I have this problem and I would like to know if my solution is correct.
Let $g(x)$ be af funtion from the positive reals to the reals satisfying $g(xy)=g(x)+g(y)$ for alle $x$ and $y$ in the positive reals. Proof that if $g$ is continous in $x=1$ then it's continous at every point.
By inserting $y=1$ we get that $g(x) = g(x) + g(1) \implies g(1) = 0$. Since $g$ is continous in $x=1$ we know that for all $\epsilon > 0$ and for all $x$ there exists a $\delta_1$ such that $|g(x)| < \epsilon$ whenever $|x-1| < \delta_1$. Since this holds for all $x$ we can substitute $\frac{x}{a}$ for $x$. Thus we get that $|g(\frac{x}{a})|<\epsilon$ whenever $|\frac{x}{a}-1| < \delta_1$. Rewriting $|\frac{x}{a}-1| < \delta_1 \implies |x-a| < |a|\delta_1=\delta_2$. Thus whenever $|x-a| < \delta_2$ then $|g(\frac{x}{a})|<\epsilon \implies |g(a \cdot \frac{x}{a})-g(a)| < \epsilon \implies |g(x)-g(a)| < \epsilon$.