## Continuous Cauchy functional equation

Posted April 15th, 2010 by Structure

We shall prove that continuous solutions of equation

are functions g(x)=ax where a=g(1).

We have g(0)=g(0+0)=g(0)+g(0) so

as g(0)=g(x+(-x))=g(x)+g(-x)=0

so g(-x)=-g(x)

induction

g(2x)=g(x+x)=g(x)+g(x)=2g(x)

g(nx)=g((n-1)x+x)=g((n-1)x)+g(x)=(n-1)g(x)+g(x)=ng(x);

as

so

From (v) we have

Now let there is a sequence such as

As g is continuous

Now we want to solve in the class of continuous function equation

let

Then

Let

So

We know that g(u)=au;

so

Then