Hölder Regularity of Subelliptic Pseudo-differential Operators

Pengfei Guan Department of Mathematics

Princeton University

Current Address: Department of Mathematics

McMaster University Hamilton, Ontario L8S 4K1

Canada

Abstract

We consider Hölder regularity of subelliptic pseudo-differential operators. Our main results are the following.

1. In general, Hölder regularity does not hold for subelliptic operators. We construct a class of second order subelliptic differential operators in R^{n} (n ≥ 3) for which Hölder regularity breaks down. Moreover, for any s > 0, if n large, there exists f ∈ Λ_{loc}^{s}(R^{n}), any distribution solution u is not even locally bounded.

2. On the other hand, we prove the Hölder regularity for subelliptic pseudo-differential operators in diagonal form. We also establish the Hölder regularity for any subelliptic pseudo-differential operators in two variables which satisfy Nirenberg-Treves condition (P).

3. As an application, we obtain the optimal Hölder regularity results for the subelliptic oblique derivative problem.

Introduction.

Since Hans Lewy’s celebrated example of a non-solvable operator, much important work has been done on the study of solvability and hypoellipticity of pseudo-differential operators. (e.g. [H1], [NT1],[NT2],[T],[BF],[E1],[H3],etc.) The importance of study of the pseudo-differential operators is also illustrated by its connection to other problems ([H2],[FK]).

The following condition (Ψ) and condition (P) introduced by Nirenberg and Treves are fundamental.

Condition (Ψ):

Let p(x, ξ) be the principal symbol of P, given any point (x_{0}, ξ^{0}) ∈ Ω × {R^{n} \ {O}}, and such that there exists some z ∈ C \ {O},

p(x_{0}, ξ^{0}) = 0, d_{ξ}Re(zp)(x_{0}, ξ^{0}) ≠ 0

the function Im(zp)(x, ξ), restricted to the bicharacteristics curve of Re(zp)(x, ξ) through (x_{0}, ξ^{0}) can never change sign from - to + when one moves in the positive direction.

Condition (P):

The function Im(zp)(x, ξ) in above does not change sign in the bicharacteristic curve of re(zp)(x, ξ) through (x_{0}, ξ^{0}).

Condition (P) is necessary and sufficient for local solvability of differential operators. ([NT2],[BF]). Condition (Ψ) is necessary for local solvability of pseudo-differential operator P and for hypoellipticity of P̄. ([M]). For the subellipticity, a very strong result of Egorov characterizes all subelliptic operators. A pseudo-differential operator P is subelliptic if and only if P̄ satisfies condition (Ψ) and the function Imzp vanishes only finite order along the bicharacteristic curves of Rezp. If k is the smallest number such that the order of vanishing is less than or equal to k, then we can take δ = \frac{1}{k+1}, this is sharp. Egorov’s original proof ([E1]) of the sufficiency does not seem to be very rigorous. The first complete proof for differential operators was given by Treves ([T]). Hörmander ([H3]) later supplied the missing part in the proof of [E1]. (see also [F]).

In this paper, we study Hölder regularity of subelliptic pseudo-differential operators. It is well-known that if P is an elliptic pseudo-differential operator of order m, then

u ∈ D′(Ω), P u ∈ Λ_{comp}^{α}(Ω)→u ∈ Λ_{loc}^{α+m}(Ω),

where Λ^{α} is Hölder space with index α > 0. It was expected that subelliptic pseudo-differential operators would have the similar property. Surprisingly, this is not true in general, even for subelliptic differential operators. In the following theorem we give a class of differential operators in R^{n}(n ≥ 3) which does not behave well in Hölder spaces.

Theorem 1.1 Let P = \frac{∂^{2}}{∂t^{2}} − (\frac{∂^{2}}{∂x_{1}^{2}} + … + \frac{∂^{2}}{∂x_{n}^{2}} ) − it^{2k}(\frac{∂^{2}}{∂x_{1}^{2}} + … + \frac{∂^{2}}{∂x_{n}^{2}} ), k ≥ 1, n ≥ 2, P is a

subelliptic with gain δ = \frac{1}{2k+1}. When \frac{n−1}{2} − \frac{n}{2(k+1)} > 0, there is a function f ∈ Λ_{comp}^{\frac{n−1}{2} − \frac{n}{2(k+1)}} (R^{n+1}), and for any u ∈ D′(R^{n+1}) with P u = f, then u ∉ Λ_{loc}^{1+ϵ}(R^{n+1}) for any ϵ > 0. If \frac{n−5}{2} − \frac{n}{2(k+1)} > 0

there is f ∈ Λ_{comp}^{\frac{n−5}{2} − \frac{n}{2(k+1)}} (R^{n+1}), any distribution u of P u = f is not locally bounded.

Corollary: For any p > 2, β > 0, there exists n, s, such that there exists f ∈ L_{s+β,comp}^{p}(R^{n+1}), any distribution solution of P u = f is not in L_{s,loc}^{p}(R^{n+1}).

These are strong indications of the difference of behaviors of subelliptic operators in Sobolev spaces and in Hölder or L_{s}^{p}(p ≠ 2) spaces. Fourier integral operators are indispensable in the study of subelliptic operators in L^{2} space, but they do not preserve Hölder or L^{p} spaces (p ≠ 2). This may be the main obstacle. It becomes more obvious as we show that the diagonalized operators have good regularity properties in Hölder space. In (2.1) we define a diagonal operator and we obtain the following result.

Theorem 4.1: Let L be a diagonal operator.

(a) If L satisfies condition P, then for any f ∈ Λ_{comp}^{α}, α > 0, there is u ∈ Λ_{loc}^{α+δ}, Lu = f. And any solution u, Lu = f is in Λ_{loc}^{α+δ}.

(b) If L satisfies condition (Ψ), under some smoothness assumption on some null set of L, then for any f ∈ Λ_{comp}^{α}, α > 0, there exist u ∈ Λ_{loc}^{α+δ}, Lu = f.

(c) If L̄ satisfies condition (Ψ), under some smoothness assumption on some null set of L, then

u ∈ D′, Lu ∈ Λ_{comp}^{α}→u ∈ Λ_{loc}^{α+δ} for α > 0

Unlike in the higher dimensional case, in R^{2}, we prove:

Theorem 4.2: If P is a subelliptic pseudo-differential operator of order m in two variables, if P satisfies condition (P), then

u ∈ D′, P u ∈ Λ^{α}→u ∈ Λ^{α+m−1+δ} for α > 0,

the gain δ is same as in Egorov’s theorem.

Our analysis for diagonal operators also leads to a regularity result for the oblique derivative problem. The oblique derivative problem consists of the following boundary value problem:

\{ \begin{matrix}P u = f & on Ω \\ lu = g & on ∂Ω \end{matrix}

where P = a_{ij}(x)\frac{∂^{2}}{∂x_{i}∂x_{j}} +b_{j}(x)\frac{∂}{∂x_{j}} is a second order real elliptic differential operator, Ω is a bounded domain in R^{n+1}, l is a real vector field. The oblique derivative problem has been studied in recent years by many people (e.g. [H2],[EK],[NS], [S],[W]). For more references related to the oblique derivative problem we refer to [W].

If we write l = T + a(x)n⃗, where T is a tangential vector field to ∂Ω, n⃗ is the outward normal vector field and if a(x) is nowhere vanishing, the problem is an elliptic boundary value problem.([ADN]). In general, the problem is not well-posed. The solution of the problem depends on the behavior of a(x) near the null set. We assume ∑_{j=0}^{k} |T^{j}a(x)| > 0. We have following regularity result:

Theorem 4.3: (a) If a(x) never changes sign from - to + in the positive direction of T, let z = {x ∈ ∂Ω|_{a(x)=0}, a(x) changes sign from + to - in the positive direction of T at x}. If z ≠ ∅, we assume it is a closed submanifold in ∂Ω, and T is transversal to it. Then, if u is a solution of the oblique derivative problem with f ∈ Λ^{α}(Ω̄), g ∈ Λ^{β}(∂Ω), which satisfy a finite number of compatibility conditions, then u ∈ Λ^{γ}(Ω̄) where γ = min(α + 2, β + \frac{1}{k+1}).

(b) If −a(x) satisfies the assumptions in (a), i.e., a(x) is ”emergent type”,and we assume z ≠ ∅, then the following problem

\{ \begin{matrix}P u = f & on Ω \\ lu|_{∂Ω} = g \\ u|_{z} = 0 \end{matrix}

with f ∈ Λ^{α}(Ω̄) and g ∈ Λ^{β}(∂Ω) satisfying a finite necessary linear compatibility conditions, has a solution u ∈ Λ^{γ}(Ω̄) with γ = min(α + 2, β + \frac{1}{k+1}).

The regularity result in the theorem is optimal. We have a full gain with respect to f just like in elliptic boundary value problem ([ADN]), and the gain with respect to g is also sharp. This is in contrast to the ¯∂-Neuman problem ([K],[FK]), which can-not be obtained fully from f. The full gain in L^{P} space with respect to f for l ”emergent type” was proved in [S].

The organization of this paper is as follows: In chapter 1, we study the counterexample. We decompose P into two first order pseudo-differential operators and construct the fundamental solution. Then we can read off the high singularity from the fundamental solution. Chapter 2, and chapter 3 are devoted to study diagonal operators. We construct the parametrix in chapter 2, and give the basic estimates in chapter 3. The regularity results for subelliptic operators in two variables and for the oblique derivative problem appear in chapter 4.

This paper is a minor revision of the author’s Ph.D. dissertation at Princeton University. I am grateful to my thesis adviser, Professor J. J. Kohn, for sharing with me his profound insight and providing with constant encouragement. I am indebated to Professor E. M. Stein for many helpful discussions. I would also like to thank Professors C. Fefferman and F. Treves for valuable comments.

Chapter 1

The Counterexample

Let P = \frac{∂^{2}}{∂t^{2}} − (\frac{∂^{2}}{∂x_{1}^{2}} + … + \frac{∂^{2}}{∂x_{n}^{2}} ) − it^{2k}(\frac{∂^{2}}{∂x_{1}^{2}} + … + \frac{∂^{2}}{∂x_{n}^{2}} ), where k is a positive integer. P is a second order differential operator in R^{n+1}. P is of principal type because its real part is a wave operator. The symbol of P is

p(t, x, τ, ξ) = −(τ^{2} − |ξ|^{2}) + it^{2k}|ξ|^{2}

where τ ∈ R^{1}, ξ = (ξ_{1}, …, ξ_{n}) ∈ R^{n}. Solve the system of equations:

\{ \begin{matrix}\frac{dt}{ds} = −2τ \\ \frac{dτ}{ds} = 0 \\ \frac{dx_{j}}{ds} = 2ξ_{j} \\ \frac{dξ_{j}}{ds} = 0 \end{matrix}  (1.1)

The bicharacteristic curve of Rep through a point (t_{0}, x_{0}, τ^{0}, ξ^{0}) ∈ R^{n+1} × {R^{n+1} \ {0}} is

\{ \begin{matrix}t(s) = t_{0} − 2τ^{0}s \\ x(s) = x_{0} + 2sξ^{0} \\ τ(s) = τ^{0} \\ ξ(s) = ξ^{0} \end{matrix}

The null set of p(t, x, τ, ξ) is {(t, x, τ, ξ)|t = 0, |τ| = |ξ| ≠ 0}. So at a point (t_{0}, x_{0}, τ^{0}, ξ^{0}) where p vanishes, the restriction of Imp on the bicharacteristic curve of Rep is

Imp = 4^{k}|τ^{0}|^{2k}|ξ^{0}|^{2k}s^{2k}

with |τ^{0}| = |ξ^{0}| ≠ 0. This verifies that P satisfies condition (P). By the Egorov-Treves theorem, P is a subelliptic operator with gain δ = \frac{1}{k+1}.

To study the solution of P in Hölder spaces,we decompose P to two first order operators.

Lemma 1.1 Let R(t, |D_{x}|) be a pseudo-differential operator in the x variables with t as a parameter, with symbol

R̃ = k ∫_{−∞}^{t} e^{i ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ|ξ|} \frac{t^{′2k−1}}{\sqrt{1 + it^{′2k}}} |ξ|dt′  (1.2)

then we have

|\frac{∂^{α}}{∂t^{α}} \frac{∂^{γ}}{∂x^{γ}} \frac{∂^{β}}{∂ξ^{β}} R̃| ≤ C_{αβγ}|ξ|^{−|β|} for|ξ| ≥ 1.  (1.3)

Proof: R̃(t, x, ξ) is independent of x, we need only to estimate

|\frac{∂^{α}}{∂t^{α}} \frac{∂^{γ}}{∂ξ^{γ}} R̃(t, x, ξ)| ≤ C_{αβ}|ξ|^{−|β|}

because Rei ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ < 0 for t′ < t, given any N, for |ξ| > 1. Integrate by parts the expression for R̃ N+1 times. We get

R̃ = f_{0}(t) + f_{1}(t)\frac{1}{|ξ|} + … + f_{N}(t)\frac{1}{|ξ|^{N}} + \frac{1}{|ξ|^{N}} ∫_{−∞}^{t} f_{N+1}(t′)e^{i ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ|ξ|}dt′

where f_{j}(t) are smooth functions of t, f_{0}(t) = \frac{kt^{2k−1}}{i(1+it^{2k})} and |f_{j}^{(l)}(t)| ≤ \frac{C_{jl}}{1+t^{2}} , j = 1, 2, ..., N + 1, l =

0, 1, 2, . . .. Apply \frac{∂^{α}}{∂t^{α}} \frac{∂^{β}}{∂ξ^{β}} on both sides of above equation, if N > α + 2|β|, then |\frac{∂^{α}}{∂t^{α}} \frac{∂^{β}}{∂ξ^{β}} R̃| ≤ C_{αβ}|ξ|^{−|β|}.

Proposition 1.1 Let P_{1} = \frac{∂}{∂t} − i\sqrt{1 + it^{2k}}|D_{x}|, P_{2} = \frac{∂}{∂t} + i\sqrt{1 + it^{2k}}|D_{x}| + R(t, |D_{X}|), then P = P_{1}P_{2}. Here we denote |D_{x}| as a pseudo-differential operator with symbol |ξ| in x variables.

Proof:

P̃(\frac{∂}{∂t}, ξ, t) = \frac{∂^{2}}{∂t^{2}} + (1 + it^{2k})|ξ|^{2}
= (\frac{∂}{∂t} − i\sqrt{1 + it^{2k}}|ξ|)(\frac{∂}{∂t} + i\sqrt{1 + it^{2k}}|ξ| + R̃(t, |ξ|))

Because the coefficients of above two operators are independent of x, this proves P = P_{1}P_{2}.

From now on we will concentrate on P_{1}. We are going to construct the fundamental solution of P_{1}. Notice that P_{1} is a subelliptic. Suppose f ∈ C_{0}^{∞}(R^{n+1}), and P_{1}u = f. Take the Fourier transform in the x variables. The equation changes to

\frac{∂}{∂t}ũ(t, ξ) − i\sqrt{1 + it^{2k}}|ξ|ũ(t, ξ) = f̃(t, ξ)

Solve above ordinary equation. We get

ũ(t, ξ) = ∫_{−∞}^{t} e^{i ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ|ξ|}f̃(t′, ξ)dt′

It is important to notice that Rei ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ|ξ| ≤ 0. The inverse Fourier transform takes us back to

u(t, x) = C ∫_{R^{N}} ∫_{−∞}^{t} e^{ixξ+i ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ}|ξ|f̃(t′, ξ)dt′dξ
= C ∫_{R^{n}} ∫_{−∞}^{t} ∫_{R^{n}} e^{i(x−x′)ξ+i ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ|ξ|}f(t′, x′)dx′dt′dξ

We observe that above integral operator is actually a zero-order Fourier integral operator with a regular complex phase function. It gives a distribution solution of P u = f. This also gives a fundamental solution of P. It is a Fourier integral distribution

K(t, t′, x, x′) = C ∫_{R^{n}} e^{i(x−x′)·ξ+i ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ|ξ|}.  (1.4)

Our next task is to study K(t, t′, x, x′).

Lemma 1.2 Let a ≠ 0, b > 0 be constants, and Ima > 0. We have

∫_{0}^{2π} \frac{1}{(a + bcosθ)^{2}} dθ = 2π \frac{a}{(\sqrt{a^{2} − b^{2}})^{3}}  (1.5)

∫_{0}^{2π} \frac{sin^{n−2}θ}{(a + bcosθ)^{n}} dθ = \{ \begin{matrix}C_{n} \frac{a^{n−1}}{b^{n−2}(a^{2}−b^{2})^{\frac{n+1}{2}}} + ∑_{j≤n,h<n−2,n≤h+j} \frac{C_{jh}a^{j+h−n}}{b^{h}(a^{2}−b^{2})^{j/2}} , n even \\ C′_{n} \frac{a^{\frac{n−3}{2}}}{b^{n−2}} (_{\frac{1}{(a+b)^{\frac{n+1}{2}}}} − _{\frac{1}{(a−b)^{\frac{n+1}{2}}}} ) \\ +∑_{j≤ \frac{n−1}{2} ,h≤n−2,n≤h+j} \frac{C′_{jh}a^{j+h−n}}{b^{h}} (\frac{1}{(a+b)^{j}} − \frac{1}{(a−b)^{j}}), n ≥ 3, odd \end{matrix} (1.6)

where the C_{n}, C_{jh}, C′_{n}, C′_{jh} are constants independent of a, b, and C_{n} ≠ 0, C′_{n} ≠ 0.

Proof: It suffices to prove the first equation for b = 1. Let α = −a−\sqrt{a^{2} − 1}, β = −a+\sqrt{a^{2} − 1}, because Ima > 0, |α| > 1, |β| < 1, αβ = 1. α, β are roots of z^{2} + 2a + 1.

∫_{0}^{2π} \frac{1}{(cosθ + a)^{2}} = 4i ∫_{|z|=1} \frac{1}{z(z + z^{−1} + 2a)^{2}} dz
= 4i ∫_{|z|=1} \frac{z}{(z^{2} + 1 + 2az)^{2}}

Use residue formula

∫_{0}^{2π} \frac{1}{(cosθ + a)^{2}} dθ = 8π(\frac{z}{(z − α)^{2}} )′|_{z=β}
= −8πi\frac{β + α}{(β − α)^{3}}
= 2π \frac{a}{(\sqrt{a^{2} − 1})^{3}}

As for the second formula, when n is odd

∫_{0}^{π} \frac{sin^{n−2}θ}{(a + bcosθ)^{n}} dθ = ∫_{−1}^{1} \frac{(1 − s^{2})^{\frac{n−3}{2}}}{(a + bs)^{n}} ds
= −\frac{n − 3}{2nb^{2}} ∫_{−1}^{1} \frac{(1 − s^{2})^{\frac{n−5}{2}}}{(a + bs)^{n−2}} ds − \frac{(n − 3)a}{2nb^{2}} ∫ \frac{(1 − t^{2})^{\frac{n−5}{2}}}{(a + bs)^{n−1}} ds

repeating the process, we will reach

∫_{−1} \frac{(1 − s^{2})^{\frac{n−3}{2}}}{(a + bs)^{n}} ds = C′′ \frac{a^{\frac{n−3}{2}}}{b^{n−3}} ∫_{−1}^{1} \frac{1}{(a + b)^{n− \frac{n−3}{2}}} ds + ∑_{j≤ \frac{n+1}{2} ,n≤j+h,h≤n−3} C′′_{jh} \frac{a^{j+h−n}}{b^{h}} ∫_{−1}^{1} \frac{1}{(a + bs)^{j}} ds

This will give the equation for n odd. When n is even, n ≥ 4, do the same computation as for n odd. But in the final stage we read

∫_{0}^{π} \frac{sin^{n−2}θdθ}{(a + bcosθ)^{n}} dθ = ∫_{−1}^{1} \frac{(1 − s^{2})^{\frac{n−3}{2}}}{(a + bs)^{n}} ds
= C′′ \frac{a^{\frac{n−4}{2}}}{b^{n−4}} ∫_{−1}^{1} \frac{(1 − s^{2})^{1/2}}{(a + bs)^{\frac{n+4}{2}}} ds  (1.7)

+ ∑_{4≤j≤ \frac{n+4}{2} ,h≤n−4,n≤j+h} C′′_{jh} \frac{a^{j+h−n}}{b^{h}} ∫_{−1}^{1} \frac{(1 − s^{2})^{1/2}}{(a + bs)^{j}} ds  (1.8)

We need to compute ∫_{−1}^{1} \frac{(1−s^{2})^{1/2}}{(a+bs)^{l}} ds, for l ≥ 4

∫_{−1}^{1} \frac{\sqrt{1 − s^{2}}}{(a + bs)^{l}} dθ = ∫_{0}^{π} \frac{sin^{2}θ}{(a + bcosθ)^{l}} dθ
= \frac{1}{2} ∫_{0}^{2π} \frac{sin^{2}θ}{(a + bcosθ)^{l}} dθ
= \frac{1}{2i} ∫_{|z|=1} \frac{(\frac{z−z^{−1}}{zi})^{2}}{z(a + b\frac{z+z^{−1}}{2} )^{l}} dz

= \frac{i}{8} ∫_{|z|=1} \frac{z^{l−3}(z^{2} − 1)^{2}}{2az + bz^{2} + b)^{l}} dz
= \frac{i}{8b^{l}} ∫_{|z|=1} \frac{z^{l−3}(z^{2} − 1)^{2}}{(2a_{1}z + z^{2} + 1)^{l}} dz

where a_{1} = \frac{a}{b}. Let α = −a_{1} − \sqrt{a_{1}^{2} − 1}, β = −a_{1} + \sqrt{a_{1}^{2} − 1}, α, β are roots of z^{2} + 1 + 2a_{1}z, and αβ = 1, using the residue formula again.

∫_{−1}^{1} \frac{\sqrt{1 − s^{2}}}{(a + bs)^{l}} = C \frac{1}{8b^{l}} (\frac{z^{l−3}(z^{2} − 1)^{2}}{(z − l)^{l}})^{(l−1)}|_{z=β}

= C \frac{(−1)^{l−1}i(2l − 3)!}{8b^{l}(l − 1)!}\{\frac{(2l − 2)β^{l−3}(β^{2} − 1)^{2}}{(β − α)^{2l−1}} − \frac{4β^{l−2}(β^{2} − 1)}{(β − α)^{2l−2}}
+ ∑ C_{m,h}^{j} \frac{β^{l−3−h}(β^{2} − 1)^{j}}{(β − α)^{l+m}} \}
= C \frac{(−1)^{l−1}i(2l − 3)!}{8b^{l}(l − 1)!}\{(2l − 6)\frac{β^{l−1}}{(β − α)^{2l−3}}
+ ∑_{0≤m≤l−4} C_{m} \frac{β^{m+2}}{(β − α)^{l+k}} \}
= C̃ \frac{i(2l − 3)!}{8b^{2}(l − 1)!}\{\frac{(2l − 6)a^{l−1}}{(\sqrt{a^{2} − b^{2}})^{2l−3}} + \frac{1}{b^{2}} ∑_{0≤m≤l−4} C̃_{m} \frac{a^{m+2}}{(\sqrt{a^{2} − b^{2}})^{l+m}} \}

C ≠ 0, C̃ ≠ 0. Put the above into (1.7), we conclude the proof of lemma 1.2. Proposition 1.2

K(t, t′, x, x′) = \{ \begin{matrix}−2π \frac{∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ}{[(∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ)^{2}−|x−x′|^{2}]^{3/2}} \\ when n = 2; \\ C_{n} \frac{(∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ)^{n−1}}{|x−x′|^{n−2}((∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ)^{2}−|x−x′|^{2})^{\frac{n+1}{2}}} \\ + ∑_{j≤n,h≤n−2,n≤h+j} \frac{C_{jh}(∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ)^{j+h−n}}{|x−x′|^{h}((∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ)^{2}−|x−x′|^{2})^{j/2}} \\ when n ≥ 4, even; \\ C_{n}^{\frac{(∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ)^{\frac{n−3}{2}}}{|x−x′|^{n−2}}} (_{\frac{1}{(∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ+|x−x′|)^{\frac{n+1}{2}}}} − _{\frac{1}{(∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ−|x−x′|)^{\frac{n+1}{2}}}} ) \\ + ∑_{j≤ \frac{n−1}{2} ,h≤n−2,n≤h+j} ^{\frac{C′_{jh}(∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ)^{j+h−n}}{|x−x′|}} (_{\frac{1}{(∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ+|x−x′|)^{\frac{n+1}{2}}}} \\ −_{\frac{1}{(∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ−|x−x′|)^{\frac{n+1}{2}}}} ) \\ when n ≥ 3, odd; \end{matrix}

Proof:

K(x, x′, t, t′) = C ∫_{R^{n}} e^{i(x−x′)ξ+|ξ|i ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ}dξ
= C ∫_{|ξ|=1} ∫_{0}^{∞} r^{n−1}e^{i(x−x′)ξ·r+ri ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ}drdσ_{ξ}
= C′ ∫_{|ξ|=1} \frac{1}{((x − x′) · ξ + ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ)^{n}} dσ_{ξ}
= C′′ ∫ \frac{sin^{n−2} θ}{(|x − x′|cosθ + ∫_{t′}^{t} \sqrt{1 + iθ^{2k}}dθ)^{n}} dθ

Applying Lemma 1.2, we prove the Proposition. Lemma 1.3

| ∫_{−A}^{0} ∫_{|x′|<|t′|} \frac{(∫_{t′}^{0} \sqrt{1 + iθ^{2k}}dθ)^{n−1}}{|(∫_{t′}^{0} \sqrt{1 + iθ^{2k}}dθ)^{2} − |x′|^{2}|^{1+ \frac{n}{(k+1)2}} |x′|^{n−2}} dx′dt′| = ∞

for any constant A > 0.

Proof:

∫_{t′}^{0} \sqrt{1 + iθ^{2k}}dθ = −t′ − \frac{it^{′2k+1}}{2(2k + 1)} + O(|t′|^{4k+1})

(∫_{t′}^{0} \sqrt{1 + iθ^{2k}}dθ)^{2} = t^{′2} + \frac{it^{′2k+2}}{(2k + 1)} + O(|t′|^{4k+2})

|(∫_{t′}^{0} \sqrt{1 + iθ^{2k}}dθ)^{2} − |x′|^{2}| = \sqrt{|t^{′2} − |x′|^{2}|^{2} + |t′|^{4k+4}} + O(|x′|^{2}|t′|^{4k+2}) (∫_{t′}^{0} \sqrt{1 + iθ^{2k}}dθ)^{n−1} = (−1)^{n−1}t^{′n−1} + O(|t′|^{n+2k−1})

| ∫_{−A}^{0} ∫_{|x′|<|t′|} \frac{(∫_{t′}^{0} \sqrt{1 + iθ^{2k}}dθ)^{n−1}}{|(∫_{t′}^{0} \sqrt{1 + iθ^{2k}}dθ)^{2} − |x′|^{2}|^{1+ \frac{n}{(k+1)2}} |x′|^{n−2}} dx′dt′|

≥ \frac{1}{2} ∫_{−A}^{0} ∫_{0}^{|t′|} \frac{|t′|^{n−1}r}{(t′^{2} − r^{2} + |t′|^{2k+2})^{1+ \frac{n}{2(k+1)}}} drdt′
≥ \frac{1}{2} ∫_{−A}^{0} \frac{|t′|^{n−1}}{|t′|^{n}} dt′
= \frac{1}{2} ∫_{−A}^{0} \frac{1}{|t′|}dt′
= ∞

Finally, we are ready to prove the following: Theorem 1.1 The subelliptic differential operator P (k ≥ 1) in R^{n+1}(n ≥ 2) does not have the regularity gain in Hölder spaces as in Sobolev spaces. More precisely, when \frac{n−1}{2} − \frac{n}{2(k+1)} > 0, there is a function f in the Hölder space Λ_{comp}^{\frac{n−1}{2} − \frac{n}{2(k+1)}} (R^{n+1}) such that if u is a solution of P u = f then u ∉ Λ_{loc}^{1+ϵ}(R^{n+1}) for any ϵ > 0. If \frac{n−5}{2} − \frac{n}{2(k+1)} > 0, there is a function g in Hölder space Λ_{comp}^{\frac{n−5}{2} − \frac{n}{2(k+1)}} (R^{n+1}) such that if v is a solution of P v = g, then v is not locally bounded.

Proof: First, let h(t, x) = x^{2} − (∫_{t}^{0} \sqrt{1 + iθ^{2k}}dθ)^{2}, w(t, x) ∈ C_{0}^{∞}(R^{n+1}), w=1 in a small

neighborhood of the origin. Set f = w \frac{h^{\frac{n+1}{2}}}{|h|^{1+ \frac{n}{2(k+1)}}} , then f ∈ Λ_{comp}^{\frac{n+1}{2} −1− \frac{n}{2(k+1)}} . By Proposition 1.2

and Lemma 1.3, we know Kf is unbounded near the origin. If u is a solution of P u = f, then P_{1} · (P_{2}u − Kf) = 0. P_{1} is subelliptic, so P_{2}u − Kf ∈ C^{∞}. That is P_{2}u is unbounded near the origin. By Lemma 1.1, P_{2} maps Λ^{α}→Λ^{α−1} for α > 1. This proves the first part of the theorem. For the second part, we claim Δ_{x}u is unbounded near the origin. If Δ_{x}u ∈ L^{∞}, by the equation \frac{∂^{2}u}{∂t^{2}} = f +(1+it^{2k})Δ_{x}u, so \frac{∂^{2}u}{∂t^{2}} ∈ L^{∞}. Then (\frac{∂^{2}}{∂t^{2}} +Δ_{x})u ∈ L^{∞}. But the Laplacian \frac{∂^{2}}{∂t^{2}} +Δ_{x} is elliptic in R^{n+1}. This would give u ∈ Λ_{loc}^{2}(R^{n+1}). This is a contradiction. Now let g = Δ_{x}f, v = Δ_{x}u. Because the coefficients of P are independent of x,P Δ_{x}u = P v = Δ_{x}f = g.

Remark: If we make a little more careful analysis, when \frac{n−1}{2} − \frac{n}{2(k+1)} = 0, (this only happens for n = 2, k = 1), we can conclude that there exists a function f ∈ L_{comp}^{∞}(R^{n}), if u is a solution of P u = f, then u ∉ Λ_{loc}^{1+ϵ}(R^{n+1}) for any ϵ > 0. Also, P is locally solvable, so for f, g in Theorem 1.1, u, v always exist. Our solution u is in H^{\frac{n+1}{2} − \frac{n}{2(k+1)} + \frac{1}{2k+1}} by the Egorov-Treves theorem.

Corollary 1.1 For any p > 2, β > 0, there exist k, n, and s, such that f ∈ L_{s+β,comp}^{p}(R^{n}) any solution u of P u = f is not in L_{s,loc}^{p}(R^{n+1}). That is, P does not have similar regularity property in L^{p} spaces as in L^{2} spaces.

Proof: Let s = \frac{n+2}{p}, choose k, n large, such that \frac{n−5}{2} − \frac{n}{2(k+1)} > \frac{n+2}{p} + β. This is possible because p > 2. The corollary now follows from Theorem 1.1 and the Sobolev embedding theorem.

Construction of The Parametrix

In this chapter and chapter 3, we study diagonal pseudo-differential operators in R^{n+1} of the form

L = \frac{∂}{∂x_{n+1}} + a(x_{n+1}, x)Q  (2.1)

where x_{n+1} ∈ R, x ∈ R^{n}, Q is a first order pseudo-differential operator in x variables with real symbol and x_{n+1} as a parameter. For the oblique derivative problem, we will also deal with Q of

σ(Q) = \sqrt{∑_{j,k=1}^{n+1} Q_{jk}ξ_{j}ξ_{k}}  (2.2)

where a(x_{n+1}, x), Q_{jk}(x_{n+1}, x) are real smooth functions, the matrix {Q_{jk}}_{j,k}^{n+1,n+1} is semi-positive definite and the submatrix {Q_{jk}}_{j,k}^{n,n} is positive definite. Thus there exists a constant λ such that

λ^{−1} ∑_{j=1}^{n} ξ_{j}^{2} ≤ ∑_{j,k=1}^{n} Q_{jk}ξ_{j}ξ_{k} ≤ λ ∑_{j=1}^{n} ξ_{j}^{2}  (2.3)

for all ξ_{1}, …, ξ_{n} and (x, x_{n+1}) ∈ R^{n+1}. We call L in (2.1) a diagonal operator.

If a(x_{n+1}, x) is nowhere vanishing, L is elliptic and the regularity result is well known. We deal with a(x_{n+1}, x) that vanishes somewhere. We will only concentrate in a neighborhood Ω of some point (x^{0}, x_{n+1}^{0}) where a(x^{0}, x_{n+1}^{0}) = 0. The basic assumption is:

There exist k integer, C > 0 constant, such that

∑_{j=0}^{k} |\frac{∂^{j}}{∂x_{n+1}^{j}} a(x_{n+1}, x)| > C  (2.4)

We consider the following three cases:

(i) a(x_{n+1}, x) never changes sign.

(ii) a(x_{n+1}, x) never changes sign from + to - in the positive direction of \frac{∂}{∂x_{n+1}} . Let

z^{+} = {(x, x_{n+1}) ∈ Ω|a(x_{n+1}, x) = 0, a(x_{n+1}, x)

changes sign from - to + in the positive direction of \frac{∂}{∂x_{n+1}} \}

If z^{+} ≠ ∅, we assume z^{+} is a closed submanifold, \frac{∂}{∂x_{n+1}} is transversal to it.

(iii) −a(x_{n+1}, x) satisfies condition (ii). We denote

z^{−} = {(x_{n+1}, x) ∈ Ω|a(x_{n+1}, x) = 0, a(x_{n+1}, x)
changes sign from + to - at positive direction of \frac{∂}{∂x_{n+1}} \}

First, we want to transfer L in (2.2) to a form as in (2.1). If z^{+} ≠ ∅ in (ii) or z^{−} ≠ ∅ in (iii), we can assume they are {x_{n+1} = 0}. This is so because we can make local coordinate change:

y_{j} = x_{j} j = 1, …, n,
y_{n+1} = x_{n+1} − φ(x).

Under the transformation, z^{+} or z^{−} will be straightened to \{y_{n+1} = 0\}, and \frac{∂}{∂x_{n+1}} = \frac{∂}{∂y_{n+1}} , Q is changed to Q′ which still satisfies (2.3). We may assume that we are working on a neighborhood of O. Next, we want to decompose L to L′L_{0}, where L_{0} is a elliptic zero-order pseudo-differential operator, and

L′ = \frac{∂}{∂x_{n+1}} + a′(x_{n+1}, x)Q′  (2.5)

where the symbol of Q′ is

σ(Q′) = \sqrt{∑_{j,k=1}^{n} Q′_{jk}ξ_{j}ξ_{k}}  (2.6)

and {Q′_{jk}}_{j,k}^{n,n} satisfies (2.3). It can be done as follows:

σ(L) = iξ_{n+1} + a(x_{n+1}, x)\sqrt{Q_{n+1,n+1}ξ_{n+1}^{2} + 2(∑_{j=1}^{n} Q_{n+1,j}ξ_{j})ξ_{n+1} + ∑_{j,k=1}^{n} Q_{jk}ξ_{j}ξ_{k}}

Because a is small, we have \frac{∂}{∂x_{n+1}} σ(L) ≠ 0. Use the implicit function theorem, in a cone of {|ξ_{n+1}| ≤ C ∑_{j=1}^{n} ξ_{j}^{2}},

σ(L) = [iξ_{n+1} + b(x_{n+1}, x, ξ_{1}, …, ξ_{n})] · g(x_{n+1}, x, ξ_{1}, …, ξ_{n+1})

where b is homogeneous of degree 1, g is homogeneous of degree zero, and g ≠ 0 for (ξ_{1}, …, ξ_{n+1}) ∈ R^{n+1} \ {0} on that cone. Now −ib is a root of

τ^{2} + a^{2}(Q_{n+1,n+1}τ^{2} + 2(∑_{j=1}^{n} Q_{n+1,j}ξ_{j})τ + ∑_{j,k=1}^{n} Q_{jk}ξ_{j}ξ_{k})

so

−ib = \frac{−a^{2}(∑_{j=1}^{n} Q_{n+1,j}ξ_{j}) ± ia\sqrt{(∑_{j,k=1}^{n} Q_{jk}ξ_{j}ξ_{k})(1 + a^{2}Q_{n+1,n+1}) − a^{2}(∑_{j=1}^{n} Q_{n+1,j}ξ_{j})^{2}}}{1 + a^{2}}

Now let

σ(L_{1}) = iξ_{n+1}+\frac{−ia^{2}(∑_{j=1}^{n} Q_{n+1,j}ξ_{j}) ∓ a\sqrt{(∑_{j,k=1}^{n} Q_{jk}ξ_{j}ξ_{k})(1 + a^{2}Q_{n+1,n+1}) − a^{2}(∑_{j=1}^{n} Q_{n+1,j}ξ_{j})^{2}}}{1 + a^{2}}

(2.7)

If a is small, then the quadratic term inside the square root is positive definite. We make the final local coordinate change: Set y_{n+1} = x_{n+1}, for l = 1, …, n, let y_{l} be the solution of the linear differential equation

(\frac{∂}{∂x_{n+1}} − \frac{a^{2}}{1 + a^{2}} ∑_{j=1}^{n} Q_{n+1,j} \frac{∂}{∂x_{j}})y_{l} = 0

which equals x_{1} on the hyperplane x_{n+1} = 0. This is a coordinate change used in [NT1]. To see it is a coordinate change we notice the Jacobian has determinant 1 at 0. In terms of new coordinates (y_{1}, …, y_{n+1}), we have

L_{1} = \frac{∂}{∂y_{n+1}} ∓ \frac{a}{1 + a^{2}} Q_{1}  (2.8)

where σ(Q_{1}) is \sqrt{∑_{j,k=1}^{n} Q′_{jk}ξ_{j}ξ_{k}}, with \{Q′_{jk}\} satisfies (2.3). We contend that it satisfies condition (i) or (ii). Then the sign in front of the second term in (2.8) is +, because by [NT2], condition (Ψ) is invariant under change of coordinates and multiplication of non-vanishing terms. Also, L is elliptic on any cone {ξ_{n+1}^{2} > C(ξ_{1}^{2} + … + ξ_{n}^{2})}. We need only to concentrate on {ξ_{n+1}^{2} < C(ξ_{1}^{2} + … + ξ_{n}^{2})}. In that cone, we have simplified L to form in (2.5).

We are going to construct a right parametrix for L in the form (2.1) that satisfies condition (i) or (ii). We may assume a ≥ 0 in the condition (i) by changing x_{n+1} to −x_{n+1}. We also specify x_{n+1} = t and assume that the domain Ω is (−1, 1) × B_{n}. Let ω ∈ C_{0}^{∞}(Ω),and ω = 1 on [−\frac{1}{2}, \frac{1}{2}] × \frac{1}{2}B_{n}. We define

K_{0}f(t, x, ξ) = ∫_{T}^{t} ω(t, x)e^{ix·ξ−A(t,t′,x)Q(t,x,ξ)}χ(t′)f̃(t′, ξ)dt′  (2.9)

where χ(t) ∈ C_{0}^{∞}((−1, 1)), χ(t) = 1 on [−\frac{1}{2}, \frac{1}{2}], T = −1 when a satisfies condition(i), T = 0 when a satisfies condition(ii), and

A(t, t′, x) = ∫_{t′}^{t} a(θ, x)dθ  (2.10)

First of all

\frac{∂}{∂t}(K_{0}f)(t, x, ξ) = χ(t)f̃(t, ξ) − a(t, x)Q(t, x, ξ)K_{0}f(t, x, ξ)

−∫_{T}^{t} (\frac{∂}{∂t}Q(t, x, ξ))A(t, t′, x)ω(t, x)χ(t′)e^{ixξ−A(t,t′,x)Q(t,x,ξ)}f̃(t′, ξ)dt  (2.11)

Define

Kf(t, x) = (2π)^{−n}_{∫R_{n}}e^{ix·ξ}K_{0}f(t, x, ξ)dξ  (2.12)

Following lemma appearing as Lemma C.1 in [T]:

Lemma 2.1 There is a constant C_{d} > 0, depending only on the integer d ≥ 0 such that, for all polynomials g in one variable, of degree less than or equal to d, with complex coefficients, and all real number t, t′

|g^{(l)}(t)||t′ − t|^{j+1} ≤ C_{d}| ∫_{t}^{t′} |g(s)|ds|, j = 0, …, d.  (2.13)

We refer to [T] for a proof.

By Malgrange preparation theorem, there is an open neighborhood V of the origin in R^{n+1}, as well as two real-valued smooth function a_{1}(t, x), a_{2}(t, x), defined on V, such that

a(t, x) = a_{1}(t, x)a_{2}(t, x)  (2.14)
a_{2}(t, x) > C > 0 for all (t, x) ∈ V
a_{1}(t, x) = t^{k} + a_{1}(x)t^{k−1} + … + a_{k}(x)

where the a_{j}(x) are real-valued and smooth in V, a_{j}(0) = 0. We may still assume V = [−2, 2]×2B_{n}.

Lemma 2.2 If a(θ, x) keeps the same sign in (t′, t), then

|A(t, t′, x)| ≥ C|a_{1}^{(j)}(t, x)||t − t′|^{j+1}, j = 0, …, k  (2.15)

And for |ξ| = 1

|grad_{x}(A(t, t′, x) · Q(t, x, ξ))|^{k+1} ≤ C|A(t, t′, x)|Q(t, x, ξ)  (2.16)

Proof: (2.15) is a direct consequence of Lemma 2.1. Especially

|A(t, t′, x)| ≥ C|t − t′|^{k+1}

This gives (2.16).

We have the following asymptotic expansion in the standard sense of symbols:

a(t, x)Q(t, x, D_{x})(w(t, x)e^{ix·ξ−A(t,t′,x)Q(t,x,ξ)})
∼ e^{ix·ξ} ∑_{q∈N^{n}} \frac{(−1)^{q}}{q!} a(t, x)Q^{(q)}(t, x, ξ)D_{x}^{q}(w(t, x)e^{−A(t,t′,x)Q(t,x,ξ)})  (2.17)

(2.17) appeared in [T] for a satisfying condition (i). A general asymptotic lemma was in [NT] labeled as Lemma 5.1. If the hypothesis (5.3), loc. cit., is replaced by

|gradw_{1}|^{k+1} ≤ Rew_{1}  (2.18)

the whole proof of lemma in [NT] will still hold, except we may take N′ = −1 + sup(0, (k + 1)(m + n + M + M′)) in lemma. (2.16) guarantees (2.18), so asymptotic formula (2.17) holds.

For N large, let

rN^{′}(t, t′, x, ξ) = ∑_{1≤|q|≤N} \frac{(−1)^{q}}{q!} a(t, x)Q^{(q)}(t, x, ξ)D_{x}^{q}(w(t, x)e^{−A(t,t′,x)Q(t,x,ξ)})  (2.19)

r_{N}(t, t′, x, ξ) = A(t, t′, x)(\frac{∂}{∂t}Q(t, x, ξ)e^{−A(t,t′,x)Q(t,x,ξ)}) + r′_{N}(t, t′, x, ξ)  (2.20)

What (2.17) states is that, given any integer N′, N′′ > 0, there is an N > 0 such that all derivatives with respect to (t, t′, x) of order less than or equal to N′′ of the function

|ξ|^{N′}{a(t, x)Q(t, x, D_{x})(w(t, x)e^{ix·ξ−A(t,t′,x)Q(t,x,ξ)})
−e^{ix·ξ}[w(t, x)a(t, x)Q(t, x, ξ)e^{−A(t,t′,x)Q(t,x,ξ)} − r′_{N}(t, t′, x, ξ)]}  (2.21)

are bounded independently of x, t, t′ and ξ, |ξ| > 1. In view of this we reach the following conclusion:

LKf(x, t) = w(x, t)f(x, t) + (2π)^{−n}_{∫R_{n}}∫_{T}^{t} e^{ix·ξ}r_{N}(t, t′, x, ξ)f̃(t′, ξ)dt′dξ
+ _{∫R_{n}}∫_{T}^{t} S_{N}(t, t′, x, ξ)f̃(t′, ξ)dt′dξ  (2.22)

By (2.21), the last term, if we denote S_{N}f, is very smooth, say, if f ∈ Λ^{α}, then S_{N}f ∈ Λ^{α+100}.

Let

R_{N}f = (2π)^{−n}_{∫R_{n}}∫_{T}^{t} r_{N}(t, t′, x, ξ)f̃(t′, ξ)dt′dξ  (2.23)

We restate (2.22) in the form

LK = I + R_{N} + S_{N}  (2.24)

In the next chapter, we will investigate K and R_{N}. The Fourier distribution kernels of K, R_{N} are

K(t, t′, x, x′) = (2π)^{−n}w_{∫R_{n}}e^{i(x−x′)ξ−A(t,t′,x)Q(t,x,ξ)}dξ  (2.25)
R_{N}(t, t′, x, x′) = (2π)^{−n}w_{∫R_{n}}r_{N}(t, t′, x, ξ)e^{i(x−x′)ξ}dξ  (2.26)

It is important to notice that we have chosen T in that way, so that a(θ, x) keeps the same sign on (t′, t), for any t′ ∈ (T, t). Also A(t, t′, x) > 0 for any t′ ∈ (T, t). This is crucial in the construction of K and R_{N}.

As a final remark, our construction of the parametrix K is similar to what Treves used in [T] for a satisfying condition (i). In [NS] Stein and Nagel constructed a parametrix for L when a = t which produces solutions orthogonal to the null space of L. They developed a symbolic calculus and proved Hölder and L^{p} regularities for L. Smith in [S] generalized their construction to the case where a satisfies condition (ii) and proved L^{p} regularity for L. The difference between the two constructions is that while their parametrix gives orthogonal solutions , our parametrix gives solutions with initial values on z^{+} for a satisfying condition (ii) for the oblique derivative problem.

Estimate of The Parametrix

We carry on the basic estimates for operators K, R_{N} constructed in Chapter 2. To avoid complication with index, we always refer to C as a universal constant. Assume A(t, t′, x) > 0. We have

Lemma 3.1 If g(t, t′, x, ξ) is smooth, m is a non-negative integer, 0 ≤ β < 1, let

I = ∫_{|ξ|=1} \frac{g(t, t′, x, ξ)}{(i(x − x′) · ξ − (A(t, t′, x))Q(t, x, ξ))^{m+β}} dσ_{ξ}

where dσ_{ξ} is the standard volume element in S^{n−1} ⊂ R^{n}. Then

|I| ≤ C\{\frac{1}{(|x − x′| + A(t, t′, x))^{m+β}} \}  (3.1)

where C depends only on g, Q and M.

Proof: For n = 1, (3.1) is obvious. The following is a proof for n ≥ 2. We may assume |x − x′| = 1 in I, A(t, t′, x) < 1. We need to show |I| < C. Choose sphere coordinates (θ_{1}, …, θ_{n−1}) with (x − x′) · ξ = cosθ_{1}, 0 ≤ θ_{1} ≤ π, lets = \frac{θ_{1}}{Q}, for θ_{1} small. Make use of Taylor expansion for g in s variable, then the problem is reduced to estimate

II = ∫_{−1}^{1} \frac{s^{k}}{(is + A(t, t′, x))^{m+β}} ds.

for k < m + β. Because A(t, t′, x) > 0, this is easy by applying integration by parts. Therfore we conclude the proof. Lemma 3.1 is in more general setting than the one in [G], we thank professor E.M. Stein for the remark on the improvement.

Proposition 3.1 (a). There is constant C > 0, independent of t, x, for any ϵ > 0

| ∫_{T}^{t} ∫_{|x−x′|<ϵ} |K(t, t′, x, x′)|dx′dt′| ≤ Cϵ^{δ}  (3.2)
|∫_{T}^{t}_{∫B_{n}}|K(t, t′, x + Δx, x′) − K(t, t′, x, x′)|dx′dt′| ≤ C|Δx|^{δ}  (3.3)

if a(t, x)a(t + ϵ, x) ≥ 0,

| ∫_{t}^{t+ϵ} _{∫B_{n}}|K(t, t′, x, x′)|dx′dt′| ≤ Cϵ|lnϵ|  (3.4)

if a(t, x)a(t + Δt, x) ≥ 0,

|∫_{T}^{t}_{∫B_{n}}|K(t + Δt, t′, x, x′) − K(t, t′, x, x′)|dx′dt′| ≤ C|Δt||ln|Δt||  (3.5)

(b). ∃C_{N} > 0, independent of t, x, ∀ϵ > 0,

| ∫_{T}^{t} ∫_{|x−x′|<ϵ} |R_{N}(t, t′, x, x′)|dx′dt′| ≤ Cϵ^{δ}  (3.6)
| ∫_{T}^{t} ∫_{|x−x′|<ϵ} |R_{N}(t, t′, x + Δx, x′)|dx′dt′| ≤ Cϵ^{δ}  (3.7)

if a(t, x)a(t + ϵ, x) ≥ 0,

| ∫_{t}^{t+ϵ} ∫_{B_{n}} |R_{N}(t, t′, x, x′)|dx′dt′| ≤ Cϵ|lnϵ|  (3.8)

if a(t, x)a(t + Δt, x) ≥ 0,

|∫_{T}^{t}_{∫B_{n}}|R_{N}(t + Δt, t′, x, x′) − R_{N}(t, t′, x, x′)|dx′dt′| ≤ C|Δt||ln|Δt||  (3.9)

where δ = \frac{1}{k+1}.

Proof: In the expression of (2.25).

K(t, t′, x, x′) = (2π)^{−n}w(t, x)_{∫R_{n}}e^{i(x−x′)ξ−A(t, t′, x)Q(t,x,ξ)}dξ

Because A(t, t′, x) > 0 for t′ ∈ (t, t + ϵ) if a(t, x)a(t + ϵ, x) ≥ 0, (3.5) and (3.6) follow from Lemma 3.1:

| ∫_{T}^{t} ∫_{|x−x′|<ϵ} |K(t, t′, x, x′)|dx′dt′|
≤ C\{| ∫_{T}^{t} ∫_{|x−x′|<ϵ} \frac{1}{(|x − x′| + A(t, t′, x))^{n}} dx′dt′\}

≤ C\{| ∫_{T}^{t} ∫_{0}^{ϵ} \frac{1}{r + A(t, t′, x)}drdt′
≤ C\{| ∫_{T}^{t} ∫_{0}^{ϵ} \frac{1}{r + |t − t′|^{k+1}} drdt′|\}
≤ C{| ∫_{T}^{t}(ln(ϵ + |t − t′|^{k+1}) − ln|t − t′|^{k+1})dt′|}
≤ C\{ln(1 + ϵ) + ∫_{0}^{1} \frac{ϵ}{ϵ + s^{k+1}} ds\}
≤ C\{2^{k+1}ϵ∫_{0}^{1} \frac{1}{(ϵ^{1/k+1} + s)^{k+1}} ds\}
≤ Cϵ^{\frac{1}{k+1}} .
| ∫_{t}^{t+ϵ} ∫_{B_{n}} |K(t, t′, x, x′)|dx′dt′|
≤ C\{| ∫_{t}^{t+ϵ} ∫_{B_{n}} \frac{1}{(|x − x′| + A(t, t′, x))^{n}} dx′dt′|\}
≤ C\{| ∫_{t}^{t+ϵ} ∫_{0}^{1} \frac{1}{r + A(t, t′, x)}dx′dt′|\}
≤ C\{| ∫_{t}^{t+ϵ} ∫_{0}^{1} \frac{1}{τ + |t − t′|^{k+1}} dx′dt′|\}
≤ C| ∫_{t}^{t+ϵ} ln|t − t′|^{k+1}dt′|
≤ Cϵ|lnϵ|

As for (3.6),

|∫_{T}^{t}_{∫B_{n}}|K(t, t′, x, x′) − K(t, t′, x + Δx, x′)|dx′dt′|

≤ {|∫_{T}^{t} ∫_{|x−x′|<4λ|Δx|} |K(t, t′, x, x′) − K(t, t′, x + Δx, x′)|dx′dt′|

+|∫_{T}^{t} ∫_{|x−x′|>2λ|Δx|} |K(t, t′, x, x′) − K(t, t′x + Δx, x′)|dx′dt′|}

= I_{1} + I_{2}

By (3.5), I_{1} ≤ |Δx|^{δ}. To estimate I_{2}, for Δx fixed, let

F (t, t′, x, x′, s) = K(t, t′, x + sΔx, x′),

F is a smooth function on |x − x′| ≥ 2λ|Δx| ≥ 2|Δx|.

K(t, t′, x + Δx, x′) − K(t, t′, x, x′) = ∫_{0}^{1} \frac{d}{ds}F ds

\frac{d}{ds}F = _{∫R_{n}}[iΔx · ξ − Δx · ▽_{x}A(t, t′, x + sΔx)Q(t, x + sΔx, ξ)
− A(t, t′, x + sΔx)Δx · ▽_{x}Q(t, x + sΔx, ξ)]
e^{i(x+sΔx−x′)ξ−A(t,t′,x+sΔx,ξ)}dξ

By Lemma 3.1

|\frac{dF}{ds} | ≤ C|Δx| \frac{1}{(|x − x′| + A(t, t′, x + sΔx))^{n+1}}

So

|I_{2}| ≤ |∫_{T}^{t} ∫_{|x−x′|>2λ|Δx|} ∫_{0}^{1}|\frac{dF}{ds} |dsdx′dt′|

≤ C|Δx|\{|∫_{0}^{1}∫_{T}^{t} ∫_{|x−x′|>2λ|Δx|} \frac{1}{(|x − x′| + A(t, t′, x + sΔx))^{n+1}} dx′dt′ds|\}
≤ C|Δx|\{|∫_{0}^{1}∫_{T}^{t} ∫_{2|Δx|}^{1} \frac{1}{(r + A(t, t′, x + sΔx))^{2}} drdt′ds|\}
≤ C|Δx|\{|∫_{0}^{1}∫_{T}^{t} ∫_{2|Δx|}^{1} \frac{1}{(r + |t − t′|^{k+1})^{2}} drdt′ds|\}
≤ C|Δx|\{|∫_{T}^{t} \frac{1}{2|Δx| + |t − t′|^{k+1}} dt′|\}
≤ C|Δx|2^{k+1}\{|∫_{T}^{t} \frac{1}{((2|Δx|)^{\frac{1}{k+1}} + |t − t′|)^{k+1}} dt′\}
≤ C|Δx|^{\frac{1}{k+1}}

Now for (3.8): If a(t, x)a(t + Δt, x) ≥ 0, then a(θ, x) keeps the sign in (T, t), and A(t, t′, x) > 0 for t′ ∈ (T, t). If |t − T | ≤ 2|Δt|, (3.8) follows from (3.7). We may assume t − T > 2|Δt| > 0.

|∫_{T}^{t}_{∫B_{n}}|K(t, t′, x, x′) − K(t + Δt, t′, x, x′)|dx′dt′|
≤ {∫_{|t−t′|<2|Δt|} … + ∫_{|t−t′+Δt|<2|Δt|} … + ∫ |t − t′| > 2|Δt|, |t − t′ + Δt| > 2|Δt| …}
= II_{1} + II_{2} + II_{3}

By (3.7), (II_{1}+II_{2}) ≤ C|Δt|^{δ} |ln|Δt| |. For II_{3}, let Δt be fixed, G(t, t′x, x′, s) = K(t+sΔt, t′, x, x′). Then G is smooth on |t − t′| > 2|Δt|, |t − t′ + Δt| > 2|Δt|, |x − x′| ≥ |Δt|, 0 ≤ s ≤ 1.

K(t + sΔt, t′, x, x′) − K(t, t′, x, x′) = ∫_{0}^{1} \frac{d}{ds}Gds

\frac{d}{ds}G = Δt_{∫R_{n}}[−a(t, +sΔt, x)Q(t + sΔt, x, ξ) − A(t + sΔt, t, x)\frac{∂}{∂t}Q(t + sΔt, x, ξ)]
e^{i(x−x′)ξ−A(t+sΔt,t′,x)Q(t+sΔt,x,ξ)}dξ

By Lemma 3.1

|\frac{d}{ds}G| ≤ C|Δt|\{\frac{(|a(t + sΔt, x)| + A(t + sΔt, t′, x))}{(|x − x′| + A(t + sΔt, t′, x))^{n+1}}\}

|II_{3}| ≤ C|Δt|∫_{T}^{t−2Δt} _{∫B_{n}} \frac{|a(t + sΔt, x)| + A(t + sΔt, t′, x)}{((|x − x′| + A(t + sΔt, t′, x))^{n+1}}dx′dt′\}
≤ C|Δt|\{∫_{0}^{1} ∫_{T}^{t−2Δt} ∫_{0}^{1} \frac{|a(t + sΔt, x)| + A(t + sΔt, t′, x)}{(r + A(t + sΔt, t′, x))^{2}}drdsdt′\}
≤ C|Δt|\{∫_{0}^{1} ∫_{T}^{t−2Δt} |lnA(t + sΔt, t′, x)|dsdt′ + ∫_{0}^{1} ∫_{T}^{t−2Δt} \frac{|a(t + sΔt, x)|}{A(t + sΔt, t′, x)}dt′ds
≤ C|Δt|\{∫_{0}^{1} ∫_{T}^{t−2Δt} |ln|t + sΔt − t′|^{k+1}|dsdt′ + ∫_{0}^{1} ∫_{T}^{t−2Δt} \frac{1}{t + sΔt − t′} dt′ds\}
≤ C|Δt|\{∫_{T}^{t−2Δt} \frac{1}{t − t′} dt′\}
≤ C|Δt||ln|Δt||

The proof of (a) is completed. As for (b), from (2.19) and (2.20), r_{N} is sum of terms like

A(t, t′, x)h(t, t′, x, ξ)e^{−A(t, t′, x)Q(t,x,ξ)}

and

a(t, x)▽_{x}A(t, t′, x)h(t, t′, x, ξ)e^{−A(t, t′, x)Q(t,x,ξ)}

where h is homogeneous degree one with respect to ξ. This is because D^{q} brings down (Q(t, x, ξ))^{|q|}, and Q^{(q)}(t, x, ξ) is of homogeneous degree 1 − |q|. From this

R(t, t′, x, x′) ∼ ∫_{|ξ|=1} \frac{A(t, t′, x)g}{(i(x − x′)ξ − A(t, t′, x))^{n+1}} dσ_{ξ}

or ∫_{|ξ|=1} \frac{a(t, x)▽_{x}A(t, t′, x)g}{(i(x − x′)ξ − A(t, t′, x))^{n+1}} dσ_{ξ}

where g is as in Lemma 3.1. We notice the appearance of A(t, t′, x) and a(t, x)▽_{x}A(t, t′, x), and also by Lemma 2.2,

|a(t, x)▽_{x}A(t, t′, x)| ≤ C|a(t, x)||t − t′| ≤ CA(t, t′, x)

Use this fact and Lemma 3.1, then the argument in the proof of (a) can be completely adopted to prove (b). We will not repeat it here.

Remark: Our proof for part (a) in Proposition 3.1 will work for any

K′(t, t′, x, x′) = ω_{∫R_{n}}g(t, t′x, x′, ξ)e^{i(x−x′)ξ−A(t,t′,x)Q(t,x,ξ)}dξ

with g smooth and homogeneous of degree zero with respect to ξ. Next we want to study K ∘ \frac{∂}{∂t}, K ∘ (aQ), R_{N} ∘ \frac{∂}{∂t}, and R_{N} ∘ (aQ). We will prove in Chapter 4 that they preserve Hölder spaces. But first we deal with K ∘ \frac{∂}{∂t} ∘ |D_{x}|^{−σ}, K ∘ (aQ) ∘ |D_{x}|^{−σ}, R_{N} ∘ \frac{∂}{∂t} ∘ |D_{x}|^{−σ}, and R_{N} ∘ (aQ) ∘ |D_{x}|^{−σ}, for σ > 0. We define

(K ∘ \frac{∂}{∂t} ∘ |D_{x}|^{−σ}f)(t, x) = ∫_{ξ} ∫_{T}^{t} ω(t, x)|ξ|^{−σ}e^{ix·ξ−A(t,t′,x)Q(t,x,ξ)}χ(t′) ˜\frac{∂}{∂t′} f(t′, ξ)dt′dξ  (3.10)

(K ∘ aQ ∘ |D_{x}|^{−σ}f)(t, x) = ∫_{ξ} ∫_{T}^{t} ω(t, x)|ξ|^{−σ}a(t′x)Q(t′, x, ξ)e^{ix·ξ−A(t,t′,x)Q(t,x,ξ)}χ(t′)f̃(t′, ξ)dt′dξ

(3.11)

If f ∈ Λ^{α+1}, α > 0, integrate by parts, then (3.10) becomes

(K ∘ \frac{∂}{∂t} ∘ |D_{x}|^{−σ}f)(t, x) = − ∫_{ξ} ∫_{T}^{t} ω(t, x)|ξ|^{−σ}a(t′, x)Q(t′, x, ξ)e^{ix·ξ−A(t,t′,x)Q(t,x,ξ)}χ(t′)f̃(t′, ξ)dt′dξ
+ |D_{x}|^{−σ}f(t, x)

Let

H_{σ}(t, t′x, x′) = ∫_{ξ} |ξ|^{−σ}a(t′, x)Q(t′, x, ξ)e^{i(x−x′)·ξ−A(t,t′,x)Q(t,x,ξ)}dξ  (3.12)
H_{σ}^{∗}(t, t′x, x′) = ∫_{ξ} |ξ|^{−σ}a(t′, x)Q(t′, x′, ξ)e^{i(x−x′)·ξ−A(t,t′,x)Q(t′,x,ξ)}dξ  (3.13)

They are the distribution kernels for K ∘ \frac{∂}{∂t} ∘ |D_{x}|^{−σ} − |D_{x}|^{−σ}, K ∘ (aQ) ∘ |D_{x}|^{−σ} respectively. Similarly we define R_{N} ∘ \frac{∂}{∂t} ∘ |D_{x}|^{−σ}, and R_{N} ∘ (aQ) ∘ |D_{x}|^{−σ}. and their corresponding distribution kernels J_{σ}(t, t′, x, x′) and J_{σ}^{∗}(t, t′, x, x′).

Proposition 3.2 There exists C_{σ} > 0, independent of t, x, for any ϵ > 0 ,

|∫_{T}^{t} ∫_{|x−x′|<ϵ} |H_{σ}(t, t′, x, x′)|dx′dt′| ≤ Cϵ^{σ}  (3.14)
|∫_{T}^{t}_{∫B_{n}}|H_{σ}(t, t′, x + Δx, x′) − H_{σ}(t, t′, x, x′)|dx′dt′| ≤ C|Δx|^{σ}  (3.15)

If a(t, x)a(t + ϵ, x) ≥ 0

| ∫_{t}^{t+ϵ} _{∫B_{n}}|H_{σ}(t, t′, x, x′)|dx′dt′| ≤ Cϵ^{σ}  (3.16)

If a(t, x)a(t + Δt, x) ≥ 0

|∫_{T}^{t}_{∫B_{n}}|H_{σ}(t + Δt, t′, x, x′) − H_{σ}(t, t′, x, x′)|dx′dt′| ≤ C|Δt|^{σ}  (3.17)

And the same estimates hold for H^{∗}, J, J^{∗}.

Proof: It just repeats the work done in the proof of Proposition 3.1. We give a simple proof of (3.14), the others can be handled in the same fashion.

By Lemma 3.1

∫_{T}^{t} ∫_{|x−x′|<ϵ} |H_{σ}(t, t′, x, x′)|dx′dt′

≤ C\{|∫_{T}^{t} ∫_{|x − x′|<λϵ} \frac{|a(t, x)|}{(|x − x′| + A(t, t′, x))^{n+1−σ}} dx′dt′|
≤ C\{|∫_{T}^{t} ∫_{0}^{ϵ} \frac{|a(t, x)|}{(r + A(t, t′, x))^{2−σ}} dt′dr|\}
≤ C{|∫_{T}^{t} ((A(t, t′, x))^{−1+σ} − (ϵ + A(t, t′, x))^{−1+σ})|a(t′, x)dt′|}
= C{|∫_{T}^{t} ((A(t, t′, x))^{−1+σ} − (ϵ + A(t, t′, x))^{−1+σ})dA(t, t′, x)|}
≤ C|ϵ|^{σ}

As a conclusion of this chapter, we give a brief discussion of L which satisfies condition (iii) in Chapter 2. Let L = \frac{∂}{∂t} + aQ, and ^{t}L be the formal transpose of L. Then ^{t}L satisfies condition(ii) in Chapter 2. Let ^{t}K,^{t} R_{N},^{t} S_{N} be the operators constructed in Chapter 2 for ^{t}L. We have

tL^{t}K = I_{d} +^{t} R_{N} +^{t} S_{N}.

Now, let K, R_{N}, S_{N} be the formal transposes of ^{t}K,^{t} R_{N} and ^{t}S_{N} respectively. Define

K^{+}f(t, x) = ∫_{t}^{1} _{∫B_{n}}^{t}K(t, t′, x, x′)f(t′, x′)dx′dt′ for t ≥ 0

K^{−}f(t, x) = − ∫_{−1}^{t} _{∫B_{n}}^{t}K(t, t′, x, x′)f(t′, x′)dx′dt′ for t ≤ 0
R_{N}^{+}f(t, x) = ∫_{t}^{1} _{∫B_{n}}^{t}R_{N}f(t′, x′)dx′dt′ for t ≥ 0
R_{N}^{−}f(t, x) = − ∫_{−1}^{t} _{∫B_{n}}^{t}R_{N}f(t′, x′)dx′dt′ for t ≤ 0

and

S_{N}^{+}f(t, x) = ∫_{t}^{1} _{∫B_{n}}^{t}S_{N}f(t′, x′)dx′dt′ for t ≥ 0

S_{N}^{−}f(t, x) = − ∫_{−1}^{t} _{∫B_{n}}^{t}S_{N}f(t′, x′)dx′dt′ for t ≤ 0

K, R_{N} and S_{N} would have jump at t = 0, so it is not well defined. But, we have broken equations

K^{+}L = I + R_{N}^{+} + S_{N}^{+} for t ≥ 0  (3.18)
K^{−}L = I + R_{N}^{−} + S_{N}^{−} for t ≤ 0  (3.19)

Also the similar estimates for K^{+}, K^{−}, R_{N}^{+}, R_{N}^{−} hold as for K and R_{N} in Proposition 3.1 and 3.2, except we specify for t ≥ 0 or t ≤ 0 separately. S_{N}^{+}, S_{N}^{−} are very smooth on R_{+}^{n+1} = {t ≥ 0} and R_{−}^{n+1} = {t ≤ 0} respectively. As we will show in Chapter 4 these properties will be enough to establish Hölder regularity of L which satisfies condition(iii).

The Regularity Results

In this chapter we will prove Hölder regularity of subelliptic operators in two variables and of the oblique derivative problem. First, we prove the regularity results for operators constructed in the last two chapters. We use the argument in [RS] to prove the following:

Proposition 4.1 The operator K and R_{N} are bounded linear maps

Λ_{comp}^{α}(Ω)→Λ_{loc}^{α+δ}(Ω)  (4.1)

and

L_{comp}^{∞}(Ω)→Λ_{loc}^{δ}(Ω)  (4.2)

Proof:

Kf(t + Δt, x) − Kf(t, x)
= ∫_{T}^{t} ∫_{B_{n}} K(t + Δt, t′, x, x′)f(t′, x′)dx′dt′
− ∫_{T}^{t} ∫_{B_{n}} K(t, t′, x, x′)f(t′, x′)dx′dt′
+ ∫_{t}^{t+Δt} ∫_{B_{n}} K(t, t′, x, x′)f(t′, x′)dx′dt′

If a(t, x)a(t + Δt, x) ≥ 0, then by Proposition 3.1

|Kf(t + Δ, x) − Kf(t, x)| ≤ C|Δt| |ln|Δt|| ||f||_{L^{∞}}
≤ C|Δt|^{δ}||f||_{L^{∞}}

If a(t, x)a(t + Δt, x) < 0, this means t < T < t + Δt, or t + Δt < T < t. In any case, |t − T | < |Δt|, |t + Δt − T | < |Δt|, by Proposition 3.1 again,

|Kf(t + Δ, x) − Kf(t, x)| ≤ Kf(t + Δt, x) + Kf(t, x)
≤ C|Δt||ln|Δt||||f||_{L^{∞}}
≤ C|Δt|^{δ}||f||_{L^{∞}}

As for |Kf(t, x + Δx) − Kf(t, x)| ≤ C|Δx^{δ}|||f||_{L^{∞}} it is a direct consequence of Proposition 3.1. This shows (4.2) for K. For R argument is the same.

Now, let T denote either K or R, Let f be a function with compact support. Define

u(X, z) = ∫_{R^{n+1}} P_{z}(r)f(X − r)dr

where

X = (t, x) ∈ R^{n+1}, P_{z}(r) = \frac{C_{n+1}z}{(|r|^{2} + z^{2})^{\frac{n+1}{2}}}

Then, f ∈ Λ^{α}, 0 < α < 1 if and only if

||\frac{∂}{∂z} u(X, z)||_{L^{∞}} ≤ Az^{−1+α}. (see [ST])

Set f_{z}(X) = u(X, z), then

||f_{ϵ}(X) − f(X)||_{L^{∞}} ≤ Cϵ^{α} if f ∈ Λ^{α}, 0 < α < 1

f − f_{ϵ} = − ∫_{0}^{ϵ} \frac{∂f_{z}}{∂z} dz

||\frac{∂f_{z}}{∂z} ||_{L^{∞}} ≤ Cz^{α−1}

||\frac{∂f_{z}}{∂t} ||_{L^{∞}} ≤ Cz^{α−1}

|||D_{x}|f_{z}||_{L^{∞}} ≤ Cz^{α−1}

||\frac{∂f_{z}}{∂x_{j}} ||_{L^{∞}} ≤ Cz^{α−1} for j = 1, …, n.

Pick two cut-off function w_{1}, w_{2} ∈ C_{0}^{∞}(Ω), write T′ = w_{1}T w_{2}, F = w_{1}T (w_{2}f) = T′f and F_{ϵ} = w_{1}T (w_{2}f_{2}) = T′f_{ϵ}. By Proposition 3.1, we have

||F − F_{ϵ}||_{Λ^{δ}} ≤ C||f − f_{ϵ}||_{L^{∞}} ≤ Cϵ^{α}  (4.3)
||F_{ϵ}||_{Λ^{δ+1}} ≤ C(||F_{ϵ}||_{Λ^{δ}} + ||\frac{∂}{∂t}F_{ϵ}||_{Λ^{δ}} + ∑_{j+1}^{n} ||\frac{∂}{∂x_{j}} F_{ϵ}||_{Λ^{δ}})  (4.4)

Observation 1:

\frac{∂}{∂t}Kf = b_{1}(t, x)f + R̃_{1}f + K̃_{1}|D_{x}|f

\frac{∂}{∂x_{j}} Kf = R̃_{2}f + K(\frac{∂}{∂x_{j}} f)

\frac{∂}{∂t}Kf = b_{2}(t, x)f + R̃_{3}f + R̃_{4}|D_{x}|f + K̃_{3}f

\frac{∂}{∂x_{j}} Kf = R̃_{5}f + R(\frac{∂}{∂x_{j}} f)

where b_{1}, b_{2} are smooth functions, K̃_{j}, R̃_{l} are operators like K and R. The observation can be checked directly from (2.9),(2.12),(2.19) and(2.20).

The inequalities

|||D_{x}|f_{ϵ}||_{L^{∞}} ≤ Cϵ^{α−1}

and

||\frac{∂}{∂x_{j}} f_{ϵ}||_{L^{∞}} ≤ Cϵ^{α−1}

implying

||\frac{∂}{∂t}F_{ϵ}||_{Λ^{δ}} ≤ C(||f_{ϵ}||_{Λ^{δ}} + ϵ^{α−1})
≤ C(||Δf_{ϵ}||_{Λ^{δ}} + ϵ^{α−1})
≤ Cϵ^{α−1}

Also

||\frac{∂}{∂t}F_{ϵ}||_{Λ^{δ}} ≤ Cϵ^{α−1}

by (4.4),

||F_{ϵ}||_{Λ^{δ+1}} ≤ Cϵ^{α−1}  (4.5)

We have δ + α < 2, because 0 < α < 1, 0 < δ < 1. To show F ∈ Λ^{δ+α}, it suffices to estimate

|Δ_{h}^{2}F | ≤ C|h|^{δ+α} for h small

where

Δ_{h}^{2}F = F (X + h) + F (X − h) − 2F (X)  (4.6)

However, in general ||Δ^{2}G||_{L^{∞}} ≤ |h|^{β}||G||_{Λ^{β}}, if 0 < β < 2. Therefore

||Δ_{h}^{2}F ||_{L^{∞}} ≤ C(|h|^{δ}ϵ^{α} + |h|^{δ+1}ϵ^{α−1})

by (4.3),(4.5) and (4.6). We take therefore ϵ = |h| and get ||Δ_{h}^{2}F ||_{L^{∞}} ≤ C|h|^{α+δ}. This proves that F ∈ Λ^{α+δ} when 0 < α < 1. To prove the proposition for l < α < l + 1, we use the identities in observation 1 l times and reduce problem to the case 0 < α < 1.

Thus the proposition is proved for α > 0 but integers. The case when α is an integer follows by the standard interpolation theorems for the Λ^{α} spaces(see [C]). So the proof is complete.

Corresponding to Proposition 3.2, we have the following:

Proposition 4.2 For σ > 0 fixed, the operators (K · \frac{∂}{∂t}) · |D_{x}|^{−δ}, (K · aQ) · |D_{x}|^{−δ}, (R_{N} · \frac{∂}{∂t}) · |D_{x}|^{−δ}, (R_{N} · aQ) · |D_{x}|^{−δ} are bounded linear maps on

Λ_{comp}^{α}→Λ^{α+σ} for α > 1

Proof: The whole proof of Proposition 4.1 can be adopted here with out change, because a similar observation for operators in this proposition holds as in Proposition 4.1.

Proposition 4.3 The operators (K · \frac{∂}{∂t}), (K ·aQ), (R_{N} · \frac{∂}{∂t}), (R_{N} ·aQ) are bounded linear operators on

Λ_{comp}^{α}→Λ_{loc}^{α} for α > 1

Proof: It follows from Proposition 4.2 and the fact that, for any 0 < σ < 1 fixed, |D_{x}|^{σ} is a bounded operator Λ^{α+σ}→Λ^{α}.

Remark: For operators K^{+}, K^{−}, R_{N}^{+} and R_{N}^{−}, we have similar results as in Proposition 4.1 and 4.2, except that the underlined spaces will be Λ^{α}(R_{+}^{n+1}) and Λ^{α}(R_{−}^{n+1}) respectively, because similar estimates hold for these operators as pointed out at the end of Chapter 3.

Theorem 4.1 Let L = \frac{∂}{∂x_{n+1}} +a(x_{n+1}, x)Q on R^{n+1} as in (2.1), with Q satisfying (2.2) and (2.3). Then, the following results hold locally:

(a). If L satisfies condition (i) in Chapter 2, then for any f ∈ Λ_{comp}^{α}, α > 0 there exists u ∈ Λ_{loc}^{α+δ}, Lu = f and any solution u, Lu = f, is in Λ_{loc}^{α+δ}.

(b). If L satisfies condition(ii), and z^{+} ≠ ∅, then for any f ∈ Λ_{comp}^{α}, α > 0 there exists u ∈ Λ_{loc}^{α+δ}, Lu = f.

(c). If L satisfies condition (iii), and z_{−} ≠ ∅, then

Lu ∈ Λ_{comp}^{α}⇒u ∈ Λ_{loc}^{α+δ} for α > 0

Proof: As we pointed out in Chapter 2, L is elliptic on any cone {ξ_{n+1}^{2} > c(ξ_{1}^{2} + … + ξ_{n}^{2})}. If c is suitably small, L = L_{0}L_{1} on the cone {ξ_{n+1}^{2} < c(ξ_{1}^{2} + … + ξ_{n}^{2})}, and L_{1} is in the form (2.5), L_{0} elliptic on the cone. Let 1 ≥ ρ(ξ) ≥ 0 be a homogeneous degree zero function, smooth on R^{n+1} \ \{O\}, with suppρ ⊂ \{ξ_{n+1}^{2} < c(ξ_{1}^{2} + … + ξ_{n}^{2})\}, and ρ = 1 on \{ξ_{n+1}^{2} < \frac{c}{2}(ξ_{1}^{2} + … + ξ_{n}^{2})\}. Let ρ_{1}(ξ) = 1 − ρ(ξ), and P_{0}, P_{1} be pseudo-differential operators with symbols ρ, ρ_{1}, respectively, then P_{0} + P_{1} ≡ Imod(S^{−∞}). Pick ω ∈ C_{0}^{∞}, ω = 1 in neighborhood of the origin. For f ∈ Λ_{comp}^{α}, we define f_{0} = wP_{0}f, f_{1} = wP_{1}f. Then f_{0}, f_{1} ∈ Λ_{comp}^{α} because L is elliptic on \{ξ_{n+1}^{2} ≥ \frac{c}{2}(ξ_{1}^{2}+…+ξ_{n}^{2})\}. So there exists u_{1} such that Lu_{1} = f_{1}. For f_{0}, use the operators K, R_{N}, S_{N} and (2.24). We have , for N′ > 0 integer

L_{1}K(∑_{j=0}^{N′}(−1)^{j}(R_{N} + S_{N})^{j}) = I + (−1)^{N′}(R_{N} + S_{N})^{N′}  (4.7)

Let v_{N′} = (K ∑_{j=0}^{N′}(−1)^{j}(R_{N} + S_{N})^{j})L_{0}^{−1}f_{0}, we have

L_{1}v_{N′} = L_{0}^{−1}f_{0} + (−1)^{N′}(R_{N} + S_{N})^{N′}L_{0}^{−1}f_{0}  (4.8)

Because L is locally solvable, if necessary, we shrink the domain so there exists w_{N′} ∈ H_{loc}^{N′δ+α}, L_{1}w_{N′} = (−1)^{N′+1}(R_{N} + S_{N})^{N′}L_{0}^{−1}f_{0}. If N′ is large, w_{N′} ∈ Λ_{loc}^{α+δ}. Then u_{0} = v_{N′} + w_{N′} is a solution of L_{1}u_{0} = L_{0}^{−1}f_{0}, so Lu_{0} = f_{0} + g, where g is a smooth function. Combining, we have

Lu′ = f + g′

where u′ = u_{0} + u_{1}, g′ smooth. Finally, let v ∈ Λ^{α+δ} be a solution of lv = −g′, then u = u′ + v is a solution of

Lu = f, u ∈ Λ_{loc}^{α+δ}.

This proves (b) and the first part of (a). Since L is subelliptic in (a), any solution of Lu = 0 is smooth. This gives the second part of (a). We show (c) in three steps:

\underline{Step 1}: If f ∈ Λ_{comp}^{α}, Lu = f and u is continuous, then u ∈ Λ^{α+δ}. Let α+δ < 1, from the remark after Proposition 4.3 and the subellipticity of L, u ∈ Λ_{loc}^{α+δ}(R_{+}^{n+1} ∩ Ω) and u ∈ Λ_{loc}^{α+δ}(R_{−}^{n+1} ∩ Ω). By the assumption, u is continuous, so u ∈ Λ_{loc}^{α+δ}(R^{n+1} ∩ Ω). If α + δ ≥ 1, from what we have just proved, u ∈ Λ^{1−ϵ} for any ϵ > 0 because Λ^{α} ⊂ Λ^{1−δ−ϵ}. We then have [(I − δ)^{1/4}, L]u ∈ Λ^{1/2−ϵ}, and

L((I − Δ)^{1/4}u) = (I − Δ)^{1/4}f − [(I − Δ)^{1/4}, L]u  (4.9)

and (I − Δ)^{1/4}u ∈ Λ^{1/2−ϵ}, so it is continuous. This gives (I − Δ)^{1/4}u ∈ Λ^{1/2−ϵ+δ}, so u ∈ Λ^{1+δ−ϵ}. Replace (I −Δ)^{1/4} by (I −Δ)^{1/4+δ/2} in (4.9), we will get (I −Δ)^{1/4+δ/2}u ∈ Λ^{1/2+δ−ϵ}, ⇒u ∈ Λ^{1−ϵ+lδ}. Repeat the process, we get u ∈ Λ^{1−ϵ+(l+1)δ} if 1 − ϵ + lδ < α + δ. So after a finite number of times, we get u ∈ Λ^{α+δ}.

\underline{Step 2}: If L is as in Step 1, R is a zero-order pseudo-differential operator, for any ϵ > 0, if f ∈ Λ^{α}, v ∈ Λ^{ϵ}, with (L + R)v = f, then v ∈ Λ^{α+δ}. First we know L + R is subelliptic with the same gain δ. If f ∈ Λ^{α}, v ∈ Λ^{ϵ}, (L + R)v = f, so Lv = f + Rv, Rv ∈ Λ^{ϵ}, by step 1, v ∈ Λ^{δ+ϵ} if δ + ϵ < α + δ. So Rv ∈ Λ^{ϵ+δ}, v satisfies the same equation Lv = f + Rv, by step 1, v ∈ Λ^{ϵ+2δ} if ϵ + 2δ < α + δ. Repeat the argument, then v ∈ Λ^{α+δ}.

\underline{Step 3}: If f ∈ Λ^{α}, α > 0, Lu = f, then u ∈ Λ^{α+δ}. If f ∈ Λ^{α}, α ≥ n/2, then u ∈ H^{n/2+δ}, by the Sobolev embedding theorem u is continuous, from Step 1, u ∈ Λ^{α+δ}. This proves (c) for α ≥ n/2.

For any α > 0, f ∈ Λ^{α}, Lu = f, by subellipticity, u ∈ H^{α+δ}, so (I − Δ)^{−n/4}u ∈ H^{n/2+α+δ}, ⇒(I − Δ)^{−n/4}u ∈ Λ^{δ}.L((I − Δ)^{−n/4}u) + ([(I − Δ)^{−n/4}, L](I − Δ)^{n/4}((I − Δ)^{−n/4}u) = (I − Δ)^{−n/4}f, L_{0} = [(I − Δ)^{−n/4}, L](I − Δ)^{n/4} is zero-degree. From Step 2, (I − Δ)^{−n/4}u ∈ Λ^{n/2+α+δ}, ⇒u ∈ Λ^{α+δ}.

In Chapter 1, we constructed a counterexample in R^{n}, n ≥ 3, for which the Hölder regularity failed to hold. For subelliptic operators in two variables, we have the following:

Theorem 4.2 If P is a pseudo-differential operator of order m in two variables and P is of principal type and satisfies condition (P), then P u ∈ Λ^{α}⇒u ∈ Λ^{α+m−1+δ}. The gain δ here is equal to the gain in Sobolev space.

Proof: Let P (x, ξ) be the principal symbol of P. If P (x, ξ) never vanishes P is elliptic, the wellknown result. Suppose P (x_{0}, ξ^{0}) = 0 for some unit vector ξ^{0}. We may assume x_{0} = 0, \frac{∂P (x_{0},ξ^{0})}{∂ξ_{1}} ≠ 0. There is a smooth function λ(x, ξ_{2}) in U × X, where U is a neighborhood of the origin in R^{2}, and X is an open cone which contains ξ^{0}, with λ(0, ξ^{0}) = ξ_{1}^{0} and such that

P (x, ξ) = (ξ_{1} − λ(x, ξ_{2})) · q(x, ξ)  (4.10)

with q ≠ 0 in U × X. So q is elliptic in U × X of order m − 1.

Let L_{1} be the pseudo-differential operator of first order with symbol ξ_{1} − λ(x, ξ_{2}). We write λ(x, ξ_{2}) = β(x, ξ_{2}) + ir(x, ξ_{2}), β, r are real functions. L_{1} is elliptic on the cone |ξ_{1}| > C|ξ_{2}| for C large. We need only to consider X_{+} = {|ξ_{1}| < Xξ_{2}} and X_{−} = {|ξ_{1}| < −Cξ_{2}}. We may assume that ξ^{0} ∈ X_{+}. Because λ is homogeneous of degree 1 with respect to ξ_{2}, so on X_{+}, β(x, ξ_{2}) = β(x)ξ_{2} and r(x, ξ_{2}) = r(x)ξ_{2}, the symbol of L_{1} is now ξ_{1} − β(x)ξ_{2} − ir(x)ξ_{2}. As we have done in Chapter 2, we make a coordinate change to eliminate the β term. We take y_{1} = x_{1}, and let y_{2} be a solution of

\frac{∂y_{2}}{∂x_{1}} − β(x)\frac{∂y_{2}}{∂x_{2}} = 0

which equals x_{2} on x_{1} = 0. This is a coordinate change as we pointed out in chapter 2 as long as we restrict it to a neighborhood of the origin. Under this transformation, ξ_{1} − β(x)ξ_{2} − ir(x)ξ_{2} changes to

η_{1} − ir(x)\frac{∂y_{2}}{∂x_{2}} η_{2} = η_{1} + ia(y)η_{2}

So L_{1} = \frac{∂}{∂y_{1}} + ia(y)\frac{∂}{∂y_{2}} . By[NT1], condition (P) is invariant under changes of coordinates and multiplication of non-vanishing factors. So a satisfies condition (ii) in Chapter 2. Because |ξ| = 1 is a compact set, we can pick finite open cones Γ_{j}, j = 1, …, l, which cover |ξ| = 1, and on those cones, P is a composition of L_{j} and Q_{j}, with L_{j} first order diagonalizable and Q_{j} is m − 1 order elliptic on Γ_{j}. Use partition of unity and let ρ_{j} ≥ 0, ρ_{j} ∈ C_{0}^{∞}(Γ_{j}), ρ_{j} is homogeneous of degree zero, and ∑_{j=1}^{l} ρ_{j} = 1. Let ρ_{j}(D) be the pseudo-differential operators with ρ_{j}(ξ) as symbol, and

f_{j} = Q_{j}^{−1}ρ_{j}(D)P u  (4.11)

By Theorem 4.1, there exists v_{j} ∈ Λ^{α+m−1+δ}

L_{j}v_{j} = f_{j}, j = 1, …, l  (4.12)

combining together, we have

P (∑_{j=1}^{l} v_{j}) = P u + g  (4.13)

where g is a smooth function. So

P (∑_{j=1}^{l} v_{j} − u) = g  (4.14)

But P is subelliptic, so ∑_{j=1}^{l} v_{j} − u ∈ C^{∞}, that is u ∈ Λ^{α+m−1+δ}.

Finally, as an application of our analysis of diagonal pseudo-differential operators, we prove a regularity result for the subelliptic oblique derivative problem. The so-called oblique derivative problem is to solve the following boundary value problem:

\{ \begin{matrix}P u = f & on Ω \\ lu|_{∂Ω} = g \end{matrix}  (4.15)

where Ω is a bounded open set in R^{n+1}, P is a second order real elliptic differential operator

P = a_{ij} \frac{∂}{∂x_{i}} \frac{∂}{∂x_{j}} + b_{j} \frac{∂}{∂x_{j}} ,  (4.16)

and l is a nowhere vanishing real vector field. We assume ∂Ω is smooth. In general, the problem (4.15) is not well-posed. We refer to [E2] for a more detailed discussion. In order to have a solution, it is necessary that f and g satisfy some compatibility conditions (see [EK] or [W]). This is similar to the elliptic boundary value problem ([ADN]).

To state our next theorem, we write

l = T + a(x)n⃗  (4.17)

where T is a tangential vector field to ∂Ω, and n⃗ is the outward normal vector field. We assume that a(x) vanishes only for finite order less than k along the integral curves of T, k is an integer. Let

z^{+} = {x ∈ ∂Ω|a(x) = 0,

a(x)changes sign from - to + in the positive direction of T at x}

z^{−} = {x ∈ ∂Ω|a(x) = 0,

a(x)changes sign from + to - in the positive direction of T at x} Theorem 4.3 (a). If a(x) never changes sign from - to + in the positive direction of T and if z^{−} ≠ ∅, we assume it is a closed submanifold in ∂Ω, and T is transversal to z^{−}. If u is a solution of (4.15) with f ∈ Λ^{α}(Ω̄), g ∈ Λ^{β}(∂Ω), which satisfy a finite necessary linear compatibility conditions, then u ∈ Λ^{γ}(Ω̄), where γ = min(α + 2, β + \frac{1}{k+1}).

(b). If a(x) never changes sign from + to - in the positive direction of T, suppose z^{+} ≠ ∅, and z^{+} is a submanifold in ∂Ω, and T is transversal to z^{+}, then the following boundary value problem

\{ \begin{matrix}P u = f & in Ω \\ lu|_{∂Ω} = g \\ u|_{z^{+}} = 0 \end{matrix}  (4.18)

with f ∈ Λ^{α}(Ω̄)andg ∈ Λ^{β}(∂Ω) satisfying finite necessary linear compatibility conditions, has a solution u ∈ Λ^{γ}(Ω̄), γ = min(α + 2, β + \frac{1}{k+1}).

Proof: By [H2], the problem can be reduced to a pseudo-differential equation problem on the boundary. Let E be a parametrix defined on R^{n+1} for P and let f^{0} be the extension of f on R^{n+1} with f^{0}(x) = 0 for x ∉ Ω̄. Let v(x) = Ef^{0}(x), we have

\{ \begin{matrix}P (u(x) − v(x)) = 0 & on Ω \\ l(u(x) − v(x)) = g(x) − lEf^{0} & on ∂Ω \end{matrix}  (4.19)

If we can solve

lw = g(x) − lEf^{0} on ∂Ω  (4.20)

Using the Poisson operator of P, we can extend w to Ω, then u = v + w would be a solution of (4.15). They are equivalent. Use partion of unity, in a local coordinate chart (4.20) is

\frac{∂w}{∂x_{n+1}} + a(x)Qw = g(x) − \frac{∂}{∂x_{n+1}} (Ef^{0}) + a(x)Q(Ef^{0})  (4.21)

where Q is a pseudo-differential operator with symbol σ(Q) = \sqrt{∑_{j,k}^{n+1} Q_{jk}ξ_{j}ξ_{k}}, and the matrix {Q_{jk}} is positive definite. If a(x) satisfies the hypothesis in (b), by Theorem 4.1, there is a local solution w of (4.21), w ∈ Λ^{γ}, γ = min(α + 1 + \frac{1}{k+1}, β + \frac{1}{k+1}). We claim that we can take γ = min(α + 2, β + \frac{1}{k+1}). If we go back to the proof of Theorem 4.1, the approximate solution we constructed is v_{N′} = (K ∑ j = 0^{N′}(−1)^{j}(R_{N} + S_{N})^{j})L_{0}^{−1}f_{0}, in this case now,

f = g(x) − \frac{∂}{∂t}(Ef^{0}) + a(x)Q(Ef^{0})  (4.22)

By Proposition 4.3, K ∘ \frac{∂}{∂t}, K ∘ (aQ), R_{N} ∘ (aQ) preserve Hölder spaces, and S_{N} is a very smooth operator, so we contend that v_{N′} ∈ Λ^{γ}, γ = min(α + 2, β + \frac{1}{k+1}). We proved that the local solution w ∈ Λ^{γ}, γ = min(α + 2, β + \frac{1}{k+1}). combining the local solutions together, we have a global w satisfying

lw = g(x) − lEf^{0} + h(x) on ∂Ω  (4.23)

where h(x) ∈ C^{∞}(∂Ω). If we go back to the original (4.18), we have a solution u′ ∈ Λ^{γ}(Ω̄) of

\{ \begin{matrix}P u′ = f & on Ω \\ lu′ = g + h & on ∂Ω \\ u′|_{z^{+}} = h_{1} \end{matrix}

where h_{1} ∈ Λ^{γ}. Now, by [W], the following problem

\{ \begin{matrix}P u′′ = 0 & on Ω \\ lu′′ = h & on ∂Ω \\ u′′|_{z^{+}} = h_{1} \end{matrix}

has a solution u′′ ∈ Λ^{γ}(Ω̄), because h ∈ C^{∞}(∂Ω) and f = 0 ∈ C^{∞}(Ω̄). Finally, u = u′ − u′′ is a solution of (4.18) with u ∈ Λ^{γ}(Ω̄), γ = min(α + 2, β + \frac{1}{k+1}).

As for (a), if the solution u of (4.15) exists, we will have a corresponding w, satisfying (4.21) in any local coordinate chart. If a(x) satisfies condition (i) of chapter 2, in the local coordinate chart we can use the same argument in the proof with the help of Proposition 4.3. We then can get w ∈ Λ^{γ} in that coordinate chart. If a(x) satisfies condition (iii) there, from lw = g(x) − lEf, and Theorem 4.1, w ∈ Λ^{γ}, γ = min(α+1+ \frac{1}{k+1}, β + \frac{1}{k+1}). But from w +R_{N}^{+}w +S_{N}^{+}w = K^{+}(g(x)−lEf) on R_{+}^{n+1} and w + R_{N}^{−}w + S_{N}^{−}w = K^{−}(g(x) − lEf) on R_{−}^{n+1} and the fact K^{±}lEf ∈ Λ_{loc}^{α+2}(R_{±}^{n+1}), we conclude that w ∈ Λ^{γ}, γ = min(α + 2, β + \frac{1}{k+1}), by using the same argument in the proof of Theorem 4.1 . So the original solution u of (4.15) is in Λ^{γ}(Ω̄) with γ = min(α + 2, β + \frac{1}{k+1}).

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