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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 $|g(\frac{x}{a})|<\epsilon \implies |g(a \cdot \frac{x}{a})-g(a)| < \epsilon \implies |g(x)-g(a)| < \epsilon$.

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Pepesadgeomegalul
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  • Sounds like a very good test to me. But the only thing that causes me noise is the fact that $x$ you take it in a neighborhood around zero, i. e., you need $x$ to be a real number, right? – Eduardo Maza Mar 15 '21 at 17:19
  • I'm struggling to find an example of such a function other than $g(x) = 0$ (which is of course continuous) ? – Tom Collinge Mar 15 '21 at 20:09
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    g(x)=log(x) satifies the condition. – Pepesadgeomegalul Mar 15 '21 at 20:22
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    @TomCollinge log function has the property $\log(xy) = \log x + \log y$. The question says specifically $f:\mathbb R^{+}\to \mathbb R$ so $g(0)$ need not exists. (If it did then $g(a)+ g(0) = g(0)$ and $g(a) = 0$ for all $a$). – fleablood Mar 15 '21 at 20:23
  • " i. e., you need x to be a real number, right?" we need $x$ to be a positive real number. – fleablood Mar 15 '21 at 20:25
  • Your proof looks good to me and is what I would have done. I haven't gone through it with a fine tooth comb but I don't see any obvious logical problems. – fleablood Mar 15 '21 at 22:12
  • I'm pretty sure (but I'm learned never to say these things for certain without backup) that we can prove the only such continuous $g$ exists if $g(x)=\log_b x$ for some $b> 0;b\ne 1$ or $g(x) = 0$... If $g$ need not be continuous at $x=1$ we can maybe come up with a funky alternative. – fleablood Mar 15 '21 at 22:16
  • @fleablood: yes you are correct. With continuity the only solution is $k\log x$ where $k$ is any fixed real number. I have the proof on mathse but still searching. – Paramanand Singh Mar 16 '21 at 05:41
  • @fleablood: search complete, see https://math.stackexchange.com/a/2091337/72031 – Paramanand Singh Mar 16 '21 at 05:44

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