Differentiable Proof 2

thaopanda · #1 $Old$ November 1st, 2009, 07:20 PM

Suppose that f: R $\rightarrow$ R is differentiable and that for every a,b $\in$ R there holds f (a + b) = f(a) + f(b).

Prove that f '(x) = f '(0) for all x $\in$ R. What kind of function is f?

umgambit · #2 $Old$ November 1st, 2009, 07:26 PM

Start by noticing that f(0)=0. Indeed, putting a=b=0, one gets f(0)=2f(0), hence f(0)=0.
Now put a=x, b=h:

f(x+h)-f(x)=f(h)=f(h)-f(0)
Divide by h to get
(f(x+h)-f(x))/h=(f(h)-f(0))/h

Take the limit h->0 to finally get
f'(x)=f'(0).

This means that f'(x)=a is a constant: f(x)=cx+d, but d must be zero in order to the first relation to hold.

Hence f(x)=cx. Those are called linear functions.