Macroeconomics/Math Review

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.

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 ${\displaystyle \rho _{max}(x,y)=\max _{1\leq i\leq l}x_{i}-y_{i}}$,

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.

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 ${\displaystyle n>N_{\epsilon }}$,

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?

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

• ${\displaystyle (\mathbb {R} ,\rho _{E})}$ 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.

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.

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

Now we state the 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.

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.