ВУЗ:
Составители:
Рубрика:
a = 1 a = −1
ϕ(x) = χ
[0,1)
(x) ϕ(x) = χ
[0,1)
(x 1)
a = 0 [1/2, 1)
ϕ
ϕ(x) = (1/2)χ
[0,1)
(x/2) {ϕ(·k) | k ∈ Z
+
}
ϕ(x) = ϕ(x 1)
0 < |a| < 1
ϕ(x) = (1/2)χ
[0,1)
(x/2)(1 + a
∞
X
j=0
b
j
w
2
j+1
−1
(x/2)), x ∈ R
+
.
ϕ
L
2
(R
+
)
Ω
j
(
ϕ
) ≤ C|b|
j
, j ∈ N,
ϕ(x) =
(1 + a − b)/2 + bϕ(2x) 0 ≤ x < 1,
(1 − a + b)/2 − bϕ(2x −2) 1 ≤ x ≤ 2.
|b| < 1/2 {2
j/2
ψ(2
j
· k) | j ∈ Z, k ∈ Z
+
}
ψ
ϕ
L
q
(R
+
), 1 < q < ∞
p = 2 n = 3
m(ω) =
1, ω ∈ [0, 1/4) ∪ [3/8, 1/2),
b, ω ∈ [1/4, 3/8),
0, ω ∈ [1/2, 3/4) ∪ [7/8, 1),
β, ω ∈ [3/4, 7/8),
0 ≤ |b| < 1 |β| =
p
1 − |b|
2
m
E = [0, 1/2) ∪[3/4, 1) ∪ [3/2, 7/4)
ϕ(x) =
1
4
χ
[0,1)
(x/4) [1 + w
1
(x/4) + bw
2
(x/4) + w
3
(x/4) + βw
6
(x/4)] . (16)
ϕ
ϕ(x) =
7
X
k=0
c
k
ϕ(2x k)
c
0
= (3 + b + β)/4, c
1
= (3 + b − β)/4, c
2
= c
6
= (1 − b − β)/4,
Ïðè a = 1 è a = −1 ðåøåíèÿìè óðàâíåíèÿ (15) ÿâëÿþòñÿ ôóíêöèÿ Õàà- ðà ϕ(x) = χ[0,1) (x) è ñìåùåííàÿ ôóíêöèÿ Õààðà ϕ(x) = χ[0,1) (x 1) ñîîò- âåòñòâåííî.  ñëó÷àå a = 0 èíòåðâàë [1/2, 1) ÿâëÿåòñÿ áëîêèðîâàííûì ìíî- æåñòâîì äëÿ ìàñêè óðàâíåíèÿ (15), ôóíêöèÿ ϕ îïðåäåëÿåòñÿ ïî ôîðìóëå ϕ(x) = (1/2)χ[0,1) (x/2) è ñèñòåìà {ϕ(· k) | k ∈ Z+ } ëèíåéíî çàâèñèìà (ëåãêî âèäåòü, ÷òî ϕ(x) = ϕ(x 1)). Ïóñòü òåïåðü 0 < | a| < 1. Òîãäà èç ôîðìóëû (14) ïîëó÷àåòñÿ ìàñøòàáè- ðóþùàÿ ôóíêöèÿ Ëýíãà: ∞ X ϕ(x) = (1/2)χ[0,1) (x/2)(1 + a bj w2j+1 −1 (x/2)), x ∈ R+ . j=0 Ýòà ôóíêöèÿ ϕ ãåíåðèðóåò 2-ÊÌÀ â L2 (R+ ), óäîâëåòâîðÿåò íåðàâåíñòâó Ωj (ϕ) ≤ C| b|j , j ∈ N, è îáëàäàåò ôðàêòàëüíûì ñâîéñòâîì: äëÿ 0 ≤ x < 1, (1 + a − b)/2 + bϕ(2x) ϕ(x) = (1 − a + b)/2 − bϕ(2x − 2) äëÿ 1 ≤ x ≤ 2. Êðîìå òîãî, ïðè óñëîâèè | b| < 1/2 ñèñòåìà {2j/2 ψ(2 j · k) | j ∈ Z, k ∈ Z+ }, ãäå âåéâëåò ψ îïðåäåëåí ïî ϕ ñ ïîìîùüþ ôîðìóëû (2), ÿâëÿåòñÿ áåçóñëîâíûì áàçèñîì âî âñåõ ïðîñòðàíñòâàõ Lq (R+ ), 1 < q < ∞. 4.10. Ïóñòü p = 2, n = 3 è 1, ω ∈ [0, 1/4) ∪ [3/8, 1/2), b, ω ∈ [1/4, 3/8), m(ω) = 0, ω ∈ [1/2, 3/4) ∪ [7/8, 1), β, ω ∈ [3/4, 7/8), ãäå 0 ≤ | b| < 1, |β| = 1 − | b|2 . Äëÿ ìàñêè m ìîäèôèöèðîâàííîå óñëîâèå p Êîýíà âûïîëíåíî íà ìíîæåñòâå E = [0, 1/2) ∪ [3/4, 1) ∪ [3/2, 7/4). Ïîëüçóÿñü (14), ïîëó÷àåì ìàñøòàáèðóþùóþ ôóíêöèþ 1 ϕ(x) = χ[0,1) (x/4) [1 + w1 (x/4) + bw2 (x/4) + w3 (x/4) + βw6 (x/4)] . (16) 4 Ôóíêöèÿ ϕ óäîâëåòâîðÿåò óðàâíåíèþ 7 X ϕ(x) = ck ϕ(2x k) k=0 ñ êîýôôèöèåíòàìè c0 = (3 + b + β)/4, c1 = (3 + b − β)/4, c2 = c6 = (1 − b − β)/4, 101
Страницы
- « первая
- ‹ предыдущая
- …
- 99
- 100
- 101
- 102
- 103
- …
- следующая ›
- последняя »