I. Basic math.
 II. Pricing and Hedging.
 III. Explicit techniques.
 IV. Data Analysis.
 V. Implementation tools.
 1 Finite differences.
 2 Gauss-Hermite Integration.
 3 Asymptotic expansions.
 4 Monte-Carlo.
 5 Convex Analysis.
 A. Basic concepts of convex analysis.
 B. Caratheodory's theorem.
 C. Relative interior.
 D. Recession cone.
 E. Intersection of nested convex sets.
 F. Preservation of closeness under linear transformation.
 G. Weierstrass Theorem.
 H. Local minima of convex function.
 I. Projection on convex set.
 J. Existence of solution of convex optimization problem.
 K. Partial minimization of convex functions.
 L. Hyperplanes and separation.
 M. Nonvertical separation.
 N. Minimal common and maximal crossing points.
 O. Minimax theory.
 Q. Polar cones.
 R. Polyhedral cones.
 S. Extreme points.
 T. Directional derivative and subdifferential.
 U. Feasible direction cone, tangent cone and normal cone.
 V. Optimality conditions.
 W. Lagrange multipliers for equality constraints.
 X. Fritz John optimality conditions.
 Y. Pseudonormality.
 Z. Lagrangian duality.
 [. Conjugate duality.
 VI. Basic Math II.
 VII. Implementation tools II.
 VIII. Bibliography
 Notation. Index. Contents.

## Extreme points.

efinition

For a nonempty convex set the point is an extreme point if there is no two points such that . We denote the set of all extreme points.

Proposition

(Krein-Milman theorem). Let be a nonempty convex set. Then

1. For a hyperplane that contains in one of its closed halfspaces

2. If is closed then

3. If is compact then

Proof

(1). Assume the contrary: . Then there must be . There are three cases:

a. . Since and this means that does not contain in one of its halfspaces.

b. . Then .

c. . Impossible because and .

Proof

(2). If then any candidate to be in may be translated in both directions along any while remaining in . Hence, . We have for the same reason.

If then for any point there is a direction such that the line hits the relative boundary of at some point . By proposition ( Proper separation 1 ) there is a properly separating hyperplane at that point. By closedness of the set is not empty and . We reduced the dimensionality of our proof. Because of the part (1) of the proposition we can complete this proof by induction in the number of dimensions.

Proof

(3). We prove the statement by induction in the number of dimensions of . For the statement is trivial. Assume that it is true in . Let and , . There is a line that passes through such that and . There are properly seprating hyperplanes and at points and . By applying the statement in , Then by (1), i=1,2. Hence, .

Proposition

(Extreme points of polyhedral set 1). Let be a polyhedral set. According to the proposition ( Minkowski-Weyl representation ) We have

Proof

According to the proposition ( Minkowski-Weyl representation ) any point has the representation , , . An extreme point may not have a non zero -part because it would contradict the definition of the extreme point. The also cannot be a convex combination of . Therefore, the only possibility is .

Proposition

(Extreme points of polyhedral set 2). Let be a polyhedral subset of . Then

1. Let has the form and denote then

2. Let has the form and denote then iff has the maximal rank (all columns are linearly independent) and .

3. Let has the form and denote then iff has the maximal rank (all columns are linearly independent) and .

Proof

(1). We state that iff for any direction vector such that for some and any small one of the conditions is violated. If then no can be orthogonal to all and then one of the conditions is violated. Hence,

Conversely, if there is a that is orthogonal to all then for such if . Hence, .

Proof

(2). We apply the part (1) of the proposition. In context of the part (1) the is represented by where the is the coordinate basis. Therefore, the for such situation has the form Note that is given, hence, the condition above is not restrictive. The set contains linearly independent vectors. Let . We cannot state that according to (1), for we need to have at least independent vectors among . Indeed, some of the might be in the linear span of the . Hence, we need to exclude the projection on . For we need to have where the is the projection. The original matrix has columns in total. To establish the last equality it is enough to form a matrix from the columns , remove the columns that correspond to and check that the remaining matrix has the maximal rank .

Proof

The proof of (3) is the same as the proof of (2).

Proposition

Let be a closed convex set with at least one extreme point. A convex function that attains a maximum over attains the maximum at some extreme point of .

Proof

Let be a segment . Note that is open. A convex function that attains its maximum at is constant on . Such statement follows directly from the definition of convexity.

The proof is based on the above statement and the theorem ( Proper separation 1 ). Let be a point where the maximum is attained. By the above statement either is constant on or . In the former case we are done. In the latter case there is a properly separating hyperplane . Since we have . If then we are done: the is an extreme point. Otherwise we observe that we reduce the dimension of the proof by switching the consideration from to .

 Notation. Index. Contents.