Do a bunch more work on the intro/background.

diff --git a/doc/README.rst b/doc/README.rst
index 2a01957e8ea5f3773019b7019fc2604383dd2075..6d00dee5e77396cbee8d844108ef55c14ff74312 100644 (file)
--- a/doc/README.rst
+++ b/doc/README.rst
@@ -114,6 +114,102 @@ is linear, the set :math:`K` is a symmetric cone in :math:`V`, and the
 points :math:`e_{1}` and :math:`e_{2}` belong to
 :math:`\operatorname{int}\left(K\right)`.
 
+**Definition.**  The strategy sets for our linear game are
+
+.. math::
+  \begin{aligned}
+    \Delta_{1}\left(L,K,e_{1},e_{2}\right) &=
+      \left\lbrace
+        x \in K\ \middle|\ \left\langle x,e_{2} \right\rangle = 1
+      \right\rbrace\\
+    \Delta_{2}\left(L,K,e_{1},e_{2}\right) &=
+      \left\lbrace
+        y \in K\ \middle|\ \left\langle y,e_{1} \right\rangle = 1
+      \right\rbrace.
+  \end{aligned}
+
+Since :math:`e_{1},e_{2} \in \operatorname{int}\left(K\right)`, these
+are bases for :math:`K`. We will usually omit the arguments and write
+:math:`\Delta_{i}` to mean :math:`\Delta_{i}\left(L,K,e_{1},e_{2}\right)`.
+
+To play the game :math:`\left(L,K,e_{1},e_{2}\right)`, the first
+player chooses an :math:`x \in \Delta_{1}`, and the second player
+independently chooses a :math:`y \in \Delta_{2}`. This completes the
+turn, and the payoffs are determined by applying the payoff operator
+:math:`\left(x,y\right) \mapsto \left\langle L\left(x\right), y
+\right\rangle`. The payoff to the first player is :math:`\left\langle
+L\left(x\right), y \right\rangle`, and since we want the game to be
+zero-sum, the payoff to the second player is :math:`-\left\langle
+L\left(x\right), y \right\rangle`.
+
+The payoff operator is continuous in both arguments because it is
+bilinear and the ambient space is finite-dimensional. We constructed
+the strategy sets :math:`\Delta_{1}` and :math:`\Delta_{2}` to be
+compact and convex; as a result, Karlin's [Karlin]_ general min-max
+Theorem 1.5.1, guarantees the existence of optimal strategies for both
+players.
+
+**Definition.** A pair :math:`\left(\bar{x},\bar{y}\right) \in
+\Delta_{1} \times \Delta_{2}` is an *optimal pair* for the game
+:math:`\left(L,K,e_{1},e_{2}\right)` if it satisfies the *saddle-point
+inequality*,
+
+.. math::
+  \left\langle L\left(x\right),\bar{y} \right\rangle
+  \le
+  \left\langle L\left( \bar{x}\right), \bar{y} \right\rangle
+  \le
+  \left\langle L\left(\bar{x}\right),y \right\rangle
+  \text{ for all }
+  \left(x,y\right) \in \Delta_{1} \times \Delta_{2}.
+
+At an optimal pair, neither player can unilaterally increase his
+payoff by changing his strategy. The value :math:`\left\langle L
+\left( \bar{x} \right) , \bar{y} \right\rangle` is unique (by the same
+min-max theorem); it is shared by all optimal pairs. There exists at
+least one optimal pair :math:`\left(\bar{x},\bar{y}\right)` of the
+game :math:`\left(L,K,e_{1},e_{2}\right)` and its *value* is
+:math:`v\left(L,K,e_{1},e_{2}\right) = \left\langle
+L\left(\bar{x}\right), \bar{y} \right\rangle`.
+
+Thanks to Karlin [Karlin]_, we have an equivalent characterization of
+a game's value that does not require us to have a particular optimal
+pair in mind,
+
+.. math::
+  v\left( L,K,e_{1},e_{2} \right)
+  =
+  \underset{x \in \Delta_{1}}{\max}\
+  \underset{y\in \Delta_{2}}{\min}\
+  \left\langle L\left(x\right),y \right\rangle
+  =
+  \underset{y\in \Delta_{2}}{\min}\
+  \underset{x \in \Delta_{1}}{\max}\
+  \left\langle L\left(x\right),y \right\rangle.
+
+Linear games reduce to two-person zero-sum matrix games in the
+right setting.
+
+**Example.** If :math:`K = \mathbb{R}^{n}_{+}` in :math:`V =
+\mathbb{R}^{n}` and :math:`e_{1} = e_{2} =
+\left(1,1,\ldots,1\right)^{T} \in \operatorname{int}\left(K\right)`,
+then :math:`\Delta_{1} = \Delta_{2} = \Delta`. For any :math:`L \in
+\mathbb{R}^{n \times n}`, the linear game :math:`\left(
+L,K,e_{2},e_{2} \right)` is a two-person zero-sum matrix game. Its
+payoff is :math:`\left(x,y\right) \mapsto y^{T}Lx`, and its value is
+
+.. math::
+  v\left( L,K,e_{1},e_{2} \right)
+  =
+  \underset{x \in \Delta}{\max}\
+  \underset{y\in \Delta}{\min}\
+  \left( y^{T}Lx \right)
+  =
+  \underset{y\in \Delta}{\min}\
+  \underset{x \in \Delta}{\max}\
+  \left( y^{T}Lx \right).
+
+
 References
 ----------