Find all function $f \colon \mathbb R \to \mathbb R$ such that $f(xf(y)) + f(yf(x)) = 2xy, \forall x, y \in \mathbb R$.

Question

Find all function $f \colon \mathbb R \to \mathbb R$ such that $$\large f(xf(y)) + f(yf(x)) = 2xy, \forall x, y \in \mathbb R$$

Me, thinking about things without noticing the lesson: I wonder what is the hardest problems about function equations?

My maths teacher: Let's move on the next problem which you can see on the board there, Thành Đạt!

30 minutes later...

Everyone in class: Well, we can calculate the value _$f(0)$ and prove that $f(xf(x))$ equals to-

That one kid who is certainly better than everyone (and is studying advanced calculus while you haven't started on the basics yet): Also, it can also be proven that $f(1) = \pm 1$.

(Everyone knows that.)

My maths teacher: Actually, this problem has only been proposed yesterday so there hasn't been a solution yet. Let us work this out together.

After 1 hour of figuring out what we are doing with our lives...

My maths teacher: That there's the answer, $f(x) = \pm 1, \forall x \in \mathbb R$, and don't write that down in the actual test if there is a problem in a competition that is similar to this. I've taken a shortcut just because-

Me: I want to go home.

My maths teacher: Thank you for finishing my sentence but please don't interrupt me, again.

So I have provided the solution below, please help me with coming up with a shorter solution.

Answer 1

It is evident that $f(0) = 0$

Let $P(x, y)$ be the assertion of $(x, y)$ in $f(xf(y)) + f(yf(x)) = 2xy$.

For $P(x, x)$, we have that $f(xf(x)) = x^2, \forall x \in \mathbb R$.

$u$ and $v$ are numbers such that $f(u) = f(v)$. For $P(u, v)$, we have that $$2uv = f(uf(v)) + f(vf(u)) = f(uf(u)) + f(vf(v)) = u^2 + v^2 \implies u = v$$

$\implies f$ is an injective function.

Replacing $x$ by $-x$ into $f(xf(x)) = x^2$, we have that $f(-xf(-x)) = x^2, \forall x \in \mathbb R$

$$f(xf(x)) = f(-xf(-x)) \iff xf(x) = -xf(-x) \implies -f(x) = f(-x), \forall x \in \mathbb R, x \ne 0$$

Besides, $-f(0) = f(-0) (= 0) \implies f$ is an odd function.

For $P(1, 1)$ and $P(f(1), f(1))$, we have that $f^2(1) = 1$ and $f(f(1)f^2(1)) = [f(1)]^2$

$$ \implies f(f(1) \cdot 1) = [f(1)]^2 \iff 1 = [f(1)]^2 \iff f(1) = \pm 1$$

In the case of $f(1) = 1$, for $P\left(x, \dfrac{1}{x}f\left(\dfrac{1}{x}\right)\right)$, what would happen is $$\begin{align} &f\left(xf\left(\frac{1}{x}f\left(\frac{1}{x}\right)\right)\right) + f\left(\frac{1}{x}f\left(\frac{1}{x}\right)f(x)\right) = 2f\left(\frac{1}{x}\right)\\ \iff &f\left(x \cdot \frac{1}{x^2}\right) + f\left(\frac{1}{x}f\left(\frac{1}{x}\right)f(x)\right) = 2f\left(\frac{1}{x}\right)\\ \iff &f\left(\frac{1}{x}f\left(\frac{1}{x}\right)f(x)\right) = f\left(\frac{1}{x}\right) \iff \frac{1}{x}f\left(\frac{1}{x}\right)f(x) = \frac{1}{x}\\ \iff &f\left(\frac{1}{x}\right) = \dfrac{1}{f(x)}, \forall x \in \mathbb R, x \ne 0 \end{align}$$

Furthermore, $f^2\left(\dfrac{1}{x}\right) = f\left(\dfrac{1}{f(x)}\right) = \dfrac{1}{f^2(x)}, \forall x \in \mathbb R, x \ne 0$.

For $P(x, 1)$ and $P\left(\dfrac{1}{x}, 1\right)$, we can, respectively, see that $f(x) + f^2(x) = 2x, \forall x \in \mathbb R, x \ne 0$ and $f\left(\dfrac{1}{x}\right) + f^2\left(\dfrac{1}{x}\right) = \dfrac{2}{x} \iff \dfrac{1}{f(x)} + \dfrac{1}{f^2(x)} = \dfrac{2}{x} \iff \dfrac{f(x) + f^2(x)}{f(x)f^2(x)} = \dfrac{2}{x}$

$ \iff \dfrac{2x}{f(x)f^2(x)} = \dfrac{2}{x} \iff f(x)f^2(x) = x^2, \forall x \in \mathbb R, x \ne 0$

$\iff f(x) = f^2(x) = x, \forall x \in \mathbb R, x \ne 0$

However, $f(0) = 0 \implies f(x) = x, \forall x \in \mathbb R$

In the case of $f(1) = -1$, for $P\left(-\dfrac{1}{x}, -xf(-x)\right)$, what would happen is $$\begin{align} &f\left(-\dfrac{1}{x}f\left(-xf(-x)\right)\right) + f\left(-xf(-x)f\left(-\dfrac{1}{x}\right)\right) = 2f(-x)\\ \iff &f\left(-\dfrac{1}{x} \cdot x^2\right) + f\left(-xf(-x)f\left(-\dfrac{1}{x}\right)\right) = 2f(-x)\\ \iff &f\left(-xf(-x)f\left(-\dfrac{1}{x}\right)\right) = f\left(-x\right) \iff -xf(-x)f\left(-\dfrac{1}{x}\right) = -x\\ \iff &f\left(-\frac{1}{x}\right) = \dfrac{1}{-f(x)}, \forall x \in \mathbb R, x \ne 0 \end{align}$$

Furthermore, $f^2\left(\dfrac{1}{x}\right) = f\left(-f\left(-\dfrac{1}{x}\right)\right) = f\left(-\dfrac{1}{-f(x)}\right) = \dfrac{1}{f^2(x)}, \forall x \in \mathbb R, x \ne 0$.

For $P(x, 1)$ and $P\left(\dfrac{1}{x}, 1\right)$, we can, respectively, see that $f(-x) + f^2(x) = 2x, \forall x \in \mathbb R, x \ne 0$ and $f\left(-\dfrac{1}{x}\right) + f^2\left(\dfrac{1}{x}\right) = \dfrac{2}{x} \iff \dfrac{1}{f(-x)} + \dfrac{1}{f^2(x)} = \dfrac{2}{x} \iff \dfrac{f(-x) + f^2(x)}{f(x)f^2(x)} = \dfrac{2}{x}$

$ \iff \dfrac{2x}{f(-x)f^2(x)} = \dfrac{2}{x} \iff f(-x)f^2(x) = x^2, \forall x \in \mathbb R, x \ne 0$

$\iff f(-x) = f^2(x) = x \iff f(x) = -f^2(x) = -x, \forall x \in \mathbb R, x \ne 0$

However, $f(0) = 0 \implies f(x) = -x, \forall x \in \mathbb R$