(USA Winter TST 2012) Determine all functions $f:\mathbb{R}\to \mathbb{R}$ so that for every pair of real numbers $x$ and $y, f(x+y^2) = f(x)+|yf(y)|$
I've managed to come up with the work below. How can I solve this question?
Obviously the zero function works. If $f$ is constant, then $f$ must be identically equal to zero, so assume f is nonconstant. Suppose $f(x) = ax+b$ for some real constants a and b. Then the given equation simplifies to $a(x+y^2) +b = ax+b + |y(ay+b)| \Rightarrow ay^2 = |ay^2 + by|$ for all real numbers $y$. In particular, we must have $a > 0$ ($f$ is not constant by assumption and if $a < 0$, then the LHS is negative while the RHS is nonnegative for any nonzero y). For $y$ large enough, $ay^2 + by > 0$ and so the RHS equals $ay^2 + by$ for infinitely many y. Equating the coefficients (possible since (single-variable) polynomials that agree on infinitely many points are identically equal), we get that $b=0.$ Hence $f(x) = ax$ is a solution for any $a > 0.$ Now suppose $f$ is not linear. It could be useful to determine whether such a function is surjective and/or injective, as some useful equations can usually be derived from these results. It could also be useful to determine whether $f$ is odd or even or whether there exists some $x$ with $f(x) = 0$ (the latter can be useful if one cannot show $f$ is surjective easily for instance). Substituting $x=0$ gives $f(y^2) = f(0) + |yf(y)|.$ Substituting $x=-y^2$ gives $f(0) = f(-y^2)+|yf(y)|.$ Then $f(0) = f(y^2) - |yf(y)| = f(-y^2) + |yf(y)|$ for all real numbers y, which implies that for all real y, $f(y^2)-f(-y^2) = 2|yf(y)|$ for all real numbers $y$. If f is odd, then this implies that $f(y^2) = |yf(y)|$ for all real numbers $y$, so $f(x)\ge 0$ whenever $x\ge 0$ and $f(x) \leq 0$ whenever $x<0.$ Substituting $-y$ for $y$ in the original equation yields $f(x)+|-yf(-y)| = f(x+y^2) = f(x)+|yf(y)|$. Hence $|f(y)| = |f(-y)|$ for all real numbers $y\neq 0.$ Suppose there exists $y\neq 0$ so that $f(y) = 0.$ Then $f(x+y^2) = f(x)$ for all real numbers x, so inductively $f(x + ny^2) = f(x)$ for all $n \in \mathbb{Z}, x \in \mathbb{R}.$