Difference between revisions of "Additive function"

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(Definition: addition operation should not be called additive function)
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==Definition==
 
==Definition==
Here {{M|(X,+_X:X\times X\rightarrow Y)}} (which we'll denote {{M|X}} and {{M|+_X}})  denotes a set endowed with a binary operation called addition.
+
Here {{M|(X,+_X:X\times X\rightarrow X)}} (which we'll denote {{M|X}} and {{M|+_X}})  denotes a set endowed with a binary operation called addition.
  
 
The same goes for {{M|(Y,+_Y:Y\times Y\rightarrow Y)}}.
 
The same goes for {{M|(Y,+_Y:Y\times Y\rightarrow Y)}}.

Revision as of 07:49, 19 March 2016

An additive function is a homomorphism that preserves the operation of addition in place on the structure in question.

In group theory (because there's only one operation) it is usually just called a "group homomorphism"

Definition

Here [ilmath](X,+_X:X\times X\rightarrow X)[/ilmath] (which we'll denote [ilmath]X[/ilmath] and [ilmath]+_X[/ilmath]) denotes a set endowed with a binary operation called addition.

The same goes for [ilmath](Y,+_Y:Y\times Y\rightarrow Y)[/ilmath].

A function [ilmath]f[/ilmath] is additive[1] if for [ilmath]a,b\in X[/ilmath]

[math]f(a+_Xb)=f(a)+_Yf(b)[/math]

Warning about structure

If the spaces X and Y have some sort of structure (example: Group) then some required properties follow, for example:

[math]x=x+0\implies f(x)=f(x+0)=f(x)+f(0)\implies f(x)=f(x)+0\implies f(0)=0[/math] so one must be careful!

On set functions

A set function, [ilmath]\mu[/ilmath], is called additive if[2] whenever:

  • [ilmath]A\in X[/ilmath]
  • [ilmath]B\in X[/ilmath]
  • [ilmath]A\cap B=\emptyset[/ilmath]

We have:

[math]\mu(A\cup B)=\mu(A)+\mu(B)[/math] for valued set functions (set functions that map to values)

An example would be a measure

Variations

Finitely additive

This follows by induction on the additive property above. It states that:

  • [math]f(\sum^n_{i=1}A_i)=\sum^n_{i=1}f(A_i)[/math] for additive functions
  • [math]\mu(\bigcup^n_{i=1}A_i)=\sum^n_{i=1}\mu(A_i)[/math] for valued set functions

Countably additive

This is a separate property, while given additivity we can get finite additivity we cannot get additivity, we cannot get countable additivity from just additivity.

  • [math]f(\sum^\infty_{n=1}A_n)=\sum^\infty_{n=1}f(A_n)[/math] for additive functions
  • [math]\mu(\bigcup^\infty_{n=1}A_n)=\sum^\infty_{n=1}\mu(A_n)[/math] for valued set functions

Countable additivity can imply additivity

If [math]f(0)=0[/math] or [math]\mu(\emptyset)=0[/math] then given a finite set [math]\{a_i\}_{i=1}^n[/math] we can define an infinite set [math]\{b_n\}_{n=1}^\infty[/math] by:

[math]b_i=\left\{\begin{array}a_i&\text{if }i\le n\\ 0\text{ or }\emptyset & \text{otherwise}\end{array}\right.[/math]

Thus:

  • [math]f(\sum^\infty_{n=1}b_n)= \begin{array}{lr} f(\sum^n_{i=1}a_i) \\ \sum^\infty_{n=1}f(b_n)=\sum^n_{i=1}f(a_i)+f(0)=\sum^n_{i=1}f(a_i) \end{array}[/math]
  • Or indeed [math]\mu(\sum^\infty_{n=1}b_n)= \begin{array}{lr} \mu(\sum^n_{i=1}a_i) \\ \sum^\infty_{n=1}\mu(b_n)=\sum^n_{i=1}\mu(a_i)+\mu(0)=\sum^n_{i=1}\mu(a_i) \end{array}[/math]

References

  1. http://en.wikipedia.org/w/index.php?title=Additive_function&oldid=630245379
  2. Halmos - p30 - Measure Theory - Springer - Graduate Texts in Mathematics (18)