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Printable PDF file
I. Basic math.
II. Pricing and Hedging.
III. Explicit techniques.
IV. Data Analysis.
V. Implementation tools.
VI. Basic Math II.
VII. Implementation tools II.
1. Calculational Linear Algebra.
2. Wavelet Analysis.
A. Elementary definitions of wavelet analysis.
B. Haar functions.
C. Multiresolution analysis.
D. Orthonormal wavelet bases.
E. Discrete wavelet transform.
F. Construction of MRA from scaling filter or auxiliary function.
G. Consequences and conditions for vanishing moments of wavelets.
H. Existence of smooth compactly supported wavelets. Daubechies polynomials.
I. Semi-orthogonal wavelet bases.
a. Biorthogonal bases.
b. Riesz bases.
c. Generalized multiresolution analysis.
d. Dual generalized multiresolution analysis.
e. Dual wavelets.
f. Orthogonality across scales.
g. Biorthogonal QMF conditions.
h. Vanishing moments for biorthogonal wavelets.
i. Compactly supported smooth biorthogonal wavelets.
j. Spline functions.
k. Calculation of spline biorthogonal wavelets.
l. Symmetric biorthogonal wavelets.
J. Construction of (G)MRA and wavelets on an interval.
3. Finite element method.
4. Construction of approximation spaces.
5. Time discretization.
6. Variational inequalities.
VIII. Bibliography
Notation. Index. Contents.

Riesz bases.


efinition

(Riesz basis) A collection of functions MATH is a Riesz basis for the MATH (closure taken in $L^{2}$ ) iff

(a). MATH are linearly independent,

(b). MATH such that MATH we have

MATH (Riesz frame)

Proposition

(Biorthogonality criteria 1) Let MATH . Then MATH is biorthogonal to MATH iff MATH

Proof

is similar to the proof of the proposition ( OST property 1 ).

Proposition

(Existence of biorthogonal basis 1) If MATH has the property

MATH (Frame formula 1)
then there exists a biorthogonal basis MATH with the function $\tilde{\phi}$ given by MATH

Proof

Apply the proposition ( Biorthogonality criteria 1 ).

Proposition

(Frame property 1) If MATH satisfies the property ( Frame formula 1 ) then for MATH MATH where MATH

Proof

We calculate MATH hence MATH Make change $z=y+n$ and use MATH . MATH and we arrive to the desired conclusion.

Proposition

(Frame property 2)

1. If a function MATH satisfies the formula ( Frame formula 1 ) then there exist $A,B=const>0$ such that MATH we have

MATH (Riesz property)

2. If a function MATH has compact support and $\phi$ satisfies the formula ( Riesz property ) then it satisfies the formula ( Frame formula 1 ).

Proof

(1). For a function MATH we calculate MATH MATH We use the formula ( Property of scale and transport 4 ). MATH We continue calculation of MATH We make a change $z=y+m$ . The function MATH has period 1. MATH The function MATH has period 1 and the above expression is the Fourier coefficient of it. By proposition ( Parseval equality ) we have MATH We substitute the formula ( Frame formula 1 ): MATH so that MATH and use the following consequence of the proposition ( Frame property 1 ): MATH thus MATH

Proof

(2) According to the proposition ( Property of transport 2 ), the function MATH takes the form MATH for some finite sequence of real numbers MATH . For this reason we already have MATH To prove the part MATH it suffices to disprove existence of MATH such that MATH . This is so because MATH is periodic.

For MATH to vanish, every term MATH has to vanish. In other words, we aim to show that there cannot exist a $y^{\ast}$ such that MATH or MATH

In the part (1) of the proof we obtained the generic relationship MATH . We will reach our goal if we can find a sequence MATH : MATH where the $\hat{c}_{N}$ is Fourier transform of MATH : MATH Indeed, if such MATH exists then, according to MATH , MATH and we would have a contradiction with the formula ( Riesz property ).

Existence of such MATH is standard. We approximate MATH on $\left[ 0,1\right] $ by an $N$ -indexed sequence of periodic functions, obtain its Fourier coefficients MATH and use them in the combination MATH .

Proposition

(Frame property 3) If functions MATH satisfy the conditions

(a) $\phi$ is compactly supported,

(b) MATH is a Riesz basis for MATH ,

(c) MATH is biorthogonal to MATH .

then

(1) $\forall f\in$ MATH MATH

(2) there exist $A,B=const>0$ such that $\forall f\in$ MATH MATH

Proof

It suffices to prove (1) and (2) for MATH . Extending to the closure is a standard exercise because all involved operations are $L^{2}$ -continuous.

Let MATH then by applying MATH we get MATH hence (1).

According to the proposition ( Biorthogonality criteria 1 ), MATH and by condition (b) and proposition ( Frame property 2 )-2, MATH for some MATH From MATH we derive MATH for some MATH .

We now apply the proposition ( Frame property 1 ): MATH By the proposition ( Parseval equality ), MATH hence (2).





Notation. Index. Contents.


















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