A fundamental problem in partial differential equations is to determine the extent to which the solutions are unique. There is an extensive literature on the use of Carleman's inequalities in proving unique continuation theorems. For example, suppose that ¢J is a continuous function on the domain n ~ Rn and there exists a point Xo E n and an open set.
The aim of the thesis is to present results on a smaller family of convex weights. The effect of osculation is to localize matters to sets where the convex function is close to its linear part. The ability to sum these local estimates amounts to a proof of the covering lemma for the sets st( C).
To prove Carleman's inequalities for the exact gap, we restrict our attention to some subsets of convex functions, allowing us to prove stronger lemmas of type L 00. If the weights are sufficiently convex, the restriction to bounded domains is not necessary and the osculation argument can be used to prove the estimates in R n.
Chapter 1
Preliminaries
Some notation
Convex sets and functions
Our local Carleman inequalities will be done on sets where ¢> is close to its linear part.
Chapter 2
Convex Weights
Osculation by linear weights
We note that condition (2.1) is a necessary scaling condition, and that (2.2) spans an interval about r, because the adjoint of the operator under consideration is another operator of the same type. We start by evaluating the local oscillation.. from its linear part in the state) and applying theorem (2.2). This can be done in several ways, each reducing the Carleman inequality to a certain covering lemma for stet groups).
Covering estimates
- Intersection properties
- Covering properties
- Proof of the covering lemma
The problem here, of course, is that we require the sets to cover all n, However, even if we relax the statement of Theorem 2.3 in this way, it can still be shown to be false for coverings by arbitrary rectangles. We begin by analyzing the intersection properties of the sets st(t) in one dimension, and thus along lines in Rn, and proceed to derive some covering properties of the st(t).
The proposition now follows by assuming that b is the right endpoint of V" when k > 0, and the left endpoint when k < O. 2.3) The intersection estimate in proposition 2.2 immediately implies an estimate for the intersection of the quantities st( t) along lines in Rn. We show that when this happens, their union is contained in a set of the form St(At), for some dimensional constant A. Since a local Carleman estimate can be made (using Theorem 2.2 ) ) on such a set, this allows us to basically discard one of the two sets in this situation.
We will see that this will reduce matters to considering the case where two such sets overlap "slightly". Taking the absolute values and using the triangle inequality, we see that it suffices to show that each of the four terms on the RHS has an absolute value of S C(A)t. Our first proposition in this section will tell us that we can choose the coverage 0 = UaEJ st(t) such that the sets st(t) do not overlap too much.
At this point, we have not yet fully exploited the special structure of st(t) sets that is necessary to obtain the full power of the covering lemma. The key fact is that when the sets st(t) have a small relative intersection, as in Proposition 2.5, they can be shrunk about their barycenters by a fixed factor and made disjoint. Since the dilations have a known effect on the volume, the estimate in the covering lemma follows easily.
Let C be the constant from Lemma 1.2. 2.6) The proof in this case is based only on the convexity of the sets st(t). This case is more difficult and the set structure S:(t) will be used here. According to Lemma 1.4, we can choose a rectangle Ra with Ra C stet) c CRa, with the same barycenter as stet).
Note that the latter lies on the distance connecting Pl and pz in st(t).) Subtracting (2.13) from (2.12) and using the definitions of g and h, we obtain. To complete the proof of Theorem 2.1, we only need to combine the covering lemma with Proposition 2.1.
Chapter 3
Uniform Convexity and Related Results
Osculation estimates
The proposition now proceeds by taking the q-th roots and applying Holder's inequality to the right-hand side. Proof: The proof of this proposition is actually a composite of the proofs of Proposition 2.1 and Proposition 3.1. Now we proceed as before, dominating the integral over 0 by the sum of the integrals over st(t).
Since (s/r) ~ 1, we can bring the sum inside the integral on the right-hand side by getting To complete the proof, we simply take the power 1/s and apply Holder's inequality once more.
Uniform convexity
Proof Since is linear in the (Xk+b .. , xn) directions, it follows that there is a constant C such that. The lemma follows immediately from the left inequality, since the set where the left side is < 1 is essentially a cube in R k of side (C jt)1/2 crossed with a rectangle in the remaining n - k directions whose side lengths are bounded by the diameter of O. Let F be a bounded subset of Rn, and for an E F, let Ra be the translation of R centered at a.
Proof: We start by selecting a cover F = UaEJ Ra with the property that if Xj is the center of Raj', then axes, Lemma 3.2 applies, and so we may assume that every point in n lies in at most Cn elements of UaEJ C3 Ra. We note that Theorem 3.3 is false if it is assumed that ( x) is only convex.
When r = s = 2, this is essentially in [6], with the inclusion of an additional estimate on the gradient.
Estimates on Rn
Proof: We begin by showing that the sets st(p - 1) are essentially spheres of radius lal(2-p)/2 centered at a. This proves the claim, and since the integral on the right is clearly bounded by a constant depending only on p and n, we have the required estimate for the sum over J1•. Proof: When p = 2, the cover obtained in Lemma 3.3 extends to the cover Rn, since all sets st(t) are spheres of constant radius.
In this case, the covering lemma is particularly easy to prove, and we obtain LP(Rn ---+ Lq(Rn) estimates for the same range of exponents valid for linear weights. Using the Besicovitch covering lemma, we can choose a subgroup cover Q = Uk (st" (1) n Q) with the property that each point in Q belongs to at most 2 elements in the cover and that each element in the cover intersects at most 2 other elements. For a given f E COO(Rn), we can choose a cube Q, as in lemma 3.4, large enough to contain the support for f.
Bibliography
