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.

## Intersection of nested convex sets.

ntersection of nested closed compact sets is not empty.

Intersection of nested unbounded closed convex sets may be empty. Consider . The sets escape to infinity along the common direction of recession . However, if all directions of recession are included in a common linearity space then this cannot happen as stated in the proposition below.

Proposition

(Principal intersection result). Let be a sequence of nonempty closed convex sets, . Let and be the recession cone and linearity space of and let If and then the intersection is nonempty and for some nonempty compact set .

Proof

Starting from some the has to stop decreasing because it is a space of finite dimension. After such we have . Let us restrict attention to such .

Starting from some we must have Indeed, if this is not so then for every there is . We consider . The sets are closed and nested. Hence, a limit point of such sequence has to be in . This contradicts the condition .

We now apply the result ( Decomposition of a convex set ). Hence, Starting from some the set has no direction of recession. Hence, the sets are nested and compact. We conclude .

Proposition

(Linear intersection result). Let be a sequence of closed convex subsets of .

Let be the set given by the relationships where and .

Assume that

1. for all .

2. for all

3. where , , .

Then .

To see that the has to be linear consider and . Such and fail only the linearity requirement and the conclusion of the theorem.

Proof

If then the statement is a consequence of the ( Principal intersection result ). We exclude such case from further consideration.

We consider the case when . Since always and then there has to be a that does not belong to .

Let us take a sequence such that . Since the sets are nested it is enough to prove the statement for some subsequence.

For any we form the sum . Since and then for some the lies on the boundary of . Hence, . The is given by a finite number of linear conditions. Hence, there is some such that for infinite number of . We restrict our attention to such subsequence.

The set satisfies the conditions of the proposition within the subspace and is one dimension smaller then . Therefore, we proceed by induction in the number of dimensions of . The proposition is true for dimension 0 ( is a point). Then we assume that it is true for and prove it for using the construction above. Indeed, since the proposition holds for then the intersection is not empty for the chosen subindexing of . But hence .

Proposition

(Quadratic intersection result). Let be a sequence of subsets of given by where is a symmetric positive semidefinite matrix, is a vector, is a scalar and is a non-increasing sequence of real numbers converging to 0.

Let be a subset of of the form where the are positive semidefinite matrixes.

Let be nonempty for all .

Then the intersection is nonempty.

Proof

The elements of recession cones and linear spaces of are given by and are -independent.

If then the statement follows from the ( Principal intersection result ). Hence, we consider the situation and . If there is a then and for any we have . Note,

If then for a sufficiently large the lies in all and in and we are done.

Therefore, it remains to consider a situation when for any we have but .

The recession cone of is given by Hence, we are considering the case when for any we have and for some .

We now proceed by induction in the number of conditions . For the case that we are considering is excluded. Hence, we assume that the proposition holds for and proceed to prove that it hold for . We are interested only in adding an equation with because the all the equations with may be arranged to be in the beginning of the induction and, hence, fall into the category.

A step of the induction in proceeds in the following stages.

1. Assume that the statement holds for .

2. Let be the -equations set.

3. Let be the set holding equations of . The exclusion of a condition from makes the a bigger set. Hence, the conditions of the statement holds for the set and is not empty.

4. We take a point and a direction . In our case for the one additional equation. Hence, we can construct by taking a sufficiently large .

 Notation. Index. Contents.