Let $f$: $\Bbb R$ $\rightarrow$ $\Bbb R$ be a non-constant function such that $f(a + b) = f(a)f(b)$ for all real numbers $a$ and $b$.
Prove that if $f(x + y) = f(x)f(y)$ is continuous, then it has to be injective.
I derived a few useful properties from this function such that $f(x)\neq0;f(0) = 1 ; f(x) = \frac{1}{\mathrm f(-x)}$
In order to prove $f$ is injective, I need to prove that $f(x) = f(y)$ implies that $x = y$
I suppose that there are $x, y \in \Bbb R$ such that $f(x) = f(y)$ $$f(x) = \frac{1}{f(-y)}$$ $$f(x)f(-y) = 1$$ $$f(x - y) = 1$$ To finish the proof I need to prove $f(x) = 1$ implies $x = 0$. This is where I got stuck.
I don't know if I'm on the right track. Any suggestion is much appreciated.
Hint : show that $f(t)=f(\frac{t}{2})^2$. Deduce $f\geq 0$. Let $G=\lbrace x\in{\mathbb R} | f(x)=1\rbrace$. Then $G$ is an additive subgroup of $\mathbb R$. If $f$ is continuous it is also a closed subgroup. Also, show that $t\in G \Rightarrow \frac{t}{2} \in G$ using the above identity. Deduce that $G$ is dense if $G\neq \lbrace 0 \rbrace$. Conclude.