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Affine Functions

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data
    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]
    3. 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.
     
  2. jcsd
  3. Jul 9, 2011 #2

    micromass

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    Staff Emeritus
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    2016 Award

    Hi psholtz! :smile:

    Maybe you can start as follows:

    [tex]g(x+y)=g \left( 2 \left( \frac{1}{2}x+\frac{1}{2}y \right) \right) [/tex]

    Now, what can you do with this?
     
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