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Let $f\colon \mathbb R^+\to\mathbb R$ be a function that satisfies the following conditions: $$\tag1 \lim_{x\to 1}f(x)=0 $$ $$\tag2f(x_1)+f(x_2)=f(x_1x_2)$$ Show that $f$ is continuous in its domain.

I managed to show that $f$ is continuous at $x=1$, but I have no idea how to continue from there. Here's what I've done so far:

Because $\lim_{x\to 1}f(x)=0$, for every ϵ>0 there exists a δ>0 so that $$0<|x - 1|<δ⇒|f(x)-0|<ϵ$$

To prove continuity at $x=1$ it's enough to show that $f(1)=0$ using the condition 2): $$f(1)+f(1)=f(1 ·1)$$ $$f(1)=f(1)-f(1)$$ $$f(1)=0$$ So now we have the definition of continuity at $x=1$: $$|x - 1|<δ⇒|f(x)-f(1)|<ϵ$$

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  • $\begingroup$ Use $f(x+ε) = f(x) + f(\frac{x+ε}{x})$. $\endgroup$ – Rolf Sievers Mar 20 '15 at 12:12
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    $\begingroup$ Btw, here the abstract underlying concept is the topological group. In this case $(ℝ^+, \cdot)$ and $(ℝ, +)$. Whenever you have maps which respect the group structure you only need to verify continuity at the unit element. $\endgroup$ – Rolf Sievers Mar 20 '15 at 12:18
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    $\begingroup$ Somewhat hesitantly, but since $f(x_1)+f(x_2)=f(x_1x_2)$ is a functional equation, I have added (functional-equation)$. You can find some links to posts about this particular functional equation (and also some closely related equations) here. $\endgroup$ – Martin Sleziak Mar 20 '15 at 13:22
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    $\begingroup$ Note that you do know the function explicitly: the equation can be transformed into a Lipschitz-continuous differential equation, which pins down the solution to be unique. Hence $\ln(x_1) + \ln(x_2) = \ln(x_1\cdot x_2)$ is enough to confirm $f = \ln$. $\endgroup$ – leftaroundabout Mar 20 '15 at 17:08
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You have shown continuity at $x=1$, i.e., $$\lim_{x\to1}f(x)=f(1).$$ Consequently, for any $x_0\ne0$ $$\lim_{x\to x_0}f(x)=\lim_{x\to x_0}f\left(\frac x{x_0}\cdot x_0\right)=\lim_{x\to x_0}\left(f\left(\frac x{x_0}\right)+f(x_0)\right)=f(1)+f(x_0)=f(x_0).$$

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