$f(x-y) = f(x)f(y)$ defined from $\mathbb{R}\to\mathbb{R}$ and $f(x)\ge 0$ for all $x$. find $f(3)$

Question

The solution is given as if $x=0,$ then $f(0) \ge 0$ therefore $f(0) = 1$ and $f(3) = 1$

How?

Answer 1

Because for $y=0$ we obtain $f(x)=f(x)f(0)$, which gives $f(x)=0$ and $f(3)=0$ or $f(0)=1$.

But for $x=y$ we obtain $$f(x)^2=f(0)=1$$ and since $f(x)\geq0$, we obtain $f(x)=1$ and $f(3)=1.$

Answer 2

Putting $x=y=0$ we get $f(0)=f(0-0)=f(0)f(0)$. Then either $f(0)=0$ or $f(0)=1$.

If $f(0)=0$ then $f(x)=f(x-0)=f(x)f(0)=0$ for all $x$.

Otherwise let $f(0)=1$. Then $f(-y)=f(0-y)=f(y)$ for all $y$. Also then $1=f(0)=f(y-y)=f(y)f(-y)=f^2(y)$ for all $y$. Since $f(y)\ge 0$ for all $y$, hence $f(y)$=1 for all $y$.