Finding all continuous functions with the property that g(x + y) = g(x) + g(y)

[jdinatale]

Homework Statement

Determine all continuous functions g: R -> R such that g(x + y) = g(x) + g(y) for all x, y \in \mathbf{R}

The Attempt at a Solution

g(x) = g(x + 0) = g(x) + g(0). Hence G(0) = 0.

G(0) = g(x + -x) = g(x) + g(-x) = 0. Therefore g(x) = -g(-x).

It seems obvious that the only solutions that satisfy these properties are in the form of g(x) = \alpha x for some \alpha \in \mathbf{R}.

My issue is determining that these are the ONLY such functions. I have to somehow rule out every other possible function.

I can rule out all functions in the form of g(s) = ax + b for b \not= 0 since solutions in that form would imply that

g(s + t) = a(s + t) + b = as + at + b

and

g(s + t) = g(s) + g(t) = as + b + at + b = as + at + 2b

which is impossible. But I have to somehow rule out the infinitely many other types of possible functions.

 

[micromass]

OK, so g(0)=0. That's good.

Now, set \alpha=g(1). We want to prove now that g(x)=\alpha x. We do this in steps:

First, can you find g(2)? g(3)? In general, can you find g(n) for positive integers n??
Then can you find g(x) for integers x?? Not necessarily positive?
Then can you find g(x) for rational numbers x?

 

[Ray Vickson]

Homework Statement

Determine all continuous functions g: R -> R such that g(x + y) = g(x) + g(y) for all x, y \in \mathbf{R}

The Attempt at a Solution

g(x) = g(x + 0) = g(x) + g(0). Hence G(0) = 0.

G(0) = g(x + -x) = g(x) + g(-x) = 0. Therefore g(x) = -g(-x).

It seems obvious that the only solutions that satisfy these properties are in the form of g(x) = \alpha x for some \alpha \in \mathbf{R}.

My issue is determining that these are the ONLY such functions. I have to somehow rule out every other possible function.

I can rule out all functions in the form of g(s) = ax + b for b \not= 0 since solutions in that form would imply that

g(s + t) = a(s + t) + b = as + at + b

and

g(s + t) = g(s) + g(t) = as + b + at + b = as + at + 2b

which is impossible. But I have to somehow rule out the infinitely many other types of possible functions.

This is Cauchy's functional equation; see the Wiki article on that topic, and especially some of the cited external links.

RGV