Fun(ctional) equation #2

The condition $f(1) = 0$ is implied by the functional equation (put $x=y=1$ ). If we define the function $g(x) = f(e^x)$ , then the function $g$ satisfies the equations $g(x+y) \; = \; g(x) + g(y) \hspace{1cm} (x,y \in \mathbb{R}) \hspace{2cm} g(0) = 0\,,\,g'(0) = 1$ Thus we are assuming that $g$ is differentiable at $0$ , and hence is continuous at $0$ . Since $g(x+h) - g(x) \; = \; g(h) \; = \; g(h) - g(0) \hspace{2cm} x,h \in \mathbb{R}$ it follows that $g$ is continuous everywhere.

It is a simple matter of induction to show that $g(mx) = mg(x)$ for all $x \in \mathbb{R}$ and $m \in \mathbb{Z}$ , and since $ng\big(\tfrac{m}{n}x\big) \; = \; g(mx) \; = \; mg(x) \hspace{2cm} x \in \mathbb{R}\,,\,m,n \in \mathbb{Z}\,,\, n \neq 0$ we deduce that $g(qx) \; = \; qg(x) \hspace{2cm} q \in \mathbb{Q}\,,\,x \in \mathbb{R}$ In particular, therefore, $q(q) = qg(1)$ for $q \in \mathbb{Q}$ . Since $g$ is continuous everywhere and the rationals are dense in the reals, it follows that $g(x) = kx$ for all real $x$ , where $k = g(1)$ . Thus $k = g'(1) = 1$ , and hence $g(x) = x$ . This implies that $f(x) = \ln x$ for all $x > 0$ , making the answer $\ln1000 = \boxed{6.907755279}$ .

The condition that $f'(1) = 1$ is key, since it implies the continuity of $g$ . There are infinitely many examples of discontinuous functions $g$ on $\mathbb{R}$ satisfying the identity $g(x+y) = g(x) + g(y)$ for all $x,y$ . Equivalently, there are infinitely many examples of discontinuous functions $f$ satisfying the main functional equation.

Step 1: Differentiate equation w.r.t. $x$ and then put $x= 1 and y=x$ .

Step 2: Differentiate equation w.r.t. to $y$ and then put $y= 1$

By solving both equations simultaneously one easily gets $1= x\times dy/dx$ ,Taking $y=f(x)$ .

Solve this elementary diffrential equation.

By hit and error f(x) = lnx