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

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In summary, the conversation discusses the proof that every affine function can be expressed as f(x) = Ax + b, where b is a constant vector and A is a linear transformation. The attempt at a solution involves defining a function g(x) and proving its linearity. The conversation then delves into trying to prove the property g(x+y) = g(x) + g(y), with the suggestion to start by simplifying g(x+y) to g(2(x+y)/2).
  • #1
psholtz
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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.
 
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  • #2
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?
 

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

1. What is an affine function?

An affine function is a mathematical function that can be written in the form f(x) = ax + b, where a and b are constants. This type of function is commonly used in linear algebra and optimization problems.

2. What does it mean to prove an affine function inequality?

Proving an affine function inequality involves showing that a certain mathematical expression involving affine functions is always true. In this case, we are trying to prove that f(x + y) = f(x) + f(y) - f(0), which is known as the affine function inequality.

3. How is an affine function inequality proven?

An affine function inequality can be proven using various techniques, such as algebraic manipulation or mathematical induction. In this case, we can use the properties of affine functions and basic algebraic operations to show that the inequality holds for all values of x and y.

4. What is the significance of proving an affine function inequality?

Proving an affine function inequality can have various applications in mathematics and other fields. It can help in solving optimization problems, proving theorems, and understanding the behavior of linear systems. Additionally, it can also improve our understanding of affine functions and their properties.

5. Can an affine function inequality be generalized to higher dimensions?

Yes, the affine function inequality can be extended to higher dimensions. In this case, the expression would involve multiple variables and constants, but the same principles and techniques can be used to prove its validity.

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