# Prove that function is continuous without knowing the function explicitly

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 $x=1$: $$|x - 1|<δ⇒|f(x)-f(1)|<ϵ$$

• Use $f(x+ε) = f(x) + f(\frac{x+ε}{x})$. – Rolf Sievers Mar 20 '15 at 12:12
• 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. – Rolf Sievers Mar 20 '15 at 12:18