- #1
psholtz
- 136
- 0
Homework Statement
I'm trying to show that every affine function f can be expressed as:
[tex]f(x) = Ax + b[/tex]
where b is a constant vector, and A a linear transformation.
Here an "affine" function is one defined as possessing the property:
[tex]f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)[/tex]
provided that:
[tex]\alpha + \beta = 1[/tex]
The Attempt at a Solution
I've defined:
[tex]g(x) = f(x) - f(0)[/tex]
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:
[tex]g(0) = f(0) - f(0) = 0[/tex]
and it's easy to show that:
[tex]g(\alpha x) = f(\alpha x) - f(0)[/tex]
[tex]g(\alpha x) = f(\alpha x + (1-\alpha) \cdot 0) - f(0)[/tex]
[tex]g(\alpha x) = \alpha \cdot f(x) + (1-\alpha)\cdot f(0) - f(0)[/tex]
[tex]g(\alpha x) = \alpha \cdot f(x) - \alpha \cdot f(0)[/tex]
[tex]g(\alpha x) = \alpha \cdot \left( f(x) - f(0) \right)[/tex]
[tex]g(\alpha x) = \alpha \cdot g(x)[/tex]
But I'm having more trouble proving the property:
[tex]g(x+y) = g(x) + g(y)[/tex]
On the one hand we have:
[tex]g(x+y) = f(x+y) - f(0)[/tex]
and on the other hand we have:
[tex]g(x) + g(y) = f(x) + f(y) - 2f(0)[/tex]
so it seems that if we could prove that:
[tex]f(x + y) = f(x) + f(y) - f(0)[/tex]
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.