Carleman estimates and unique continuation property for abstract elliptic equations

Boundary Value Problems20122012:46

DOI: 10.1186/1687-2770-2012-46

Received: 26 January 2012

Accepted: 23 April 2012

Published: 23 April 2012

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Abstract

The unique continuation theorems for elliptic differential-operator equations with variable coefficients in vector-valued L p -space are investigated. The operator-valued multiplier theorems, maximal regularity properties and the Carleman estimates for the equations are employed to obtain these results. In applications the unique continuation theorems for quasielliptic partial differential equations and finite or infinite systems of elliptic equations are studied.

AMS: 34G10; 35B45; 35B60.

Keywords

Carleman estimates unique continuation embedding theorems Banach-valued function spaces differential operator equations maximal L p -regularity operator-valued Fourier multipliers interpolation of Banach spaces

1 Introduction

The aim of this article, is to present a unique continuation result for solutions of a differential inequalities of the form:
P ( x , D ) u ( x ) E V ( x ) u ( x ) E , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ1_HTML.gif
(1)
where
P ( x ; D ) u = i , j = 1 n a i j 2 u x i x j + A u + k = 1 n A k u x k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equa_HTML.gif

here a ij are real numbers, A = A (x), A k = A k (x) and V (x) are the possible linear operators in a Banach space E.

Jerison and Kenig started the theory of L p Carleman estimates for Laplace operator with potential and proved unique continuation results for elliptic constant coefficient operators in [1]. This result shows that the condition VLn/2,locis in the best possible nature. The uniform Sobolev inequalities and unique continuation results for second-order elliptic equations with constant coefficients studied in [2]. This was latter generalized to elliptic variable coefficient operators by Sogge in [3]. There were further improvement by Wolff [4] for elliptic operators with less regular coefficients and by Koch and Tataru [5] who considered the problem with gradients terms. A comprehensive introductions and historical references to Carleman estimates and unique continuation properties may be found, e.g., in [5]. Moreover, boundary value problems for differential-operator equations (DOEs) have been studied extensively by many researchers (see [618] and the references therein).

In this article, the unique continuation theorems for elliptic equations with variable operator coefficients in E-valued L p spaces are studied. We will prove that if n 1 p - 1 p 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq1_HTML.gif 1 μ = 1 p - 1 p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq2_HTML.gif, 1 p + 1 p | = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq3_HTML.gif, VL μ (R n ; L(E)), p, μ∈ (1, ∞) and u W p 2 ( R n ; E ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq4_HTML.gif satisfies (1), then u is identically zero if it vanishes in a nonempty open subset, where W p 2 ( R n ; E ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq5_HTML.gif is an E-valued Sobolev-Lions type space. We prove the Carleman estimates to obtain unique continuation. Specifically, we shall see that it suffices to show that if w ( x ) = x 1 + x 1 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq6_HTML.gif, then
e t w u L p | ( R n ; E ) C e t w L ( ε x , D ) u L p ( R n ; E ) , 1 p + 1 p | = 1 , | α | 1 t ( 1 + 1 n | α | ) e t w D α u L p ( R n ; E ) + e t w A u L p ( R n ; E ) C e t w L ( ε x , D ) u L p ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equb_HTML.gif
In the Hilbert space L2 (R n ; H), we derive the following Carleman estimate
| α | 2 t 3 2 | α | e t w D α u L 2 ( R n ; H ) + e t w A u L 2 ( R n ; H ) C e t w L 0 u L 2 ( R n ; H ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equc_HTML.gif
Any of these inequalities would follow from showing that the adjoint operator L t (x; D) = e tw L (x; D) e -tw satisfies the following relevant local Sobolev inequalities
u L p | ( R n ; E ) C L t u L p ( R n ; E ) , 1 p + 1 p | = 1 , | α | 1 t ( 1 + 1 n | α | ) D α u L p ( R n ; E ) A u L p ( R n ; E ) C L t u L p ( R n ; E ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equd_HTML.gif

uniformly to t, where L0t = e tw L0e -tw . In application, putting concrete Banach spaces instead of E and concrete operators instead of A, we obtain different results concerning to Carleman estimates and unique continuation.

2 Notations, definitions, and background

Let R and C denote the sets of real and complex numbers, respectively. Let
S φ = { ξ C , | arg ξ | φ } { 0 } , φ [ 0 , π ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Eque_HTML.gif
Let E and E1 be two Banach spaces, and L (E, E1) denotes the spaces of all bounded linear operators from E to E1. For E1 = E we denote L (E, E1) by L (E). A linear operator A is said to be a φ-positive in a Banach space E with bound M > 0 if D (A) is dense on E and
( A + ξ I ) 1 L ( E ) M ( 1 + | ξ | ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equf_HTML.gif
with λS φ , φ ∈ (0, π], I is identity operator in E. We will sometimes use A + ξ or A ξ instead of A + ξI for a scalar ξ and (A + ξI)-1 denotes the inverse of the operator A + ξI or the resolvent of operator A. It is known [19, §1.15.1] that there exist fractional powers A θ of a positive operator A and
E ( A θ ) = { u D ( A θ ) , u E ( A θ ) = A θ u E + u < , - < θ < } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equg_HTML.gif
We denote by L p (Ω; E) the space of all strongly measurable E-valued functions on Ω with the norm
u L p = u L p ( Ω ; E ) = Ω u ( x ) E p d x 1 / p , 1 p < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equh_HTML.gif

By Lp,q(Ω) and W p , q l ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq7_HTML.gif let us denoted, respectively, the (p, q)-integrable function space and Sobolev space with mixed norms, where 1 ≤ p, q < ∞, see [20].

Let E0 and E be two Banach spaces and E0 is continuously and densely embedded E.

Let l be a positive integer.

We introduce an E-valued function space W p l ( Ω ; E 0 , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq8_HTML.gif (sometimes we called it Sobolev-Lions type space) that consist of all functions uL p (Ω; E0) such that the generalized derivatives D k l u = l u x k l L p ( Ω ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq9_HTML.gif are endowed with the
u W p l ( Ω ; E 0 , E ) = u L p ( Ω ; E 0 ) + k = 1 n D k l u L p ( Ω ; E ) < , 1 p < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equi_HTML.gif

The Banach space E is called an UMD-space if the Hilbert operator ( H f ) ( x ) = lim ε 0 | x - y | > ε f ( y ) x - y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq10_HTML.gifdy is bounded in L p (R, E), p ∈ (1, ) (see e.g., [21, 22]). UMD spaces include, e.g., L p , l p spaces and Lorentz spaces L pq , p, q ∈ (1, ).

Let E1 and E2 be two Banach spaces. Let S (R n ; E) denotes a Schwartz class, i.e., the space of all E-valued rapidly decreasing smooth functions on R n . Let F and F-1denote Fourier and inverse Fourier transformations, respectively. A function Ψ ∈ C m (R n ; L (E1, E2)) is called a multiplier from L p (R n ; E1) to L q (R n ; E2) for p, q ∈ (1, ) if the map u → Ku = F-1 Ψ (ξ) Fu, uS (R n ; E1) is well defined and extends to a bounded linear operator
K : L p ( R n ; E 1 ) L q ( R n ; E 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equj_HTML.gif

We denote the set of all multipliers from L p (R n ; E1) to L q (R n ; E2) by M p q ( E 1 , E 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq11_HTML.gif. For E1 = E2 = E and q = p we denote M p q ( E 1 , E 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq11_HTML.gif by M p (E). The L p -multipliers of the Fourier transformation, and some related references, can be found in [19, § 2.2.1-§ 2.2.4]. On the other hand, Fourier multipliers in vector-valued function spaces, have been studied, e.g., in [2328].

A set KL (E1, E2) is called R-bounded [22, 23] if there is a constant C such that for all T1, T2, . . . , T m K and u1,u2, . . . , u m E1, mN
0 1 j = 1 m r j ( y ) T j u j E 2 d y C 0 1 j = 1 m r j ( y ) u j E 1 d y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equk_HTML.gif

where {r j } is a sequence of independent symmetric {-1, 1}-valued random variables on [0,1]. The smallest C for which the above estimate holds is called a R-bound of the collection K and denoted by R (K).

Let
U n = { β = ( β 1 , β 2 , , β n ) , β i { 0 , 1 } , i = 1 , 2 , , n } , ξ β = ξ 1 β 1 ξ 2 β 2 ξ n β n , | ξ β | = | ξ 1 | β 1 | ξ 2 | β 2 | ξ n | β n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equl_HTML.gif
For any r = (r1, r2, . . . , r n ), r i ∈ [0, ) the function () r , ξR n will be defined such that
( i ξ ) r = ( i ξ 1 ) r 1 ( i ξ n ) r n , ξ 1 , ξ 2 , , ξ n 0 , 0 , ξ 1 , ξ 2 , , ξ n = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equm_HTML.gif
where
( i t ) ν = | t | ν exp i π 2 sign  t , t ( - , ) , ν [ 0 , ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equn_HTML.gif
Definition 2.1. The Banach space E is said to be a space satisfying a multiplier condition with respect to p, q ∈ (1, ) (with respect to p if q = p) when for Ψ ∈ C(n)(R n ; L (E1, E2)) if the set
ξ | β | + 1 p - 1 q D β Ψ ( ξ ) : ξ R n \ 0 , β U n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equo_HTML.gif

is R-bounded, then Ψ M p q ( E 1 , E 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq12_HTML.gif.

Definition 2.2. The φ-positive operator A is said to be a R-positive in a Banach space E if there exists φ ∈ [0, π) such that the set
L A = { ξ ( A + ξ I ) - 1 : ξ S φ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equp_HTML.gif

is R-bounded.

Remark 2.1. By virtue of [29] or [30] UMD spaces satisfy the multiplier condition with respect to p ∈ (1, ).

Note that, in Hilbert spaces every norm bounded set is R-bounded. Therefore, in Hilbert spaces all positive operators are R-positive. If A is a generator of a contraction semigroup on L q , 1 ≤ q ≤ ∞ [31], A has the bounded imaginary powers with ( A i t ) L ( E ) C e ν | t | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq13_HTML.gif, ν < π 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq14_HTML.gif or if A is a generator of a semigroup with Gaussian bound in E ∈ UMD then those operators are R-positive (see e.g., [24]).

It is well known (see e.g., [32]) that any Hilbert space satisfies the multiplier condition with respect to p ∈ (1, ). By virtue of [33] Mikhlin conditions are not sufficient for operator-valued multiplier theorem. There are however, Banach spaces which are not Hilbert spaces but satisfy the multiplier condition (see Remark 2.1).

Let H k = { Ψ h M p q ( E 1 , E 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq15_HTML.gif, h = ( h 1 , h 2 , , h n ) K } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq16_HTML.gif be a collection of multipliers in M p q ( E 1 , E 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq17_HTML.gif. We say that H k is a uniform collection of multipliers if there exists a constant M > 0, independent on hK, such that
F - 1 Ψ h F u L q ( R n ; E 2 ) M u L p ( R n ; E 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equq_HTML.gif

for all hK and uS (R n ; E1).

We set
C b ( Ω ; E ) = u C ( Ω ; E ) , lim | x | u ( x )  exists . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equr_HTML.gif

In view of [17, Theorem A0], we have

Theorem 2.0. Let E1 and E2 be two UMD spaces and let
Ψ C ( n ) ( R n \ 0 ; L ( E 1 , E 2 ) ) for  p , q ( 1 , ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equs_HTML.gif
If
R ξ | β | + 1 p - 1 q D ξ β Ψ h ( ξ ) : ξ R n \ 0 , β U n K β < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equt_HTML.gif

uniformly with respect to hK then Ψ h (ξ) is a uniformly collection of multipliers from L p (R n ; E1) to L q (R n ; E2).

Let
χ = | α | + n 1 p - 1 q l , α = ( α 1 , α 2 , , α n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equu_HTML.gif

Embedding theorems in Sobolev-Lions type spaces were studied in [1318, 32, 34]. In a similar way as [17, Theorem 3] we have

Theorem 2.1. Suppose the following conditions hold:
  1. (1)

    E is a Banach space satisfying the multiplier condition with respect to p, q ∈ (1, ) and A is a R-positive operator on E;

     
  2. (2)

    l is a positive and α k are nonnegative integer numbers such that 0 ≤ μ ≤ 1 - ϰ, t and h are positive parameters.

     
Then the embedding
D α W p l ( R n ; E ( A ) , E ) L q ( R n ; E ( A 1 - χ - μ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equv_HTML.gif
is continuous and there exists a positive constant C µ such that for
u W p l ( R n ; E ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equw_HTML.gif
the uniform estimate holds
D α u L q ( R n ; E ( A 1 - χ - μ ) ) C μ h μ u W p l ( R n ; E ( A ) , E ) + h - ( 1 - μ ) u L p ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equx_HTML.gif
Moreover, for u W p l ( R n ; E ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq18_HTML.gif the following uniform estimate holds
A 1 - χ - μ u L p ( R n ; E ) C μ h μ u W p l ( R n ; E ( A ) , E ) + h - ( 1 - μ ) u L p ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equy_HTML.gif

3 Carleman estimates for DOE

Consider at first the equation with constant coefficients
L 0 u = k = 1 n D k 2 u + A u = f ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ2_HTML.gif
(2)

where D k = i k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq19_HTML.gif and A is the possible unbounded operator in a Banach space E.

Let w ( x ) = x 1 + x 1 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq20_HTML.gif and t is a positive parameter.

Remark 3.1. It is clear to see that
e t w L 0 [ e - t w u ] = L 0 t ( x , D ) u = e t w k = 1 n D k 2 ( e - t w u ) + e - t w A u = k = 1 n D k 2 u + A u + 2 t w 1 u x 1 + [ - t 2 w 1 2 + t ] u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ3_HTML.gif
(3)
where w 1 = w x 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq21_HTML.gif. Let L0t(x, ξ) is the principal operator symbol of L0t(x, D) on the domain B0, i.e.,
L 0 t ( x , ξ ) = ξ 1 2 - 2 i ξ 1 w 1 t + A + | ξ | | 2 - t 2 w 1 2 = G t ( x , ξ ) B t ( x , ξ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equz_HTML.gif
where
G t ( x , ξ ) = ξ 1 - i A + | ξ | | 2 1 2 + t w 1 , B t ( x , ξ ) = ξ 1 + i A + | ξ | | 2 1 2 - t w 1 , | ξ | | 2 = k = 2 n ξ k 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equaa_HTML.gif

Our main aim is to show the following result:

Remark 3.2. Since Q(ξ) ∈ S (φ) for all φ ∈ [0, π) due to positivity of A, the operator function A + ||2, ξR n is uniformly positive in E. So there are fractional powers of A+||2 and the operator function A + | ξ | | 2 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq22_HTML.gif is positive in E (see e.g., [19, §1. 15.1]).

First, we will prove the following result.

Theorem 3.1. Suppose A is a positive operator in a Hilbert space H. Then the following uniform Sobolev type estimate holds for the solution of Equation (3)
| α | 2 t 3 2 | α | e t w D α u L 2 ( R n ; H ) + e t w A u L 2 ( R n ; H ) C e t w L 0 u L 2 ( R n ; H ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ4_HTML.gif
(4)
By virtue of Remark 3.1 it suffices to prove the following uniform coercive estimate
| α | 2 t 3 2 | α | D α u L 2 ( R n ; H ) + A u L 2 ( R n ; H ) C L 0 t u L 2 ( R n ; H ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ5_HTML.gif
(5)

for u W 2 2 ( R n ; H ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq23_HTML.gif.

To prove the Theorem 3.1, we shall show that L0t(x, D) has a right parametrix T, with the following properties.

Lemma 3.1. For t > 0 there are functions K = K t and R = R t so that
L 0 t ( x , D ) K ( x , y ) = δ ( x - y ) + R ( x , y ) , x , y B 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ6_HTML.gif
(6)
where δ denotes the Dirac distribution. Moreover, if we let T = T t be the operator with kernel K, i.e.,
T f ( x ) = B 0 K ( x , y ) f ( y ) d y , f C 0 ( B 0 ; E ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equab_HTML.gif
and R is the operator with kernel R (x, y), then for large t > 0, the adjoint of these operators satisfy the following estimates
| α | 2 t 2 | α | D α T * f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) , A T * f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ7_HTML.gif
(7)
t 1 2 R * f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ8_HTML.gif
(8)
t 1 2 D ν R * f L 2 ( B 0 ; H ) C | α | | ν | 1 D α f L 2 ( B 0 ; H ) , 1 | ν | 2. http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ9_HTML.gif
(9)
Proof. By Remark 3.2 the operator function A + | ξ | | 2 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq24_HTML.gif is positive in E for all ξR n . Since tw1 + 1S(φ), due to positivity of A, for φ [ π 2 , π ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq25_HTML.gif the factor G t ( x , ξ ) = - i A + | ξ | | 2 1 2 + w 1 t + i ξ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq26_HTML.gif has a bounded inverse G t - 1 ( x , ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq27_HTML.giffor all ξR n , t > 0 and
G t 1 ( x , ξ ) B ( H ) C ( 1 + | t w 1 + i ξ 1 | ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ10_HTML.gif
(10)
Therefore, we call G t (x, ξ) the regular factor. Consider now the second factor
B t ( x , ξ ) = i A + | ξ | | 2 1 2 - ( w 1 t + i ξ 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equac_HTML.gif
By virtue of operator calculus and fractional powers of positive operators (see e.g., [19, §1.15.1] or [35]) we get that - [tw1 + 1] ∉ S (φ) for ξ1 = 0 and tw1 = | |, i.e., the operator B t (x, ξ) does not has an inverse, in the following set
Δ t = { ( x , ξ ) B 0 × R n : ξ 1 = 0 , | ξ | | = t w 1 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equad_HTML.gif
So we will called B t the singular factor and the set Δ t call singular set for the operator function B t . The operator B t - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq28_HTML.gif cannot be bounded in the set Δ t . Nevertheless, the operator B t - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq28_HTML.gif, and hence L 0 t - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq29_HTML.gif, can be bounded when (x, ξ) is sufficiently far from Δ t . For instance, if we define
Γ t = ( x , ξ ) B 0 × R n : | ξ | | t 4 , 4 t , | ξ 1 | t 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equae_HTML.gif
by properties of positive operators we will get the same estimate of type (10) for the singular factor B t . Hence, using this fact and the resolvent properties of positive operators we obtain the following estimate
L 0 t 1 ( x , ξ ) B ( E ) C ( 1 + | ξ | 2 + t 2 ) 1 when  ( x , ξ ) c Γ t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ11_HTML.gif
(11)

where the constant C is independent of x, ξ, t and c Γ t denotes the complement of Γ t .

Let β C 0 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq30_HTML.gif such that, β(ξ) = 0 if | ξ | 1 4 , 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq31_HTML.gif and β (ξ) = 0 near the origin. We then define
β 0 ( ξ ) = β 0 t ( ξ ) β 0 ( ξ ) = 1 - β ( | ξ | | / t ) β ( 1 - ξ 1 / t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equaf_HTML.gif
and notice that β0 (ξ) = 0 on Γ t . Hence, if we define
K 0 ( x , y ) = ( 2 π ) - n R n β 0 ( ξ ) e i ( ( x - y ) , ξ ) L 0 t - 1 ( y , ξ ) d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ12_HTML.gif
(12)
and recall (11), then by [31] it follows from standard microlocal arguments that
L 0 t ( x , D ) K 0 ( x , y ) = ( 2 π ) - n R n β 0 ( ξ ) e i ( ( x - y ) , ξ ) d ξ + R 0 t ( x , y ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equag_HTML.gif
where R0tbelongs to a bounded subset of S-1 which is independent of t. Since operator R 0 t * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq32_HTML.gif also has the same property, it follows that for all f C 0 ( B 0 ; H ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq33_HTML.gif
D ν R 0 t * f L 2 ( B 0 ; H ) C | α | | ν | 1 D α f L 2 ( B 0 ; H ) , 1 | ν | 2. http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equah_HTML.gif
By reasoning as in [31] we get that tR0tbelongs to a bounded subset of S0. So, we have the following estimate
t D ν R 0 t * f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equai_HTML.gif
Moreover, the Remark 3.2, positivity properties of A and, (11) and (12) imply that, the operator functions | α | 2 β 0 ( ξ ) t 2 - | α | ξ α L 0 t - 1 ( x , ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq34_HTML.gif and β 0 ( ξ ) A L 0 t - 1 ( x , ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq35_HTML.gif are uniformly bounded. Then, if we let T0 be the operator with kernel K0 (x, y), by using the Minkowski integral inequality and Plancherel's theorem we obtain
| α | 2 t 2 | α | D α T 0 f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) , A T 0 f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equaj_HTML.gif
For inverting L0t(x, D) on the set Γ t we will require the use of Fourier integrals with complex phase. Let β1 (ξ) = 1 - β0 (ξ). We will construct a Fourier integral operator T1 with kernel
K 1 ( x , y ) = ( 2 π ) - n R n β 1 ( ξ ) e i Φ ( x , y , ξ ) L 0 t - 1 ( y , ξ ) d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ13_HTML.gif
(13)
so that the analogs of (16) and the estimates (7)-(9) are satisfied. Since G t - 1 ( x , ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq36_HTML.gif is uniformly bounded on Γ t , we should expect to construct the phase function Φ in (13) using the factor B t (x, ξ). Specifically, we would like Φ to satisfy the following equation
B t ( x , Φ x ) = B t ( y , ξ ) , y B 0 , ( x , ξ Γ t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ14_HTML.gif
(14)
The Equation (14) leads to complex eikonal equation (i.e., a non-linear partial differential equation with complex coefficients).
( A + | Φ x | ( x , y , ξ ) | 2 ) 1 2 - [ w 1 ( x ) t + i Φ x 1 ( x , y , ξ ) ] = A + | ξ | | 2 1 2 - ( w 1 ( y ) t + i ξ 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ15_HTML.gif
(15)
Since w1 (x) = 1 + x1, w1 (y) = 1 + y1, we have
Φ = ( x - y , ξ ) + ( x 1 - y 1 ) 2 ξ 1 2 ( 1 + y 1 ) + i ( x 1 - y 1 ) 2 | ξ | | 2 ( 1 + y 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ16_HTML.gif
(16)
is a solution of (15). To use this we get
L 0 t ( x , D ) e i Φ ( x , y , ξ ) = e i Φ L 0 t ( x , Φ x ) + e i Φ 2 Φ x 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equak_HTML.gif
Next, if we set
r ( x , y , ξ ) = G t ( y , ξ ) - G t ( x , ξ ) = - i [ w 1 ( y ) - w 1 ( x ) ] t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ17_HTML.gif
(17)
then it follows from L0t(x, ξ) = G t (x, ξ)B t (x, ξ) and (14) that
L 0 t ( x , Φ x ) = L 0 t ( y , ξ ) + B t ( y , ξ ) r ( x , y , ξ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ18_HTML.gif
(18)
Consequently, (16)-(18) imply that
( 2 π ) n L 0 t ( x , D ) K 1 ( x , y ) = R n β 1 ( ξ ) e i Φ d ξ + R n β 1 ( ξ ) r ( x , y , ξ ) G t - 1 ( y , ξ ) e i Φ d ξ R n β 1 ( ξ ) A L 0 t - 1 ( y , ξ ) e i Φ d ξ + R n β 1 ( ξ ) 2 Φ x 1 2 L 0 t - 1 ( y , ξ ) e i Φ d ξ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ19_HTML.gif
(19)
By reasoning as in [3] we obtain that the first and second summands in (19) belong to a bounded subset of S0. So, we see that the equality (5) must hold. Now we let K (x, y) = K0 (x, y) + K1 (x, y) and R (x, y) = R0 (x, y) + R1 (x, y), where
R 1 ( x , y ) = R 10 ( x , y ) + R 11 ( x , y ) , R 10 ( x , y ) = R n β 1 ( ξ ) r ( x , y , ξ ) G t - 1 ( y , ξ ) e i Φ d ξ , R 11 ( x , y ) = R n β 1 ( ξ ) 2 Φ x 1 2 L 0 t - 1 ( y , ξ ) e i Φ d ξ , T 0 f ( x ) = B 0 K 0 ( x , y ) f ( y ) d y , T 1 f ( x ) = B 0 K 1 ( x , y ) f ( y ) d y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equal_HTML.gif
Due to regularity of kernels, by using of Minkowski and Hölder inequalities we get the analog estimate as (7) and (9) for the operators T0 and R10. Thus, in order to finish the proof, it suffices to show that for fL2 (B0; E) one has
| α | 2 t 2 | α | D α T 1 * f L 2 ( B 0 ; H ) + A T 1 * f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ20_HTML.gif
(20)
t 1 2 R 11 * f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ21_HTML.gif
(21)
t 1 2 | | D ν R 11 * f L 2 ( B 0 ; H ) C | α | | ν | 1 D α f L 2 ( B 0 ; H ) , 1 | ν | 2. http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ22_HTML.gif
(22)
However, since R1,1tT1, we need only to show the following
t 3 / 2 T 1 * f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ23_HTML.gif
(23)
By using the Minkowski inequalities we get
T 1 * f L 2 ( R n 1 ; E ) 1 4 1 4 B 0 K 1 * ( x , y ) f ( y ) d y | d y 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equam_HTML.gif
where K 1 * ( x , y ) = K ̄ 1 ( y , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq37_HTML.gif. The estimates (13) and (16) imply that
K 1 * ( x , y ) = ( 2 π ) - n R n - 1 e i x - y m ( x 1 , y 1 , ξ | ) d ξ | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equan_HTML.gif
where
m ( x 1 , y 1 , ξ | ) = - β 1 ( ξ ) e i ( x 1 - y 1 ) ξ 1 + ( x 1 - y 1 ) 2 ( i | ξ | | - ξ 1 ) / 2 ( 1 + x 1 ) L 0 t - 1 ( y , ξ ) d ξ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equao_HTML.gif
Consequently, it follows from Plancherel's theorem that
R n - 1 K 1 * ( x , y ) f ( y ) d y | sup ξ | | m ( x 1 , y 1 , ξ | ) | R n - 1 | f ( y ) | 2 d y | 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ24_HTML.gif
(24)
Note that for every N we have
e i [ ( x 1 - y 1 ) 2 | ξ | | / 2 ( 1 + x 1 ) ] C N [ 1 + t ( x 1 - y 1 ) 2 ] - N on supp  β 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equap_HTML.gif
Since A is a positive operator in E, we have
L 0 t 1 ( x , ξ ) B ( E ) 1 + | 2 i ξ 1 w 1 t + | ξ | 2 t 2 w 1 2 | 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equaq_HTML.gif
when - 2 i ξ 1 w 1 t + A + | ξ | 2 - t 2 w 1 2 S ( φ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq38_HTML.gif. Then by using the above estimate it not easy to check that
- β 1 ( ξ ) e i ξ 1 ( x 1 - y 1 ) - ( x 1 - y 1 ) 2 / 2 ( 1 + x 1 ) L 0 t - 1 ( y , ξ ) d ξ 1 = O ( t - 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equar_HTML.gif
i.e.,
| m ( x 1 , y 1 , ξ | ) | C t - 1 [ 1 + t ( x 1 - y 1 ) 2 ] - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equas_HTML.gif
Moreover, it is clear that
- ( 1 + t x 1 ) - 1 d x 1 = O t - 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equat_HTML.gif
Thus from (24) by using the above relations and Young's inequality we obtain the desired estimate
T 1 * f L 2 ( B 0 ; H ) C t - 1 1 + t ( x 1 - y 1 ) 2 f ( y 1 , ) L 2 d y 1 d x 1 C t - 3 / 2 f L 2 ( R n ; H ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equau_HTML.gif
Moreover, by using the estimate (10) and the resolvent properties of the positive operator A we have
A T 1 * f L 2 ( B 0 ; H ) C f L 2 ( B 0 ; H ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equav_HTML.gif

The last two estimates then, imply the estimates (20)-(22).

Proof of Theorem 3.1: The estimates (7)-(9) imply the estimate (5), i.e., we obtain the assertion of the Theorem 3.1.

4 L p -Carleman estimates and unique continuation for equation with variable coefficients

Consider the following DOE
L ( x , D ) u = i , j = 1 n a i j ( x ) D i j 2 u + A u = f ( x ) , x R n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ25_HTML.gif
(25)

where D k = i k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq39_HTML.gif and A is the possible unbounded operator in a Banach space E and a ij are

real-valued smooth functions in B ε = {xR n , |x| < ε}.

Condition 4.1. There is a positive constant γ such that i , j = 1 n a i j ( x ) ξ i ξ j γ | ξ | 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq40_HTML.gif for all ξR n , x B 0 = { x R n , | x | < 1 4 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq41_HTML.gif

The main result of the section is the following

Theorem 4.1. Let E be a Banach space satisfies the multiplier condition and A be a R-positive operator in E. Suppose the Condition 4.1 holds, n ≥ 3, p = 2 n n + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq42_HTML.gifand p' is the conjugate of p, w = x 1 + x 1 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq43_HTML.gif and a ij C (B ε ). Then for u C 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq44_HTML.gif( B ε ; E(A)) and > 0 , 1 t < 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq45_HTML.gif the following estimates are satisfied:
e t w u L p | ( R n ; E ) C e t w L ( ε x , D ) u L p ( R n ; E ) , 1 p + 1 p | = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ26_HTML.gif
(26)
α 1 t 1 + 1 n - α e t w D α u L p ( R n ; E ) + e t w A u L p ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ27_HTML.gif
(27)
C e t w L ( ε x , D ) u L p ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equaw_HTML.gif
Proof. As in the proof of Theorem 3.1, it is sufficient to prove the following estimates
v L p | ( R n ; E ) C L t ε x , D v L p ( R n ; E ) , 1 p + 1 p | = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ28_HTML.gif
(28)
| α | 1 t ( 1 + 1 n - | α | ) D α v L p ( R n ; E ) + A v L p ( R n ; E ) C L t ( ε x , D ) v L p ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ29_HTML.gif
(29)
where,
L t ε x , D = e t w L ε x , D e - t w = L ε x , D + 2 t w 1 x 1 - ( t w 1 ) 2 - t 2 , w 1 = w x 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equax_HTML.gif
Consequently, since w1 ≃ 1 on B ε , it follows that, if we let Q t (εx, D) be the differential operator whose adjoint equals
Q t * ( ε x , D ) = w 1 - 2 L ( ε x , D ) + 2 t w 1 - 1 x 1 - t 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equay_HTML.gif
then it suffices to prove the following
v L p | ( R n ; E ) C Q t ( ε x , D ) v L p ( R n ; E ) , 1 p + 1 p | = 1 , | α | t ( 1 + 1 n - | α | ) | | D α v | | L p ( R n ; E ) + | | A v | | L p ( R n ; E ) C | | Q t ( ε x , D ) v | | L p ( R n ; E ) , v C 0 ( B ε ; E ( A ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ30_HTML.gif
(30)

The desired estimates will follow if we could constrict a right operator-valued parametrix T, for Q t * (εx, D) satisfying L p estimates. these are contained in the following lemma.

Lemma 4.1. For t > 0 there are functions K = K t and R = R t , so that
Q t * ε x , D K x , y = δ x - y + R x , y , x , y B ε , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ31_HTML.gif
(31)
where δ denotes the Dirac distribution. Moreover, if we let T = T t be the operator with kernel K (x, y) and R be the operator with kernel R (x, y), then if ε and 1 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq46_HTML.gif are sufficiently small, the adjoint of these operators satisfy the following uniform estimates
T * f L p 1 ( R n ; E ) C f L p ( R n ; E ) , 1 p + 1 p | = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ32_HTML.gif
(32)
| α | 1 t ( 1 + 1 n - | α | ) D α T * f L p ( R n ; E ) C f L p ( R n ; E ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ33_HTML.gif
(33)
A T * f L p ( R n ; E ) C f L p ( R n ; E ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equaz_HTML.gif
t 1 n R * f L q R n ; E C f L q R n ; E , q = p , p | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ34_HTML.gif
(34)
t - 1 + 1 n R * f L p R n ; E C f L p R n ; E , f C 0 B ε ; E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ35_HTML.gif
(35)
Proof. The key step in the proof is to find a factorization of the operator-valued symbol Q t * ε x , ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq47_HTML.gif that will allow to microlocally invert Q t * ε x , D http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq48_HTML.gif near the set where Q t * ε x , ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq49_HTML.gif vanishes. Note that, after making a suitable choice of coordinates, it is enough to show that if L (x, D) is of the form
L x , D = D 1 2 + i , j = 2 n a i j D i D j , D j = 1 i x j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equba_HTML.gif
therefore, we can expressed Q t * ε x , ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq50_HTML.gif as
Q t * ε x , ξ = B t ε x , ξ G t ε x , ξ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ36_HTML.gif
(36)
where
B t x , ξ = w 1 - 1 ξ 1 + i A + w 1 - 1 a ε x , ξ | - t , G t x , ξ = w 1 - 1 ξ 1 - i A + w 1 - 1 a ε x , ξ | + t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbb_HTML.gif
where
a x , ξ | = i , j = 2 n a i j x ξ i ξ j . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbc_HTML.gif
The ellipticity of Q(x, D) and the positivity of the operator A, implies that the factor G t (x, ξ) never vanishes and as in the proof of Theorem 3.1 we get that
G t - 1 ε x , ξ B H C 1 + | w 1 - 1 a ε x , ξ | | 1 2 + | t + w 1 - 1 ξ 1 | - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ37_HTML.gif
(37)
x B ε , ξ R n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbd_HTML.gif
i.e., the operator function G t (εx, ξ) has uniformly bounded inverse for (x, ξ) ∈ B ε ×R n . One can only investigate the factor B t (εx, ξ). In fact, if we let
Δ t = x , ξ B ε × R n : ξ 1 = 0 , | ξ | | = t w 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Eqube_HTML.gif
then the operator function B t (x, ξ) is not invertible for (x, ξ) ∈ Δ t . Nonetheless, B t (εx, ξ) and Q t * ε x , ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq51_HTML.gif can be have a bounded inverse when (x, ξ) is sufficiently far away. For instance, if we define
Γ t = x , ξ B ε × R n : | ξ | | t 4 , 4 t , | ξ 1 | t 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbf_HTML.gif
by properties of positive operators we will get the same estimate of type (37) for the singular factor B t . Hence, we using this fact and the resolvent properties of positive operators we obtain the following estimate
Q t * - 1 ε x , ξ B E C 1 + | ξ | | + | t + w 1 - 1 ξ 1 | - 1 when x , ξ c Γ t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ38_HTML.gif
(38)
As in § 3, we can use (38) to microlocallity invert Q t * ε x , D http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq52_HTML.gif away from Γ t . To do this, we first fix β C 0 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq53_HTML.gif as in § 3. We then define
β 0 = β 0 t = 1 - β ξ | / t β 1 - ξ 1 / t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbg_HTML.gif
It is clear that β0 (ξ) = 0 on Γ t . Consequently, if we define
K 0 x , y = 2 π - n R n β 0 ξ e i x - y , ξ Q t * - 1 ε y , ξ d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ39_HTML.gif
(39)
and recall (37), then we can conclude that standard microlocal arguments give that
Q t * ε x , D K 0 x , y = 2 π - n R n β 0 ξ e i x - y , ξ d ξ + R 0 x , y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ40_HTML.gif
(40)
where R0 belongs to a bounded subset of S-1 that independent of t. Since the adjoint operator R 0 * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq54_HTML.gif also is abstract pseudodifferential operator with this property, by reasoning as in [31, Theorem 6] it follows that
R 0 * f L p R n ; E C f L p R n ; E , f C 0 B ε ; E , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ41_HTML.gif
(41)
t R 0 * f L q R n ; E C f L q R n ; E , f C 0 B ε ; E , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ42_HTML.gif
(42)
q = p , p , 1 p + 1 p = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbh_HTML.gif
Moreover, the positivity properties of A and the estimate (38) imply that the operator functions α 2 β 0 ξ t 2 - α ξ α Q t * - 1 ε x , ξ and β 0 ξ A Q t * - 1 ε x , ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq55_HTML.gif are uniformly bounded. Next, let T0 be the operator with kernel K0. Then in a similar way as in [31] we obtain that
α 1 t 2 - α D α T 0 * f L p R n ; E C f L p R n ; E , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ43_HTML.gif
(43)
A T 0 * f L p R n ; E C f L p R n ; E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbi_HTML.gif
which also the first estimate is stronger than the corresponding inequality in Lemma 4.1. Finally, since T0S-2 and 1 p - 1 p | = 2 n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq56_HTML.gif it follows from imbedding theorem in abstract Sobolev spaces [17] that
T 0 * f L p | R n ; E C f L p R n ; E , f C 0 B ε ; E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ44_HTML.gif
(44)

Thus, we have shown that the microlocal inverse corresponding to cΓ t , satisfies the desired estimates.

Let β1 (ξ) = 10 (ξ). To invert Q t * ε x , D http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq57_HTML.gif for (x, ξ) ∈ Γ t , we have to construct a Fourier integral operator T1, with kernel
K 1 x , y = 2 π - n R n β 1 ξ e i Φ x , y , ξ Q 0 t * - 1 ε y , ξ d ξ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ45_HTML.gif
(45)
such that the analogs of (39) and (32)-(35) are satisfied. For this step the factorization (36) of the symbol Q t * ε y , ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq58_HTML.gif will be used. Since the factor G t (εx, ξ) has a bounded inverse for (x, ξ) ∈ Γ t , the previous discussions show that we should try to construct the phase function in (46) using the factor B t (εx, ξ). We would like Φ (x, y, ξ) to solve the complex eikonal equation
B t ε x , Φ x = B t ε y , ξ , x , y B ε , ξ supp  β 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ46_HTML.gif
(46)
Since B t (εx, Φ x ) - B t (εy, ξ) is a scalar function (it does not depend of operator A ), by reasoning as in [3, Lemma 3.4] we get that
Φ ( x , y , ξ ) = ϕ ( x , y , ξ ) + ψ ( x , y , ξ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbj_HTML.gif
where ϕ is real and defined as
ϕ x , y , ξ = x 1 - y 1 ξ 1 + O x - y 2 ξ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbk_HTML.gif
while
ψ x , y , ξ = x 1 - y 1 ξ 1 + O x 1 - y 1 2 ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbl_HTML.gif
and
Im ψ x , y , ξ c x 1 - y 1 2 ξ , c > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ47_HTML.gif
(47)
Then we obtain from the above that
Q t * ε x , D e i Φ x , y , ξ = e i Φ Q t * ε x , Φ x + e i Φ w 1 - 2 L ε x , D Φ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbm_HTML.gif
Next, if we set
r x , y , ξ = G t ε y , ξ - G t ε x , ξ = w 1 - 1 y ξ 1 - i a ε y , ξ - w 1 - 1 x ξ 1 - i a ε x , ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ48_HTML.gif
(48)
then it follows from (36) and (48) that
e i Φ Q t * ε x , Φ x = e i Φ Q t * ε y , ξ + e i Φ B t ε y , ξ r x , y , ξ + O t - N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ49_HTML.gif
(49)
for every N when β1 (ξ) ≠ 0. Consequently, (49), (50) imply that
2 π n Q t * ε x , D K 1 x , y = β 1 ξ e i Φ d ξ + β 1 ξ r x , y , ξ G t - 1 ε y , ξ e i Φ d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbn_HTML.gif
w 1 - 2 β 1 ( ξ ) Q t * - 1 ( ε y , ξ ) ( L ( ε x , D ) Φ ) e i Φ d ξ + O ( t - N ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ50_HTML.gif
(50)
By reasoning as in Theorem 3.1 we obtain from (51) that
Q t * ( ε x , D ) K 1 ( x , y ) = ( 2 π ) - n β 1 ( ξ ) e i ( x - y , ξ ) d ξ + R 10 ( x , y ) + R 11 ( x , y ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbo_HTML.gif
where
R 11 ( x , y ) = ( 2 π ) - n w 1 - 2 β 1 ( ξ ) Q t * - 1 ( ε y , ξ ) ( L ( ε x , D ) Φ ) e i Φ d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ51_HTML.gif
(51)
while R10 belongs to a bounded subset of S-1 and tR10 belongs to a bounded subset of S0. In view of this formula, we see that if we let K (x, y) = K0 (x, y) + K1 (x, y) and R (x, y) = R0 (x, y)+R1 (x, y), where R1 = R10 +R11, then we obtain (31). Moreover, since R10 satisfies the desired estimates, we see from Minkowski inequality that, in order to finish the proof of Lemma 4.1, it suffices to show that for f C 0 ( B ε ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq59_HTML.gif
T 1 * f L p | ( R n ; E ) C f L p ( R n ; E ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ52_HTML.gif
(52)
| α | 1 t ( 1 + 1 n | α | ) D α T 1 * f L p ( R n ; E ) C f L p ( R n ; E ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ53_HTML.gif
(53)
t 1 n R 11 * f L q ( R n ; E ) C f L q ( R n ; E ) , q = p , p | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ54_HTML.gif
(54)
t 1 + 1 n R 11 * f L p ( R n ; E ) C f L p ( R n ; E ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ55_HTML.gif
(55)

where 1 p + 1 p | = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq60_HTML.gif.

To prove the above estimates we need the following prepositions for oscillatory integral in E-valued L p spaces which generalize the Carleson and Sjolin result [36].

Preposition 4.1. Let E be Banach spaces and A C 0 ( R n , L ( E ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq61_HTML.gif. Moreover, suppose Φ ∈ C satisfies | ∇Φ| ≥ γ > 0 on supp A. Then for all λ > 1 the following holds
e i λ Φ ( x ) A ( x ) d x L ( E ) C N λ - N , N = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbp_HTML.gif

where C N -depends only on γ if Φ and A (x) belong to a bounded subset of C and C (R n , L (E)) and A is supported in a fixed compact set.

Proof. Given x0 ∈ supp A. There is a direction νSn-1such that |(ν, ∇Φ)| ≥ γ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq62_HTML.gif on some ball centered at x0. Thus, by compactness, we can choose a partition of unity φ j C 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq63_HTML.gif consisting of a finite number of terms and corresponding unit vectors ν j such that j = 1 m φ j ( x ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq64_HTML.gif on supp A and | ( ν j , Φ ) | γ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq65_HTML.gif on supp φ j . For A j = φ j A it suffices to prove that for each j
e i λ Φ ( x ) A j ( x ) d x L ( E ) C N λ - N , N = 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbq_HTML.gif
After possible changing coordinates we may assume that ν j = (1, 0, . . . , 0) which means that Φ x 1 γ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq66_HTML.gif on supp φ j . If let L ( x ; D ) = Φ x 1 - 1 1 i λ x 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq67_HTML.gif, then L ( x ; D ) e i λ Φ ( x ) = e i λ Φ ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq68_HTML.gif. Consequently, if L * = x 1 1 i λ Φ x 1 - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq69_HTML.gif is a adjoint, then
e i λ Φ ( x ) A ( x ) d x = e i λ Φ ( x ) ( L * ) N A j ( x ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbr_HTML.gif

Since our assumption imply that (L*) N A j (x) = O (λ -N ), the result follows.

Preposition 4.2. Suppose Φ ∈ C is a phase function satisfying the non-degeneracy condition det 2 Φ x i x j 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq70_HTML.gifon the support of
A ( x , y ) C 0 ( R n × R n , L ( E ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbs_HTML.gif
Then for T λ f = R n e i λ Φ ( x , y ) A ( x , y ) f ( y ) d x , λ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq71_HTML.gif the following estimates hold
T λ f L p ( R n ; E ) C λ - n - 1 p f L p ( R n ; E ) , 1 p 2 , T λ f L p ( R n ; E ) C λ - n p f L p ( R n ; E ) , 1 p + 1 p = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbt_HTML.gif
Proof. In view of [3, Remark 2.1] we have
| x [ Φ ( x , y ) - Φ ( x , z ) ] | | y - z | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ56_HTML.gif
(56)
where |y - z| is small. By using a smooth partition of unity we can decompose A (x, y) into a finite number of pieces each of which has the property that (57) holds on its support. So, by (57) we can assume
| x [ Φ ( x , y ) - Φ ( x , z ) ] | C | y - z | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ57_HTML.gif
(57)
on supp A for same C > 0. To use this we notice that
T λ f 2 2 = K λ ( y , z ) f ( y ) f ̄ ( z ) d y d z , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbu_HTML.gif
where
K λ ( y , z ) = R n e i λ [ Φ ( x , y ) - Φ ( x , z ) ] A ( x , y ) Ā ( x , z ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ58_HTML.gif
(58)
Hence, by virtue of Preposition 4.1 and by (58) we obtain that
K λ ( y , z ) L ( E ) C N 1 + | λ | | y - z | - N ,  for all  N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbv_HTML.gif
Consequently, by Young's inequality, the operator with kernel K λ acts
L p ( R n ; E )  to  L p ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbw_HTML.gif
By (59) we get that
T λ f L 2 ( R n ; E ) C λ - n f L 2 ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbx_HTML.gif
Moreover, it is clear to see that
T λ f L ( R n ; E ) C λ - n f L 1 ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equby_HTML.gif

Therefore, by applying Riesz interpolation theorem for vector-valued L p spaces (see e.g., [19, § 1.18]) we get the assertion.

In a similar way as in [3, Preposition 3.6] we have.

Preposition 4.3. The kernel K1 (x, y) can be written as
K 1 ( x , y ) = j = 0 , 1 A j ( x , y ) t n - 2 e i t φ j ( x , y ) | t ( x - y ) | ( n - 2 ) / 2 | t ( x - y ) | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equbz_HTML.gif
where, for every fixed N, the operator functions A j satisfy
D α A j ( x , y ) C α ( 1 + t ( x 1 - y 1 ) 2 ) - N | x - y | - | α | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equca_HTML.gif
and moreover, the phase functions φ j are real and the property that when ε is small enough, 0 < δ ≤ ε and y1 ∈ [, ε] is fixed, the dilated functions
( x , y ) ( - 1 ) j δ - 1 φ j ( δ x , y 1 , δ y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equcb_HTML.gif
in the some fixed neighborhood of the function φ 0 ( x , y ) = | x - y | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq72_HTML.gif in the C topology. Then, the following estimates holds
| K 1 ( x , y ) | C t n - 2 ( 1 + t | x 1 - y 1 | ) - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ59_HTML.gif
(59)
Proof. By representation of K1 (x, y) and Φ (x, y, ξ) we have
K 1 ( x , y ) t n - 2 R n β 1 ( t ξ ) e i t Φ ( x , y , ξ ) Q 0 t * - 1 ( ε y , ξ ) d ξ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equcc_HTML.gif

Then, by using (36) in view of positivity of operator A, by reasoning as in [3, Preposition 3.6] we obtain the assertion.

Let us now show the end of proof of Lemma 4.1. Let η C 0 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq73_HTML.gif be supported in 1 4 , 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq74_HTML.gif such that ν  =  - η ( 2 ν s ) = 1 , s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq75_HTML.gif and set η 0 ( s ) = 1 - ν  =  - 0 η ( 2 ν s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq76_HTML.gif. Then we define kernels K1,ν, ν = 0, 1, 2, . . . , as follows
K 1 , ν = η ( t 2 - ν | x - y | ) K 1 ( x , y ) , ν > 0 η 0 ( t | x - y | ) K 1 ( x , y ) , ν = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equcd_HTML.gif
Let T1,νdenotes the operators associated to these kernels. Then, by positivity properties of the operator A and by Prepositions 4.2, 4.3 we obtain for f C 0 ( B ε ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq77_HTML.gif the following estimates
T 1 , ν * f L p ( R n ; E ) C 2 2 ν / n f L p ( R n ; E ) , 1 p + 1 p 1 = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ60_HTML.gif
(60)
T 1 , ν * f L p ( R n ; E ) C ( t 2 ν ) 1 / p t ( 1 + 1 n ) f L p ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_Equ61_HTML.gif
(61)

By summing a geometric series one sees that these estimates imply (52) and (53) for case of α = 0.

Let us first to show (60). One can check that the estimate (59) implies that the L r norm of K 10 * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq78_HTML.gif is O (tn-2t -n/r). But, if we let r = n/n - 2, it is follows from Young inequality and the fact that 1 p - 1 p = 2 n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-46/MediaObjects/13661_2012_Article_133_IEq79_HTML.gif that
T 1 , 0 * f L p | ( R n ; E ) C t n 2 t n / r f L p ( R n ; E ) = C f L p