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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

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    $\begingroup$ First, you should tell us your attempts and thoughts on the problem. $\endgroup$ – GEdgar Dec 9 '15 at 15:30
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    $\begingroup$ If you define $g:\mathbb R\to\mathbb R$ as $g(x)=f(e^x)$, then $g$ satisfies the functional equation $g(x+y)=g(x)+g(y)$, and is continuous at $x=0$. This reduces to a more common problem.. $\endgroup$ – Thomas Andrews Dec 9 '15 at 15:31
  • $\begingroup$ Why did you put this question on hold? Relax with this thing.... $\endgroup$ – Jimmy R. Dec 10 '15 at 9:43
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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 $$\lim\limits_{x\to x_0}f(x)=f(x_0).$$

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

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.

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