A function $f:\mathbb{R^+}→\mathbb{R}$ satisfies the condition $f(ab)=f(a)+f(b)$ for $a,b>0$. prove that $f$ is continuous on $\mathbb{R^+}$.

Question

A function $f:\mathbb{R^+}→\mathbb{R}$ satisfies the condition $f(ab)=f(a)+f(b)$ for $a,b>0$. If $f(x)$ is continuous at $x=1$, then prove that $f$ is continuous on $\mathbb{R^+}$.

Attempt:

I have proved $f(1)=0$, $f(1/c)=-f(c),~~ c>0$.

Consider a sequence $\{x_n\}$ in $\mathbb{R^+}$ converges to $c>0$, then $$\lim_{n\to \infty}f(x_n)=\lim_{n\to \infty}f(\frac{x_n}{c}c)=\lim_{n\to \infty}f(\frac{x_n}{c})+\lim_{n\to \infty}f(c)$$

how to proceed further to use sequential criterion of continuity to prove that $f$ is continuous on $\mathbb{R^+}$.

My aim is to use sequential criterion of continuity $x_n\to c$ and $f(x_n)\to f(c)$ implies $f$ is continuous.

Please help.

Answer 1

Using your desired approach, we have by conditinuity of scalar multiplication that $x_n/c\to c/c=1$. So, as $f$ is continuous at $1$ by hypothesis, we have, as you wrote, $$\lim_{n\to \infty}f(x_n)=\lim_{n\to \infty}f\left(\frac{x_n}{c}c\right)=\lim_{n\to \infty}f\left(\frac{x_n}{c}\right)+\lim_{n\to \infty}f(c)=f(1)+f(c)=f(c)$$ (as $f (1)=f(1\cdot 1)=f(1)+f(1)$ implies $f(1)=0$) and so $f$ is continuous at $c$ as for any sequence $\{x_n\}$ converging to $c$, we have that $f(x_n)\to f(c)$.

Answer 2

Note that $f(1) = 0$. Fix $x > 0$. We have: $$f(x+h) - f(x) = f \left( 1 + \frac{h}{x} \right)$$

We have $1 + \frac{h}{x} \to 1$ as $h \to 0$, so by continuity at $1$, $f\left( 1 + \frac{h}{x} \right) \to f(1) = 0$. Therefore $f$ is continuous at $x$.