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

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SUMMARY

The discussion focuses on proving the property of affine functions, specifically that for an affine function f, the equation f(x + y) = f(x) + f(y) - f(0) holds true. The user defines a new function g(x) = f(x) - f(0) and aims to demonstrate that g is linear. The proof hinges on establishing the relationship g(x+y) = g(x) + g(y), which leads to the conclusion that if f satisfies this property, it can be expressed in the form f(x) = Ax + b, where A is a linear transformation and b is a constant vector.

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psholtz
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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.
 
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Hi psholtz! :smile:

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?
 

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