Finding the function $f(x)$ through the given set of conditions

Question

I got terribly stuck while solving this question recently

Let $f(x)$ be a real valued function such that $$f(0)=\frac{1}{2}; \quad f(x+y) =f(x) f(a-y) +f(y) f(a-x).$$

I do not want the rigorous solution to this problem. I want a possible approach that can be used.

I tried plugging in the values of $0$, $a$, etc but couldn't find the function anyway.

How do I proceed?

EDIT: a is a constant

Answer 1

Using $f(x-y)=f(x)$ in the original equation results in $f(x+y)=2f(x)f(y)$ or $m(x+y)=m(x)m(y)$ for $m:=2f$. Because of $m(a-x)=m(x)$ (a result contained in the answer by Anurag A) and $m(a)=1$ we get $m(2x)=m(x)^2=1$. What does this imply for $f$?

Answer 2

Hint:

Assuming $a$ is fixed. Set $x=y=0$, to get $$f(0)=2f(0)f(a).$$ This implies $f(a)=1/2$. Now just set $y=0$, to get $$f(x)=f(x)f(a)+f(0)f(a-x) \implies f(x)=f(a-x).$$ Can you take it from here?