1. Introduction and statement of the main results
Testo completo
− i div SN −1
where div SN −1
u − a(x/|x|)u |x|2
e −|x|24
e −|x|24
T A,a = − i ∇ SN −1
− a(θ) = −∆ SN −1
− 2i A · ∇ SN −1
∇ SN −1
(14) V n,j (x) = |x| −αj
(15) kφk Ht
kuk Lt
N (t) = γ e m0
|∇ A u(x, t) | 2 − a(x/ |x| |x|)2
e −4(t0−t)|x|2
|u(x, t)| 2 e −4(t0 −t)|x|2
Furthermore, denoting as J 0 the finite set of indices J 0 = {(m, k) ∈ N×(N\{0}) : m− α 2k
λ −2γm0,k0
λ −2γm0,k0
e −|x−y|24t
u(x, t) = t −N2
e −|x|2 +|y|24
e iπ2
being β k as in (12), ψ k as in (10), and being J βk
defined, for every t > 0, as the completion of C c ∞ (R N \ {0}, C) with respect to the norm kuk HAt
described in [16]. From [23, Proposition 3.1] (see Lemma 3.1 in section 3) it follows that if N > 3 then H t = H t = H t 0 and the norms k · k Ht
We denote as H ⋆ t the dual space of H t and by H⋆t
|v(x)| 2 e −|x|2
r ∇ SN −1
(∇ SN −1
= t −N2
r N −1 e −r24t
+ t −N2
r N −1 e −r24t
|(∇ SN −1
t −N2
r N −1 e −r24t
= t −N2
r N −1 e −r24t
= t −N2
− t −N2
r N −1 e −r24t
r N −1 e −r24t
|(∇ SN −1
being the norms k · k Ht
u ∈H inft
|∇ A u(x) | 2 − a(x/ |x| |x|)2
|∇ A v(x) | 2 − a(x/|x|) |x|2
|∇ A v ε (x) | 2 + (1 + kAk L∞
6 ε(1 + kAk L∞
1 − (1 + kAk L∞
1 − (1 + kAk L∞
< (1 + kAk L∞
1 − (1 + kAk L∞
4 < (1 + kAk L∞
-norm. Therefore, letting ε → 0 in the above inequality, we obtain µ 1 (A, a) + (N −2) 4 2
Proof. Let u ∈ C c ∞ (R N \ {0}, C) and w = e −|x|28
∇ A w = e −|x|28
e −|x|24
e −|x|24
e −|x|24
e −|x|24
e −|x|24
6 C s t −Ns
Proposition 4.2. Let ˜ T : H → ˜ H be defined as ˜ T u(x) = e −|x|24
|u(x)| 2 e −|x|24
|∇u(x)| 2 e −|x|24
L β mk
if (m 1 , k 1 ) 6= (m 2 , k 2 ) then V m1
. Moreover, letting U n,j = e −|x|24
e −|x|24
= kV m,k k 2 L = 2 1+2βk
Lemma 5.2. Letting C 1 as in Corollary 3.4, we have that, for every j = 1, . . . , k, the function t 7→ t −2C1
H j (t), which implies that dt d t −2C1
Proof. From Lemma 5.2, the function t 7→ t −2C1
(˜u j ) t (x, t) + ∇A
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