Difference between revisions of "Additive function"

Revision as of 16:44, 13 March 2015 (view source) Alec (Talk | contribs) (Created page with "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...")		Revision as of 16:54, 13 March 2015 (view source) Alec (Talk | contribs) m (→‎Countable additivity can imply additivity) Newer edit →
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	Thus:		Thus:
	* <math>f(\sum^\infty_{n=1}b_n)=		* <math>f(\sum^\infty_{n=1}b_n)=
−	\begin{array}	+	\begin{array}{lr}
−	ff(\sum^n_{i=1}a_i) \\	+	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)		\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>		\end{array}</math>
	* Or indeed <math>\mu(\sum^\infty_{n=1}b_n)=		* Or indeed <math>\mu(\sum^\infty_{n=1}b_n)=
−	\begin{array}	+	\begin{array}{lr}
−	\mu\mu(\sum^n_{i=1}a_i) \\	+	\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)		\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>		\end{array}</math>
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−	Note that a [[Mathjax bug]] was found here.
	{{Definition|Abstract Algebra|Measure Theory}}		{{Definition|Abstract Algebra|Measure Theory}}
		+
	==References==		==References==

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Revision as of 16:54, 13 March 2015

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 $(X,+_X:X\times X\rightarrow Y)$ (which we'll denote $X$ and $+_X$) denotes a space with an additive function assigned.

The same goes for $(Y,+_Y:Y\times Y\rightarrow Y)$.

A function is additive^[1] if for $a,b\in X$

$$f(a+_Xb)=f(a)+_Yf(b)$$

Warning about structure

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

$$x=x+0\implies f(x)=f(x+0)=f(x)+f(0)\implies f(x)=f(x)+0\implies f(0)=0$$ so one must be careful!

On set functions

A set function, $\mu$, is called additive if^[2] whenever:

• $A\in X$
• $B\in X$
• $A\cap B=\emptyset$

We have:

$$\mu(A\cup B)=\mu(A)+\mu(B)$$ 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:

• $$f(\sum^n_{i=1}A_i)=\sum^n_{i=1}f(A_i)$$ for additive functions
• $$\mu(\bigcup^n_{i=1}A_i)=\sum^n_{i=1}\mu(A_i)$$ 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.

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

Countable additivity can imply additivity

If

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

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

Thus:

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

References

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