PHLST WITH ADAPTIVE TILING AND ITS APPLICATION TO ...

10 ZHIHUA ZHANG AND NAOKI SAITO
where A k is the area of Q k . From this and (17)-(19), by the maximum modulus
principle, we get
( )
‖u k −ũ nk ‖ C(Qk ) = max
1≤i≤4 ‖h(k) i −σ nk h (k)
i ‖ C
(τ ) ≤ C kn ′ −α k+s− 1
(k)
2
k
. (20)
i
From this, we have the following proposition.
Proposition 3. Let
( ) 4C
n ′ ′
k ≈ k
M
(αk −s+ 2) 1 −1 M∑
and n ′ = n ′
ε
k.
Denote
Then ‖U −U n ′‖ L p (Q) ≤ ε 4 .
Again, by (20),
U n ′(x) =
‖u k −ũ n ′
k
‖ L p (Q k ) = O
This implies the following proposition.
Proposition 4.
‖U −U n ′‖ L p (Q) = O
By Propositions 1 and 3, we obtain
M∑
ũ n ′
k
(x)χ Qk (x).
k=1
( )
(n ′ k) −α k+s− 1 2
( )
(n ′ ) −α k 0 +s− 1 2
k=1
(k = 1,...,M).
(s > 0).
Theorem 5.1. Suppose that f ∈ L p (Ω) (2 ≤ p < ∞) and Ω be a bounded domain
in R 2 . For ε > 0, Q = M ⋃
Q k is an adaptive tiling of Ω described in Section 4
which satisfies
k=1
‖f χ Ω\Q ‖ L p (Ω) < ε 2 .
Denote the global Besov index of f on Q k by α k . Let
f(x) = u k (x)+v k (x)
on Q k
be the m k −fold PHLST decomposition where m k is the maximal integer satisfying
2m−1 ≤ α k and denote
M∑
M∑
U(x) = u k (x)χ Qk (x), V(x) = v k (x)χ Qk (x).
Set
k=1
( 4Ck M
n k =
ε
k=1
) (
α k
2 −s ′ − 1 p +1 2) −1 , n =
M∑
k=1
( ) 4C
n ′ ′
k = k
M
(αk −s+ 2) 1 −1 M∑
, n ′ = n ′
ε
k,
where the constants C k and C
k ′ are stated in (13) and (18). Denote
M∑
M∑
U n ′(x) = u n ′
k
(x)χ Qk (x), V n (x) = v nk (x)χ Qk (x).
k=1
k=1
k=1
n k