Balti Tee 2024 · Ülesanne 1

Let $P(x, y)$ denote the assertion of the given functional equation. Note that $P(1,0)$ is $f(1)=f(1)+f(0)$ which implies

$$f(0)=0$$

Applying this result to $P(x,-x)$ and $P(-x, x)$ where $x \neq 0$ we get:

$$\begin{aligned} & 0=(1-\alpha) x f(x)+x f(-x) \\ & 0=(\alpha-1) x f(-x)-x f(x) \end{aligned}$$

By adding (1) and (2) and simplifying, we get $0=\alpha x f(-x)-\alpha x f(x)$ which implies

$$f(x)=f(-x)$$

for all $x \neq 0$. Since $f(0)=0=f(-0)$, we can conclude that $f$ is even. Therefore (1) simplifies to

$$0=x f(x)(2-\alpha)$$

which implies that if $\alpha \neq 2$ then $f(x)=0$ for all $x \in \mathbb{R}$. It is easy to check that this function works. Now let us consider the case $\alpha=2$. The initial functional equation becomes

$$x f(x+y)=(x+2 y) f(x)+x f(y)$$

which can be rewritten as

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

Note that the right hand side is symmetric with respect to $x$ and $y$. From this we can deduce that $x f(x+y)-(x+y) f(x)=y f(x+y)-(x+y) f(y)$ where factorizing yields

$$(x-y) f(x+y)=(x+y)(f(x)-f(y)) .$$

By replacing $y$ with $-y$ and using the fact that $f$ is even, we get

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

Taking $x=\frac{z+1}{2}$ and $y=\frac{z-1}{2}$ in both (3) and 4, we get

$$\begin{aligned} f(z) & =z\left(f\left(\frac{z+1}{2}\right)-f\left(\frac{z-1}{2}\right)\right) \\ z f(1) & =f\left(\frac{z+1}{2}\right)-f\left(\frac{z-1}{2}\right) \end{aligned}$$

respectively. Equations (5) and (6) together yield $f(z)=z \cdot z f(1)=z^{2} f(1)$ which must hold for all $z \in \mathbb{R}$.

Thus, the only possible functions that satisfy the given relation for $\alpha=2$ are $f(x)=c x^{2}$ for some real constant $c$. It is easy to check that they indeed work.

Multiplying the given equation by $y$ gives

$$x y f(x+y)=(x+\alpha y) y f(x)+x y f(y)$$

which is equivalent to

$$x y(f(x+y)-f(x)-f(y))=\alpha y^{2} f(x)$$

The left-hand side of this equation is symmetric in $x$ and $y$. Hence the right-hand side must also stay the same if we swap $x$ and $y$, i.e.,

$$\alpha y^{2} f(x)=\alpha x^{2} f(y)$$

As $\alpha \neq 0$, this implies

$$y^{2} f(x)=x^{2} f(y)$$

Setting $y=1$ in this equation immediately gives $f(x)=c x^{2}$ where $c=f(1)$. Plugging $f(x)=c x^{2}$ into the original equation gives

$$c x(x+y)^{2}=c(x+\alpha y) x^{2}+c x y^{2}$$

where terms can be rearranged to obtain

$$c x(x+y)^{2}=c x\left(x^{2}+\alpha x y+y^{2}\right)$$

If $c=0$ then (7) is satisfied. Hence for every $\alpha$, the function $f(x)=0$ is a solution. If $c \neq 0$ then (7) is satisfied if and only if $\alpha=2$. Hence in the case $\alpha=2$, all functions $f(x)=c x^{2}$ (where $c \neq 0$ ) are also solutions.