How do I finish solving $f(x)f(2y)=f(x+4y)$?

Question

I'm trying to solve this functional equation: $$f(x)f(2y)=f(x+4y)$$

The first thing I tried was to set $x=y=0$; then I get: $$f(0)f(0)=f(0)$$ which means that either $f(0)=0$ or we can divide the equation by $f(0)$ and then $f(0)=1$.

Case 1: If $f(0)=0$ then we can try to set $x=0$. Then; $$f(0)f(2y)=f(4y)$$ $$0=f(4y)$$ Which means that one of the solutions is a constant function $f(x)=0$

Case 2: If $f(0)=1$. This is where I'm stuck.

I tried to set $x=0$, then I get: $$f(2y)=f(4y)$$ I also tried to set $x=-4y$. Then I get: $$f(-4y)f(2y)=f(0)=1$$

I think that these two observations could be useful, but I don't know how to continue from here. I have guessed that another solution is $f(x)=1$ but I don't know how to show that there aren't any others as well.

Answer 1

Assuming that $ f(0)=1$, you got

for $ y\in \Bbb R $, $$f(4y)=f(2y)=f(y)=f(\frac y2)$$ $$=...=f(\frac{y}{2^n})$$ for each $ n\ge 0$.

but by continuity of $ f $ at $ 0$, $$\lim_{n\to +\infty}f(\frac{y}{2^n})=f(0)=1$$

thus $$(\forall y\in \Bbb R)\;\; f(y)=1$$

Answer 2

When solving functional equations, we aren't able to claim continuity and other such properties. We in fact don't need the continuity condition in order to solve the problem.

For the $f(0)=1$ case, we can reuse some of your claims:

(1)$$f(2y)=f(4y)$$

(2)$$f(-4y)f(2y)=1$$

We can substitute 1) into 2) to get that

(3)$$f(-4y)f(4y)=1. $$

We now plug $y=-\frac{x}{2}$ into our original equation to get another identity:

(4)$$f(x)f(-x)=f(-x).$$

We can replace $4y$ with $x$ and combine (3) with (4):

$$f(-x)=f(x)f(-x)=f(4y)f(-4y)=1.$$

We get $f(-x)=1$, and we can flip the sign of the $-x$ to get that

$f(x)=1$ for all x

Answer 3

Ryan's answer is showing how to finish the argument you are making. This answer provides alternative way to solve the equation without splitting into the cases in the first place.

Substituting $y/2$ in place of $y$, we get functional equation $f(x)f(y)=f(x+2y)$, which is little more symmetric. If we now swap $x$ and $y$, we get $f(x+2y)=f(y+2x)$. Put $y=-2x$ and observe that $f(-3x)=f(0)$. Finally, substitute $-x/3$ in place of $x$ and get $f(x)=f(0)$, hence $f(x)$ is a constant function. Let $f(x)=a$, putting into the original equation this gives $a^2=a$, so $a=0$ or $a=1$. Thus $f(x)=0$ and $f(x)=1$ are the only solutions.