such that Ω is open, F is F

151 (1996)

The Zahorski theorem is valid in Gevrey classes

by

Jean S c h m e t s (Li`ege) and Manuel V a l d i v i a (Valencia)

Abstract. Let {Ω, F, G} be a partition of R

such that Ω is open, F is F

and of the first category, and G is G

. We prove that, for every γ ∈ ]1, ∞[, there is an element of the Gevrey class Γ

which is analytic on Ω, has F as its set of defect points and has G as its set of divergence points.

1. Introduction. Let f be a real C

-function on R

. The set where f is analytic is of course an open subset of R

; denote it by Ω

. It is clear that x belongs to Ω

if and only if the radius of convergence %

(x) of the Taylor series of f at x is strictly positive and this series represents f on some neighbourhood of x.

As the set T

of x ∈ R

such that %

(x) > 0 is easily seen to be an F

-set, the set F

= T

\ Ω

is also F

and its elements x are characterized by the fact that %

(x) > 0 and that the Taylor series of f at x represents f on no neighbourhood of x. By use of a lemma of R. P. Boas ([1], p. 234), one easily sees that F

is also a first category set (cf. [3] or [4]).

Finally, one may consider the set G

= R

\ T

, a G

-set given by

%

(x) = 0.

It is clear that {Ω

, G

, F

} is a partition of R

.

The Zahorski theorem (cf. [6]) asserts conversely that for every partition {Ω, F, G} of [0, 1], where Ω is an open subset of [0, 1], F is a first category F

-subset of [0, 1] and G is a G

-subset of [0, 1], there is a real C

-function f on [0, 1] such that Ω = Ω

, F = F

and G = G

. In [3], H. Salzmann and K. Zeller have provided a shorter proof of the Zahorski theorem and in [4], J. Siciak has extended this result to R

.

1991 Mathematics Subject Classification: Primary 26E10; Secondary 26E05.

Key words and phrases: Gevrey classes, defect point, divergence point.

Research of the first author partially supported by the Belgian C.G.R.I.

Research of the second author partially supported by DGICYT PB91–0326 and the Convenio Hispano-Belga.

[149]

The purpose of this paper is to prove that, for every γ ∈ ]1, ∞[, the Zahorski theorem has a solution f belonging to the Gevrey class Γ

.

Let us recall that, for an open subset Ω of R

and for γ ∈ ]1, ∞[, the Gevrey class Γ

(Ω) is the set of f ∈ C

(Ω) for which there are constants a, b > 0 such that

kD

f k

≤ ab

(|α|!)

, ∀α ∈ N

. If Ω = R

, we simply write Γ

instead of Γ

(R

).

It is known that

(a) a function f ∈ C

(Ω) belongs to Γ

(Ω) if and only if there are constants c, d > 0 such that

kD

f k

≤ cd

|α|

, ∀α ∈ N

.

(b) the Denjoy–Carleman–Mandelbrojt result (cf. [2]) states that, for every closed ball b of R

and every γ ∈ ]1, ∞[, there is a nonzero function f ∈ Γ

with support contained in b.

In order to get an efficient way to state the results, for a real C

-function f on R

, let us call the elements of Ω

(resp. F

; G

) the analytic points (resp. the defect points; the divergence points) of f .

The purpose of this article is to prove the following result.

Theorem 1.1. For every partition {Ω, F, G} of R

, where Ω (resp. F ; G) is an open set (resp. a first category F

-set; a G

-set) and every γ ∈ ]1, ∞[, there is an element of Γ

having Ω (resp. F ; G) as its set of analytic points (resp. defect points; divergence points).

R e m a r k. It is a direct matter to check that the Zahorski theorem ex- tends to the case when R

is replaced by an open or a closed subset of R

. 2. An auxiliary result. We begin with the following easy result, where, as usual, D

(Ω) denotes the space of C

-functions on the open subset Ω of R

which have a compact support contained in Ω.

Proposition 2.1. Let Ω be a nonvoid open subset of R

. For every f ∈ C

(Ω), g ∈ D

(Ω) and λ > 0, it is well known that

h(x) = π

λ

\

Ω

f (y)g(y)e

dy

belongs to C

(R

). If moreover a

, a

, b

, b

> 0 and ζ > 1 are such that kD

f k

≤ a

b

(|α|!)

and kD

gk

≤ a

b

(|α|!)

, ∀α ∈ N

, then

kD

hk

≤ a

a

(b

+ b

)

(|α|!)

, ∀α ∈ N

.

P r o o f. As g has a compact support contained in Ω, up to extension by 0 on R

\ Ω, we may suppose that the product f g is a D

-function on R

with compact support contained in Ω. So in the definition of h(x), we may consider that we integrate on R

. Therefore integrating by parts |α| times gives

D

h(x) = π

λ

\

Rⁿ

D

(f g)(y)e

dy, hence

|D

h(x)| ≤ π

λ

\

Rⁿ

X

β≤α

 α β



|D

f (y)D

g(y)|e

dy

≤ X

β≤α

 α β



a

a

b

b

(|β|!)

(|α − β|!)

≤ a

a

(b

+ b

)

(|α|!)

.

3. Special compact covers of open subsets of R

. In the following results, we are going to use systematically the following construction and notations. Let Ω be a nonvoid open subset of R

. Then we set

Ω

= (R

\ Ω) ∪ {x ∈ R

: |x| ≥ m √

n}, ∀m ∈ N,

and denote by µ the first positive integer m for which there is at least one cube Q of the type

Y

[2

a

, 2

(a

+ 1)] with a ∈ Z

contained in Ω and such that d(Q, Ω

) > 2

√

n. Let Q

, . . . , Q

be these cubes (of course, we have p

∈ N) and set

H

=

p1

[

h=1

Q

.

Now we proceed by recursion. If the sets H

, . . . , H

are obtained, we let Q

, . . . , Q

denote all the cubes of the type

Y

[2

a

, 2

(a

+ 1)] with a ∈ Z

contained in Ω, disjoint from the interior of H

∪ . . . ∪ H

and such that d(Q, Ω

) > 2

√

n. Then we set

H

=

p

[

h=1

Q

.

At this point let us remark that

d(H

, H

) ≥ 2

, ∀r ∈ N.

Finally, we set

K

= H

∪ . . . ∪ H

, ∀r ∈ N.

It is clear that {K

: r ∈ N} is a compact cover of Ω such that K

⊂ K

for every r ∈ N.

4. The auxiliary functions v

. With this construction in mind and the notations therein, we now prove the following result.

Proposition 4.1. Let Ω be a nonvoid open subset of R

and ζ ∈ ]1, ∞[.

Then there are integers c, d ∈ N and functions v

∈ C

(R

) for r ∈ {3, 4, 5, . . .} such that

(a) supp(v

) ⊂ K

\ K

, (b) v

(R

) ⊂ [0, 1],

(c) v

(H

) = {1},

(d) kD

v

k

≤ c(2

d)

(|α|!)

, ∀α ∈ N

, for every integer r > 2.

P r o o f. Let ϕ be an element of C

(R) for which there are l, d > 0 such that

ϕ(t) > 0 if |t| < n

2

, ϕ(t) = 0 if |t| ≥ n

2

, kϕ

k

≤ ld

(s!)

, ∀s ∈ N

, \

R

ϕ(t) dt = 1.

The existence of such a ϕ is provided by the Denjoy–Carleman–Mandelbrojt theorem (cf. [2]). Then we define

ψ(x) = ϕ(x

) . . . ϕ(x

), ∀x ∈ R

.

Clearly ψ belongs to C

(R

), has compact support equal to [−n

2

, n

2

]

and satisfies

kD

ψk

≤ l

d

(|α|!)

, ∀α ∈ N

, \

Rⁿ

ψ(x) dx = 1.

Now for every integer r ∈ N, we set

L

= {x ∈ R

: d(x, H

) ≤ 2

},

ψ

(x) = 2

ψ(2

x), ∀x ∈ R

,

and note that ψ

belongs to C

(R

) and has a compact support of diameter 2

. Then we set

v

(x) = ψ

∗ χ

(x) = \

Rⁿ

ψ

(y)χ

(x − y) dy, ∀x ∈ R

.

It is well known or easy to check that the function v

belongs to C

(R

) and has the properties (a)–(c) as well as

kD

v

k

≤ \

Rⁿ

|D

ψ

| dx ≤ `(supp(ψ

))kD

ψ

k

≤ c(2

d)

(|α|!)

for every α ∈ N

if we set c = l

(n

2

)

, a constant which does not depend on r > 2 nor on α ∈ N

.

5. Approximation in Gevrey classes and consequences. In the proof of Theorem 5.2, we shall make use of the following property which results immediately from the proof of Lemma 5 of [5].

Proposition 5.1 Let r ∈ N and g ∈ D

(R

). Then, for every ε > 0, there is λ

> 0 such that, for every λ ≥ λ

, the function

h(x) = π

λ

\

Rⁿ

g(y)e

dy

belongs to C

(R

) (in fact, it is analytic on R

) and satisfies kD

h − D

gk

≤ ε if |α| ≤ r.

Theorem 5.2. Let Ω be a nonvoid open subset of R

and let ζ, γ be real numbers such that 1 < ζ < γ. Then, for every f ∈ Γ

(Ω), there is g ∈ Γ

(Ω) which is analytic on Ω and such that

kD

f − D

gk

≤ 1

s if |α| ≤ s and s ≥ 2, with K

defined as in the special compact cover of Ω.

P r o o f. Of course there are numbers a, b > 1 such that

kD

f k

≤ ab

(|α|!)

and kD

v

k

≤ a(2

b)

(|α|!)

for every integer r ≥ 3 and α ∈ N

.

Now we introduce by recursion a sequence (g

)

in C

(R

) such that kD

g

k

≤ 2

a

2

b

(|α|!)

, ∀α ∈ N

, ∀s ∈ N.

At this point, to get the functions g

, we just need to consider a strictly increasing sequence (λ

)

of ]0, ∞[ but later on we shall make more strin- gent restrictions on these positive numbers.

We start with

g

(x) = π

λ

\

Rⁿ

v

(y)f (y)e

dy, ∀x ∈ R

,

where of course v

f has been extended by 0 on R

\ Ω. By Proposition 2.1, g

belongs to C

(R

) and satisfies

kD

g

k

≤ a

3

b

(|α|!)

≤ 2

a

2

b

(|α|!)

for every α ∈ N

. Now if g

, . . . , g

are obtained, we first remark that we certainly have

D

 f −

X

g



≤ kD

f k

+ X

kD

g

k

≤ ab

(|α|!)

+ X

2

a

2

b

(|α|!)

≤ 2

a

2

b

(|α|!)

for every α ∈ N

and then check by direct use of Proposition 2.1 that the function g

defined by

g

(x) = π

λ

\

Rⁿ

v

(y)

 f (y) −

X

g

(y)



e

dy

suits our purpose.

Of course we have D

X

g

≤

X

2

a

2

b

(|α|!)

≤ 2

a

2

b

(|α|!)

for every s ∈ N and α ∈ N

.

With this majorant at our disposal, we are in a position to make a more precise (but not yet final) choice of the numbers λ

(we are free to take them larger but strictly increasing). As we have

|α|→∞

lim

2

a

2

b

(|α|!)

(|α|!)

= 0, ∀s ∈ N,

there is a strictly increasing sequence (A

)

in N such that, for every s ∈ N,

2

a

2

b

(|α|!)

≤ (|α|!)

if |α| ≥ A

.

Then we can also fix a strictly increasing sequence (B

)

in N such that sup

|α|≤As

a(2

b)

(|α|!)

≤ B

, ∀s ∈ N.

Next we introduce the following elements of D

(R

):

h

(x) =

 v

(x)f (x), ∀x ∈ Ω,

0, ∀x ∈ R

\ Ω,

and, for every integer s ≥ 2, h

(x) =

 v

(x)(f (x) − P

j=1

g

(x)), ∀x ∈ Ω,

0, ∀x ∈ R

\ Ω.

Finally, by recursive use of Proposition 5.1, we may require that the numbers λ

are such that, for every s ∈ N,

kD

h

− D

g

k

≤ (2

B

)

if |α| ≤ A

. Now we consider the series

g(x) = X

g

(x), ∀x ∈ Ω.

We first prove that g is defined and belongs to C

(Ω). Indeed, for every s ∈ N and every α ∈ N

such that |α| ≤ A

, we get

kD

h

k

≤ X

β≤α

 α β



kD

v

k

D

 f −

X

g



≤

(∗)

X

β≤α

 α β



a(2

b)

(|β|!)

(2

B

)

≤ X

β≤α

 α β



B

(2

B

)

≤ 2

2

≤ 2

(at (∗), we have used the fact that f − P

j=1

g

= h

− g

on H

). As v

(x) = 0 for every x ∈ K

, we get

kD

h

k

≤ 2

if s ∈ N and |α| ≤ A

, hence

kD

g

k

≤ kD

g

− D

h

k

+ kD

h

k

≤ (2

B

)

+ 2

≤ 2

for every s ∈ N and α ∈ N

such that |α| ≤ A

. Now it is clear that g ∈ C

(Ω).

We establish next that g satisfies the inequalities announced in the state- ment of the theorem—in fact, we are going to prove more. Consider an inte- ger s ≥ 2 and a multi-index α ∈ N

such that |α| ≤ A

—this is certainly the case if |α| ≤ s. For every x

∈ Ω \ K

, there is a first integer p ≥ 2 such that x

∈ K

. According to the last established inequality, we of course have

|D

g

(x

)| ≤ 2

, ∀r ∈ {p − 1, p, p + 1, . . .}.

As v

(H

) = {1}, we also get  

D

f (x

) − D

s+p−2

X

j=1

g

(x

)  

 = |D

h

(x

) − D

g

(x

)| ≤ 2

.

Thus

|D

f (x

) − D

g(x

)| ≤  

D

f (x

) − D

s+p−2

X

j=1

g

(x

)    +

X

|D

g

(x

)|

≤ 2

+ X

2

≤ 2

+ 2

< 1 s , and hence

kD

f − D

gk

≤ 1

s if s ≥ 2 and |α| ≤ A

.

At this step, if we proceed as in the proof of Lemma 6 of [5], we see that it is possible to select successively the numbers λ

in such a way that g is an analytic function on Ω. (This is our last refinement on the choice of λ

.)

We still have to prove that g ∈ Γ

(Ω).

Set

sup

|α|≤A2

kD

gk

/(|α|!)

= a

and consider α ∈ N

and x

∈ Ω.

On the one hand, if |α| ≤ A

and (i) if x

∈ K

, we trivially have

|D

g(x

)| ≤ kD

gk

≤ a

(|α|!)

, (ii) if x

∈ Ω \ K

, we get

|D

g(x

)| ≤ |D

f (x

)| + |D

f (x

) − D

g(x

)|

≤ ab

(|α|!)

+ 1

2 ≤ 2ab

(|α|!)

. Hence there are constants a

, b

> 0 such that

kD

gk

≤ a

b

(|α|!)

if |α| ≤ A

.

On the other hand, if |α| > A

, we first let s be the integer such that A

< |α| ≤ A

(of course s ≥ 2) and then consider the following two possibilities:

(i) if x

∈ Ω \ K

, then we have at once

|D

g(x

)| ≤ |D

f (x

)| + |D

f (x

) − D

g(x

)|

≤ ab

(|α|!)

+ 1

s + 1 ≤ 2ab

(|α|!)

,

(ii) if x

∈ K

, we have |D

g

(x

)| ≤ 2

for every r ∈ N, hence X

r=1

|D

g

(x

)| ≤ X

2

= 2

. But we also have

  D

X

g

(x

)  

 ≤ 2

a

2

b

(|β|!)

for every β ∈ N

and s ∈ N, hence

  D

X

g

(x

)  

 ≤ (|α|!)

since |α| ≥ A

. Therefore

|D

g(x

)| ≤   D

X

g

(x

)    +

X

|D

g

(x

)| ≤ 2

+ (|α|!)

≤ 2(|α|!)

. Consequently, there are constants a

, b

> 0 such that

kD

gk

≤ a

b

(|α|!)

if |α| > A

.

Corollary 5.3. For every open and nonvoid subset Ω of R

and γ ∈ ]1, ∞[, there is a function g ∈ Γ

which is

(a) analytic on Ω,

(b) identically 0 on no connected component of Ω,

(c) flat on R

\ Ω (i.e. identically 0 together with all its derivatives on R

\ Ω).

P r o o f. If Ω is connected, we choose ζ ∈ ]1, γ[ and f ∈ Γ

(Ω) with compact support contained in K

\ K

and such that kf k

> 1/2. Then Theorem 5.2 provides g ∈ Γ

(Ω) which is analytic on Ω and such that

kD

f − D

gk

≤ 1

2 if |α| ≤ 2 (which implies that g is not identically 0 on Ω) as well as

kD

gk

= kD

g − D

f k

≤ 1

s if |α| ≤ s and s ≥ 3.

It is then well known that extending g by 0 on R

\ Ω provides a solution.

If Ω has a finite number of connected components—say Ω

, . . . , Ω

— then, for every k ∈ {1, . . . , m}, there is g

∈ Γ

which is analytic on Ω

, not identically 0 on Ω

and flat on R

\ Ω

. It is then clear that g = P

k=1

g

is a solution.

As Ω always has countably many connected components, to conclude,

we just have to settle the case when {Ω

: m ∈ N} is the set of connected

components of Ω. For this purpose, fix ζ ∈ ]1, γ[. By the first part of the proof, for every m ∈ N, there is g

∈ Γ

which is analytic on Ω

, not identically 0 on Ω

, flat on R

\ Ω

and such that

kD

g

k

≤ 1

s if |α| ≤ s and s ≥ 2,

where of course K

is the rth element of the special cover of Ω

. So, for every m ∈ N, there are constants a

, b

> 1 such that

kD

g

k

≤ a

b

(|α|!)

, ∀α ∈ N

, hence there is an integer k

≥ m such that

a

b

(|α|!)

≤ (|α|!)

if |α| ≥ k

. Then we set

c

= ( sup

|α|≤k_m

a

b

(|α|!)

)

and g = X

2

c

g

.

It is clear that g is a function defined on R

which is analytic on Ω, identi- cally 0 on no connected component of Ω and identically 0 on R

\Ω. Now we prove that g belongs to C

(R

) and is flat on R

\Ω. Let x ∈ R

\Ω. For ev- ery integer k ≥ 3, there is r > 0 such that the ball b = {y ∈ R

: |x − y| ≤ r}

is disjoint from the compact sets K

, . . . , K

. For every α ∈ N

such that |α| ≤ k, this leads to

sup

x∈b

X

2

c

|D

g

|

≤ sup

 sup

m≤k

2

c

k , sup

m>k

2

c

kD

g

k



≤ sup{1/k, 2

},

hence g belongs to C

(R

) and is flat on R

\ Ω. We still have to prove that g ∈ Γ

. This is immediate: for every α ∈ N

, we have

kD

gk

= sup

m∈N

2

c

kD

g

k

≤ sup

m∈N

sup{2

, 2

(|α|!)

} ≤ (|α|!)

by consideration of the cases |α| ≤ k

and |α| > k

.

Corollary 5.4. For every γ ∈ ]1, ∞[ and nondegenerate compact in- tervals I, J of R such that J ⊂ I

, there are f ∈ C

(R) and c > 0 such that

(a) f has no divergence point, (b) f (R) ⊂ [0, 1],

(c) f (R \ I) = {0}, f (I

) ⊂ ]0, 1] and f (J) = {1},

(d) kf

k

≤ ck

, ∀k ∈ N.

P r o o f. Let I = [a

, b

] and J = [a

, b

]. We choose ζ ∈ ]1, γ[ and apply Corollary 5.3 with Ω = ]a

, a

[: there is a nonzero g ∈ Γ

which is analytic on ]a

, a

[ and such that supp(g) = [a

, a

]. Now we choose k > 0 such that g

= kg

satisfies T

R

g

(x) dx = 1 and define the function f

on R by f

(x) =

x

\

−∞

g

(t) dt, ∀x ∈ R.

It is clear that f

belongs to Γ

, is analytic on ]a

, a

[ and satisfies f

(]−∞, a

]) = {0}, f

(]a

, a

[) ⊂ ]0, 1] and f

([a

, ∞[) = {1}.

So it is clear that f

has no divergence point.

Similarly there is f

∈ Γ

which is analytic on ]b

, b

[, has no divergence point and satisfies

f

(]−∞, b

]) = {0}, f

(]b

, b

[) ⊂ ]0, 1] and f

([b

, ∞[) = {1}.

Finally, we set

f (x) = f

(x)f

(b

+ b

− x), ∀x ∈ R.

Of course f belongs to Γ

and satisfies (a)–(c). Let us establish that f also satisfies (d). As f ∈ Γ

, there are a, b > 0 such that

kf

k

≤ ab

k

, ∀k ∈ N

. Since

k→∞

lim

ab

k

k

= 0,

there is k

∈ N such that ab

k

≤ k

for every k ≥ k

; therefore there is c > 0 such that

kf

k

≤ ck

, ∀k ∈ N.

Proposition 5.5. Let γ ∈ ]1, ∞[, let p ∈ N and let I, J be nondegenerate compact intervals of R such that J ⊂ I

. Then there is m

∈ N such that, for every integer m ≥ m

, there is a function u ∈ C

(R) satisfying the following conditions:

(a) u has no divergence point, (b) supp(u) ⊂ I,

(c) ku

k

≤ 2

, ∀k ∈ {0, 1, . . . , m}, (d) ku

k

≤ 2

k

, ∀k ∈ N,

(e) for every x ∈ J, one has either

|u

(x)| ≥ 5

m

or

|u

(x)| ≥ 5

 pm + 1 p + 1



.

P r o o f. Corollary 5.4 provides a function f ∈ C

(R) and a constant c > 0 such that f has no divergence point and satisfies

f (R) ⊂ [0, c], f (R \ I) = {0}, f (J) = {1}

as well as

kf

k

≤ ck

, ∀k ∈ N.

For any m ∈ N, we can introduce

a = m

, b = (2

cm

)

, u(x) = bf (x) sin(ax), ∀x ∈ R.

It is then clear that u is a C

-function on R satisfying the conditions (a) and (b) as well as

kuk

≤ bc = (2

m

)

≤ 2

. Moreover, for every x ∈ R and k ∈ N, we have

|u

(x)| ≤ b X

 k h



|f

(x)|a

≤ bc X

 k h



h

a

≤ (2

m

)

(k

+ a)

, hence

|u

(x)| ≤ (2

m

)

(m

+ m

)

= 2

if 1 ≤ k ≤ m as well as

|u

(x)| ≤ (2

m

)

(k

+ k

)

≤ 2

k

if k > m;

i.e. u also satisfies the conditions (c) and (d).

Now we investigate (e). Let x ∈ J. Of course we have u

(x) = a

b sin(kπ/2 + ax), ∀k ∈ N.

Now, for every m ∈ N, we certainly have

sup{|sin(pmπ/2 + ax)|, |sin((pm + 1)π/2 + ax)|} ≥ 2

. So on the one hand, if |sin(pmπ/2 + ax)| ≥ 2

, we get

|u

(x)| ≥ 2

a

b = 4

( √

2c)

m

, and on the other hand, if |sin((pm + 1)π/2 + ax)| ≥ 2

, then

|u

(x)| ≥ 2

a

b = 4

( √

2c)

m

≥ 4

( √

2c)

m

≥ 4

( √ 2c)

 pm + 1 p + 1



.

To conclude, it is enough to take as m

any positive integer m

such that sup{4( √

2c)

, 4( √

2c)

} ≤ 5.

6. Characterizing the sets of divergence points. In this section, we establish the following result.

Theorem 6.1. For every γ ∈ ]1, ∞[ and every G

-subset G of R

, there is an element of Γ

having G as its set of divergence points.

P r o o f. We proceed in several steps.

S t e p 1: the numbers γ

and p

. We fix a strictly increasing sequence (γ

)

⊂ ]1, γ[ and, for every j ∈ N, denote by p

a positive integer such that p

(γ

− γ

) > γ

.

S t e p 2: some auxiliary inequalities and the numbers q

. For every r ∈ N, we certainly have

p

− 1

p

γ

> γ

> . . . > γ

> 1.

Therefore, for every j ∈ {0, . . . , r − 1}, it is a straightforward matter to check the following limits:

(1) lim

m→∞

5

2

(p

m)

1 m

= lim

m→∞

 4.5

p

m



= 0,

(2) lim

m→∞

5

2

(p

m+1)

 p

+ 1 p

m + 1



= lim

m→∞

 4.5

(p

+ 1)

(p

m + 1)



= 0, (3) lim

m→∞

5

(nr)

(p

m + 1)

1 m

= lim

m→∞

 5

nrp

(p

m)





p

m + 1 p

m



= 0,

(4) lim

m→∞

5

(nr)

(p

m + 2)

 p

+ 1 p

m + 1



= lim

m→∞

 5

nr(p

+ 1)

(p

m + 1)





p

m + 2 p

m + 1



= 0.

With these limits at our disposal, we find that, for every r ∈ N, there is

q

∈ N such that, for every integer m ≥ q

and j ∈ {1, . . . , r − 1}, we have

the following auxiliary inequalities:

(I) 4 · 2

(p

m)

< 5

m

, (II) 4 · 2

(p

m + 1)

< 5

 p

m + 1 p

+ 1



, (III) (nr)

(p

m + 1)

< 5

m

,

(IV) (nr)

(p

m+2)

< 5

 p

m + 1 p

+ 1



. S t e p 3: the sets G

, Q

, P

, I

and J

. Being a G

-subset of R

, G is equal to the intersection of a sequence (G

)

of open subsets of R

that we may suppose decreasing.

Proceeding as in the construction of the special compact cover of an open set, we find that each G

is the union of countably many compact cubes Q

that we may renumber as a sequence, say (Q

)

. Then for every l, k ∈ N, we denote by P

the compact cube in R

having the same center as Q

and

diam(Q

) as diameter. Now we arrange N

into a sequence ((l

, k

))

, set

Q

= Q

and P

= P

,

and let I

and J

for j ∈ {1, . . . , n} be the compact intervals in R such that

Q

= Y

J

and P

= Y

I

.

Of course this construction leads to J

⊂ I

for every r ∈ N and j ∈ {1, . . . , n}.

S t e p 4: the functions u

and the numbers m

. At this point, every- thing is set up to introduce the functions u

for r ∈ N and j ∈ {1, . . . , n}, as well as the sequence (m

)

of N by the following recursion.

An application of Proposition 5.5 to γ = γ

and p = p

leads to an integer m

> q

and to functions u

, . . . , u

∈ C

(R) such that, for every j ∈ {1, . . . , n},

(a) u

has no divergence point, (b) supp(u

) ⊂ I

,

(c) ku

k

≤ 2

, ∀k ∈ {0, 1, . . . , m

}, (d) ku

k

≤ 2

k

, ∀k ∈ N,

(e) for every t ∈ J

, one has either

|u

(t)| ≥ 5

m

or

|u

(t)| ≥ 5

 p

m

+ 1 p

+ 1



.

Now, for an integer r ≥ 2, if the functions u

for t ∈ {1, . . . , r − 1}

and j ∈ {1, . . . , n} and the integers m

, . . . , m

are obtained, we ap- ply Proposition 5.5 to γ = γ

and p = p

and obtain an integer m

>

sup{p

m

, q

} and functions u

, . . . , u

∈ C

(R) such that, for ev- ery j ∈ {1, . . . , n},

(a) u

has no divergence point, (b) supp(u

) ⊂ I

,

(c) ku

k

≤ 2

, ∀k ∈ {0, 1, . . . , m

}, (d) ku

k

≤ 2

k

, ∀k ∈ N,

(e) for every t ∈ J

, one has either

|u

(t)| ≥ 5

m

or

|u

(t)| ≥ 5

 p

m

+ 1 p

+ 1



. S t e p 5: the functions u

and u. Finally, for every r ∈ N, we define,

u

(x) = u

(x

) . . . u

(x

), ∀x ∈ R

, and consider the series u = P

r=1

u

. For every k ∈ N, we certainly have k ≤ m

. Therefore, for every α ∈ N

,

X

kD

u

k

≤

X

2

≤ 1;

this implies that u is a bounded C

-function on R

. Moreover, for every α ∈ N

such that |α| ≥ 1, we have

kD

uk

≤

|α|−1

X

r=1

kD

u

k

+ X

kD

u

k

≤

|α|−1

X

r=1

2

|α|

+ 1 ≤ 3

|α|

, hence u ∈ Γ

.

To conclude, we prove that G is the set of divergence points of u.

On the one hand, if x ∈ R

does not belong to G, then x 6∈ G

for some

l

, hence x 6∈ G

for all l ≥ l

. This implies that x belongs to an at most