Proving Continuity for $f(x_1 + x_2) = f(x_1)f(x_2)$ Suppose $f$ is defined on $\mathbb{R}$ and has the following properties. Its limit when $x$ approaches $0$ is $1$ and $f(x_1 + x_2) = f(x_1)f(x_2)$. Then prove that
a) $f(x)>0$ for all $x$
b) $f(rx) = (f(x))^r$ if $r$ is rational
c) If $f(1) = 1$, then $f$ is constant
d) If $f(1)>1$, then $f$ is increasing and limit of $f(x)$ when $x$ approaches infinity is infinity. And the limit of $f(x)$ when $x$ approaches infinity from the negative side is $0$.
Part (a) and (b) were pretty simple but I am having trouble with part (c) and (d)
Thanks for your help
 A: You first have to prove that $f$ is everywhere continuous.
It is continuous at $0$ because
$$
f(0)=f(0+0)=f(0)f(0)
$$
so either $f(0)=0$ or $f(0)=1$. But $f(0)=0$ would imply $f(x)=0$ for all $x$ (why?); this is ruled out by the fact that $\lim_{x\to0}f(x)=1$.
We can also say that $f(x)\ne0$, for all $x$, because $1=f(0)=f(x-x)=f(x)f(-x)$. This shows also that $f(-x)=f(x)^{-1}$.
Now we want to prove $f$ is everywhere continuous. Let $x_0\in\mathbb{R}$ and fix $\varepsilon>0$. By definition we have
$$
|f(x_0+h)-f(x_0)|=|f(x_0)f(h)-f(x_0)|=|f(x_0)|\,|f(h)-1|
$$
so it suffices to take $\delta>0$ such that …
(a) By continuity and the facts that $f(0)=1$ and $f(x)\ne0$ for all $x$, we can conclude $f(x)>0$ for all $x$, because …
(b) Prove it first for integer $r$, and then …
(c) If $x\in\mathbb{R}$, then $x=\lim_{n\to\infty}a_n$ where $a_n$ are rational numbers; then $f(x)=\lim_{n\to\infty}f(a_n)=\dots$
(d) If $x>1$ and $f(x)<1$, then by continuity you can find $r$ rational such that $r>1$ and $f(r)<1$. Then $1>f(r)=f(1)^r$ and …
A: Hint 1: To show continuity at point $x$ take any sequence $(a_i)\rightarrow 0$ and consider sequence $f(x+a_i)=f(x)f(a_i)$.
Hint 2: By b) and continuity you get $f(r)=(f(1))^r$ for all $r\in \mathbb R$.
