Proving Affine Function Inequality: f(x + y) = f(x) + f(y) - f(0)

[psholtz]

Homework Statement

I'm trying to show that every affine function f can be expressed as:

$$f(x) = Ax + b$$
where b is a constant vector, and A a linear transformation.

Here an "affine" function is one defined as possessing the property:

$$f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$$
provided that:

$$\alpha + \beta = 1$$

The Attempt at a Solution

I've defined:

$$g(x) = f(x) - f(0)$$
and the idea is to show that g(x) is linear. If so, the form of f we are trying to derive above follows easily.

It's easy to show that g maps zero onto zero:

$$g(0) = f(0) - f(0) = 0$$
and it's easy to show that:

$$g(\alpha x) = f(\alpha x) - f(0)$$
$$g(\alpha x) = f(\alpha x + (1-\alpha) \cdot 0) - f(0)$$
$$g(\alpha x) = \alpha \cdot f(x) + (1-\alpha)\cdot f(0) - f(0)$$
$$g(\alpha x) = \alpha \cdot f(x) - \alpha \cdot f(0)$$
$$g(\alpha x) = \alpha \cdot \left( f(x) - f(0) \right)$$
$$g(\alpha x) = \alpha \cdot g(x)$$
But I'm having more trouble proving the property:

$$g(x+y) = g(x) + g(y)$$
On the one hand we have:

$$g(x+y) = f(x+y) - f(0)$$
and on the other hand we have:

$$g(x) + g(y) = f(x) + f(y) - 2f(0)$$
so it seems that if we could prove that:

$$f(x + y) = f(x) + f(y) - f(0)$$
we would be done.

This relation seems to hold for various affine functions that I've tried substituting into it, but I'm having trouble proving it in general.

 

[micromass]

Hi psholtz!

Maybe you can start as follows:

$$g(x+y)=g \left( 2 \left( \frac{1}{2}x+\frac{1}{2}y \right) \right)$$

Now, what can you do with this?