Open Access Research

Carleman estimates and unique continuation property for abstract elliptic equations

Veli B Shakhmurov

Author Affiliations

Department of Electronics Engineering and Communication, Okan University, Akfirat Beldesi, Tuzla, 34959, Istanbul, Turkey

Boundary Value Problems 2012, 2012:46  doi:10.1186/1687-2770-2012-46


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/46


Received:26 January 2012
Accepted:23 April 2012
Published:23 April 2012

© 2012 Shakhmurov; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The unique continuation theorems for elliptic differential-operator equations with variable coefficients in vector-valued Lp-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 Lp-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:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M1">View MathML</a>

(1)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M2">View MathML</a>

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

Jerison and Kenig started the theory of Lp 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 V Ln/2,loc is 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 [6-18] and the references therein).

In this article, the unique continuation theorems for elliptic equations with variable operator coefficients in E-valued Lp spaces are studied. We will prove that if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M3">View MathML</a><a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M4">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M5">View MathML</a>, V Lμ (Rn; L(E)), p, μ∈ (1, ∞) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M6">View MathML</a> satisfies (1), then u is identically zero if it vanishes in a nonempty open subset, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M7">View MathML</a> 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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M8">View MathML</a>, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M9">View MathML</a>

In the Hilbert space L2 (Rn; H), we derive the following Carleman estimate

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M10">View MathML</a>

Any of these inequalities would follow from showing that the adjoint operator Lt (x; D) = etwL (x; D) e-tw satisfies the following relevant local Sobolev inequalities

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M11">View MathML</a>

uniformly to t, where L0t = etwL0e-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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M12">View MathML</a>

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 onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M13">View MathML</a>

with λ Sφ, φ ∈ (0, π], I is identity operator in E. We will sometimes use A + ξ or Aξ 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 Aθ of a positive operator A and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M14">View MathML</a>

We denote by Lp (Ω; E) the space of all strongly measurable E-valued functions on Ω with the norm

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M15">View MathML</a>

By Lp,q (Ω) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M16">View MathML</a> 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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M17">View MathML</a> (sometimes we called it Sobolev-Lions type space) that consist of all functions u Lp (Ω; E0) such that the generalized derivatives <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M18">View MathML</a> are endowed with the

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M19">View MathML</a>

The Banach space E is called an UMD-space if the Hilbert operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M20">View MathML</a>dy is bounded in Lp (R, E), p ∈ (1, ) (see e.g., [21,22]). UMD spaces include, e.g., Lp, lp spaces and Lorentz spaces Lpq, p, q ∈ (1, ).

Let E1 and E2 be two Banach spaces. Let S (Rn; E) denotes a Schwartz class, i.e., the space of all E-valued rapidly decreasing smooth functions on Rn. Let F and F-1denote Fourier and inverse Fourier transformations, respectively. A function Ψ ∈ Cm (Rn; L (E1, E2)) is called a multiplier from Lp (Rn; E1) to Lq (Rn; E2) for p, q ∈ (1, ) if the map u → Ku = F-1 Ψ (ξ) Fu, u S (Rn; E1) is well defined and extends to a bounded linear operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M21">View MathML</a>

We denote the set of all multipliers from Lp (Rn; E1) to Lq (Rn; E2) by <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M22">View MathML</a>. For E1 = E2 = E and q = p we denote <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M22">View MathML</a> by Mp (E). The Lp-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 [23-28].

A set K L (E1, E2) is called R-bounded [22,23] if there is a constant C such that for all T1, T2, . . . , Tm K and u1,u2, . . . , um E1, m N

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M23">View MathML</a>

where {rj} 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M24">View MathML</a>

For any r = (r1, r2, . . . , rn), ri ∈ [0, ) the function ()r, ξ Rn will be defined such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M25">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M26">View MathML</a>

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) (Rn; L (E1, E2)) if the set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M27">View MathML</a>

is R-bounded, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M28">View MathML</a> .

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M29">View MathML</a>

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 Lq, 1 ≤ q ≤ ∞ [31], A has the bounded imaginary powers with <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M30">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M31">View MathML</a> 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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M32">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M33">View MathML</a> be a collection of multipliers in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M34">View MathML</a> . We say that Hk is a uniform collection of multipliers if there exists a constant M > 0, independent on h K, such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M35">View MathML</a>

for all h K and u S (Rn; E1).

We set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M36">View MathML</a>

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

Theorem 2.0. Let E1 and E2 be two UMD spaces and let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M37">View MathML</a>

If

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M38">View MathML</a>

uniformly with respect to h K then Ψh (ξ) is a uniformly collection of multipliers from Lp (Rn; E1) to Lq (Rn; E2).

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M39">View MathML</a>

Embedding theorems in Sobolev-Lions type spaces were studied in [13-18,32,34]. In a similar way as [17, Theorem 3] we have

Theorem 2.1. Suppose the following conditions hold:

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

Then the embedding

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M40">View MathML</a>

is continuous and there exists a positive constant Cµ such that for

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M41">View MathML</a>

the uniform estimate holds

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M42">View MathML</a>

Moreover, for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M43">View MathML</a> the following uniform estimate holds

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M44">View MathML</a>

3 Carleman estimates for DOE

Consider at first the equation with constant coefficients

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M45">View MathML</a>

(2)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M46">View MathML</a> and A is the possible unbounded operator in a Banach space E.

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M47">View MathML</a> and t is a positive parameter.

Remark 3.1. It is clear to see that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M48">View MathML</a>

(3)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M49">View MathML</a>. Let L0t (x, ξ) is the principal operator symbol of L0t (x, D) on the domain B0, i.e.,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M50">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M51">View MathML</a>

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, ξ Rn is uniformly positive in E. So there are fractional powers of A+||2 and the operator function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M52">View MathML</a> 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)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M53">View MathML</a>

(4)

By virtue of Remark 3.1 it suffices to prove the following uniform coercive estimate

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M54">View MathML</a>

(5)

for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M55">View MathML</a>.

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 = Kt and R = Rt so that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M56">View MathML</a>

(6)

where δ denotes the Dirac distribution. Moreover, if we let T = Tt be the operator with kernel K, i.e.,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M57">View MathML</a>

and R is the operator with kernel R (x, y), then for large t > 0, the adjoint of these operators satisfy the following estimates

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M58">View MathML</a>

(7)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M59">View MathML</a>

(8)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M60">View MathML</a>

(9)

Proof. By Remark 3.2 the operator function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M61">View MathML</a> is positive in E for all ξ Rn. Since tw1 + 1 S(φ), due to positivity of A, for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M62">View MathML</a> the factor <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M63">View MathML</a> has a bounded inverse <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M64">View MathML</a>for all ξ Rn, t > 0 and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M65">View MathML</a>

(10)

Therefore, we call Gt (x, ξ) the regular factor. Consider now the second factor

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M66">View MathML</a>

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 Bt (x, ξ) does not has an inverse, in the following set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M67">View MathML</a>

So we will called Bt the singular factor and the set Δt call singular set for the operator function Bt. The operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M68">View MathML</a> cannot be bounded in the set Δt. Nevertheless, the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M68">View MathML</a>, and hence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M69">View MathML</a>, can be bounded when (x, ξ) is sufficiently far from Δt. For instance, if we define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M70">View MathML</a>

by properties of positive operators we will get the same estimate of type (10) for the singular factor Bt. Hence, using this fact and the resolvent properties of positive operators we obtain the following estimate

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M71">View MathML</a>

(11)

where the constant C is independent of x, ξ, t and cΓt denotes the complement of Γt.

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M72">View MathML</a> such that, β(ξ) = 0 if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M73">View MathML</a> and β (ξ) = 0 near the origin. We then define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M74">View MathML</a>

and notice that β0 (ξ) = 0 on Γt. Hence, if we define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M75">View MathML</a>

(12)

and recall (11), then by [31] it follows from standard microlocal arguments that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M76">View MathML</a>

where R0t belongs to a bounded subset of S-1 which is independent of t. Since operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M77">View MathML</a> also has the same property, it follows that for all <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M78">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M79">View MathML</a>

By reasoning as in [31] we get that tR0t belongs to a bounded subset of S0. So, we have the following estimate

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M80">View MathML</a>

Moreover, the Remark 3.2, positivity properties of A and, (11) and (12) imply that, the operator functions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M81">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M82">View MathML</a> 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M83">View MathML</a>

For inverting L0t (x, D) on the set Γt 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M84">View MathML</a>

(13)

so that the analogs of (16) and the estimates (7)-(9) are satisfied. Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M85">View MathML</a> is uniformly bounded on Γt, we should expect to construct the phase function Φ in (13) using the factor Bt (x, ξ). Specifically, we would like Φ to satisfy the following equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M86">View MathML</a>

(14)

The Equation (14) leads to complex eikonal equation (i.e., a non-linear partial differential equation with complex coefficients).

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M87">View MathML</a>

(15)

Since w1 (x) = 1 + x1, w1 (y) = 1 + y1, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M88">View MathML</a>

(16)

is a solution of (15). To use this we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M89">View MathML</a>

Next, if we set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M90">View MathML</a>

(17)

then it follows from L0t (x, ξ) = Gt (x, ξ)Bt (x, ξ) and (14) that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M91">View MathML</a>

(18)

Consequently, (16)-(18) imply that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M92">View MathML</a>

(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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M93">View MathML</a>

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 f L2 (B0; E) one has

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M94">View MathML</a>

(20)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M95">View MathML</a>

(21)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M96','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M96">View MathML</a>

(22)

However, since R1,1 tT1, we need only to show the following

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M97">View MathML</a>

(23)

By using the Minkowski inequalities we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M98','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M98">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M99','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M99">View MathML</a>. The estimates (13) and (16) imply that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M100">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M101','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M101">View MathML</a>

Consequently, it follows from Plancherel's theorem that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M102','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M102">View MathML</a>

(24)

Note that for every N we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M103','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M103">View MathML</a>

Since A is a positive operator in E, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M104','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M104">View MathML</a>

when <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M105','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M105">View MathML</a>. Then by using the above estimate it not easy to check that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M106','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M106">View MathML</a>

i.e.,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M107','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M107">View MathML</a>

Moreover, it is clear that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M108','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M108">View MathML</a>

Thus from (24) by using the above relations and Young's inequality we obtain the desired estimate

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M109','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M109">View MathML</a>

Moreover, by using the estimate (10) and the resolvent properties of the positive operator A we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M110','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M110">View MathML</a>

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 Lp-Carleman estimates and unique continuation for equation with variable coefficients

Consider the following DOE

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M111">View MathML</a>

(25)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M112','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M112">View MathML</a> and A is the possible unbounded operator in a Banach space E and aij are

real-valued smooth functions in Bε = {x Rn, |x| < ε}.

Condition 4.1. There is a positive constant γ such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M113','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M113">View MathML</a> for all ξ Rn, x <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M114','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M114">View MathML</a>

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 = <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M115','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M115">View MathML</a>and p' is the conjugate of p, w = <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M116','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M116">View MathML</a> and aij C(Bε). Then for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M117','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M117">View MathML</a>( Bε; E(A)) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M118','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M118">View MathML</a> the following estimates are satisfied:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M119','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M119">View MathML</a>

(26)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M120','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M120">View MathML</a>

(27)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M121','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M121">View MathML</a>

Proof. As in the proof of Theorem 3.1, it is sufficient to prove the following estimates

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M122','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M122">View MathML</a>

(28)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M123','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M123">View MathML</a>

(29)

where,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M124','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M124">View MathML</a>

Consequently, since w1 ≃ 1 on Bε, it follows that, if we let Qt (εx, D) be the differential operator whose adjoint equals

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M125','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M125">View MathML</a>

then it suffices to prove the following

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M126','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M126">View MathML</a>

(30)

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

Lemma 4.1. For t > 0 there are functions K = Kt and R = Rt, so that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M127','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M127">View MathML</a>

(31)

where δ denotes the Dirac distribution. Moreover, if we let T = Tt be the operator with kernel K (x, y) and R be the operator with kernel R (x, y), then if ε and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M128','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M128">View MathML</a> are sufficiently small, the adjoint of these operators satisfy the following uniform estimates

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M129','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M129">View MathML</a>

(32)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M130','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M130">View MathML</a>

(33)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M131','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M131">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M132','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M132">View MathML</a>

(34)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M133','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M133">View MathML</a>

(35)

Proof. The key step in the proof is to find a factorization of the operator-valued symbol <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M134','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M134">View MathML</a> that will allow to microlocally invert <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M135','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M135">View MathML</a> near the set where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M136','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M136">View MathML</a> vanishes. Note that, after making a suitable choice of coordinates, it is enough to show that if L (x, D) is of the form

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M137','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M137">View MathML</a>

therefore, we can expressed <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M138','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M138">View MathML</a> as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M139','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M139">View MathML</a>

(36)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M140','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M140">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M141','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M141">View MathML</a>

The ellipticity of Q(x, D) and the positivity of the operator A, implies that the factor Gt (x, ξ) never vanishes and as in the proof of Theorem 3.1 we get that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M142','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M142">View MathML</a>

(37)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M143','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M143">View MathML</a>

i.e., the operator function Gt (εx, ξ) has uniformly bounded inverse for (x, ξ) ∈ Bε ×Rn. One can only investigate the factor Bt (εx, ξ). In fact, if we let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M144','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M144">View MathML</a>

then the operator function Bt (x, ξ) is not invertible for (x, ξ) ∈ Δt. Nonetheless, Bt (εx, ξ) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M145','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M145">View MathML</a> can be have a bounded inverse when (x, ξ) is sufficiently far away. For instance, if we define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M146','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M146">View MathML</a>

by properties of positive operators we will get the same estimate of type (37) for the singular factor Bt. Hence, we using this fact and the resolvent properties of positive operators we obtain the following estimate

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M147','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M147">View MathML</a>

(38)

As in § 3, we can use (38) to microlocallity invert <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M148','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M148">View MathML</a> away from Γt . To do this, we first fix <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M149','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M149">View MathML</a> as in § 3. We then define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M150','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M150">View MathML</a>

It is clear that β0 (ξ) = 0 on Γt. Consequently, if we define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M151','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M151">View MathML</a>

(39)

and recall (37), then we can conclude that standard microlocal arguments give that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M152','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M152">View MathML</a>

(40)

where R0 belongs to a bounded subset of S-1 that independent of t. Since the adjoint operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M153','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M153">View MathML</a> also is abstract pseudodifferential operator with this property, by reasoning as in [31, Theorem 6] it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M154','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M154">View MathML</a>

(41)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M155','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M155">View MathML</a>

(42)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M156','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M156">View MathML</a>

Moreover, the positivity properties of A and the estimate (38) imply that the operator functions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M157','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M157">View MathML</a> are uniformly bounded. Next, let T0 be the operator with kernel K0. Then in a similar way as in [31] we obtain that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M158','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M158">View MathML</a>

(43)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M159','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M159">View MathML</a>

which also the first estimate is stronger than the corresponding inequality in Lemma 4.1. Finally, since T0 S-2 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M160','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M160">View MathML</a> it follows from imbedding theorem in abstract Sobolev spaces [17] that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M161','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M161">View MathML</a>

(44)

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

Let β1 (ξ) = 10 (ξ). To invert <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M162','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M162">View MathML</a> for (x, ξ) ∈ Γt, we have to construct a Fourier integral operator T1, with kernel

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M163','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M163">View MathML</a>

(45)

such that the analogs of (39) and (32)-(35) are satisfied. For this step the factorization (36) of the symbol <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M164','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M164">View MathML</a> will be used. Since the factor Gt (εx, ξ) has a bounded inverse for (x, ξ) ∈ Γt, the previous discussions show that we should try to construct the phase function in (46) using the factor Bt (εx, ξ). We would like Φ (x, y, ξ) to solve the complex eikonal equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M165','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M165">View MathML</a>

(46)

Since Bt (εx, Φx) - Bt (εy, ξ) is a scalar function (it does not depend of operator A ), by reasoning as in [3, Lemma 3.4] we get that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M166','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M166">View MathML</a>

where ϕ is real and defined as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M167','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M167">View MathML</a>

while

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M168','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M168">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M169','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M169">View MathML</a>

(47)

Then we obtain from the above that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M170','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M170">View MathML</a>

Next, if we set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M171','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M171">View MathML</a>

(48)

then it follows from (36) and (48) that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M172','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M172">View MathML</a>

(49)

for every N when β1 (ξ) ≠ 0. Consequently, (49), (50) imply that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M173','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M173">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M174','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M174">View MathML</a>

(50)

By reasoning as in Theorem 3.1 we obtain from (51) that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M175','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M175">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M176','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M176">View MathML</a>

(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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M177','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M177">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M178','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M178">View MathML</a>

(52)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M179','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M179">View MathML</a>

(53)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M180','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M180">View MathML</a>

(54)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M181','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M181">View MathML</a>

(55)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M182','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M182">View MathML</a>.

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

Preposition 4.1. Let E be Banach spaces and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M183','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M183">View MathML</a>. Moreover, suppose Φ ∈ Csatisfies | ∇Φ| ≥ γ > 0 on supp A. Then for all λ > 1 the following holds

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M184','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M184">View MathML</a>

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

Proof. Given x0 ∈ supp A. There is a direction ν Sn-1 such that |(ν, ∇Φ)| ≥ <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M185','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M185">View MathML</a> on some ball centered at x0. Thus, by compactness, we can choose a partition of unity <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M186','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M186">View MathML</a> consisting of a finite number of terms and corresponding unit vectors νj such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M187','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M187">View MathML</a> on supp A and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M188','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M188">View MathML</a> on supp φj. For Aj = φjA it suffices to prove that for each j

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M189','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M189">View MathML</a>

After possible changing coordinates we may assume that νj = (1, 0, . . . , 0) which means that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M190','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M190">View MathML</a> on supp φj. If let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M191','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M191">View MathML</a>, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M192','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M192">View MathML</a>. Consequently, if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M193','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M193">View MathML</a> is a adjoint, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M194','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M194">View MathML</a>

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

Preposition 4.2. Suppose Φ ∈ Cis a phase function satisfying the non-degeneracy condition det <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M195','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M195">View MathML</a>on the support of

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M196','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M196">View MathML</a>

Then for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M197','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M197">View MathML</a> the following estimates hold

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M198','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M198">View MathML</a>

Proof. In view of [3, Remark 2.1] we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M199','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M199">View MathML</a>

(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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M200','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M200">View MathML</a>

(57)

on supp A for same C > 0. To use this we notice that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M201','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M201">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M202','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M202">View MathML</a>

(58)

Hence, by virtue of Preposition 4.1 and by (58) we obtain that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M203','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M203">View MathML</a>

Consequently, by Young's inequality, the operator with kernel Kλ acts

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M204','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M204">View MathML</a>

By (59) we get that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M205','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M205">View MathML</a>

Moreover, it is clear to see that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M206','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M206">View MathML</a>

Therefore, by applying Riesz interpolation theorem for vector-valued Lp 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M207','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M207">View MathML</a>

where, for every fixed N, the operator functions Aj satisfy

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M208','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M208">View MathML</a>

and moreover, the phase functions φj are real and the property that when ε is small enough, 0 < δ ≤ ε and y1 ∈ [, ε] is fixed, the dilated functions

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M209','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M209">View MathML</a>

in the some fixed neighborhood of the function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M210','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M210">View MathML</a> in the Ctopology. Then, the following estimates holds

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M211','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M211">View MathML</a>

(59)

Proof. By representation of K1 (x, y) and Φ (x, y, ξ) we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M212','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M212">View MathML</a>

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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M213','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M213">View MathML</a> be supported in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M214','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M214">View MathML</a> such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M215','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M215">View MathML</a> and set <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M216','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M216">View MathML</a>. Then we define kernels K1,ν, ν = 0, 1, 2, . . . , as follows

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M217','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M217">View MathML</a>

Let T1,ν denotes the operators associated to these kernels. Then, by positivity properties of the operator A and by Prepositions 4.2, 4.3 we obtain for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M218','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M218">View MathML</a> the following estimates

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M219','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M219">View MathML</a>

(60)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M220','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M220">View MathML</a>

(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 Lr norm of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M221','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M221">View MathML</a> is O (tn-2t -n/r). But, if we let r = n/n - 2, it is follows from Young inequality and the fact that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M222','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M222">View MathML</a> that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M223','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M223">View MathML</a>

as desired. To prove the result for ν > 0, set <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M224','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M224">View MathML</a> and let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M225','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M225">View MathML</a> be the kernel of the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M226','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M226">View MathML</a> . Then, if we fix x1 and y1, it follows that the <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M227','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M227">View MathML</a> norm of the operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M228','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M228">View MathML</a>

equal <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M229','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M229">View MathML</a> times the norm of the dilated operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M230','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M230">View MathML</a>

where δ = 2ν t-1. By Preposition 4.3, the kernel in last integral equals the complex conjugate of

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M231','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M231">View MathML</a>

and, consequently by using the Proposition 4.2, for 0 < δ ≤ ε and for supp <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M232','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M232">View MathML</a> we obtain that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M233','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M233">View MathML</a>

This estimate implies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M234','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M234">View MathML</a>

For <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M235','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M235">View MathML</a> we set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M236','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M236">View MathML</a>

Then, the desired estimate (60) follows from the above estimate and Young's inequality. The other inequality (61), follows from a similar argument.

Preposition 4.4. The estimates (32)-(34) imply (30).

Proof. Indeed, (31) implies that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M237','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M237">View MathML</a>

and so Minkowski's inequality, (32) and (34) give that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M238','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M238">View MathML</a>

which implies that the first inequality in (30) for sufficiently large t. Moreover, in a similar way, using (32) and (33) we get (30) for α = 0. To prove (30) for |α| = 1, we use (33), (34) and obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M239','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M239">View MathML</a>

Hence, the result follows.

Now we can show the end of the proof of Theorem 4.1. Really, we obtain the estimate (30), which implies the estimates (26) and (27). That is the assertion of Theorem 4.1 is hold.

Theorem 4.2. Assume all conditions of Theorem 4.1 are satisfied, then for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M240','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M240">View MathML</a> if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M241','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M241">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M242','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M242">View MathML</a> then u is identically 0 if it vanishes in a nonempty open subset.

Proof. Suppose

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M243','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M243">View MathML</a>

(62)

in a connected open set G, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M244','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M244">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M245','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M245">View MathML</a>. Then, after the possibly change of variables, one sees that Theorem 4.2 would follow if we could show that if

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M246','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M246">View MathML</a>

(63)

then 0 ∉supp u. Moreover, by making a proper choice of geodesic coordinate system, we may assume L (x, D) as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M247','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M247">View MathML</a>

Then argue as in [29], first set uε (x) = u (εx) where ε is chosen small enough so that (26) and (27) hold for Bε. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M248','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M248">View MathML</a> be equal to one when |x| < <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M249','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M249">View MathML</a> and set Uε = ηuε. Then if Vε (x) = V (εx) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M250','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M250">View MathML</a>

which implies that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M251','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M251">View MathML</a>

(64)

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M252','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M252">View MathML</a>

If the condition (63) holds, then we can always choose δ to be small enough that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M253','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M253">View MathML</a>

and so that if C is as in (26), (27) and C0 is as in (64) then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M254','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M254">View MathML</a>

Next, (26), (27) imply

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M255','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M255">View MathML</a>

If we recall that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M256','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M256">View MathML</a>, then we see that (64) and Hölder's inequality imply

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M257','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M257">View MathML</a>

Thus, by (63) for sufficiently large t > 0 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M258','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M258">View MathML</a> we can conclude that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M259','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M259">View MathML</a>

finally, since w' (x) = 1 + x1> 0 on Bε , this forces Uε (x) = 0 for x Sδ and so 0 ∉ supp u which completes the proof.

Consider the differential operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M260','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M260">View MathML</a>

where aij are real-valued functions numbers, A = A (x), Ak = Ak (x), V (x) are the possible linear operators in a Banach space E.

By using Theorem 4.2 and perturbation theory of linear operators we obtain the following result

Theorem 4.3. Assume:

(1) all conditions of Theorem 4.1 are satisfied;

(2) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M261','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M261">View MathML</a>

Then, for Dαu Lp,loc (B0; E) if ||P (x, D) u||E ||Vu||E and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M262','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M262">View MathML</a>, then u is identically 0 if it vanishes in a nonempty open subset.

Proof. By condition (2) and by Theorem 2.1, for all ε > 0 there is a C (ε) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M263','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M263">View MathML</a>

Then, by using (29) and the above estimate we obtain the assertion.

5 Carleman estimates and unique continuation property for quasielliptic PDE

Let Ω ⊂ Rl be an open connected set with compact C2m-boundary Ω. Let us consider the BVP for the following elliptic equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M264','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M264">View MathML</a>

(65)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M265','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M265">View MathML</a>

(66)

where u = (x, y), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M266','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M266">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M267','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M267">View MathML</a>. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M268','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M268">View MathML</a>.

Let Q denotes the operator generated by the problem (64), (65).

Theorem 5.1. Let the following conditions be satisfied;

(1) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M269','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M269">View MathML</a> for each |α| = 2m and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M270','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M270">View MathML</a> for each |α| = k < 2m with rk ≥ q and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M271','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M271">View MathML</a>

(2) bC2m-mj (Ω) for each j, β and mj < 2m, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M272','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M272">View MathML</a>, for |β| = mj, y|∂G, where σ = (σ1, σ2, . . . , σn) ∈ Rm is a normal to ∂G ;

(3) for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M273','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M273">View MathML</a> let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M274','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M274">View MathML</a>

(4) for each y0 Ω local BVP in local coordinates corresponding to y0

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M275','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M275">View MathML</a>

has a unique solution ϑ C0 (R+) for all h = (h1, h2, . . . , hm) ∈ Rm, and for ξ1 Rl-1 with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M276','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M276">View MathML</a>

(5) Condition 4.1 holds, aij C(Bε), n ≥ 3, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M277','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M277">View MathML</a>and p' is the conjugate of p and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M278','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M278">View MathML</a>

(6) dk L(Rn × Ω).

Then:

(a) for sufficiently large b > 0, t ≥ t0 and for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M279','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M279">View MathML</a> the Carleman type estimate

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M280','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M280">View MathML</a>

holds for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M281','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M281">View MathML</a>.

(b) for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M282','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M282">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M283','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M283">View MathML</a> the differential inequality

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M284','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M284">View MathML</a>

has a unique continuation property.

Proof. Let E = Lq (Ω). Consider the following operator A which is defined by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M285','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M285">View MathML</a>

For x Rn also consider operators

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M286','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M286">View MathML</a>

The problem (5.1), (5.2) can be rewritten in the form (4.1), where u (x) = u (x, .), f (x) = f (x, .) are functions with values in E = Lq (Ω). Then by virtue of [24, Theorems 3.6 and 8.2] the (1) condition of Theorem 4.1 is satisfied. Moreover, by using the embedding <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M287','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M287">View MathML</a> and interpolation properties of Sobolev spaces (see e.g., [19, §4]) we get that there is ε > 0 and a continuous function C (ε) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M288','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M288">View MathML</a>

Due to positive of the operator A, then we obtain that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M289','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M289">View MathML</a>

Then it is easy to get from the above estimate that (2) condition of the Theorem 4.3 is satisfied. By virtue of (5) condition, (2) condition of the Theorem 4.1 is fulfilled too. Hence, by virtue of Theorems 4.1 and 4.3 we obtain the assertions.

6 Carleman estimates and unique continuation property for infinite systems of elliptic equations

Consider the following infinity systems of PDE

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M290','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M290">View MathML</a>

(67)

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M291','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M291">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M292','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M292">View MathML</a>

Let O denotes the operator generated by the problem (66).

Theorem 6.1. Let the following conditions are satisfied:

(1) ak Cb (Rn), ak (x) ≠ 0, x Rn, k = 1, 2, . . . , n and the Condition 4.1 holds;

(2) there are 0 < ν < <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M293','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M293">View MathML</a> such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M294','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M294">View MathML</a>

a.e. for x Rn.

Then:

(a) for sufficiently large b > 0, t ≥ t0 and for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M295','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M295">View MathML</a> the Carleman type estimate

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M296','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M296">View MathML</a>

holds for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M297','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M297">View MathML</a>.

(b) for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M298','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M298">View MathML</a>and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M299','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M299">View MathML</a> the differential inequality

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M300','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M300">View MathML</a>

has a unique continuation property.

Proof. Let E = lq and A, Ak (x) be infinite matrices, such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M301','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/46/mathml/M301">View MathML</a>

It is clear to see that this operator A is R-positive in lq and all other conditions of Theorems 4.1 and 4.3 are hold. Therefore, by virtue of Theorems 4.1 and 4.3 we obtain the assertions.

Competing interests

The author declares that they have no competing interests.

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