Find all $f:R-R$ such that $f(x^2+y+f(y))=2y+f(x)^2$
Problem
Source: Bulgaria TST 2003 P2
Tags: algebra, functional equation, algebra proposed
RaleD
11.09.2012 21:50
we calculate possible values of $f(0)$
put 0, 0 $f(f(0))=f(0)^2$
put 0, f(0) $f(f(0)+f(f(0)))=2f(0)+f(0)^2$
put f(0), 0 $f(f(0)^2+f(0))=f(f(0))^2=f(0)^4 $ (1)
from (1) we have f(0)=0 or 1<f(0)<2. Suppose second f(0)=t, because f is surjective for some $x_1$ $f(x_1)=0$ . If we put $x_1$ , 0 in given equation (call it *) we have $f(x_1^2 +t)=0$ now we see there is arbitrary big positive number$x_i$ such it's image is 0. But if put $y>0$ , and then x such $x^2 + y + f(y)=x_i$ we get contradiction.
We have $f(x^2)=f(x)^2$ (2) if $f(x_i)=0$ then also $f(x_i^2)=0$. if $x_i$ isn't zero then there are at least two positive numbers such their image is zero. let $x_2 < x_1$ be some two such numbers, put $y=x_2$ and $x_1= x^2 + x_2$ in (*) and we get contradiction. So f(x)=0 if and only if x=0.
Suppose $f(a)=f(b)$ if we put $a, \frac{- f(a)^2}{2} $ and then $b, \frac{- f(a)^2}{2} $ we can conclude $a^2 = b^2$. Let $m>0, m^2=a, -m^2=b$, and put $m, -m^2$ in (*) . We have $ f(f(-m^2))= f(m)^2-2m^2 $ and $f(f(m^2))=f(f(m))^2$ we have $f(m)^2-2m^2> 0$. Now $f( -m^2 + f(m^2))=-2m^2 <0 $ but $ f(m^2)> 2m^2 $ meaning that image of positive is less than zero; contradiction with (2).
We have f is injective so because $ f(y + f(y))=2y$ we have $ y+f(y)$ is bijective. (it is obviously injective, any number $p$ we can get by $y= p/2$ because of injection of f. Now replace $z=y+f(y)$ in * and we get $f(x^2+z)=f(x^2)+f(z)$ and by Cauchy ($f(x)$ is positive when $x$ is positive) and checking we get $f(x)=x$.
It may contain mistake(s).
hyperbolictangent
12.09.2012 01:35
Your proof of injectivity is very nice, however I think it is not so easy to go from $f(x^2 + z) = f(x^2) + f(z)$ to $f(x) = x$.
$f(x^2 + z) = f(x^2) + f(z)$ for all $x, z$. Also, since $f$ is injective and $f(x)^2 = f(x^2)$, $f(x) = -f(-x)$ for all $x$. Then for $a, b < 0$, \[f(a + b) =-f((-a) + (-b)) = -[f(-a) + f(-b)] = [-f(-a)] + [-f(-b)] = \] \[ f(a) + f(b)\] So $f(a + b) = f(a) + f(b)$ for all $a, b$.
While this is Cauchy's functional equation, we are working in the reals, so this is not enough to say that all solutions are of the form $f(x) = xf(1)$. However, we know that for $a \ge 0$, $f(a) = f(\sqrt{a})^2 \ge 0$ which is one of the ways to ensure all solutions are linear. It is trivial from here to check that $f(x) = x$ is the only one that works.
Also, here is an easier proof that $f(0) = 0$ and that it is the only zero of $f$.
Since $f$ is surjective (look at the right side of the equation), $f$ has a zero, call it $z$. Also, since $f(x^2) = f(x)^2$, $f(-z) = 0$ as well. Plugging in $(x, y) = (0, z)$ gives \[f(0 + z + f(z)) = f(z) = 0 = 2z + f(0)^2\] while on the other hand $(0, -z)$ gives \[f(0 - z + f(-z)) = f(-z) = 0 = -2z + f(0)^2\] meaning $2z = -2z$ and $z = 0$.
cmtappu96
03.01.2013 07:21
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=464139&p=2601448&hilit=imo+tst#p2601448
ryuzaki
20.01.2013 18:15
RaleD , you have to prove the monotonicity , first
toto1234567890
04.05.2015 17:45
There is a real number $ a $ . Find all $ f:\mathbb{R^{+}} \cup$ {0} $ \rightarrow \mathbb{R^{+}}\cup $ {0} that satisfie $ f(x^a+y+f(y))=ay+f(x)^a $ .