Find all solutions to $f\left(x^2+xf(y)\right)=xf(x+y)$

Question

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$f\left(x^2+xf(y)\right)=xf(x+y)$$ for all $x,y\in\mathbb{R}$.

This is somewhat related to this question, but with an $xf(y)$ term instead of $yf(x)$. I've managed to get $f(0)=0$, $f\left(x^2\right)=xf(x)$ and $f$ is odd, but not much more. I think that $f(x)=x$ and $f(x)=0$ are the only solutions, but I'm not sure.

Answer 1

The function defined by $f(x) = 0$ for all $x \in \mathbb{R}$ is a solution to the functional equation.

Suppose that $f$ is some other solution. Then there is a real number $c$ such that $f(c) \neq 0$.

For any $r \in \mathbb{R}$, let $x = \frac{r}{f(c)}$, and let $y = c - x$. The functional equation then gives us that $$ f \left( x^2 + xf(y) \right) = xf(x + y) = xf(c) = \frac{r}{f(c)} \cdot f(c) = r $$ showing that $r \in f(\mathbb{R})$. i.e. The function is surjective.

As you noted, we have that $f(0) = 0$, which follows by taking $x = y = 0$ in the functional equation.

Now let $x = -f(y)$ in the functional equation. We obtain that $$ 0 = f(0) = f \left( f(y)^2 - f(y) f(y) \right) = -f(y) f(y - f(y)) $$ for all $y \in \mathbb{R}$.

Then for any $y \in \mathbb{R}$, we have that if $f(y) = 0$, then $f(y - f(y)) = f(y) = 0$, and if $f(y) \neq 0$, then $-f(y) f(y - f(y)) = 0$ implies that $f(y - f(y)) = 0$. We thus have that $$ f(y - f(y)) = 0 $$ for all $y \in \mathbb{R}$.

Also as noted by you, we have that $f(x^2) = xf(x)$ by taking $y = 0$ in the functional equation.

Now for any real number $x$, we showed earlier that there is some $y \in \mathbb{R}$ such that $f(y) = x$.

We then have that $$ xf(x) = f(y) f(f(y)) = f(f(y)^2) = f \left( f(y)^2 + f(y) f \left( y - f(y) \right) \right) $$ using the fact that $f(y - f(y)) = 0$. The functional equation then gives us that $$ xf(x) = f(y) f \left( f(y) + y - f(y) \right) = f(y) f(y) = x^2 $$ holds for all $x \in \mathbb{R}$. For $x = 0$, we already have that $f(x) = 0 = x$. If $x \neq 0$, then we can divide the previous relation by $x$ to find again that $f(x) = x$. We thus conclude that $f(x) = x$ for all $x \in \mathbb{R}$.

Answer 2

Since you've allready know that, I'll continue from $f(0) = 0$ and $xf(x)=f(x^2)$

first, see that for every $x\ne0$: $$-xf(-x)=f\left(\left(-x\right)^2\right)=f(x^2)=xf(x)$$ $$f(x)=-f(-x)$$ Now, for every $a>0$: $$f(a)=\sqrt{a}f(\sqrt{a})$$ which means: $$f(a)=a^{\frac{1}{2}}f(a^{\frac{1}{2}})=a^{\frac{1}{2}}\cdot a^{\frac{1}{4}}f(a^{\frac{1}{4}})=a^{\frac{1}{2}}\cdot a^{\frac{1}{4}}\cdot a^{\frac{1}{8}}f(a^{\frac{1}{8}})=\dots=a^{\frac{2^n-1}{2^n}}f(a^{\frac{1}{2^n}})$$ therefore: $$f(a)=\lim_{n\rightarrow\infty}{a^{\frac{2^n-1}{2^n}}f(a^{\frac{1}{2^n}})} = \lim_{n\rightarrow\infty}{a^{\frac{2^n-1}{2^n}}}\cdot\lim_{n\rightarrow\infty}{f(a^{\frac{1}{2^n}})}=af\left(\lim_{n\rightarrow\infty}{a^{\frac{1}{2^n}}}\right)=af(1)$$ which means basicly that theres $m\in\Bbb{R}$ such that for every $a>0$, $f(a)=ma$. But now, for every $a<0$: $$f(a)=-f(-a)=-m\cdot(-a)=ma$$ therefore $f(a)=ma$ for every $x\in\Bbb{R}$