to me, each word is important in maths statements.
And "f is bounded" is not equivalent to "f is lowerbounded over $\mathbb R^+$ "

With this last sentence, I do agree.

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#10 h Nov 28, 2022, 9:30 PM

Y by

Alternative ending:
After we establish that $2x = f(x) + f(f(x))$ we set $a_n = n_{th}$ iteration of $f$ . Then, we get that
$a_n + a_{n-1} = a_{n-2}$ and solving the recurrence we find that $a_n = l + m(-2)^n$ .
But $f$ is lowerbounded in positive reals and therefore we must have $m=0$ and $a_n = l$ which is constant.
Thus, $f(f(x))=f(x)$ and due to surjectivity we have $f(x)=x$ in positive reals.
Now choose $x$ large enough s.t. $x^2 + y + f(y) > 0$ and from the initial relation we get that $f(y)=y, \forall y \in \mathbb{R}$

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