Let $f$ be a function such that $f(ab)=f(a)+f(b) $ with $f(1)=0$ and derivative of $ f$ at $1 $ is $1$.

Question

Let $f$ be a function such that $f(ab)=f(a)+f(b)$ with $f(1)=0$ and derivative of $f$ at $1$ is $1$

How can I show that $f$ is continuous on every positive number and

derivative of $f$ is $\frac{1}{x}$?

Answer 1

For any fixed $x>0$, we have $$ f(x+h)-f(x)=f(x(1+h/x))-f(x)=f(1+h/x)=h\frac{f(1+h/x)-f(1)}{h}. $$ As $h\to 0$, $\frac{f(1+h/x)-f(1)}{h}\to f'(1)=1$ and $h\to 0$, so the RHS above goes to $0$ as $h\to 0$. This implies that $f$ is continuous at any $x>0$. The above equalities also imply $$ \frac{f(x+h)-f(x)}{h}=\frac{1}{x}\frac{(f+h/x)-f(1)}{h/x}. $$ Letting $h\to 0$, the LHS goes to $f'(x)$ and the RHS goes to $\frac{1}{x}f'(1)=\frac{1}{x}$.

Answer 2

For the second part of your question, note that $f(x\cdot (1+\epsilon)) = f(x) + f(1 + \epsilon)$. Thus the following is equal in the limit to the derivative at $x$:

$\lim_{\epsilon \to 0} \frac{f(1+\epsilon)}{x \cdot \epsilon} = \frac{1}{x} \lim_{\epsilon \to 0}\frac{f(1+\epsilon)}{\epsilon} = \frac{1}{x} \cdot f'(1) = \frac{1}{x}$

As a side note, this also resolves the first part of the question: letting $\epsilon \to 0$ we have $x \cdot (1 + \epsilon) \to x$, and $f(x \cdot (1 + \epsilon)) = f(x) + f(1+\epsilon) \to f(x) + f(1) = f(x) + 0 = f(x)$. Hence $f$ is continuous.

Answer 3

$f(x/x)=f(x(1/x))=f(1)=f(x)+f(1/x)=0$, $f(x)=-f(1/x)$, $(x/y)=f(x)-f(y)$

$f'(x_0)=lim_{x\rightarrow x_0}{{f(x)-f(x_0)}\over{x-x_0}}=lim_{x\rightarrow x_0}{{f(x/x_0)}\over{x-x_0}}=(1/x_0)lim_{x\rightarrow x_0}{{f(x/x_0)}\over{x/x_0-1}}=(1/x_0)lim_{x\rightarrow 1}{{f(x)-f(1)}\over{x-1}}=1/x_0$. $f$ is derivable, thus continue.