Functional equation $f(xy)=f(x)+f(y)$ and continuity

Question

Prove that if $f:(0,\infty)→\mathbb{R}$ satisfying $f(xy)=f(x)+f(y)$, and if $f$ is continuous at $x=1$, then $f$ is continuous for $x>0$.

I let $x=1$ and I find that $f(x)=f(x)+f(1)$ which implies that $f(1)=0$. So, $\lim_{x\to1}f(x)=0$, but how can I use this to prove continuity of $f$ for every $x \in \mathbb R$?

Any help would appreciated. Thanks

Answer 1

Give $x_0>0$, $$f(x)-f(x_0)=f\left(x_0\cdot\frac{x}{x_0}\right)-f(x_0)=f\left(\frac{x}{x_0}\right),$$ by $f$ is continuous at $x=1$, when $x\to x_0$, $\frac{x}{x_0}\to1$, then $$\lim\limits_{x\to x_0}f(x)=f(x_0).$$

Answer 2

$f(a^+)=\lim_{\epsilon \rightarrow 0}f(a(1+\epsilon))=\lim_{\epsilon \rightarrow 0}(f(a)+f(1+\epsilon))=f(a)+f(1)$

We have used from continuity of $f$ at $x=1$ in $\lim_{\epsilon \rightarrow 0}f(1+\epsilon)=f(1)$. Now notice $f(1)=0$, since $f(x)=f(x\times1)=f(x)+f(1)$ for all valid $x$, so we can say $f(a^+)=f(a)$.

With the same reasoning you can say $f(a^-)=f(a)$, so your function is continuous for positive values.