0

Find all real functions $f(x)$ $\mathbb{R} \to \mathbb{R}$ such that $x^2f(yf(x)) = y^2f(x)f(f(x))$

Obviously, let $y=0$ you instantly get, $f(0) = 0$.

Also, a relation is:

$f(yf(x)) = \frac{y^2 f(x) f(f(x))}{x^2}$ and set $y = f(y)$ to get:

$f(f(y) f(x)) = \frac{f(y)^2 f(x) f(f(x))}{x^2}$ where $x \ne 0$.

Replacing $y$ with $x$ will not change anything in the LHS so,

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

Set the RHS of both equal to get:

$f(y) f(f(x))/x^2 = f(x) f(f(y))/y^2$

Gather all the $f(x)$ on the LHS we can get:

$f(f(x))/f(x)x^2 = f(f(y))/f(y) y^2 = k$

As we can set the RHS to a constant $k$.

Going back, we get:

$f(f(x)) = kx^2 f(x)$.

But that is all I can really get.

Trivially, with $y=1$ you can get $f(x) = x^2$ but what about other options?

Amad27
  • 10,465

1 Answers1

2

When $y=1$ you get $x^2f(f(x))=f(x)f(f(x))$. So $x^2=f(x)$ or $f(f(x))=0$.

Show that if $f(f(x))=0$ for some non-zero $x$ then $f(x)=0$ for all $x$.

This gives two functions, $f(x)=0$ and $f(x)=x^2$.

So, if $f(f(x_0))=0$ for some $x_0\neq0$, we first show that $f(x_0)=0$.

This is because if $f(x_0)\neq 0$ then we can pick $y=x_0/f(x_0),x=x_0$ and get:

$$x_0^2f(x_0)=y^2f(x_0)f(f(x_0))=0$$

So $f(x_0)=0$.

Now, if $f(x_1)\neq 0$ for any $x_1$, we let $y=x_0/f(x_1)$ and $x=x_1$ and we get:

$$0=x_1^2 f(x_0)=y^2f(x_1)f(f(x_1))$$

So either $f(x_1)=0$ or $f(f(x_1))=0$. But we've already shown that the latter implies the former.

So we've shown that the function $f$ must be one of $x^2$ or $0$.

Thomas Andrews
  • 177,126