N.M. Tri
ON THE GEVREY ANALYTICITY OF SOLUTIONS OF SEMILINEAR
PERTURBATIONS OF POWERS OF THE MIZOHATA OPERATOR
Abstract. We consider the Gevrey hypoellipticity of nonlinear equations with mul- tiple characteristics, consisting of powers of the Mizohata operator and a Gevrey analytic nonlinear perturbation. The main theorem states that under some con- ditions on the perturbation every Hs solution (with s ≥ s0) of the equation is a Gevrey function. To prove the result we do not use the technique of cut-off functions, instead we have to control the behavior of the solutions through a fun- damental solution. We combine some ideas of Grushin and Friedman.
1. Introduction
Since a Hilbert’s conjecture the analyticity of solutions of partial differential equations has at- tracted considerable interest of specialists. The conjecture stated that every solution of an an- alytic nonlinear elliptic equation is analytic (see [1]). This conjecture caused considerable re- search by several authors. In 1904 Bernstein solved the problem for second order nonlinear elliptic equations in two variables [2]. For equations with an arbitrary number of variables the problem was established by Hopf [3], Giraud [4]. For a general elliptic system it was proved by Petrowskii [5], Morrey [6], and Friedman [7]. The last author also obtained results on the Gevrey analyticity of solutions. At present the results on the analyticity and Gevrey analyticity of solutions of linear partial differential equations have gone beyond the frame of the elliptic theory. A linear partial differential equation is called Gevrey hypoelliptic if every its solution is Gevrey analytic (in this context we can assume a solution in a very large space such as the space of distributions), provided the data is Gevrey analytic. Recently the class of linear Gevrey hypoelliptic equations with multiple characteristics has drawn much interest of mathematicians (see [8], and therein references). In this paper we consider the Gevrey analyticity of solutions in the Sobolev spaces of the nonlinear equation with multiple characteristics:
M2khu+ϕ x1,x2,u, . . . ,∂l1+l2u
∂lx11∂lx22
!
l1+l2≤(h−1)
=0,
where u = u(x1,x2), M2k = ∂∂x
1 +i x2k1 ∂x∂
2, the Mizohata operator inR2. The Gevrey hy- poellipticity for the equation was obtained in [9] in the linear case. We will prove a theorem for a so-called Grushin type perturbationϕ(see the notations in §2 below). The analyticity of
∗Dedicated to my mother on her sixtieth birthday.
37
C∞solutions was considered in [10]. The paper is organized as follows. In §2 we introduce notations used in the paper and state some auxiliary lemmas. In §3 we state the main theorem and prove a theorem on C∞regularity of the solutions. In §4 we establish the Gevrey analyticity of the solutions.
2. Some notations and lemmas
First let us define the function
F2kh(x1,x2,y1,y2)= 1 2π(h−1)!
(x1−y1)h−1 x12k+1−y2k+11
2k+1 +i(x2−y2) .
For j=1, . . . ,h−1 we have (see [11]) M2kj F2kh = 1
2π(h− j−1)!
(x1−y1)h−j−1 x12k+1−y12k+1
2k+1 +i(x2−y2)
and M2kh F2kh =δ(x−y) ,
where x=(x1,x2), y=(y1,y2)andδ(·)is the Dirac function. We will denote bya bounded domain inR2with piece-wise smooth boundary. We now state a lemma on the representation formula. Its proof can be found in [10].
LEMMA4.1. Assume that u∈Ch()¯ then we have
(1)
u(x)= Z
(−1)hF2kh(x,y)M2khu(y)d y1d y2
+ Z
∂
h−1X
j=0
(−1)jM2kj u M2kh−j−1F2kh(x,y)
(n1+i y12kn2)ds.
For reason of convenience we shall use the following Heaviside function θ (z)=
1 if z≥0, 0 if z<0.
We then have the following formuladtdmm(zn)=n. . . (n−m+1)θ (n−m+1)zn−m. We will use∂1α,∂2β,γ∂α,β,∂1yα ,∂2yβ andγ∂α,βy instead of∂∂xαα
1
, ∂β
∂x2β, x1γ ∂α+β
∂x1α∂x2β, ∂y∂αα 1
, ∂β
∂y2β and yγ1 ∂α+β
∂y1α∂y2β, respectively. Throughout the paper we use the following notation n!=
n! if n≥1,
1 if n≤0, and Cn=
Cn if n≥1, 1 if n≤0.
Furthermore all constants Ciwhich appear in the paper are taken such that they are greater than 1. Put h(2k+1)=r0. For any integer r≥0 let0rdenote the set of pair of multi-indices(α, β) such that0r =0r1∪0r2where
0r1= {(α, β):α≤r0,2α+β≤r}, 0r2= {(α, β):α≥r0, α+β≤r−r0}.
For a pair(α, β)we denote by(α, β)∗ the minimum of r such that (α, β) ∈ 0r. For any nonnegative integer t we define the following sieves
4t = {(α, β, γ ):α+β≤t,2kt≥γ≥α+(2k+1)β−t}, 40t = {(α, β, γ ):(α, β, γ )∈4t, γ =α+(2k+1)β−t}. Later on we will use the following properties of4t:
4tcontains a finite number of elements (less than 2k(t+1)3elements).
If(α, β, γ )∈4t,β≥1,γ ≥1, then(α, β−1, γ−1)∈4t−1.
For every(α, β, γ )∈ 4t,(α0, β0, γ0)∈4t0 we can expressγ∂α,β(γ0∂α0,β0)as a linear combination ofγ0∂α0,β0, where(α0, β0, γ0)∈4t+t0.
For every (α, β, γ ) ∈ 4t, (α1, β1) ∈ 0r and a nonnegative integer m we can rewrite γ−m∂α+α1−m,β+β1 asγ−m∂α2,β2(∂1α3∂2β3)where(α2, β2, γ−m)∈4tand(α3, β3)∈0r−m (see [12]).
For any nonnegative integer r let us define the norm
|u, |r = max (α1,β1)∈0r
∂1α1∂2β1u,
+ max
(α1,β1)∈0r α1≥1,β1≥1
max x∈ ¯ ∂1h
∂1α1∂2β1u(x), where|w, | =P
(α,β,γ )∈4h−1maxx∈ ¯|γ∂α,βw(x)|.
For any nonnegative integer l letHlloc()denote the space of all u such that for any compact K inwe haveP
(α,β,γ )∈4lkγ∂α,βukL2(K)<∞. We note the following properties ofHlloc():
Hlocl ()⊂Hlloc()where Hlocl ()stands for the standard Sobolev spaces.
H4k+2loc ()⊂Hloc2 ()⊂C().
If u∈Hlloc()and(α, β, γ )∈4t, t≤l thenγ∂α,βu∈Hl−tloc().
The following lemma is due to Grushin (see [12])
LEMMA4.2. Assume that u∈D0()and M2khu∈Hlloc()then u∈Hl+hloc().
Next we define a space generalizing the space of analytic functions (see for example [7]).
Let Ln and L¯n be two sequences of positive numbers, satisfying the monotonicity condition i
n
LiLn−i ≤ ALn(i = 1,2. . .; n = 1,2. . .), where A is a positive constant. A function F(x, v), defined for x = (x1,x2)and forv = (v1, . . . , vµ)in aµ-dimensional open set E , is said to belong to the class C{Ln−a;| ¯Ln−a;E}(a is an integer) if and only if F(x, v)is infinitely differentiable and to every pair of compact subsets0⊂and E0⊂E there corre- spond constants A1and A2such that for x∈0andv∈E0
∂j+kF(x, v)
∂x1j1∂x2j2∂vk11. . . ∂vµkµ
≤A1A2j+kLj−aL¯k−a
j1+ j2= j,X
ki =k; j,k=0,1,2. . .
.
We use the notation L−i = 1 (i = 1,2, . . .). If F(x, v) = f(x), we simply write f(x) ∈ C{Ln−a;}. Note that C{n!;}, (C{n!s;}) is the space of all analytic functions (s-Gevrey functions), respectively, in. Extensive treatments of non-quasi analytic functions (in particular, the Gevrey functions) can be found in [13, 14]. Finally we would like to mention the following lemma of Friedman [7].
LEMMA4.3. There exists a constant C1such that if g(z)is a positive monotone decreasing function, defined in the interval 0≤z≤1 and satisfying
g(z)≤ 1
812kg z 1−6k n
!!
+ C
zn−r0−1 (n≥r0+2,C>0) , then g(z) <CC1/zn−r0−1.
3. The main theorem
We will consider the following problem
(2) M2khu+ϕ(x1,x2,u, . . . ,γ∂α,βu)(α,β,γ )∈4h−1=0 in . We now state our main
THEOREM4.1. Let s≥4k2+6k+h+1. Assume that u is aHsloc()solution of the equation (2) andϕ ∈C{Ln−a−2;|Ln−a−2;E}for every a ∈ [0,r0]. Then u ∈ C{Ln−r0−2;}. In particular, ifϕis a s-Gevrey function then so is u.
The proof of this theorem consists of Theorem 4.2 and Theorem 4.3.
THEOREM4.2. Let s ≥ 4k2+6k+h+1. Assume that u is aHsloc()solution of the equation (2) andϕ∈C∞. Then u is a C∞()function.
Proof. Note that M2kh is elliptic onR2, except the line(x1,x2)= (0,x2). Therefore from the elliptic theory we have already u∈C∞(∩R2\(0,x2)).
LEMMA4.4. Let s ≥ 4k2+6k+h+1. Assume that u ∈ Hsloc()andϕ ∈ C∞then ϕ(x1,x2,u, . . . ,γ∂α,βu)(α,β,γ )∈4h−1∈Hs−h+1loc ().
Proof. It is sufficient to prove thatγ1∂α1,β1ϕ(x1,x2,u, . . . ,γ∂α,βu)∈ L2loc()for every(α1, β1, γ1)∈4s−h+1. Let us denote(u, . . . ,γ∂α,βu)(α,β,γ )∈4h−1by(w1, w2, . . . , wµ)withµ≤ 2kh3. Since s ≥ 4k2+6k+h+1 it follows thatw1, . . . , wµ ∈ C(). It is easy to verify that∂1α1∂2β1ϕ(x1,x2,u, . . . ,γ∂α,βu)is a linear combination with positive coefficients of terms of the form
(3) ∂kϕ
∂x1k1∂x2k2∂w1k3. . . ∂wµkµ+2 Yµ
j=1 Y
(α1,j,β1,j)
∂1α1,j∂2β1,jwj
ζ (α1,j,β1,j) ,
where k=k1+k2+ · · · +kµ+2≤α1+β1;ζ (α1,j, β1,j)may be a multivalued function of α1,j, β1,j;α1,j, β1,jmay be a multivalued function of j , and
X
j
α1,j·ζ (α1,j, β1,j) ≤ α1, X
j
β1,j·ζ (α1,j, β1,j) ≤ β1.
Hence xγ11∂1α1∂2β1ϕ(x1,x2,u, . . . ,γ∂α,βu)is a linear combination with positive coefficients of terms of the form
∂kϕ
∂x1k1∂x2k2∂w1k3. . . ∂wkµµ+2 x1γ1
Yµ
j=1 Y
(α1,j,β1,j)
∂1α1,j∂2β1,jwj
ζ (α1,j,β1,j) .
Therefore the theorem is proved if we can show this general terms are in L2loc(). If all ζ (α1,j, β1,j)vanish then it is immediate that∂kϕ/∂x1k1∂x2k2∂w1k3. . . ∂wµkµ+2 ∈ C, sinceϕ ∈ C∞,w1, . . . , wµ ∈C(). Then we can assume that there exists at least one ofζ (α1,j, β1,j) that differs from 0. Choose j0such that there existsα1,j0, β1,j0withζ (α1,j0, β1,j0)≥1 and
α1,j0+(2k+1)β1,j0 = max
j=1,...,µ ζ (α1,j,β1,j)≥1
α1,j+(2k+1)β1,j.
Consider the following possibilities
I)ζ (α1,j0, β1,j0)≥2. We then haveα1,j+β1,j ≤l−(h−1)−(4k+2). Indeed, if j 6= j0 andα1,j+β1,j >l−(h−1)−(4k+2)thenα1,j0+β1,j0 ≥2k. Therefore
l−(h−1)−(4k+2) < α1,j+β1,j≤α1,j+(2k+1)β1,j
≤α1,j0+(2k+1)β1,j0 ≤(2k+1)(α1,j0+β1,j0)
≤2k(2k+1) . Thus l< (2k+2)(2k+1)+(h−1), a contradiction.
If j= j0andα1,j0+β1,j0 >l−(h−1)−(4k+2)then we have
l−(h−1)≥α1+β1≥2(α1,j0+β1,j0) >2(l−(h−1)−(4k+2)) . Therefore l< (h+1)+4(2k+1), a contradiction.
Next define
γ (α1,j, β1,j)=max{0, α1,j+(2k+1)β1,j+(h−1)+(4k+2)−l}.
We claim thatγ (α1,j, β1,j)≤2k(l−(h−1)−(4k+2)). Indeed, if j6= j0andγ (α1,j, β1,j) >
2k(l−(h−1)−(4k+2))then
(2k+1)(l−(h−1)) ≥ α1+(2k+1)β1
≥ (α1,j+2α1,j0)+(2k+1)(β1,j0+2β1,j0)
> 3(2k+1)(l−(h−1)−(4k+2)) . Thus l< (h−1)+3(2k+1), a contradiction.
If j= j0andγ (α1,j0, β1,j0) >2k(l−(h−1)−(4k+2))then it follows that (2k+1)(l−(h−1))≥α1+(2k+1)β1≥2(α1,j0+(2k+1)β1,j0)
>2(2k+1)(l−(h−1)−(4k+2)) . Thus l< (h−1)+4(2k+1), a contradiction.
From all above arguments we deduce that(α1,j, β1,j, γ (α1,j, β1,j))∈4l−(h−1)−(4k+2). Next we claim thatP
γ (α1,j, β1,j)ζ (α1,j, β1,j)≤γ1. Indeed, ifP
γ (α1,j, β1,j)ζ (α1,j, β1,j) >
γ1then we deduce that
α1+(2k+1)β1−2(l−(h−1)−(4k+2))≥X
γ (α1,j, β1,j)ζ (α1,j, β1,j) > γ1
≥α1+(2k+1)β1−(l−(h−1)) . Therefore l< (h−1)+4(2k+1), a contradiction.
Now we have xγ11
Yµ
j=1 Y
(α1,j,β1,j)
∂1α1,j∂2β1,jwjζ (α1,j,β1,j)
=x1γ1 Yµ
j=1 Y
(α1,j,β1,j)
x1γ (α1,j,β1,j)∂1α1,j∂2β1,jwjζ (α1,j,β1,j)
∈C()
since xγ (α1 1,j,β1,j)∂1α1,j∂2β1,jwj ∈H4k+2loc ()⊂C().
II)ζ (α1,j0, β1,j0)=1 andζ (α1,j, β1,j)=0 for j6= j0. We have
x1γ1 Yµ
j=1 Y
(α1,j,β1,j)
∂1α1,j∂2β1,jwj
ζ (α1,j,β1,j)
=xγ11∂1α1,j0∂2β1,j0wj0 ∈L2loc() .
III)ζ (α1,j0, β1,j0)=1 and there exists j16=j0such thatζ (α1,j1, β1,j1)6=0. Define
¯
γ (α1,j0, β1,j0)=max{0, α1,j0+(2k+1)β1,j0+(h−1)−l}.
As in part I) we can prove(α1,j, β1,j, γ (α1,j, β1,j)) ∈ 4l−(h−1)−(4k+2) for j 6= j0 and (α1,j0, β1,j0,γ (α¯ 1,j0, β1,j0))∈ 4l−(h−1). Therefore xγ (α1 1,j,β1,j)∂1α1,j∂2β1,jwj ∈ H4k+2loc ()
⊂C()for j6= j0and x1γ (α¯ 1,j0,β1,j0)∂1α1,j0∂2β1,j0wj0∈L2loc(). We also haveP
j6=j0γ (α1,j, β1,j)ζ (α1,j, β1,j)+ ¯γ (α1,j0, β1,j0)≤γ1as in part I). Now the desired result follows from the decomposition of the general terms.
(continuing the proof of Theorem 4.2) u∈ Hlocl (), l≥4k2+6k+h+1H⇒u∈Hlloc(), l≥4k2+6k+h+1H⇒ϕ(x1,x2,u, . . . ,γ∂α,βu)∈Hl−h+1loc ()(by Lemma 4.4). Therefore by Lemma 4.2 we deduce that u ∈Hl+1loc(). Repeat the argument again and again we finally arrive at u∈Hl+tloc()for every positive t , i.e.
u∈ ∩lHlloc()=C∞() .
REMARK4.1. Theorem 4.2 can be extended to the so-called Grushin type operators (non- linear version).
4. Gevrey analyticity of the solutions
PROPOSITION4.1. Assume that
ϕ(x1,x2,u, . . . ,γ∂α,βu)(α,β,γ )∈4h−1∈C{Ln−a−2;|Ln−a−2;E}
for every a∈[0,r0]. Then there exist constantsHe0,He1,C2,C3such that for every H0≥ He0, H1≥He1, H1≥C2H02r0+3if
|u, |q≤ H0H1(q−r0−2)Lq−r0−2, 0≤q≤N+1, r0+2≤N then
max x∈ ¯
∂1α1∂2β1ϕ(x1,x2,u, . . . ,γ∂α,βu)
≤C3H0H1N−r0−1LN−r0−1 for every(α1, β1)∈0N+1.
Proof. ∂1α1∂2β1ϕ(x1,x2,u, . . . ,γ∂α,βu)as in Lemma 4.4 is a linear combination with positive coefficients of terms of the form
∂kϕ
∂xk11∂xk22∂wk13. . . ∂wkµµ+2 Yµ
j=1 Y
(α1,j,β1,j)
∂1α1,j∂2β1,jwj
ζ (α1,j,β1,j) .
Substitutingwjby one of the termsγ∂α,βu with(α, β, γ )∈4h−1we obtain
∂1α1,j∂2β1,j(γ∂α,βu)= α1,j X
m=0 m
α1,j
γ· · ·(γ−m+1)θ (γ−m+1)(γ−m)∂α+α1,j−m,β+β1,ju.
We can decompose(γ−m)∂α+α1,j−m,β+β1,ju into(γ−m)∂α2,β2
∂1α3∂2β3u
with(α2, β2, γ − m)∈4h−1and(α3, β3)∈0(α1,j,β1,j)∗−m. Put S = N+1−α1−β1. Define R =r0−S.
It is easy to see that R is a nonnegative integer and less than r0. Sinceα1,j ≤ α1we deduce that(α1,j, β1,j)∈0(α1,j+β1,j+S). Using the monotonicity condition on Lnand the inductive assumption we have
m α1,j
γ· · ·(γ−m+1)θ (γ−m+1)(γ−m)∂α+α1,j−m,β+β1,ju
≤ m
α1,j
γ· · ·(γ−m+1)θ (γ−m+1)H0H1α1,j+β1,j−m−R−2Lα1,j+β1,j−m−R−2
≤C4H0H1α1,j+β1,j−m−R−2Lα1,j+β1,j−R−2. Therefore we deduce that
∂1α1,j∂2β1,j(γ∂α,βu)
≤
α1,j X
m=0 m
α1,j
γ· · ·(γ−m+1)θ (γ−m+1)(γ−m)∂α+α1,j−m,β+β1,ju
≤C5H0H1α1,j+β1,j−R−2Lα1,j+β1,j−R−2.
and the general terms can be estimated by Yµ
j=1
C5H0H1α1,j+β1,j−R−2Lα1,j+β1,j−R−2.
Sinceϕ∈C{Ln−a;|Ln−a;E}, there exist constants C6,C7such that
∂kϕ
∂x1q1∂x2q2∂wr11. . . ∂wrµµ
≤C6C7q−RLq−R−2C7rLr−R−2 (q=q1+q2,r =r1+ · · · +rµ) . Now we take X(ξ, v)=X1(v)·X2(ξ )where
X1(v)=C6 Xp
i=0
C8iLi−R−2vi
i ! , X2(ξ )= Xp
i=0
C7i−RLi−R−2ξi
i ! .
and
v(ξ )=C5H0 Xp
i=1
H1i−R−2Li−R−2ξi
i ! .
By comparing terms of the form (3) with the corresponding terms indξdppX(ξ, v)it follows that
∂1α1∂2β1ϕ(x1,x2,u, . . . ,γ∂α,βu) x=x
0
≤ dp dξpX(ξ, v)
ξ=0.
Next we introduce the following notation: v(ξ ) h(ξ )if and only ifv(j)(0) ≤ h(j)(0)for 1≤ j≤ p. It is not difficult to check that there exists a constant C9(independent of p) such that
v2(ξ )(C5H0)2C9 Xp
i=2
H1i−R−3Li−R−3ξi (i−1)! . And by induction we have
vj(ξ )(C5H0)jC9j−1 Xp
i=j
H1i−j−R−1Li−j−R−1ξi (i−j+1)! .
Next, it is easy to verify that X1(0) = 0, d Xdξ1(ξ )
ξ=0 ≤ C5C6C8H0, and djX2(ξ ) dξj
ξ=0 = C7j−RLj−R−2.
We now compute djX1(ξ ) dξj
ξ=0 when 2 ≤ j ≤ p. If 2 ≤ j ≤ 2R + 4 we can al- ways choose a constant C10 such that djX1(ξ )
dξj
ξ=0 ≤ C10H0H1j−R−2Lj−R−2, provided H1≥(C5C8C9H0)2r0+3.
If j≥2R+5, we have
(4)
djX1(ξ ) dξj
ξ=0
≤C6
R+2X
i=1
Ci8(C5H0)iC9i−1H1j−i−R−1Lj−i−R−1j ! i !(j−i+1)!
+ j−R−2
X
i=R+3
C8i(C5H0)iCi−19 H1j−i−R−1Li−R−2Lj−i−R−1j ! i !(j−i+1)!
+ Xj
i=j−R−1
C8i(C5H0)iC9i−1Li−R−2j ! i !(j−i+1)!
.
The first sum in (4) is estimated by
(5) C10H0H1j−R−2Lj−R−2
provided H1≥(C5C8C9H0)2r0+3as for 2≤ j≤2R+4.
By using the monotonicity condition on Lnthe second sum in (4) is estimated by (6)
C6 j−R−2X
i=R+3
Ci8(C5H0)iC9i−1H1j−i−R−1Li−R−2Lj−i−R−1j ! i !(j−i+1)!
≤ C11H0H1j−R−2j !Lj−2R−3 (j−2R−3)!
j−R−2X
i=R+3
1 i· · ·(i−R−1)
1
(j−i+1)· · ·(j−i−R)
≤C12H0H1j−R−2Lj−R−2, provided H1≥C5C8C9H0. For the third sum we see that
(7)
C6 Xj
i=j−R−1
Ci8(C5H0)iC9i−1Li−R−2j ! i !(j−i+1)!
≤C5C6C8H0H1j−R−2j ! Xj
i=j−R−1
Li−R−2 i !(j−i+1)!
≤C13H0H1j−R−2Lj−R−2, if H1≥(C5C8C9H0)2.
By (4), (5), (6), (7) and taking H1≥(C5C7C8C9H0)2r0+3=C2H02r0+3we obtain dpX(ξ, v)
dξp
ξ=0
≤C14 Xp
j=0 j
p
H0H1j−R−2Lj−R−2C7p−j−RLp−j−R−2
≤C15H0H1p−R−2Lp−R−2. Hence
max x∈ ¯
∂1α1∂2β1ϕ(x1,x2,u, . . . ,γ∂α,βu)
≤C15H0H1p−R−2Lp−R−2
≤C3H0H1N−r0−1LN−r0−1.
REMARK4.2. The constantsHe0,He1,C2,C3depend onincreasingly in the sense that with the sameHe0,He1,C2,C3Proposition 4.1 remains valid if we substituteby any0⊂.
COROLLARY4.1. Under the same hypotheses of Proposition 4.1 with q≤N+1 replaced by q≤N , then
max x∈ ¯
∂1α1∂2β1ϕ(x1,x2,u, . . . ,γ∂α,βu) ≤C3
|u, |N+1+H0H1N−r0−1LN−r0−1
for every(α1, β1)∈0N+1.
Proof. Indeed, as in the proof of Proposition 4.1 all typical terms, except ∂w∂ϕ
j0∂1α1∂2β1wj0 can be estimated by| · |N. Replacingwj0by one ofγ∂α,βu, we have
∂ϕ
∂wj0∂1α1∂2β1wj0= ∂ϕ
∂wj0 γ∂α+α1,β+β1u + ∂ϕ
∂wj0 α1 X
m=1 m
α1
γ· · ·(γ−m+1)θ (γ−m+1)(γ−m)∂α+α1−m,β+β1u. The first summand is estimated by C3|u, |N+1. The second sum is majorized as in Proposition 4.1.
THEOREM 4.3. Let u be a C∞ solution of the equation (2) and ϕ ∈ C{Ln−a−2;|Ln−a−2;E} for every a ∈ [0,r0]. Then u ∈ C{Ln−r0−2;}. In particu- lar, ifϕis a s-Gevrey function then so is u.
Proof. It suffices to consider the case(0,0)∈. Let us define a distanceρ((y1,y2), (x1,x2))= max
|x12k+1−y12k+1|
2k+1 ,|x2−y2|
. For two sets S1,S2the distance between them is defined as ρ(S1,S2)=infx∈S1,y∈S2ρ(x,y). Let VT be the cube with edges of size (in theρmetric) 2T , which are parallel to the coordinate axes and centered at(0,0). Denote by VδTthe subcube which is homothetic with VT and such that the distance between its boundary and the boundary of VT isδ. We shall prove by induction that if T is small enough then there exist constants H0,H1 with H1≥C2H02r0+3such that
(8)
u,VδT
n≤H0 for 0≤n≤max{r0+2,6k+1}
and (9)
u,VδT
n≤H0
H1 δ
n−r0−2
Ln−r0−2 for n≥max{r0+2,6k+1}
andδsufficiently small. Hence the Gevrey analyticity of u follows. (8) follows easily from the C∞smoothness assumption on u. Assume that (9) holds for n = N . We shall prove it for n= N+1. Putδ0=δ(1−1/N). Fix x ∈VδT and then defineσ =ρ(x, ∂VT)andσ˜ =σ/N.
Let Vσ˜(x)denote the cube with center at(x)and edges of length 2˜σ which are parallel to the co- ordinate axes, and Sσ˜(x)the boundary of Vσ˜(x). Let E1,E3(E2,E4) be edges of Sσ˜(x)which are parallel to O x1(O x2) respectively. We have to estimate maxx∈VT
δ
γ∂α,β
∂1α1∂2β1u(x)