Solve functional equation
$$f(a)f(b)=\frac{1}{2}f(a+b)+\frac{1}{2} f(max(|b-a|,a)),(*)$$ where the following conditions are satisfied:
1)$f:\mathbb{R}_{+}\cup \{0\}\rightarrow [0,1] ,$
2)$f(0)=1,$
3)$\lim\limits_{x\rightarrow +\infty} f(x)=0,$
4)$f$ is continuous and nonincreasing.
$f$ can be interpreted as probability, that nonnegative random absolute continuous variable $X$ satisfy:
$$f(a)=P(X>a).$$
EDIT: What is the solution of (*), when we don't have conditions 1)-4), but we know that:
1)' $f:\mathbb{R}\rightarrow \mathbb{R} ,$
2)' $f(x)=f(-x)$ for all $x \in \mathbb{R} ,$
3)' $f$ is continuous.
1) Let $x\geq 0$, and put $a=b=x$. You get $$(*)\,\,\,f(2x)=f(x)(2f(x)-1)$$
2) Suppose that $f(x)>0$ for all $x$. From (*), we get that $2f(x)>1$ for all $x$. This contradict $f(x)\to 0$ as $x\to +\infty$.
3) Hence there exists $u$ such that $f(u)=0$. By the fact that $f$ is non increasing and continuous, we get (as $f(0)=1$) that there exists a $m>0$ such that $f(x)=0$ for $x\geq m$, and $f(x)>0$ if $x<m$.
4) Let $x$ such that $\displaystyle m>x>\frac{m}{2}$. We have $2x>m$, hence $f(2x)=0=f(x)(2f(x)-1)$. But $f(x)>0$, hence $\displaystyle f(x)=\frac{1}{2}$. So $f(x)\to 1/2$, as $x\to m$, $x<m$. As $f(m)=0$, we have again a contradiction with the continuity of $f$ at $m$.
Perhaps the hypothesis are not correct ?
EDIT: I use the new hypothesis.
a) Let $a=x\geq 0$, and $0\leq b\leq 2x$. We have $|b-x|\leq x$, hence we get $$(1)\,\, 2f(x)f(b)=f(x+b)+f(x)$$
b) Put $x=y$, $b=y$ in (1)($y\geq 0$) $$f(2y)=f(y)(2f(y)-1)$$
c) Put $x=2y$, $b=y$ in (1) :
$$f(3y)=f(2y)(2f(y)-1)=(2f(y)-1)^2 f(y)$$
d) Put $x=y$, $b=2y$ in (1): $$f(3y)=f(y)(2f(2y)-1)=(2(2f(y)-1)f(y)-1)f(y)$$
e) Suppose now that $f(z)$ ($z\geq 0$) is not zero. By c) and d), we get $1=(2f(z)-1)$. Hence $f(z)=1$. So the function $f$ can take only the values $0$ and $1$ on $]0,+\infty[$, thus by continuity $f$ is constant, and it is easy to finish.