How can I build a fixed point theorem argument in pure strategies? To begin with, I am recalling the Banach Fixed Point Theorem.
Let $(X,d)$ be a non-empty complete metric space with a contraction mapping $T:X\to X$. Then $T$ admits a unique fixed-point $x^*$ in $X$ i.e. $T(x^*) = x^*$.
In order to prove that the combinantion of the best response strategies of $N \geqslant 2$ agents consitutes a Nash Equilibrium, we need to use a fixed point theorem argument. Is this necessary only in cases where the agents employ mixed strategies or this holds in the case where they also follow pure strategies? I am searcing for some guidance to use the Banch fixed point theorem in guadratic utility functions when agents act by submitting pure strategies. How can I build this argument? What confuses me the most is that the mapping $T:X\to X$ is about the strategies isn't it?
I can provide more details of my problem if you wish so. I would also be glad if you could answer with great details, since I do not have any idea about the topic
There is no apriori reason for the Best Response map to be a contraction in general. Here's a simple example (since Battle of Sexes has been my go-to for the past few days):
$$
\begin{array}{c|lcr}
\text{Player1/Player 2$\rightarrow$} &amp; \text{F} &amp; \text{T} \\
\hline
\text{F} &amp; 3,1 &amp; 0,0 \\
\text{T} &amp; 0,0 &amp; 1,3 \\
\end{array}
$$
Denote the strategy of player 1: $x = Pr(T)$ and that of player 2 by: $y=Pr(T)$. The best response of player 1  is a map $BR_1:[0,1]\rightarrow[0,1]$ defined by:
$
BR_1(y) = \begin{cases}
0 &amp; \text{ if } y&lt;\frac{3}{4}\\
[0,1] &amp; \text{ if } y = \frac{3}{4}\\
1 &amp; \text{ if } y&gt;\frac{3}{4}\\ 
\end{cases}
$
Similarly, the best response of player 2 is a map $BR_2:[0,1]\rightarrow[0,1]$ defined by:
$
BR_2(x) = \begin{cases}
0 &amp; \text{ if } x&lt;\frac{1}{4}\\
[0,1] &amp; \text{ if } x = \frac{1}{4}\\
1 &amp; \text{ if } x&gt;\frac{1}{4}\\ 
\end{cases}
$
Define the Best Response profile $BR:[0,1]^2\Rightarrow[0,1]^2$ as $BR(y,x) \equiv \big(BR_1(y), BR_2(x)\big)$. A Nash equilibrium of this game is a fixed point of $BR$.
This, however, need not be a contraction map. Take two points on the domain: $a = (0.8,0.5)$ and $b = (0.5,0.2)$. Its easy to see that $BR(0.8,0.5) =(1,1)$ and $BR(0.5,0.2)=(0,0)$. Thus the Euclidean distance $d\big(BR(a),BR(b)\big) = \sqrt{2}$ whereas $d\big(a,b\big) = 0.3\sqrt{2}$. It is clear that $\not\exists$ $k&gt;0$ such that $d\big(BR(a), BR(b)\big)\leq k d\big(a,b\big)$.