This problem is from my textbook:
Given : $f:\mathbb R\to\mathbb R$
Solve this functional equation :
$f(x) = f(x + y^2 + f(y))$
I think this function is just a simple constant, so I try all my best to prove this, but no result. I'm not sure this way is wrong, or not. Can give me a hint ?
Thanks :)
On
Let $f(0)=k$. Then you get $f(x) = f(x + k)$ implying that $f$ is periodic with period $k$. So, its enough to define $f : [0,k) \to \mathbb{R}$.
On
Let $\Pi$ be the group of all periods of $f$. Then for all $y \in \mathbb{R}$ we have $ y^2+f(y) \in \Pi$. Let $x \in \Pi$. Then $\Pi \ni (y+x)^2+f(x+y)=(y+x)^2+f(y)$. $\Pi$ is a group, so $2xy+x^2=(y+x)^2+f(y)-y^2-f(y) \in \Pi$. Thus if there exists nonzero $x \in \Pi$ then $\Pi=\mathbb{R}$, and $f$ is constant. If $\Pi=\{0\}$ then for all $y$ we have $y^2+f(y) \in \Pi$ so $f(y)=-y^2$.
If $f$ is differentiable, we can hold $x$ constant and differentiate the functional equation with respect to $y$. This leads pretty quickly to the result that the constant functions and $-x^2$ are the only differentiable solutions to the equation.
If $f$ is allowed not to be differentiable, some very different approach is needed (assuming the problem is even tractable, which it might not be).
ETA: Let $f$ be a piecewise continuous solution to the equation. Suppose there is some $y$ such that $f(y) \neq y^2$; then $f$ is periodic with period $y^2+f(y)$. As $f$ is non-constant, periodic, and piecewise continuous, it has a discrete set of periods; that is to say, the set $\{y^2+f(y)|y \in \Bbb{R}\}$ is discrete. Since $f$ is piecewise continuous, so is $y^2+f(y)$; thus $y^2+f(y)$ is piecewise constant, and a multiple of the minimum period.
This implies that (unless $f$ is constant or equal to $-y^2$) $f$ is periodic with some minimal period $P$, and also everywhere equal to $-y^2+nP$ for some integer $n$. But this is absurd; no two of the functions $-y^2+nP$ are horizontal translates of each other, nor are any of them periodic themselves.