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

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 $\frac{1}{x}$?

## marked as duplicate by user223391, G-man, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 20 '15 at 8:11

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 $\frac{1}{x}f'(1)=\frac{1}{x}$.
$f(x/x)=f(x(1/x))=f(1)=f(x)+f(1/x)=0$, $f(x)=-f(1/x)$, $(x/y)=f(x)-f(y)$
$f'(x_0)=lim_{x\rightarrow x_0}{{f(x)-f(x_0)}\over{x-x_0}}=lim_{x\rightarrow x_0}{{f(x/x_0)}\over{x-x_0}}=(1/x_0)lim_{x\rightarrow x_0}{{f(x/x_0)}\over{x/x_0-1}}=(1/x_0)lim_{x\rightarrow 1}{{f(x)-f(1)}\over{x-1}}=1/x_0$. $f$ is derivable, thus continue.
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.