# Macroeconomics/Math Review

## Introduction

We have a Bellman equation and first we want to know if there exists a value function that satisfies the equation and second we want to know the properties of such a solution. In order to answer the question we will define a mapping which maps a function to another function, and a fixed point of the mapping is to be a solution. The mapping we discussed is a mapping on the set of functions, which is a bit abstract. So today we will look at the math review.

So first we consider a set, ${\displaystyle S\subset \mathbb {R} ^{l}}$ , For us what it will be relevant to describe a sort of distance between any two points in a set. We will use the concept of a metric.

### Metric

A metric is a function ${\displaystyle \rho :S\times S\to \mathbb {R} }$  with the properties that it is non-negative, ${\displaystyle \rho (x,y)\geq 0}$ , symmetric, ${\displaystyle \rho (x,y)=\rho (y,x)}$ , and satisfies the triangle inequality,${\displaystyle \rho (x,z)\leq \rho (x,y)+\rho (y,z)}$ ,

A common metric is euclidean distance, ${\displaystyle \rho _{E}(x,y)={\sqrt {\sum _{i=1}^{l}(x_{i}-y_{i})^{2}}}}$ , Another is ${\displaystyle \rho _{max}(x,y)=\max _{1\leq i\leq l}x_{i}-y_{i}}$ ,

### Space

A space, is a set of objects equipped with some general properties and structure

We may be interested in a metric space, a space with a metric such as, ${\displaystyle (S,\rho )}$  where ${\displaystyle S}$  is the set of all bounded rational functions, and ${\displaystyle \rho }$  is some distance function. Once we have a metric space we can discuss convergence and continuity.

### convergence

A sequence, ${\displaystyle \{x_{i}\}\subset S}$ , converges to ${\displaystyle x}$ , ${\displaystyle x_{i}\rightarrow x}$ , if ${\displaystyle \forall \epsilon >0,\exists N_{\epsilon }}$  s.t. ${\displaystyle \rho (x_{n},y_{n})<\epsilon }$  for ${\displaystyle n>N_{\epsilon }}$ ,

### Cauchy sequence

A sequence , ${\displaystyle \{x_{i}\}\subset S}$ , is called a Cauchy sequence if ${\displaystyle \forall \epsilon >0,\exists N_{\epsilon }s.t.\rho (x_{n},y_{m})<\epsilon }$  for ${\displaystyle n,m>N_{\epsilon }}$ ,

Question: does every Cauchy sequence converge?

### Completeness

The metric space, ${\displaystyle (S,\rho )}$  is complete if every Cauchy sequence converges.

### examples of completeness

• ${\displaystyle (\mathbb {R} ,\rho _{E})}$  is complete.
• ${\displaystyle ((0,1),\rho _{E})}$  is not complete. Proof: let ${\displaystyle x_{n}={\frac {1}{n+2}}}$ , So ${\displaystyle \{x_{n}\}}$  os Cauchy, but does not converge to a point in our set ${\displaystyle (0,1)}$ ,
• ${\displaystyle ([0,1],\rho _{E})}$  is complete. Are all closed sets complete? A closed subspace of a complete space is complete.
• ${\displaystyle (\{0,1,2\},\rho _{E})}$  is complete.

## Contraction Mapping

A mappting ${\displaystyle T:S\rightarrow S}$  is a contraction mapping on a metric space, ${\displaystyle (S,\rho )}$ , if ${\displaystyle \exists o\geq \beta <1}$  such that ${\displaystyle \rho (tx,Ty)\leq \beta \rho (x,y)\forall x,y\in S}$ , Sometimes we write ${\displaystyle T(x)}$  instead of ${\displaystyle TX}$ ,

This means that any two points in our set, ${\displaystyle S}$ , are mapped such that after the mapping the distance between the points shrinks.

### examples of contraction mapping

• ${\displaystyle Tx=.9x}$  is a contraction mapping on ${\displaystyle [(0,1],\rho _{E})}$ ,

Now we state the contraction mapping theorem.

### Contraction mapping theorem

If ${\displaystyle (S,\rho )}$  is complete and ${\displaystyle T:s\rightarrow S}$  is a contraction mapping, then ${\displaystyle \exists !x^{*}}$  with ${\displaystyle Tx^{*}=x^{*}}$ ,

We will prove this theorem for a general metric space later on. However, we must remember that it is necessary for this proof that the space be complete.

Let us now look at a criteria to verify that a mapping is a contraction mapping.

### Contraction Mapping criteria

For ${\displaystyle S\subset \mathbb {R} ^{l}}$  and ${\displaystyle \rho =\rho _{E}}$ , Let ${\displaystyle T:S\rightarrow S}$  satisfy the following two conditions:

• (M, monotonic condition)${\displaystyle \forall x=(x_{1},x_{2},\ldots ,x_{l})\in S}$  and ${\displaystyle y=(y_{1},y_{2},\ldots ,y_{l})\in S}$ , and ${\displaystyle T=(T_{1},T_{2},\ldots ,T_{l}}$ , if ${\displaystyle x_{i}\geq y_{i}\Leftrightarrow x\geq y}$  then ${\displaystyle T_{i}x\geq T_{i}y\Leftrightarrow Tx\geq Ty}$ ,
• (D, discout condition) ${\displaystyle \forall x=(x_{1},x_{2},\ldots ,x_{l})\in S}$ , for ${\displaystyle {\underline {\vec {a}}}=(a,a,\ldots ,a)}$ , ${\displaystyle T_{i}(x_{1}+a,x_{2}+a,\ldots ,x_{l}+a)\leq T_{i}(x_{1},x_{2},\ldots ,x_{l})+\beta a\forall i\Leftrightarrow T(x+{\underline {a}})\leq TX+\beta {\underline {a}}}$ .

Then ${\displaystyle T}$  is a contraction mapping.