Microglobal regularity and the global wavefront set

Abstract

In this paper, we begin the study of regularity of partial differential equations in the space of global $L^q$ Gevrey functions, recently introduced in Adwan et al. (J Geom Anal 27(3):1874–1913, 2017) and Hoepfner and Raich (Indiana Univ Math J, forthcoming) and in a generalized and new function space called the space of global $L^q$ Denjoy–Carleman functions. We develop a wedge approach similar to Bony’s theorem (Bony in Séminaire Goulaouic–Schwartz (1976/1977), Équations aux dérivées partielles et analyse fonctionnelle, Exp No 3. Centre Math, École Polytech, Palaiseau, 1977) and prove three main theorems. The first establishes the existence of boundary values of continuous functions on a wedge. Next, we borrow the FBI transform approach from Hoepfner and Raich (forthcoming) to define global wavefront sets and prove a relationship between the inclusion of a direction in the global wavefront set and the existence of boundary values of sums of weighted $L^p$ functions defined in wedges. The final result is an application in which we prove a global version of a classical result: namely, the relationship between the global characteristic set of a partial differential operator P and the microglobal wavefront sets of u and Pu.

Appendices

Appendix A: On the sequence $M=\left( M_{j}\right)$

Definition A.1

Let $M=\left( M_{j}\right)$ be a sequence of positive real numbers satisfying the following properties:

$\left( \mathbf{Initial conditions }\right)$

$$\begin{aligned} M_{0}=M_{1}=1. \end{aligned}$$

(A.1)

$\left( \mathbf{Strong non-quasianalyticity }\right)$ There exists a constant $A>1$ such that for all $p=1,2,\ldots ,$ we have

$$\begin{aligned} \sum _{j=p}^{\infty }\frac{M_{j}}{M_{j+1}}\le Ap\frac{M_{p}}{M_{p+1}}. \end{aligned}$$

(A.2)

$\left( \mathbf{Strong logarithmic convexity }\right)$ For some fixed $A>0$ and for any r,  with $0\le r<1/A,$ if we set $P_{j}=M_{j}/\left( j!\right) ^{r}$, then

$$\begin{aligned} \text {the sequence }\left( \frac{P_{j}}{jP_{j-1}}\right) \text { is increasing.} \end{aligned}$$

(A.3)

$\left( \mathbf{Stability under ultradifferential operators }\right)$ There are constants $A>1$ and $H>1,$ independent of n, such that for all $n=1,2,3,\ldots ,$ we have

$$\begin{aligned} M_{n}\le AH^{n}\underset{0\le j\le n}{\min }M_{j}M_{n-j}. \end{aligned}$$

(A.4)

1.1 A.1. Some consequences

We refer to the paper [27] for consequences of the conditions listed in Definition A.1. For instance, condition (A.3) implies: (i) the (usual) logarithmic convexity condition: For all $j=1,2,3,\ldots$

$$\begin{aligned} M_{j}^{2}\le M_{j-1}M_{j+1}; \end{aligned}$$

(A.5)

(ii) for all $0\le j\le n,$

$$\begin{aligned} \left( {\begin{array}{c}n\\ j\end{array}}\right) M_{j}M_{n-j}\le M_{n} \end{aligned}$$

(A.6)

and (iii)

$$\begin{aligned} \text {the sequence }\ \ \left( \frac{M_{j}}{j!}\right) ^{1/j}\ \ \text { is increasing.} \end{aligned}$$

(A.7)

Condition (A.6) insures that the class $C^{M}\left( U\right)$ is invariant under composition and, in particular, that for all $0\le j\le n,$

$$\begin{aligned} M_{j}M_{n-j}\le M_{n}. \end{aligned}$$

(A.8)

The condition (A.4) implies the (usual) Stability under differential operators condition; i.e., There are constants $A>1$ and $H>1,$ independent of n and j, such that for all $1\le j\le n,$ we have

$$\begin{aligned} M_{n}\le AH^{n-1}M_{j}M_{n-j}. \end{aligned}$$

(A.9)

We will often replace $AH^{n-1}$ with $C^{n}$.

If the sequence M satisfies conditions (A.1) and (A.3), then it satisfies the following condition: for all $n=1,2,3,\ldots$

$$\begin{aligned} M_{n}\ge n! \end{aligned}$$

(A.10)

Condition (A.10) insures that every analytic function belongs to the class $C^{M}$.

1.2 A.2. Associated functions

Definition A.2

For each sequence $\left( M_{j}\right)$ of positive numbers we define its associated function$M\left( t\right)$ on $\left( 0,\infty \right)$ by

$$\begin{aligned} M\left( t\right) =\underset{j}{\sup }\log \frac{t^{j}}{M_{j}}. \end{aligned}$$

(A.11)

For the reader who is interested in learning more about associated functions and how each of the conditions which we impose on the sequence can be written in terms of the associated function, we recommend the paper by Komatsu [27]. In particular, it is not difficult to show that if $\left( M_{j}\right)$ satisfies conditions (A.1) and (A.10), then for all $t>0$,

$$\begin{aligned} \log t\le M\left( t\right) \le t. \end{aligned}$$

(A.12)

Appendix B: Some estimates

Lemma B.1

(See [25]) If the sequence $M=(M_j)_{j\in {\mathbb {N}}}$ satisfies (A.4) and (A.8),  then for each $\theta >0$ and $k,r, \ell \in {\mathbb {N}}$ such that $k\ge r\ge 0$ we have

$$\begin{aligned} t^r M_{k-r} \le A \frac{ H^{\ell r} }{\theta ^{ r}} M_k\, e^{\frac{1}{2^\ell }M(\theta t)}, \quad \text {for all}\ t>0 \end{aligned}$$

(B.1)

where A and H are given by (A.4).

Proof

We first note that property (A.4) is equivalent to (see [27, Proposition 3.6])

$$\begin{aligned} M\left( \frac{t}{H} \right) \le \frac{1}{2} M(t) + \log {\sqrt{A}} \end{aligned}$$

(B.2)

and this implies that for every $\ell \in {\mathbb {N}}$, the following inequality holds true

$$\begin{aligned} M\left( \frac{t}{H^\ell } \right) \le \frac{1}{2^\ell } M(t) + \log {\sqrt{A}}\sum _{j=0}^{\ell -1}\frac{1}{2^j} \le \frac{1}{2^\ell } M(t) + 2\log {\sqrt{A}}. \end{aligned}$$

(B.3)

Thus if $A>0$ and $H>0$ are given by (A.4), and $\theta >0$, $k,r,\ell \in {\mathbb {N}}$ are chosen such that $k\ge r$ then it follows from (A.8), (A.11) and (B.3) respectively that

$$\begin{aligned} t^r M_{k-r}&\le \frac{H^{r\ell }}{\theta ^{r}} M_{k} \frac{(\frac{\theta t}{H^\ell })^r}{M_r} \le \frac{H^{r\ell }}{\theta ^r} M_k e^{M((\theta t)/H^\ell )} \nonumber \\&\le \frac{H^{r\ell }}{\theta ^r} M_k \exp \{\tfrac{1}{2^\ell } M(\theta t) + 2\log {\sqrt{A}}\} \nonumber \\&=A \frac{H^{r\ell }}{\theta ^r} M_k e^{\tfrac{1}{2^\ell } M(\theta t) } \end{aligned}$$

(B.4)

as we wished to prove. $\square$

Proposition B.2

Let $k\in {\mathbb {N}}_0$. Then

• 1.
  $$\begin{aligned} \frac{d^{2k}}{dx^{2k}} e^{-ax^2} = e^{-ax^2} \sum _{j=0}^k (-1)^{k+j} a^{k+j}x^{2j} b_{2k,j} \end{aligned}$$

  and

  $$\begin{aligned} \frac{d^{2k+1}}{dx^{2k+1}} e^{-ax^2} = e^{-ax^2} \sum _{j=0}^k (-1)^{k+j+1} a^{k+j+1}x^{2j+1} b_{2k+1,j} \end{aligned}$$

  for some constants $b_{2k,j}, b_{2k+1,j} >0$.

• 2. The constants $b_{2k,j}, b_{2k+1,j}$ satisfy the following (recursion) relations.

  • (i) $b_{2k+1,j} = 2 b_{2k,j} + 2(j+1)b_{2k,j+1};$

  • (ii) $b_{2k+2,j} = 2 b_{2k+1,j-1} +(2j+1) b_{2k+1,j}$

  with the understanding that $b_{2k,j} = 0$ if $j \le -1$ or $j \ge \ell +1$ and $b_{2k+1,j} =0$ if $j \le -1$ or $j \ge \ell +1$.

• 3. The constants $b_{2k,j}$ and $b_{2k+1,j}$ have the following upper bounds : 

  • (i) $b_{2k,k} = 2^{2k}$ and $b_{2k+1,k} = 2^{2k+1}$

  • (ii) There exist constants $A,C>0$ so that

    $$\begin{aligned} b_{2k,j} \le C A^k k^{k-j} \quad \text {and}\quad b_{2k+1,j} \le C A^k k^{k-j}. \end{aligned}$$

Proof

The proofs of 1. and 2. follow easily from induction. The only interesting part is 3., and this will follow from a counting argument. The number $b_{2k+1,j}$ is the coefficient of the term (up to a sign and a power of a) $x^{2j+1}e^{-ax^2}$. Viewing the coefficient $b_{2k+1,j}$ are part of tree, the parents of $b_{2k+1,j}$ are $b_{2k,j}$ and $b_{2k,j+1}$ since

$$\begin{aligned} \frac{d}{dx} e^{-ax^2} x^{2\ell } = -2a e^{-ax^2}x^{2\ell +1} + 2j e^{-ax^2}x^{2\ell -1}. \end{aligned}$$

We will call $b_{2k,j}$ the left parent of $b_{2k+1,j}$ and $b_{2k,j}$ is the right child of $b_{2k+1,j}$. Similarly, we will call $b_{2k,j+1}$ the right parent of $b_{2k+1,j}$ and $b_{2k+1,j}$ the left child of $b_{2k,j+1}$. To pass from the left parent to the child, the term is multiplied by $-2a$, a doubling of the coefficient and an increase of the power of x. The pass from the right parent to the child, the polynomial term of $e^{-ax^2}x^{2j+2}$ is differentiated in x and consequently the child inherits a $(2j+2) b_{2k,j+1}$ summand. Visually, a tree looks like

$$\begin{aligned} \begin{matrix} b_{1,0} &{} &{} &{} \\ b_{2,0} &{}\quad b_{2,1}&{} &{} \\ &{}\quad b_{3,0}&{}\quad b_{3,1} &{} &{} \\ &{}\quad b_{4,0} &{}\quad b_{4,1}&{}\quad b_{4,2} &{} \\ &{} &{}\quad b_{5,0} &{}\quad b_{5,1}&{}\quad b_{5,2} &{} \\ &{} &{}\quad b_{6,0} &{}\quad b_{6,1} &{}\quad b_{6,2} &{}\quad b_{6,3} \\ &{} &{} &{}\quad b_{7,0} &{}\quad b_{7,1} &{}\quad b_{7,2} &{}\quad b_{7,3} \\ &{} &{} &{} &{}\quad \vdots \end{matrix}. \end{aligned}$$

For example, the right child of $b_{3,1}$ is $b_{4,2}$ (down and to the right) and the left child is $b_{4,1}$ (straight down). The key to understand the combinatorics is that in order to have a nonzero right parent and hence a factorially growing term, the power of the polynomial piece must be bigger than (in this case) $2j+1$. Observe that if we trace through the tree to get to the $(2k+1)$st row, then it is always the case that for any path

$$\begin{aligned} 2k = \#\text { right children} + \# \text { left children} \end{aligned}$$

and to arrive at a nonzero term at the $(2k+1)$st row, at least half of the children must be right children. Next, to arrive at $b_{2k+1,0}$, exactly half of the children must be right children and half left children while to arrive at $b_{2k+1,1}$, we need an additional right child and consequently one less left child. As a result, to arrive at $b_{2k+1,j}$, it follows there must be $j+k$ right children and $k-j$ left children in the path. Consequently,

$$\begin{aligned} 2j = \# \text { right children} - \# \text { left children}. \end{aligned}$$

The number of left children produce the factorially growing terms, and hence 3.ii. follows as $k-j$ left children mean the factorial contribution to the size of $b_{2k+1,j}$ is $k^{k-j}$. It follows from this observation that the only way to arrive at $b_{2k,k}$ or $b_{2k+1,k}$ is to follow the path of all right children, hence 3.i. follows. The argument to bound the size of $b_{2k,j}$ is similar. $\square$

Corollary B.3

There exist constants $C,A>0$ so that

• 1.
  $$\begin{aligned} \left| \frac{d^{2k}}{dx^{2k}} e^{-ax^2} \right| \le C e^{-ax^2}A^k a^k \sum _{j=0}^k a^j x^{2j} k^{k-j} \end{aligned}$$

  and

  $$\begin{aligned} \left| \frac{d^{2k+1}}{dx^{2k+1}} e^{-ax^2}\right| \le e^{-ax^2}A^k a^{k+1} \sum _{j=0}^k a^j x^{2j+1} k^{k-j} \end{aligned}$$

  for all $k\in {\mathbb {N}}_0$ and $a>0$.

• 2. If,  in addition,  $0 \le \ell \le d,$ then there exist constants $C_d,A>0$ so that

  $$\begin{aligned} \left\| x^\ell \frac{d^k}{dx^k} e^{-ax^2} \right\| _{L^p({{\mathbb {R}}})} \le C_d A^k a^{\frac{k}{2} - \frac{1}{2p}-\frac{\ell }{2}} k^{\frac{k}{2}}. \end{aligned}$$

Proof

Part 1. of the corollary follows immediately from Proposition B.2. For the second piece, we estimate that when k is even,

$$\begin{aligned} \left\| x^\ell \frac{d^{2k}}{dx^{2k}} e^{-ax^2} \right\| _{L^p({{\mathbb {R}}})}&\le C A^k a^k \sum _{j=0}^k a^j k^{k-j} \big \Vert x^{\ell +2j}e^{-ax^2} \big \Vert _{L^p({{\mathbb {R}}})} \\&= C A^k a^k \sum _{j=0}^k a^j k^{k-j} \left( \int _{{\mathbb {R}}}x^{p(\ell +2j)}e^{-ap x^2}\, dx\right) ^{1/p} \\&= C A^k a^k \sum _{j=0}^k a^j k^{k-j} \left( \frac{\Gamma (\frac{1}{2} + jp + \frac{\ell p}{2})}{2 (ap)^{\frac{1}{2} + jp + \frac{\ell p}{2}}}\right) ^{1/p} \\&= C A^k a^k \sum _{j=0}^k a^j k^{k-j} \frac{\Gamma (\frac{1}{2} + jp + \frac{\ell p}{2})^{1/p}}{2 (ap)^{\frac{1}{2p} + j + \frac{\ell }{2}}}. \end{aligned}$$

By Stirling’s Formula, there exist constants $C_0,A_0>0$ (which may grow from line to line) so that

$$\begin{aligned} \frac{\Gamma (\frac{1}{2} + jp + \frac{\ell p}{2})^{1/p}}{p^{\frac{1}{2p} + j + \frac{\ell }{2}}} \le C_0 \frac{(jp + \frac{\ell p}{2})^{\frac{1}{2p} + j + \frac{\ell }{2}}}{p^{\frac{1}{2p} + j + \frac{\ell }{2}}} \le C_0 A_0^j j^j \le C_0 A_0^j k^j. \end{aligned}$$

Similarly,

$$\begin{aligned} \Big \Vert x^\ell \frac{d^{2k+1}}{dx^{2k+1}} e^{-ax^2} \Big \Vert _{L^p({{\mathbb {R}}})}&\le C A^k a^k \sum _{j=0}^k a^j k^{k-j} \big \Vert x^{\ell +2j}e^{-ax^2} \big \Vert _{L^p({{\mathbb {R}}})} \\&= C A^k a^k \sum _{j=0}^k a^j k^{k-j} \frac{\Gamma ( \frac{1}{2} + p(\frac{1}{2} + j + \frac{\ell p}{2}))^{1/p}}{2 (a p)^{\frac{1}{2p} +(\frac{1}{2} + j + \frac{\ell }{2})}} \le C A_0^k k^k. \end{aligned}$$

$\square$

Corollary B.4

Let $0 \le \ell \le d,$$a>0$ and $|y|\le 1,$ then there exist constants $C_d,A>0$ so that

$$\begin{aligned} \Big \Vert (x+iy)^\ell \partial _x^{k} \big \{e^{-a [x+iy]^2}\big \}\Big \Vert _{L^p_x({{\mathbb {R}}})} \le C_d \,e^{ay^2} \,A^k \sum _{\begin{array}{c} k_1+k_2=k\\ 0\le \ell \le d \end{array}}{k\atopwithdelims ()k_1} a^{\frac{k_1}{2} +k_2 - \frac{1}{2p}-\frac{\ell }{2}}\,\, k_1^{\frac{k_1}{2}}. \end{aligned}$$

(B.5)

Proof

First we note that, since $\ell \le d$ and $|y|\le 1$ we have

$$\begin{aligned} \big |(x+iy)^\ell \big |&=\Big | \sum _{j=0}^{\ell } {\ell \atopwithdelims ()j}x^j (iy)^{\ell -j} \Big | \le C_d \sum _{\ell =0}^{d} |x|^\ell . \end{aligned}$$

(B.6)

Also, using Leibniz rule, the derivative of the complex exponential can be written as

$$\begin{aligned} \partial _x^{k} \big \{e^{-a [x+iy]^2}\big \}&= \sum _{k_1+k_2=k}{k\atopwithdelims ()k_1} \big (\partial _x^{k_1} \big \{e^{-a(x^2-y^2)}\big \}\big ) \big ( \partial _x^{k_2} \big \{e^{-2ai xy}\big \}\big ) \nonumber \\&= e^{ay^2}\sum _{k_1+k_2=k}{k\atopwithdelims ()k_1} \big (\partial _x^{k_1} \big \{e^{-ax^2}\big \}\big ) \big ( -2aiy\big )^{k_2} e^{-2ai xy}, \end{aligned}$$

(B.7)

which, recalling that $|y|\le 1$, can be easily estimated by

$$\begin{aligned} \Big |\partial _x^{k} \big \{e^{-a [x+iy]^2}\big \} \Big |&\le e^{ay^2}\sum _{k_1+k_2=k}{k\atopwithdelims ()k_1} \big |\partial _x^{k_1} \big \{e^{-ax^2}\big \}\big | \big | 2a\big |^{k_2}. \end{aligned}$$

(B.8)

The proof now is a consequence of (B.6), (B.8) and Corollary B.3. $\square$