Thailand MO $f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)$

Question

This is a functional equation question from Thailand MO 2023.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfies the equation $$f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)$$ for all $x,y \in \mathbb R$. Find all $f$.

$\textbf{My attempt: }$ By letting $x=y=0$, we have that $f(0)^3 =0 \implies f(0)=0$. Setting $y=x\neq 0$, we have that $0 = x^2f(x)$, so $f(x) = 0$ is a solution. By inspection, I also notice that $f(x) = x$ and $f(x) = -x$ are solutions. I've also tried manipulating two expressions

\begin{alignat}{2} f(x)f(y_1)f(x-y_1) &= x^2f(y_1) - {y_1}^2f(x)\\ f(x)f(y_2)f(x-y_2) &= x^2f(y_2) - {y_2}^2f(x)\\ \implies f(y_1)\left[f(x - y_1) - f(x -y_2)\right] &= {y_2}^2-{y_1}^2 \end{alignat} This doesn't seem helpful and I'm not sure where to go from here. I suspect these are the only solutions? Help would be greatly appreciated!

Answer 1

Suppose $P(x,y): f(x)f(y)f(x-y) = x^2f(y)-y^2f(x)$

As you noted, $f(0)=0$. From $P(y,x)$ and $P(x,y)$, one gets $f(x)f(y)\cdot(f(x-y)+f(y-x))=0$

Case $1$: $f(x)=0$

Case $2$: $f(x)\neq 0$.

$f(x-y)=-f(y-x)$. Taking $t \to x-y$ shows $f$ is odd, since $f(t)=-f(-t)$. Then, $P(x,-x)$ shows $f(2x)\cdot f(x) =2x^2$ and $P(2x,x)$ shows $2\cdot f(x)=f(2x)$. This gives $f^2(x)=x^2$ and that $f(x)=x$ or $f(x)=-x$. This can mean that the functions be entirely $x$, or $-x$, or perhaps be a function which is $x$ for some interval and $-x$ for some other.

But the third possibility leads to contradictions. Take $a,b \neq 0$. By way of contradiction, suppose $f(a)=a$ while $f(b)=-b$. $f(a-b)$ may be $a-b$ or $b-a$. In either case, we get from $P(a,b)$ that either $a=0$ or $b=0$, which contradicts our initial supposition. This shows the one $f$ can't be $x$ and $-x$ for different $x$. So we get two solutions $f(x)=x\ \ \forall x$ or $f(x)=-x \ \ \forall x$, where $f(x)\neq 0 \implies x\neq 0$.

Similarly, we can show that $f(x)$ can't be some function which is $0$ on some interval and non-zero (ie $\pm x$) on some other. Take $a,b\neq 0$ such that $f(a)=0$ and $f(b)=\pm b$. Then $P(a,b): a^2(\pm b) =0$ but this contradicts that $a,b$ are non-zero.

Hence, $f(x)=0 \ \forall x\lor f(x)=x\ \ \forall x \lor f(x)=-x \ \forall x$ are the solutions. Note that we can claim this for the case $2$ solutions since $f(0)$ is already known to be $0$.

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NOTE: I have elaborated this post to present a better idea of the solution.

Answer 2

Swapping $x$ and $y$, for any $x$ and $y$ we have:

$f(y)f(x)f(y-x)=y^2f(x)-x^2f(y)=-f(x)f(y)f(x-y) \Rightarrow f(x)=0~or~f(y-x)=-f(x-y) $.

Assuming a nontrivial solution, this means that $f(x)$ is an odd function. Now set $y=-x$:

$f(x)f(-x)f\left(x-(-x)\right)=x^2f(-x)-(-x)^2f(x)\Rightarrow f^2(x)f(2x)=2x^2f(x) \Rightarrow f(x)f(2x)=2x^2$ (I).

Now, consider $x\rightarrow 2x$ and $y\rightarrow x$ in the equation. We will have

$f(2x)f(x)f(2x-x)=(2x)^2f(x)-x^2f(2x)\Rightarrow f(2x)f^2(x)=x^2\left( 4f(x)-f(2x)\right)$ (II).

From (I) we have $f(2x)=\frac{2x^2}{f(x)} \Rightarrow f(2x)f^2(x)=2x^2f(x)$. (III)

(II) & (III): $2x^2f(x) = x^2\left( 4f(x)-f(2x)\right) \Rightarrow f(2x)=2f(x)$. Plug this into equation (I): $2f^2(x)=2x^2$.

Therefore, $f(x)=\pm x$. The only thing left to show is that $f(x)$ cannot be equal to $x$ for some x and equal to $-x$ for the rest. We can prove this by contradiction. Assume $f(x) =x$ and $f(y) = -y$. Then, we have

$x(-y)f(x-y)=x^2(-y)-y^2x \Rightarrow f(x-y)=x+y$. Note that f(x-y) must be equal to either $x-y$ or $y-x$.

Case 1: $f(x-y)=x-y=x+y \Rightarrow y=0$. Case 2: $f(x-y)=y-x=x+y \Rightarrow x=0$.

This completes the proof.

Answer 3

Suppose that $f(x) = 0$ for some $x \neq 0$. Then, substituting in any value of $y$ shows that $0 = x^2f(y)$, so $f(y)=0$ for all $y \in \mathbb R$. Meanwhile, letting $y=x=0$ gives $f(0)^3=0$, so $f(0)=0$.

From now on, we assume that $f(x) \neq 0$ for all $x\neq 0$. Knowing this, let's define the function $$g(x) =\begin{cases} 0 & x=0 \\ \frac{x^2}{f(x)} & x \neq 0 \end{cases}.$$

Note that for $x,y \neq 0$, dividing both sides of the functional equation by $f(x)f(y)$, $$ f(x-y) = \frac{x^2}{f(x)} - \frac{y^2}{f(y)} = g(x)-g(y). $$ Now we can play a game similar to that played for the Cauchy functional equation. Note that for any $x \neq 0$, $$ f(2x) = g(3x) - g(x) = (g(3x) - g(2x)) + (g(2x) - g(x)) = 2f(x). $$

Then, note that for any $x \neq 0$, $$ g(2x) = \frac{4x^2}{f(2x)} = \frac{2x^2}{f(x)} = 2g(x). $$

But then, $f(x) = g(2x) - g(x) = g(x)$ for all $x \neq 0$. Thus, by definition of $g$, we see that $f(x) = \pm x$ for all $x \neq 0$.

Also, this implies that $f(x-y) = g(x)-g(y) = f(x)-f(y)$ for all $x,y \neq 0$. This is in the realm of Cauchy's functional equation and similar arguments to those used in that analysis yield that

• $f$ is additive i.e. $f(rx) = rf(x)$ for all positive rational numbers $r$ and $x \neq 0$.

• $f$ is continuous at $0$, since $|f(x)| \leq |x|$ for all $x \neq 0$.

Under these conditions, $f$ is uniquely either $+x$ or $-x$, as is well known in that theory. Finally, both these solutions work, and we obtain the complete set of solutions as $f(x) = 0$ and $f(x) = \pm x$ for all $x \neq 0$ and a fixed choice of sign.

Answer 4

Leaving the discontinuous case,The function might be of Two types,Either continuous Differentiable or continuous non-Differentiable

Now either the Function is Constant function or It is a non-constant function

For Function to be constant,substitute $f(x)=k$ in original functional equation

$k^3=k^3-k^3$

Hence $k=0$,Which means $f(x)=0$ is the only function when it is a constant Differentiable function

Now Let $f(x)$ be non-constant Differentiable function

The given functional equation is as follows

$$f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)\tag1$$

Now replace $y\to x-y$ in given functional equation

$$f(x)f(x-y)f(y) = x^2f(x-y) - (x-y)^2f(x)=x^2f(x-y) - x^2f(x)+2xyf(x)-y^2f(x)\tag2$$

Hence

$$f(x)f(x-y)f(y)=x^2f(x-y) - x^2f(x)+2xyf(x)-y^2f(x)\tag3$$

From $(1)$ and $(3)$

$$x^2f(y)=x^2f(x-y)-x^2f(x)+2xyf(x)\tag4$$

Consider $x,y \in (\mathbb R-{{0}})$

From equation $4$, It follows

$$\frac{f(y)}{y}=-\left(\frac{f(x-y)-f(x)}{y}\right)+\frac{2}{x}f(x)\tag 5$$

Let $\lim \limits_{y \to 0}\frac{f(y)}{y}=t $

Apply limit $y\to 0$ on both sides of equation $5$

$$\lim \limits_{y \to 0}\frac{f(y)}{y}=-\lim \limits_{y \to 0}\left(\frac{f(x-y)-f(x)}{y}\right)+\frac{2}{x}f(x)\tag 6$$

It follows

$$t=-f'(x)+\frac{2}{x}f(x)\tag 7$$

It is A Linear first order differential equation

Solving it yields

$$f(x)=tx+cx^2\tag 8$$

Where $c$ is constant of integration

Note that Using original functional equation and replacing $x \to 2x$ and $y \to x$

We get $f(2x)=2f(x)$

Using this condition on our solved function,We get $c=0$

Hence $$f(x)=tx$$

Now taking $y\to1$ and $x\to x$in original functional equation and using $f(x)=tx$ We get an essential condition that $t^3=t$, hence $t=1$ or $t=-1$

Hence $f(x)=0$ or $f(x)=x$ or $f(x)=-x$ are the solutions to given functional equation

Note:This is unintentionally big solution as I started Typing it on mobile before it got any other answers,Hence Thought of completing and posting it anyways