PHLST WITH ADAPTIVE TILING AND ITS APPLICATION TO ...

8 ZHIHUA ZHANG AND NAOKI SAITO
∑
Proposition 1. For ε > 0, take n = M n k , where n k is stated in (13). Let V(x)
k=1
and V n (x) be stated in (10) and (11), respectively. Then
‖V −V n ‖ L p (Q) ≤ ε 4 .
From the construction of V n , we know that V n (x) is a piecewise sine polynomial
which is determined by n coefficients. From (13), we get the approximation order
of V(x) by piecewise sine polynomials V n (x).
Proposition 2. Let V(x) and V n (x) be stated in (10) and (11). Then the following
estimate holds:
(
‖V −V n ‖ Lp (Q) = O n −α k 0
)
2 +s′ + 1 p −1 2
(2 ≤ p < ∞),
where α k0 is stated in (9) and s ′ > 0 is an arbitrarily small number.
Proof. By (13), we get
n ≤
M∑
( 4Ck M
k=1
ε
k=1
) (
α k
2 −s ′ − 1 p +1 2) −1 .
By (9): α k0 = min{α 1 ,...,α M }, we have
α
( k0 ( 4
2
n ≤
ε) −s′ − 1 p +1 2 )−1 ∑ M (
)
(C k M) (α k
2 −s ′ − 1 p +1 2 )−1 J k (ε) , (14)
where J k (ε) = ( 4
ε
we have
k=1
(
αk
(
) (
α k
2 −s ′ − 1 p +1 2) −1 αk0
− 2 −s′ − 1 p +1 2
) −1
. By p > 2 and 0 < α k0 ≤ α k ,
) −1 (
αk0
−
2 −s′ − 1 p + 1 ) −1
≤ 0.
2
2 −s′ − 1 p + 1 2
So J k (ε) ≤ 1. From this, we get
M∑
M∑
(C k M) (α k
2 −s ′ − 1 p +1 2) −1 J k (ε) ≤ (C k M) ( α k
2 −s ′ − 1 p +1 2 ) −1 = O(1),
where the constant in the term “O” is independent of ε. Again, by (14),
⎛
(
n = O⎝
4
ε) ( α ) k0 −1
⎞
2 −s′ − 1 p +1 2
⎠.
k=1
This implies that
ε
(
4 = O n −α k 0
)
2 +s′ + 1 p −1 2
. (15)
By (10), (11), and (12), we get
M∑
‖V −V n ‖ L p (Q) ≤ ‖vk ∗ −σ nk (vk)‖ ∗ L p (Q k ) ≤ ε 4 .
From this and (15), we have
Proposition 2 is proved.
‖V −V n ‖ L p (Q) = O
k=1
(
n −α k 0
)
2 +s′ + 1 p −1 2
(s ′ > 0).