Partial Differential Equations - Modelling and ... - ResearchGate
Partial Differential Equations - Modelling and ... - ResearchGate Partial Differential Equations - Modelling and ... - ResearchGate
264 Y. Achdou The spaces H s (R) are Hilbert spaces, with the inner product and norm: ∫ √ (w 1 ,w 2 ) H s (R) = (1 + ξ 2 ) s ŵ 1 (ξ)ŵ 2 (ξ)dξ, ‖w‖ H s (R) = (w, w) H s (R). R We refer to [Ada75] for the properties of the spaces H s (R). If s is a nonnegative integer, we define the semi-norm ( s∑ ∥ |v| Hs (R) = d l v ∥∥∥ 2 ∥dy l l=1 If s>0 is not an integer, we define |v| Hs (R) by m ∥ |v| 2 H s (R) = ∑ d l v ∥∥∥ 2 ∥dy l l=1 where m is the integer part of s. L 2 (R) ∫ ∫ + R R L 2 (R) ) 1 2 . ( ) 2 d m v dy (y) − dm v m dy (z) m |y − z| 1+2s , Some Weighted Sobolev Spaces on R + Let L 2 (R + ) be the Hilbert space of square integrable functions on R + ,endowed with the norm ‖v‖ L2 (R +) = ( ∫ R + v(x) 2 dx) 1 2 and the inner product (v, w) L 2 (R +) = ∫ R + v(x)w(x)dx. LetV 1 be the weighted Sobolev space { V 1 = v ∈ L 2 (R + ), x ∂v } ∂x ∈ L2 (R + ) , (19) which is a Hilbert space with the norm ( ∥ ∥∥∥ ‖v‖ V 1 = ‖v‖ 2 L 2 (R + +) x ∂v ∂x∥ 2 L 2 (R +) ) 1 2 . (20) It is proved in [AP05a] that D(R + ) is a dense subspace of V 1 , and that the following Poincaré inequality is true: for all v ∈ V 1 , ‖v‖ L2 (R +) ≤ 2 ∥ x dv ∥ . (21) dx ∥ L2 (R +) Thus the semi-norm |·| V 1: |v| V 1 = ‖x dv dx ‖ L 2 (R +) is a norm equivalent to ‖·‖ V 1. For a function v defined on R + , call ṽ the function defined on R by ( y ) ṽ(y) =v(exp(y)) exp . (22) 2 By using the change of variable y =log(x), it can be seen that the mapping v ↦→ ṽ is a topological isomorphism from L 2 (R + )ontoL 2 (R), and from V 1 onto H 1 (R). This leads to defining the space V s ,fors ∈ R, by:
Calibration of Lévy Processes with American Options 265 V s = {v :ṽ ∈ H s (R)}, (23) which is a Hilbert space with the norm ‖v‖ V s = ‖ṽ‖ Hs (R). Using the interpolation theorem given, e.g., in [Ada75, Theorem 7.17], one can prove that if 0 0, we introduce the semi-norm |v| V s = |ṽ| Hs (R). Proposition 1. Let s be a real number such that 1 2 0 such that for all x ∈ [1, +∞), √ x|v(x)| ≤C‖v‖V s. (24) 2.3 The Integro-Differential Operator The Integral Operator We study the integral operator B defined in (13). Let ψ be a measurable, non-negative and essentially bounded function defined on R, andα be a real number, 0 ≤ α 0in afixedneighborhoodofz = 0. This assumption is a little restrictive, since, for example, a logarithmic singularity of k will be ruled out. Most of the results below hold without the last assumption on ψ. Proposition 2. Assume that z ↦→ ψ(z)max ( e 2z , 1 ) is a bounded function. If α =0assume, furthermore, that ∫ −1 ψ(z) −∞ |z| dz < +∞. Then, for each s ∈ R, (i) if α> 1 2 , then the operator B is continuous from V s to V s−2α , (ii) if α< 1 2 , then the operator B is continuous from V s to V s−1 , (iii) if α = 1 2 , then the operator B is continuous from V s to V s−1−ε , for any ε>0. Remark 2. As a consequence of Proposition 2, if 1 2
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Calibration of Lévy Processes with American Options 265<br />
V s = {v :ṽ ∈ H s (R)}, (23)<br />
which is a Hilbert space with the norm ‖v‖ V s = ‖ṽ‖ Hs (R). Using the interpolation<br />
theorem given, e.g., in [Ada75, Theorem 7.17], one can prove that if<br />
0 0, we introduce the semi-norm |v| V s = |ṽ| Hs (R).<br />
Proposition 1. Let s be a real number such that 1 2<br />
0 such<br />
that for all x ∈ [1, +∞),<br />
√ x|v(x)| ≤C‖v‖V s. (24)<br />
2.3 The Integro-<strong>Differential</strong> Operator<br />
The Integral Operator<br />
We study the integral operator B defined in (13). Let ψ be a measurable,<br />
non-negative <strong>and</strong> essentially bounded function defined on R, <strong>and</strong>α be a real<br />
number, 0 ≤ α 0in<br />
afixedneighborhoodofz = 0. This assumption is a little restrictive, since, for<br />
example, a logarithmic singularity of k will be ruled out. Most of the results<br />
below hold without the last assumption on ψ.<br />
Proposition 2. Assume that z ↦→ ψ(z)max ( e 2z , 1 ) is a bounded function. If<br />
α =0assume, furthermore, that ∫ −1 ψ(z)<br />
−∞ |z|<br />
dz < +∞. Then, for each s ∈ R,<br />
(i) if α> 1 2 , then the operator B is continuous from V s to V s−2α ,<br />
(ii) if α< 1 2 , then the operator B is continuous from V s to V s−1 ,<br />
(iii) if α = 1 2 , then the operator B is continuous from V s to V s−1−ε , for any<br />
ε>0.<br />
Remark 2. As a consequence of Proposition 2, if 1 2<br />
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