EQUALITIES IN ALGEBRAS OF GENERALIZED FUNCTIONS 1 ...

EQUALITIES IN ALGEBRAS OF GENERALIZED FUNCTIONS ´ DIMITRIS SCARPALEZOS, VINCENT VALMORIN STEVAN PILIPOVIC,

Abstract. Distributional and strong equalities in Colombeau algebra G of generalized functions are compared. Also it is done for G ∞ . Moreover a positive answer to a question of M. Oberguggenberger is given: A generalized function, invariant under all translations, is a generalized constant. AMS Mathematical Subject Classification(2000): 46F30, 46F05, 46S10. Key words and phrases: Algebra of generalized functions.

1. Introduction The algebra of Colombeau generalized functions G(Ω) is adequate for the analysis of various classes of non linear partial differential equations (cf. [1], [2], [3], [7]) or even for linear partial differential equations with singular coefficients which do not have solutions in the setting of the distribution theory (cf. [5] for the linear theory of Colombeau generalized functions). This algebra has many surprising properties, for example, an element can have generalized point values equal zero at any point of Ω but not being equal zero ([4]). Thus an analogy with the classical function theory is always far from trivial. In this paper we prove some new structural properties of Colombeau generalized functions. Our main tool is the use of Baire theorem as well as the use of parametrix. Our main results are the following. a) A generalized function invariant under all translations is a generalized constant (Theorem 6). RIt was a conjecture of M. Oberguggenberger. ¯ for every ρ ∈ C ∞ (Ω), then f = 0 b) If f ∈ G ∞ (Ω) and Ω f ρdt = 0, in C 0 (Theorem 5). Thus, the distribution equality of elements in G ∞ (Ω) implies their strong equality. c) Let f, g ∈ G(Ω). Though Z Z ¯ for every ρ ∈ C ∞ (Ω), (∗) f ρdt = gρdt, in C, 0 Ω

Ω

does not imply f = g in G(Ω), it is true if (∗) holds for every ρ ∈ C0k (Ω), for some k (Theorem 4). d) Let k ∈ N0 . If f ∗ ρ ∈ G ∞ (Ω) for every ρ ∈ C0k (Ω), then f ∈ G ∞ (Ω) (Theorem 3). e) If the integral of an f ∈ G(Ω) is equal to zero over any (Lebesgue) measurable set or every generalized interval, then f = 0 (Theorems 1, 2). 1

2

´ DIMITRIS SCARPALEZOS, VINCENT VALMORIN STEVAN PILIPOVIC,

Basic notions. We will recall some known facts of the theory of Colombeau generalized functions (cf. [1]-[7]). Denote by C k (Ω) the space of continuous functions on Ω with continuous derivatives up to the order k ∈ N0 ; C0k (Ω) denotes its subspace consisting of elements supported by a compact subset of Ω. We will use the notation |φ|k = sup{|φ(i) (x)|; x ∈ Ω, |i| ≤ k} and pK,k (φ) = sup{|φ(i) (x)|; x ∈ K, |i| ≤ k}, φ ∈ C k (Ω), where K ⊂⊂ Ω. In general, A ⊂⊂ B means that the closure of A is a compact subset of the interior of B. Recall, if E is a vector space on C (or R) with an increasing sequence of seminorms µn , n ∈ N, then the set of moderate nets EM (E), respectively of null nets N (E), consists of nets (Rε )ε∈(0,1) ∈ E (0,1) with the property (∀n ∈ N) (∃a ∈ R) (µn (Rε ) = O(εa )), respectively, (∀n ∈ N) (∀b ∈ R) (µn (Rε ) = O(εb )) (O is the Landau symbol). If E = Cd (or E = Rd ) and the seminorms are equal to the absolute value, then the corresponding spaces are denoted by EM (d) and N0 (d); the d− dimensional Colombeau space of generalized ¯ d = EM (d)/N0 (d). The definition of R ¯d complex numbers is defined by C is the same. In the case d = 1 one has generalized complex numbers of ¯ = EM /N0 . C ¯ is a ring, not a field (as R). ¯ Colombeau, C d Let Ω be an open set in R . If E = E(Ω) is the space of smooth functions with the sequence of seminorms µν = pKν ,ν , ν ∈ N0 , where (Kν )ν is an increasing sequence of compact sets exhausting Ω, then the above definition gives algebras EM (Ω), N (Ω) (the latter is an ideal) and as a quotient, Colombeau algebra G(Ω). It is a differential algebra with the differentiation (α) f (α) = [(fε )ε ](α) = [(fε )ε ]. The embedding of the Schwartz distribution space E 0 (Ω) is realized through the sheaf homomorphism E 0 (Ω) 3 f 7→ [f ∗ φε |Ω ] ∈ G(Ω), where a fixed net of mollifiers (φε )ε is defined by φε = ε−d φ(·/ε), R ε < 1, φ is aR rapidly decreasing smooth function (element of S(Rd )), Rd φ(t)dt = mn 1 1, Rd tm φ(t)dt = 0, m ∈ Nn0 , |m| > 0. (tm = tm 1 ...tn and |m| = m1 + ... + mn .) The extended sheaf homomorphism gives the embedding of D0 (Ω) into G(Ω). We will use the term strong equality for Rthe equality in G while the distribution equality of f, g ∈ G(Ω) means that ( Ω (fε (t)−gε (t))φ(t)dt)ε ∈ N0 (Ω), for every φ ∈ C0∞ (Ω). Recall ([7]), G ∞ (Ω) is a subalgebra of G(Ω) consisting of elements f with a representative (fε )ε such that (∀K ⊂⊂ Ω) (∃a ∈ R) (∀α ∈ Nd0 ) (sup{|fε(α) (x)|; x ∈ K}) = O(εa )). One can say, in a rather unprecise manner, that the use of G ∞ within G is similar to the use of C ∞ within the space of Schwartz distributions D0 .

EQUALITIES IN GENERALIZED ALGEBRAS

3

Let A be a subset of Cd . Then ([4]), ¯ d ; (∃ repr.(xε )ε of x)(∀ε ∈ (0, 1))(xε ∈ A)}. A˜ = {x ∈ C We embed A into A˜ by x 7→ (x)ε , where (x)ε is the corresponding constant ¯ d (R˜d = R ¯ d ). Recall, net. Clearly, C˜d = C ˜ (∃ repr.(xε )ε of x A˜c = {˜ x ∈ A; ˜)(∃K ⊂⊂ A)(∃η > 0)(∀ε ∈ (0, η))(xε ∈ K)}. We say that x ˜ and (xε )ε in the above definition are supported by K, or just compactly supported. 2. Integration We start with a simple remark. Let Ω be an open set of Rd . Let (Kn )n be an increasing sequence of compact sets of Ω, whose union equals Ω, such that the interior of Kn is non-empty, Kn is included in the interior of Kn+1 and the Lebesgue measure of the boundary of Kn equals zero, n ∈ N. Denote by κn a function in C0∞ (Ω) such that κn equals one on Kn and supp κn ⊂⊂ Kn+1 , n ∈ N. We have f = 0 in G(Ω) if and only if f κn = 0 in G(Rd ) for every n ∈ N. ˜ n ) the set of all Lebesgue measurable subsets Let n ∈ N. Denote by M(K ˜ n ), of Kn supplied with the pseudometric d(A, B) = m(A∆B), A, B ∈ M(K where m is the Lebesgue measure and ∆ is the symmetric difference of sets. ˜ n ) will be said to be equivalent if m(A∆B) = Elements A and B of M(K 0. The corresponding metric space of classes of equivalences is denoted as M(Kn ). If f ∈ G(Ω), f = [(fε )ε ] and A¯ ∈ M(Kn ) is the equivalence class of ˜ n ), we define the integral of f over A¯ by A ∈ M(K Z Z f dx = [( fε dx)ε ]. ¯ A A R ¯ It is an element ofR C. Note, if m(B) = 0, then [( B fε dx)ε ] = 0 and if (fε )ε ∈ N (Ω), then [( A fε dx)ε ] = 0. One can easily verify that M(Kn ) is a complete metric space and thus the Baire theorem can be applied. We will use the notation B(x0 , r) for a closed ball and L(x0 , r) for an open ball with the center at x0 and radius r > 0. TheoremR 1. Let (fε )ε ∈ EM (Ω) such that for every n ∈ N and every A ∈ ˜ n ), ( fε (t)dt)ε ∈ N0 . Then (fε )ε ∈ N (Ω). M(K A Proof. By the definition, we know that there exists a < 0 such that sup | grad fε (t)| ≤ εa , ε ≤ ε0 , ε0 < 1. t∈Kn

(grad denotes the gradient). Assume that (fε )ε ∈ / N (Ω). This implies that ˜ c and there exist q > 0, a compactly supported generalized point (yε )ε ∈ Ω

4


a strictly decreasing sequence (εν )ν , tending to zero, such that |fεν (yεν )| > εqν , ν ∈ N. This implies that fεν (yεν ) > εqν or fεν (yεν ) < −εqν for infinitely many indices ν. We continue with the assumption fεν (yεν ) > εqν , ν ∈ N, since, in the second case, the proof can be done in a similar way. Let p be chosen so that q + dq − da < p. This assumption will be clear at the end of the proof. With this n and p, fix l ∈ N and consider M(Kn ). Define Z ¯ fε (t)dt| ≤ εp , ε ≤ 1/l} (A ∈ A). Flp,n = {A¯ ∈ M(Kn ); | A

This is a closed subset of M(Kn ). We have ∞ [ Flp,n = M(Kn ). l=1

By the Baire theorem, there exists lp ∈ N such that Flp,n has a non-void p p,n ¯ interior. This implies that there is A0 ∈ Flp and δ > 0 such that for every ¯ ≤ δ, it follows A¯ ∈ M(Kn ), with m(A) ¯ A0 ∈ A¯0 , A0 ∪ A ∈ F p,n , d(A0 ∪ A, A0 ) < δ, A ∈ A, lp

and thus

Z

fε (t)dt| ≤ εp , ε ≤ 1/l.

| A0 ∪A

This implies Z Z | fε (t)dt| ≤ | A

A0

Z fε (t)dt| + |

fε (t)dt| ≤ 2εp , ε ≤ 1/l.

A0 ∪A

Let r > 0 be determined so that m(B(0, r)) < δ and, without loosing on generality, we assume that the balls B(yν , r), ν ∈ N, are subsets of Kn . Let t satisfy |t−yεν | ≤ εq−a ν /2 < r. It follows (by the mean-value theorem) that fεν (t) ≥ εqν /2, ν ∈ N. By the integration over the ball B(yν , εq−a /2), we obtain Z fεν (t)dt ≥ Cεq+dq−da , ν ∈ N. ν Bν

This gives a contradiction since we have chosen p such that q + dq − da < p. Generalized sets. Let A = (Aε )ε ∈ P (Ω)(0,1) such that Aε are elements of the Borel σ− algebra B(Ω), ε ∈ (0, 1), and (with the abbreviation meas for the measure) (mε )ε ∈ EM , where mε = meas (Aε ), ε ∈ (0, 1). We call (Aε )ε a measurable-moderate net and the whole family of such nets is denoted as MS M .


5

Let (Aε )ε , (Bε )ε ∈ MS M . We say that they are similar if (rε )ε ∈ N0 , where rε = meas (Aε ∆Bε ), ε ∈ (0, 1). This is an equivalence relation and we denote by SG the corresponding quotient space. Its elements are called generalized sets and the elements of MS M equivalent to (∅)ε are called zero-measurable families. We define the union and the intersection of generalized sets A and B as classes determined by (Aε ∪ Bε )ε and (Aε ∩ Bε )ε , respectively. ˜ We define a ≤ 0 if there exists a representative (aε )ε of a such Let a ∈ R. that (∀n > 0)(∃εn ∈ (0, 1))(∀ε < εn )(aε − εn ≤ 0). ˜ Define a Let a ˜, ˜b ∈ R. ˜ ≤ ˜b if a ˜ − ˜b ≤ 0. We refere to [6] for the notions of positive (and positive definite) generalized constants and functions. Let ˜ c . The generalized interval is defined by a ˜, ˜b ∈ R ˜ c; a [˜ a, ˜b] = {˜ x∈R ˜≤x ˜ ≤ ˜b)}. ˜ d as products of such one- dimensional We define generalized d-intervals of R ˜ c if generalized intervals. We say that a generalized d-interval [˜ a, ˜b] lies in Ω ˜ ˜ c. x ˜ ∈ [˜ a, b] implies x ˜∈Ω Theorem 2. If the integral of an f ∈ G(Ω) over every generalized d-interval in Ω is zero, then f equals zero. Proof. Recall [4], f = 0 if (fε (xε ))ε ∈ N0 for every generalized point ˜ c . Assume that x = [(xε )ε ] ∈ Ω ˜ c and [˜ ˜ c. a ˜, t˜ ∈ Ω a, t˜] = [˜ a1 , t˜1 ] × ... × [˜ ad , t˜d ] ⊂ Ω By the differentiation, one obtains Z t1 ,ε Z td ,ε ( ... fε (s1 , ..., sd )ds1 ...dsd )ε ∈ N0 ⇒ (fε (tε ))ε ∈ N0 . a1 ,ε

ad ,ε

However, an analogous result does not hold for intervals as can be seen by the following example: Example 1. Let θ ∈ C0∞ (R) be an odd function supported by [−1, 1] and let Rb φε (t) = θ( t−ε ε ), ε ≤ 1. Then f = [(φε )ε ] satisfies f 6= 0 but a f (t)dt = 0 for every interval [a, b] of R. 3. Regularity results and equality by duality Recall the well know parametrix formula ([8], VI (22)): For a given k ∈ N0 and a compact set K ⊂⊂ Rd containing zero in its interior, there exist m ∈ N0 , ρ ∈ C0k (Rd ) and θ ∈ C0∞ (Rd ) such that supp ρ, supp θ ⊂ K and δ = ∆m ρ + θ, which gives f = ∆m (f ∗ ρ) + f ∗ θ, f ∈ D0 (Rd ). A direct consequence of (3.1) is the following assertion.

(3.1)

6


Theorem 3. Let f ∈ G(Ω). If there exists k ∈ N0 such that for every φ ∈ C0k (Ω) f ∗ φ ∈ G ∞ , then f ∈ G ∞ . (The converse assertion is obvious.) Remark. It is clear that the assertion of the theorem holds if we assume that (f ∗ ρ) ∈ G ∞ and f ∗ θ ∈ G ∞ only for θ and ρ from (3.1). It is well known that the strong equality and the distribution equality in G are not equivalent, i.e., there exists f ∈ G(Rd ), f 6= 0, such that Z ¯ for all φ ∈ C ∞ (Rd ). f φdt = 0, in C, 0 Rd

Thus the next theorem becomes important. We give two proofs by the use of parametrix and by an elementary construction. Theorem 4. Let f = [(fε )ε ] ∈ G(Ω) such that there exists some k ∈ N0 , such that Z k fε (t)ρ(t)dt)ε ∈ N0 ). (3.2) (∀ρ ∈ C0 (Ω)) (( Rd

Then f = 0. Proof 1. By the remark at the beginning of the previous section, it is enough to prove the assertion with Ω = Rd . Note that (3.2) is equivalent to: (∀φ ∈ C0k (Rd ))(∀x ∈ Rd )((fε ∗ φ(x))ε ∈ N0 ). Suppose that supp ρ, supp θ ⊂ L(0, r0 ) (ρ and θ are from (3.1)). Let r > r0 and p ∈ N be fixed. We consider {ε−p fε ; ε < 1} as a set of functionals over the Banach space C0k (B(0, 2r)). By the hypothesis, this family is weakly bounded. By the Banach-Steinhauss theorem it follows that there exists C > 0 such that Z | ε−p fε (x)φ(x)dx| ≤ C|φ|k , φ ∈ C0k (B(0, 2r)), ε < 1. Rd

This implies |ε−p fε ∗ ψ(x)| ≤ C|ψ(· − x)|k , x ∈ B(0, r) ψ ∈ C0k (B(0, r)), ε < 1. Thus, for a fixed ψ ∈ C0k (B(0, r)) there holds: For every p ∈ N, sup ε−p |fε ∗ ψ(x)| = O(1). x∈B(0,r)

This implies (fε ∗ ψ)ε ∈ N (L(0, r)). Now by (3.1), it follows that (fε )ε ∈ N (L(0, r)). The above procedure can be done for every r > r0 . This implies that (fε )ε ∈ N (Rd ). Proof 2. Denote by C0k (Ω0 ) the cloasure of C0k (Ω0 ) in C0k (Ω) with respect to the norm | · |k , where Ω0 ⊂⊂ Ω. By the hypothesis, for every p > 0 and ρ ∈ C0k (Ω0 ), Z lim ε−p fε (x)ρ(x)dx = 0. ε→0

Rd


7

Without loosing on generality, assume that f is real valued. Let Ω00 be 00 an open subset such that Ω00 ⊂⊂ Ω0 . Denote by xm ε a point of Ω such m that |fε (xε )| = maxt∈Ω¯ 00 |fε (t)|, ε < 1. We will prove that f|Ω00 6= 0, i.e. (fε (xm / N0 , will bring a contradiction. The assumption (fε (xm / N0 ε ))ε ∈ ε ))ε ∈ implies that there exists b > 0 and a subsequence (εn )n tending to zero such b m b that fεn (xm εn ) ≥ εn , n ∈ N (or fεn (xεn ) ≤ −εn , n ∈ N which does not change the proof after multiplication with −1). By assumption, there exists N ∈ N, such that |grad fε (x)| ≤ ε−N , x ∈ 0 Ω , ε ∈ (0, 1). Let r¯n > 0 be determined so that B(xm ¯n ) ⊂ Ω0 , n ∈ N. εn , r Put m rn = min{¯ rn , ε N n fεn (xεn )}, n ∈ N. We will construct a sequence of non-negative functions ρεn ∈ C0k (Ω0 ), n ∈ N, with the following properties: supp ρεn ⊂ B(xm εn , rn ), n ∈ N, 1 ρεn (x) = 1 if |x − xm εn | ≤ rn , n ∈ N. 2 Let θ ∈ C0∞ (Rd ) satisfy the following properties: It is non-negative and θ(x) = 1, |x| ≤ 1/2, θ(x) = 0, |x| ≥ 1. Let Cj = sup{|θ(j) (x)|; x ∈ Rd }, |j| ≤ k. Then ρεn (x) =

1 x − xm εn θ( ), x ∈ Rd , n ∈ N, C1 rn

satisfy quoted conditions. Let p > (k + d)b + kN be fixed. Consider the family of continuous linear operators Z p −p p k 0 fε (x)ρ(x)dx, ε ∈ (0, 1). Sε : C0 (Ω ) → C, ρ 7→ Sε (ρ) = ε Rd

By the Banach-Steinhauss theorem, this family is a bounded family of continuous operators i.e. there exists Ap > 0 such that Z Ap |ρ|k ≥ ε−p | fε (x)ρ(x)dx|, ε ∈ (0, 1). Rd rn Let x ∈ B(xm εn , 2 ). We have m m fεn (x) ≥ fεn (xm εn ) − |x − xεn ||grad fεn (xεn )| ≥

Let C =

1 C1

fεn (xm εn ) , n ∈ N. 2

max{Cj , |j| ≤ k}. Since |ρεn |k ≤ C(rn )−k , n ∈ N,

(3.3) implies CAp (rn )−k ≥

Z Rd

ε−p n fεn (t)ρεn (t)dt

(3.3)


8

≥ ε−p n

Z B(xm εn ,rn /2)

fεn (xm εn )dx, n ∈ N.

This implies (with a new constant depending of C and the dimension d) m d (rn )−k Ap ≥ Cε−p n (fεn (xεn )) , n ∈ N.

This gives (p−kN )/(k+d) εbn ≤ sup{|fεn (x)|; x ∈ Ω0 } ≤ fεn (xm , n ∈ N. εn ) ≤ Cεn

Now we see that the choice of p implies the contradiction. The proof is completed. The next result is also a surprising one. It shows that the equality in the distribution sense of f and g implies their equality in G if both are elements of G ∞ . Theorem 5. RLet Ω denote an open set of Rd and f ∈ G ∞ (Ω). Then f = 0 if and only if Ω f ϕ = 0 for all ϕ ∈ C0∞ (Ω). ∞ (Ω) be a representative of f . Let K denote a Proof. Let (fε )ε ∈ EM 0 compact subset of Ω and κ ∈ C0∞ (Ω) such that κ|K0 = 1. Put K = supp κ. There exists s > 0 such for every m ∈ N there exists ε0 ∈ N such that

pK,m+d (κfε ) ≤ ε−s , ε > ε0 . Next, fix a positive number q and set ∂x = (∂/∂x1 )...(∂/∂xd ). Consider a net of distributions Sε , ε < 1, defined on E(Ω) by Z Sε (ϕ) = ∂x (κfε )(x)ϕ(x)dx, φ ∈ E(Ω), Ω

{ε−k S

and set Bk = ε ; ε < 1}, k > 0. The hypothesis on f implies that Bk is a bounded set in E 0 (Ω) for every k > 0. It follows, by the uniform boundedness property, that Bk is equicontinuous, thus, bounded on a neighborhood of zero in E(Ω) for every k > 0. Fix k > 0. There exist r > 0, a compact set L in Ω, which may be assume to contain K, and m ∈ N such that: if ϕ ∈ U = {ϕ ∈ E(Ω); pL,m (ϕ) ≤ r} and T ∈ Bk , then |T (ϕ)| ≤ 1. Put r∂x (κfε ) , ε < 1. ε + pK,m+d (κfε ) Since supp (κfε ) ⊂ K, we have pL,m (∂x (κfε )) = pK,m (∂x (κfε )) and ϕε =

pL,m (ϕε ) =

rpK,m (∂x (κfε )) < r. ε + pK,m+d (κfε )

Thus, ϕε ∈ U , and it follows |ε−k Sn (∂x ϕε )| ≤ 1, ε < 1. Taking k = 2q + s and using pK,m+d (κfε ) ≤ ε−s , ε < ε0 , we find a constant C such that Z |∂x (κfε )(x)|2 |ε−2q−s Sε (∂x ϕε )| = rε−2q−s dx ≥ Ω ε + pK,m+d (κfε )


C −1 ε−2q

9

Z

|∂x (κfε )(x)|2 dx, ε < ε0 . Ω p Using the inequality pK,0 (u) ≤ mes(K)k∂x ukL2 (Ω) , for u ∈ C0∞ (Ω), we find p p pK0 ,0 (fε ) ≤ mes(K)k∂x (κfε )kL2 (Ω) ≤ Cmes(K)εq , ε < ε0 . Now let α ∈ Nd . By the assumption, we have Z Z α |α| ∂ α fε (t)ϕ(t)dt)ε ∈ N0 , ϕ ∈ C0∞ (Ω). ( fε (t)∂ ϕ(t)dt)ε = ((−1) Ω

Ω

The same arguments as above show that we also have pK0 ,0 (∂ α fε ) = O(εq ) as ε → 0. The theorem is proved. In the next theorem we give a positive answer to a question of M. Oberguggenberger. Theorem 6. Let f ∈ G(Rd ) such that f (· + h) − f (·) = 0, for any h ∈ Rd . ¯ such that f = C. Then there is a C ∈ C We will present two proofs. The second one uses arguments of the parametrix which gives possibilities for some other formulations of the assertion. Proof 1. Let (fε )ε be a representative of f and R > 0. We will show that for every p ∈ N, ε−p |fε (t) − fε (0)| → 0, as ε → 0.

sup t∈B(0,R)

This is a characterization of a constant generalized function. Let δ > 0. The hypothesis is equivalent to: (∀R ∈ N)(∀p ∈ N)(∀x ∈ Rd )(∃lx ∈ N)(∀ε < 1/lx ) sup

ε−p |fε (x + t) − fε (t)| ≤ 1.

t∈B(0,R+δ)

Let l, p ∈ N. Consider the set Fp,l = {x ∈ Rd ; (ε < 1/l) ⇒ (ε−p

|fε (x + t) − fε (t)| ≤ 1)}.

sup t∈B(0,R+δ)

S d We have that Fp,l are closed sets and that ∞ l=1 Fp,l = R . By the Baire theorem, there exist l ∈ N and a ball B(x0 , r) such that the interior of Fp,l contains this ball. Denote r1 = inf{r, δ}. If x ∈ B(0, R) and h ∈ B(0, r1 ), then for ε < 1/l, sup ε−p |fε (h + t) − fε (t)| t∈B(0,R)

≤

sup

ε

−p

|fε (h + x0 + t) − fε (h + t)| +

t∈B(0,R)

≤

sup t∈B(0,R+δ)

sup

ε−p |fε (h + x0 + t) − fε (t)|

t∈B(0,R)

ε−p |fε (x0 + t) − fε (t)| +

sup t∈B(0,R+δ)

ε−p |fε (h + x0 + t) − fε (t)| ≤ 2.


10

Let n ∈ N such that n − 1 > R/r1 . For any point x in the ball we can find n points xi , i = 1, ..., n, (also in the ball) such that x1 = 0, xn = x and d(xi , xi+1 ) < r1 . We have |fε (x) − fε (0)| ≤ |fε (x2 ) − fε (x1 )| + |fε (x3 ) − fε (x2 )| +... + |fε (xn ) − fε (xn−1 )|, ε < 1/l0 . This implies ε−p |fε (t) − fε (0)| ≤ 2n, ε < 1/l0 ,

sup t∈B(0,R)

and the assertion of the theorem is proved. ˇ where ρ and θ are from (3.1) (φ(x) ˇ Proof 2. Let ρ1 = ρˇ, θ1 = θ, = d φ(−x), x ∈ R ). Let m be as in (3.1), as well. The assumption of the theorem (the translation invarianvce) implies that for any h ∈ Rd , Z Z m fε (t)(θ1 (t + h) − θ1 (t)dt)ε ∈ N0 . ∆ fε (t)(ρ1 (t + h) − ρ1 (t))dt)ε , ( ( Rd

Rd

We will prove (∃(Cε )ε ∈ EM )(∀p ∈ N)(∃r > 0)(∃z ∈ Rd )

(3.4)

−p

sup{ε |fε (t) − Cε |; t ∈ B(z, r)} = O(1). By the translation invariance of f , this implies (∃(Cε )ε ∈ EM )(∀p ∈ N)(∃r > 0)(∀x ∈ Rd ) sup{ε−p |fε (t) − Cε |; t ∈ B(x, r)} = O(1). Further on, by the compactness arguments one obtains (∃(Cε )ε ∈ EM )(∀p ∈ N)(∀K ⊂⊂ Rd ) sup{ε−p |fε (t) − Cε |; t ∈ K} = O(1). This proves the theorem. So let us prove (3.4). By the parametrix, we have fε (−x) = ∆m (fε ∗ ρ)(−x) + fε ∗ θ(−x), x ∈ Rd , ε < 1. ˇ By the assumption, for all p ∈ N, (Recall, ρ1 = ρˇ, θ1 = θ.) Z ε−p ∆m fε (t)(ρ1 (t + x) − ρ1 (t))dt = O(1), x ∈ Rd .

(3.5)

Rd

Fix p ∈ N. For every l ∈ N, denote by Aρ1 ,l the space of all x ∈ Rd with the property Z −p ε ∆m fε (t)(ρ1 (t + x) − ρ1 (t))dt ≤ 1 if ε < 1/l. Rd

S d This is a closed set, ∞ l=1 Aρ1 ,l = R , and there exists a ball L(x0 , r) and l0 such that L(x0 , r) ⊂ Aρ1 ,l0 . Thus, for x ∈ L(x0 , r), we have Z ε−p ∆m fε (t)(ρ1 (t + x) − ρ1 (t))dt ≤ 1. (3.6) Rd


11

Denote by Aθ1 ,l , l ∈ N, the set of all x ∈ B(x0 , r/2) with the property Z fε (t)(θ1 (t + x) − θ1 (t))dt ≤ 1. ε−p Rd

The use of the Baire theorem, again, implies that there exist r1 ≤ r/2, z ∈ L(x0 , r/2) and l1 > l0 such that Z −p fε (t)(θ1 (t + x) − θ1 (t))dt ≤ 1, ε < 1/l1 , x ∈ L(z, r1 ). (3.7) ε Rd

Put

Z

Z

m

fε (t)θ1 (t)dt, ε < 1.

∆ fε (t)ρ1 (t)dt +

Cε =

Rd

Rd

By (3.5), (3.6) and (3.7) we have that for ε < 1/l1 and x ∈ L(z, r1 ) it follows Z Z m −p fε (t)θ1 (t)dt| ∆ fε (t)ρ1 (t)dt − ε |fε (−x) − Rd

Rd

≤ ε−p |

Z

∆m fε (t)(ρ1 (t + x) − ρ1 (t))dt| + |

Rd

Z fε (t)(θ1 (t + x) − θ1 (t))dt|, Rd

and thus, ε−p |fε (−x) − Cε | ≤ 2, ε < 1/l1 , x ∈ L(z, r1 ). This finishes the proof of (3.4) and the proof of the theorem. Acknowledgement Work supported by START-project Y237 of the Austrian Science Fund and by the MNZZS of Serbia 144016 References [1] Biagioni, H. A.: A Nonlinear Theory of Generalized Functions, Springer-Verlag, Berlin-Heidelberg-New York, 1990. [2] J. F. Colombeau, J. F.: Elementary Introduction in New Generalized Functions, North Holland, 1985. [3] Grosser, M., Kunzinger, M., Obrguggenberger, M., Steinbauer, R.: Geometric Generalized Functions with Applications to General Relativity, Kluwer, 2001. [4] Kunzinger, M., Oberguggenberger, M.: Characterization of Colombeau generalized functions by their pointvalues, Math. Nachr. 203 (1999), 147-157. [5] Nedeljkov, M., Pilipović, S., Scarpalezos, D.: The linear theory of Colombeau generalized functions, Pitman Res. Not. Math. 259, Longman Sci. Tech., Harlow/New York, 1998. [6] Oberguggenberger, M., Pilipović, S., Scarpalezos, D.: Positivity and positive definiteness in generalized function algebras, Preprint. [7] Oberguggenberger, M.: Multiplication of Distributions and Applications to Partial Differential Equations, Pitman. Res. Not. Math. 259, Longman Sci. Techn. Essex, 1992. [8] Schwartz, L.: Théorie des distributions, Hermann, Paris, 1966. á 4, Institute of Mathematics, University of Novi Sad, Trg D. Obradovic 21000 Novi Sad, Yugoslavia e

´matiques, Universite Paris 7, 2 place Jussieu, Paris 5 =, U.F.R. de Mathe 75005, France

12


´ des Antilles et de la Guyane, De ´partement Math-Info, Campus Universite ` Pitre Cedex, France de Fouillole: 97159 Pointe a