Siberian Mathematical Journal, Vol. 47, No. 3, pp. 505–516, 2006 c 2006 Nadzhafov A. M. Original Russian Text Copyright 

EMBEDDING THEOREMS IN THE SOBOLEV–MORREY TYPE l SPACES Sp,a,κ,τ W (G) WITH DOMINANT MIXED DERIVATIVES A. M. Nadzhafov

UDC 517.518

Abstract: We construct Sobolev–Morrey type spaces with dominant mixed derivatives and, using the obtained integral representation, prove some embedding theorems in these spaces. Keywords: Sobolev–Morrey type space with dominant mixed derivative, flexible horn, integral representation, embedding theorem, H¨ older condition

The fact that some mixed derivatives Dν f cannot be estimated in terms of the derivatives of f entering the definition of the norm of Wpl and Hpl leads to the necessity of consideration of the function spaces of another type in which the key role is played by mixed derivatives. The function spaces Spl W (l ∈ Nn ) and Spl H with dominant mixed derivative were introduced and studied by S. M. Nikol ski˘ı [1] and later by A. D. Dzhabrailov [2]; the spaces Spl W were extended to the case when l = (l1 , l2 , . . . , ln ), where lj ≥ 0 cannot be an integer. l W (G) (l ∈ Nn ; p ∈ [1; ∞)n ; In this article we introduce the Sobolev–Morrey type space Sp,a,κ,τ a ∈ [0, 1]n ; κ ∈ (0, ∞)n ; τ ∈ [1, ∞]) with dominant mixed derivatives. This space is constructed on the basis of the Sobolev space Spl W with dominant mixed derivative and the Morrey type space Lp,a,κ,τ (G). In the case when G ⊂ Rn satisfies the flexible horn condition we obtain an integral representation and study some properties of the functions in the space under consideration from the view point of embedding theory. en = {1, 2, . . . , n}, e ⊆ en ; l = (l1 , l2 , . . . , ln ), lj > 0 are Suppose that G ⊂ Rnis some domain,  integers (j ∈ en ), and le = l1e , l2e , . . . , lne , where lje = lj for j ∈ e and lje = 0 for j ∈ en \e = e . Now, let be

e

f (x) dx = ae

 bj j∈e a

 dxj f (x);

j

i.e., integration is carried out only with respect to the variables xj whose indices belong to e. Henceforth we write p ≤ q, where p = (p1 , . . . , pn ) and q = (q1 , . . . , qn ), if pj ≤ qj (j ∈ en ); in particular, the notation 1 ≤ p ≤ ∞ means that 1 ≤ pj ≤ ∞ (j ∈ en ), where 1 = (1, . . . , 1) and ∞ = (∞, . . . , ∞). Given h ∈ (0, ∞)n , for each x ∈ G consider the vector-function ρ(v) = ρ(v, x) = (ρ1 (v1 , x), ρ2 (v2 , x), . . . , ρn (vn , x)), 0 ≤ vj ≤ hj (j ∈ en ), where ρj (0, x) = 0 for all j ∈ en , the functions ρj (vj , x) are absolutely continuous in uj on [0, hj ] and |ρj (vj , x)| ≤ 1 for almost all vj ∈ [0, hj ], with ρj = ∂v∂ j ρj (vj , x), j ∈ en . Given θ ∈ (0, 1]n , each of the  [ρ(v, x) + vθI] and x + V (x, θ), where I = [−1, 1]n , vθI = {(v1 θ1 y1 , . . . , vn θn yn ) : sets V (x, θ) = 0
y ∈ I}, is called a flexible horn and the point x is called the vertex of the flexible horn x + V (x, θ). We suppose that x + V (x, θ) ⊂ G. In the case v1 = · · · = vn = v, ρ(v, x) = ρ(v λ , x), and θ = (θλ1 , . . . , θλn ), θ ∈ (0, 1]n , the set [ρ(v λ , x) + v λ θλ I] V (x, θ) = V (x, λ, θ) = 0
is the flexible λ-horn introduced by O. V. Besov (see [3]). Baku. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 47, No. 3, pp. 613–625, May–June, 2006. Original article submitted March 30, 2005. c 2006 Springer Science+Business Media, Inc. 0037-4466/06/4703–0505 

505

l Denote by Sp,a,κ,τ W (G) the space of locally summable functions f on G having the generalized e l derivatives D f (e ⊆ en ) on G with the finite norm

e Dl f p,a,κ,τ ;G , f Sp,a,κ,τ l W (G) = e⊆en

where  v01 f p,a,κ,τ ;G = f Lp,a,κ,τ (G) = sup

x∈G

···

v0n 

0

0

j∈en

−

[vj ]1

f Lp,a,κ (G) = f p,a,κ;G = f p,a,κ,∞;G = sup

κj aj pj



x∈G j∈en vj >0







f p,Gvκ (x) =









p1

n

Gvκ2 (x2 ) 2



1/τ ,

j∈en

−

[vj ]1

|f (y)| dy1

... Gvκn (xn )

τ  dv j f p,Gvκ (x) vj κj aj pj

 p2 p1

,

 f p,Gvκ (x) ,

 dy2 . . .



pn pn−1

 dyn

1 pn

,

Gvκ1 (x1 ) 1

 κ Gvκ (x) = G ∩ Ivκ (x), Ivκ (x) = y : |yj − xj | < 12 vj j , j ∈ en , [vj ]1 = min{1, vj }, j ∈ en , and l (v01 , . . . , v0n ) = v0 is a fixed positive vector. For τ = ∞ the space Sp,a,κ,τ W (G) coincides with l the Sobolev–Morrey space Sp,a,κ W (G) with dominant mixed derivatives which was studied in [4]; i.e., l l Sp,a,κ,∞ W (G) ≡ Sp,a,κ W (G). l W (G). Observe some properties of Lp,a,κ,τ (G) and Sp,a,κ,τ 1. The following embeddings hold: l l Sp,a,κ,τ W (G) → Sp,a,κ W (G) → Spl W (G);

Lp,a,κ,τ (G) → Lp,a,κ (G) → Lp (G), i.e.,

f p,G ≤ f p,a,κ;G ≤ C f p,a,κ,τ ;G , f Spl W (G) ≤ f Sp,a,κ l l W (G) ≤ C f Sp,a,κ,τ W (G) . l W (G) are complete. 2. Lp,a,κ,τ (G) and Sp,a,κ,τ 3. For every real c > 0

f p,a,cκ,τ ;G =

1 c

1 τ

f p,a,κ,τ ;G

and f Sp,a,cκ,τ l W (G) =

1 1

cτ

f Sp,a,κ,τ l W (G) .

4. The following hold: (a) f p,0,κ,∞;G = f p,G and f p,1,κ,τ ;G ≥ f ∞,G ; (b) f S l l W (G) . W (G) = f Spl W (G) and f S l W (G) ≥ f S∞ p,0,κ,∞

5. If G is a bounded domain, pj ≤ qj ,

1−bj qj

p,1,κ,τ

≤

1−aj pj

(j ∈ en ), and 1 ≤ τ1 ≤ τ2 ≤ ∞ then

Lq,b,κ,τ1 (G) → Lp,a,κ,τ2 (G). We now construct an integral representation for studying the properties of functions in Spl W (G)   defined in n-dimensional domains and satisfying the flexible horn condition. Let 1e = δ1e , δ2e , . . . , δne , where δje = 1 for j ∈ e and δje = 0 for j ∈ e . We suppose that f ∈ Lloc (G) has all needed weak derivatives on G. Introduce the average of f as follows:     y ρ(v, x) −1 fv (x) = dy, (1) vj f (x + y)Ω , v v j∈en

506

Rn

 where Ω(y, z) = j∈en ωj (yj , zj ) is the kernel introduced by O. V. Besov [3, p. 77] and ωj is defined in [3, p. 77, formula (23)]. The average of (1) is constructed from the values of f at the points x + y ∈ x + V (x, θ) ⊂ G. Let ε = (ε1 , ε2 , . . . , εn ), 0 < εj < hj (j ∈ en ). Then the following equality is valid: fε (x) =



|1e |

he

(−1)

e⊆en

e

Dv1 fve +he (x) dv e .

(2)

εe

Differentiating with respect to vj , j ∈ e, and using [3, p. 77, formulas (26) and (27)], we obtain     ∂ e e −1 1e |1e | Dv fve +he (x) = fve +he (x) = (−1) hj vj−2 Ke(k +1 ) ∂vj  j∈en



Rn j∈en

j∈e

 y ρ(v e + h , x)  e e × e , , ρ (v + h , x) f (x + y) dy, v + he v e + he e

where Ke (x, y, z) =



ωj (xj , yj )

j∈e



(3)

ζj (xj , yj , zj ) ∈ C0∞ (Rn × Rn × Rn ),

j∈e

ζj is defined in [3, p. 77, formula (27)], and  (α) (α) Ke(α) (x, y, z) dx = 0 for all y, z, α such that |α| > 0. Ke (x, y, z) = Dx Ke (x, y, z), Rn

By (2), from (3) we derive



fε (x) =

e⊆en j∈e



y × e , v + he

h−1 j

 he , x)

ρ(v e

+ e v + he

he  εe j∈e

vj−2



Ke(k

e +1e )

Rn

 , ρ (v + h , x) f (x + y) dydv e . 

e

e

(4)

Then, in view of the Remark on Lemma 5.2 of [3], we have the following: if f ∈ Lloc (G) and 1 ≤ p < ∞ then fε (x) → f (x) as εj → 0 (j ∈ en ); moreover, for p > 1 we have fε (x) → f (x) for almost all x ∈ G by the Remark on Theorem 1.7 of [3]. Then it follows from (4) that f (x) = 

 ×

Me



|le |

(−1)

j∈e

e⊆en

y , e v + he

ρ(v e



 he , x)

+ e v + he

h−1 j

he  0e j∈e

−2+lj

vj

 e , ρ (v + h , x) Dl f (x + y) dydv e , 

e

e

(5)

Rn e e e Dxk +1 −l Ke (x, y, z);

where Me (x, y, z) = moreover, lj ≤ kj for j ∈ e, since the numbers kj in the kernel Ω can be chosen arbitrarily large. Suppose that ρj (vj , x) and ρj (vj , x) as functions of (vj , x) are locally summable on (0, hj ] × U , j ∈ en , where U ⊂ G is an open set. Let ν = (ν1 , ν2 , . . . , νn ) ∈ N0n ; moreover, lj ≤ νj + kj for j ∈ e and lj ≤ νj for j ∈ e . Applying the differentiation operation Dxν to both sides of (4) (and moving the differentiation onto the kernel in the summands on the right-hand side), we obtain fε(ν) (x) =  × Rn

Me(ν)





|ν|+|le |

(−1)

 j∈e

e⊆en

h−1 j

he  εe j∈e

−2−νj +lj

vj

 y ρ(v e + h , x)  e e e , , ρ (v + h , x) Dl f (x + y) dydv e . v e + he v e + he e

(6)

507

Show now that if μj = lj − νj > 0,

j ∈ en ,

(7)

then the weak derivative Dν f ∈ Lp (G) exists on G. First, establish that fε(ν) − fη(ν) → 0

as 0 < εj < ηj → 0, j ∈ en ,

(8)

in Lloc (U ). Let F ⊂ U be a compact set. Then F + T I ⊂ U for some T > 0. Put   M (ν) (x) = max max Me(ν) (x, y, z). e⊆en y,z∈I

By Minkowski’s inequality, for sufficiently small ε and h = η we have

 μj   (ν) e fε − fη(ν)  ≤C ηj Dl f 1,F +T I . 1,F +T I e⊆en j∈e

From here and (7) we obtain (8). Suppose that Dν f exists on G; i.e., fv(ν) (x) = Dν f (x) for x + cvI ⊂ G with some c = (c1 , . . . , cn ) > 0. Pass to the limit in (6) as εj → 0 (j ∈ en ), observing that the limit exists in the sense of Lloc (U ) by (7) and almost everywhere on U by the relation fv (x) → f (x) as vj → 0 (j ∈ en ) applied to Dν f . Then the equality ν

D f (x) =  ×

Me(ν)

Rn





(−1)

j∈e

e⊆en

y , e v + he



|ν|+|le |

ρ(v e

 he , x)

+ e v + he

−1−νj hj

he  0e j∈e

−2−νj +lj

vj

 e , ρ (v + h , x) Dl f (x + y) dydv e 

e

e

(9)

holds for almost all x ∈ U . Recall that the flexible horn x+V (x, θ) is the support of the representation (9) (ν) for x ∈ U . We can assume that the kernels Me and Me satisfy the following relations for all α and β:   α Dx Me (x, y, z) dx = 0, Dxβ Me(ν) (x, y, z) dx = 0 for all e ⊆ en . To prove the main theorems, we need some auxiliary inequalities given in the lemmas below. Choose ϕ(·, y, z) ∈ C0∞ so as to have

Put V =



 0
S(ϕ) = supp ϕ ⊂ I1 = {y : |yj | < 1/2, j ∈ en }.   y y : ve +h ∈ S(ϕ) . Clearly, V ⊂ Ih = {x : |xj | < 12 hj , j ∈ en }. Let U be an open e 

set contained in the domain G; henceforth we always assume that U + V ⊂ G. Put Ghκ (x) = (U + Ihκ ) ∩ G. Ghκ (U ) = x∈U

Obviously, if 0 < κ and h ≤ 1 then Ih ⊂ Ihκ and thereby U + V ⊂ Ghκ (U ) = Q. 508

Lemma 1. Let 1 ≤ p ≤ q ≤ r ≤ ∞, 0 < κ ≤ 1, 0 < v, η ≤ h ≤ 1, 0 < γ < ∞, 1 ≤ τ ≤ ∞, ν = (ν1 , . . . , νn ), νj ≥ 0 are integers (j ∈ en ), Φ ∈ Lp,a,κ,τ (G), and   1 1 , − εj = lj − νj − (1 − κj aj ) pj qj

Aeη (x)



=

Aeη,h (x)

j∈e

=

e

η  −1−νj hj 0e j∈e

 j∈e

−1−νj hj



  y ρ(v e + he , x)  e e ϕ e , , ρ (v + h , x) Φ(x + y) dy, (10) v + he v e + he



dvj

2+νj −lj

vj

Rn

he  η e j∈e

dvj



2+νj −lj

vj

Rn



  y ρ(v e + he , x)  e e ϕ e , , ρ (v + h , x) Φ(x + y) dy. (11) v + he v e + he

Then sup Aeη q,Uγ κ (¯x) x ¯∈U



≤ C1 Φ p,a,κ,τ ;Q

j∈e

−νj −(1−κj aj )( p1 − q1 ) j

hj

⎧  εj hj ⎪ ⎪ ⎪ j∈e ⎪ κj aj ⎪ ⎨   h q log ηjj × [γj ]1 j j∈e ⎪ ⎪ j∈en ⎪  εj ⎪ ⎪ ηj ⎩

 Here Uγ κ (¯ x) = x : |xj − x ¯j | < and h.

1 κj 2 γj ,

j ∈ en



 j∈en

sup Aeη,h q,Uγ κ (¯x) ≤ C2 Φ p,a,κ,τ ;Q

x ¯∈U

j

 j∈e

κj aj qj

[γj ]1

 j∈e

ε

ηj j

(εj > 0);

(12)

−νj −(1−κj aj )( p1 − q1 ) j

hj

j

for εj > 0, for εj = 0,

(13)

for εj < 0.

j∈e

and C1 and C2 are constants independent of Φ, γ , η ,

Proof. First assume that p1 = p2 = · · · = pn = p, q1 = q2 = · · · = qn = q, and r1 = r2 = · · · = rn = r. Applying the generalized Minkowski inequality, we deduce  e Aη  q,U

x) γ κ (¯

≤C

 j∈e

−1−νj

η

hj

e 

F (·, v e + he ) q,Uγ κ (¯x)

0e



dvj 2+νj −lj j∈e vj

(14)

for every x ¯ ∈ U , where 

F (x, v e + he ) =



 ϕ Rn

  y ρ(v e + he , x)  e e , , ρ (v + h , x) Φ(x + y) dy. v e + he v e + he



older’s inequality (q ≤ r) we obtain Estimate the norm F (·, v e + he ) q,Uρκ (¯x) . From H¨ 



F (·, v e + he ) q,Uγ κ (¯x) ≤ F (·, v e + he ) r,Uγ κ (¯x)

 j∈en

( 1 − r1 )κj

γj q

.

(15) 509

Let χ be the characteristic function of S(ϕ). Using the fact that 1 ≤ p ≤ r ≤ ∞, s ≤ r (since 1 1 1 s = 1 − p + r ), and 1

|Φϕ| = (|Φ|p |ϕ|s )1/r (|Φ|p χ) p

− 1r

1

1

(|ϕ|s ) s − r ,

and applying again H¨ older’s inequality ( 1r + ( p1 − 1r ) + ( 1s − 1r ) = 1), for |F | we have 

|F (x, v e + he )|

  s 1/r    ρ(v e + he , x)  e y e |Φ(x + y)| ϕ e , ρ (v + h , x)  dy ,  e e e v +h v +h



p

≤ Rn





|Φ(x + y)|p χ(y : (v e + he )) dy

×

1/p−1/r

Rn

   s 1/s−1/r    ρ(v e + he , x)  e y e   dy × , , ρ (v + h , x) . ϕ     v e + he v e + he Rn

Hence, 

e

e

F (·, v + h ) r,Uγ κ (¯x) ≤  

1/r  |Φ(x + y)| dx p

× sup y∈V

sup x ¯∈Uγ κ (¯ x)

Rn

Uγ κ (¯ x)

 |Φ(x + y)| χ p

Rn

y e v + he



1/p−1/r dy

  s 1/s    ρ(v e + he , x)  e y e ϕ , ρ (v + h , x)  dy . ,   e e e e v +h v +h

(16)

Obviously, if 0 < κ and v ≤ 1 then Qve +he (x) ⊂ Q(vκ )e +(hκ )e (x). For every x ∈ U we have 

 |Φ(x + y)| χ p

Rn



y e v + he

|Φ(y)|p dy ≤ Φ p,a,κ;Q

≤  (x) (v κ )e +(hκ )e



p



|Φ(x + y)| dx ≤ Uγ κ (¯ x)

dy  j∈e

Q

For y ∈ V



κ aj

hj j

 j∈e

|Φ(x)|p dx ≤ Φ pp,a,κ;Q

Qγ κ (¯ x+y)

κ a

vj j j .

 j∈en

κ a

[γj ]1 j j ,

s         ρ(v e + he , x)  e y e ϕ  dy = , , ρ (v + h , x) hj vj Φ ss .     v e + he v e + he j∈e

Rn

(17)

(18)

(19)

j∈e

It follows from (15)–(19) that 

F (·, v e + he ) q,Uγ κ (¯x) ≤ C Φ p,a,κ;Q ×

 j∈e

510

1−(1−κj aj )( p1 − r1 )

vj

 j∈en

 j∈e

κj ( 1q − 1r )

γj

1−(1−κj aj )( p1 − 1r )

hj



j∈en

κj aj r

[γj ]1

.

(20)

Using the inequality · p,a,κ;G ≤ C · p,a,κ,τ ;G and successively applying (20) in each single variable, we obtain the following inequality for the vectors p = (p1 , . . . , pn ), q = (q1 , . . . , qn ), and r = (r1 , . . . , rn ): 



F (·, v e + he ) q,Uγ κ (¯x) ≤ C1 Φ p,a,κ,τ ;Q ×

 j∈e

1−(1−κj aj )( p1 j

vj

− r1 j

)

 j∈en

κj ( q1 j

γj

j∈e

1−(1−κj aj )( p1 − r1 ) j

hj



− r1 ) j

j∈en

κj aj rj

[γj ]1

j

.

Inserting this inequality in (14), we see that  e Aη  q,U ×

 j∈en

≤ C2 Φ p,a,κ,τ ;Q

x) γ κ (¯

 j∈e

κj ( q1 j

γj

− r1 j

)

 j∈en

a κj r j j

[γj ]1

 j∈e

−νj −(1−κj aj )( p1 − r1 ) j

hj

j

lj −νj −(1−κj )( p1 − r1 ) j

ηj

j

.

(21)

Using rj = qj (j ∈ en ) in (21), we come to (12). Similarly, we prove (13). Corollary 1. Putting r = ∞ for 0 < γ ≤ 1 and r = q for γ > 1 in (20), we obtain sup F q,Uγ κ (¯x) ≤ C Φ p,a,κ;Q

x ¯∈U

 j∈en

κj 1q

[γj ]1

;

i.e., F q,b,κ;U ≤ C Φ p,a,κ;Q , where b ∈ [0, 1]n , whence we arrive at the inequality F q,b,κ,τ2 ;U ≤ C Φ p,a,κ,τ1 ;Q

(22)

for 1 ≤ τ1 ≤ τ2 ≤ ∞. Lemma 2. Suppose that 1 ≤ p ≤ q < ∞, 0 < κ ≤ 1, 0 < h ≤ 1, ν = (ν1 , . . . , νn ), νj ≥ 0 are integers (j ∈ en ), 1 ≤ τ1 ≤ τ2 ≤ ∞, εj > 0, and εj,0 = lj − νj − (1 − κj aj )

1 . pj

Then the following estimate is valid for every function Aeh (x) defined by (10):  e A  h q,b,κ,τ2 ;U ≤ C Φ p,a,κ,τ1 ;Q ,

(23)

where b = (b1 , . . . , bn ) and bj are arbitrary numbers satisfying the inequalities 0 ≤ bj ≤ 1 if εj,0 > 0 for j ∈ e, 0 ≤ bj < 1 if εj,0 = 0 for j ∈ e and 0 ≤ bj ≤ aj for j ∈ e ; εj,0 qj (1 − aj ) εj qj (1 − aj ) 0 ≤ bj < 1 + = aj + 1 − κj aj 1 − κj aj where C is a constant independent of Φ. Proof. Suppose first that 0 < γ < h. Then    e A  ≤ Aeγ q,U h q,U κ (¯ x) γ

x) γ κ (¯

(24)

if εj < 0 for j ∈ e,

  + Aeγh q,U

x) γ κ (¯

.

(25) 511

By (12) (ηj = γj , j ∈ en , and τ = ∞), we have  e Aγ q,U

x) γ κ (¯

≤ C1 Φ p,a,κ;Q

 j∈e

a

κj q j j

γj

 j∈e

a

εj +κj q j j

γj

,

(26)

where C1 is a constant independent of Φ and γ . Now, using the generalized Minkowski inequality and (13) (ηj = γj , j ∈ en , and τ = ∞), from (11) we obtain  e  A  ≤ C2 Φ p,a,κ;Q ψ(γ, h; r), (27) γ,h q,U κ (¯ x) γ

where C2 is a constant independent of Φ and γ and ψ(γ, h; r) =

 j∈en

δ (r ) γj j j

he  γ e j∈e

ε (rj )−1

vj j

dv e ,

  1 1 . εj (rj ) = lj − νj − (1 − κj aj ) − pj rj

κj κj − (1 − aj ), δj (rj ) = qj rj

After that we choose rj , qj ≤ rj ≤ ∞ (j ∈ en ) so that the exponent of γj in (27) be maximal. To this end, note that δj (rj ) is monotone increasing and εj (rj ) is monotone decreasing on [qj , ∞]; moreover, εj (qj ) = εj and εj (∞) = εj,0 (j ∈ en ). Consider the cases εj,0 ≥ 0 and εj,0 < 0. If εj,0 ≥ 0 then the exponent of γj in (27) is maximal for κ rj = ∞ and δj (∞) = qjj . In this case κj ⎧  qj  1  εj,0  εj,0 ⎪ ⎪ γj hj − γj ) if εj,0 > 0, ⎪ εj,0 ( ⎨ ψ(γ, h; r) =

j∈en

j∈e

κj qj

⎪  ⎪ ⎪ ⎩ γj j∈en

 j∈e

j∈e

log

j∈e

hj γj

if εj,0 = 0.

Let εj,0 < 0. Since εj (qj ) = εj > 0 and εj (∞) < 0, we have εj (rj,0 ) = 0 for some rj,0 , qj < rj,0 < ∞. Calculations show that the best estimate in this case is gained if rj = rj,0 in ψ(γ, h; r). Then  δj (rj,0 )  hj ψ(γ, h; r0 ) = γj log , (28) γj j∈en

j∈e

    εj,0 qj (1 − aj ) κj κj εj qj (1 − aj ) 1+ = aj + δj (rj,0 ) = qj 1 − κj aj qj 1 − κj aj

where

(here 1 − κj aj > 0, since εj > 0 and εj,0 < 0). Noting that εj +

κj aj 1 ≥ κj qj qj

from (25)–(28) we infer

 e A 

h q,Uγ κ (¯ x)

for εj,0 ≥ 0,

κj aj ≥ δj (εj,0 ) qj

for εj,0 < 0,

⎧ aj  κj qj  κj q1j  εj,0 ⎪ ⎪ ⎪ γj γj hj ⎪ ⎪ ⎪ j∈e j∈e j∈e ⎪ ⎪ ⎪ ⎨  κj aj  κj 1  q q h γj j γj j log γjj ≤ C3 Φ p,a,κ;Q ⎪  j∈e j∈e j∈e ⎪ ⎪ ⎪ aj ⎪ ⎪ ⎪  κj qj  δj (rj,0 )  h ⎪ ⎪ γj γj log γjj ⎩ j∈e

512

εj +

j∈e

j∈e

for εj,0 > 0, for εj,0 = 0, for εj,0 < 0.

(29)

Now, let γj ≥ hj (j ∈ en ). Applying again (12), we obtain ⎧ aj aj  κj qj  εj +κj qj ⎪ ⎪ γj γj ⎨  e j∈e j∈e A  ≤ C Φ 4 p,a,κ;Q h q,Uγ κ (¯ x)  εj ⎪ ⎪ hj ⎩

for hj ≤ γj < 1, (30) for γj > 1.

j∈e

It follows from (29) and (30) that  e A  h q,U

x) γ κ (¯

≤ C5 Φ p,a,κ;Q

b



κj qj

j∈en

j

[γj ]1

(31)

for arbitrary x ¯ ∈ U and γj , 0 < γj < ∞ (j ∈ en ), where b = (b1 , . . . , bn ), bj is an arbitrary number ¯. satisfying (24), and the constant C5 is independent of Φ, γ , and x The last inequality is equivalent to the inequality  e A  h q,b,κ,U ≤ C Φ p,a,κ;Q , and so

 e A  h q,b,κ,τ2 ;U ≤ C Φ p,a,κ,τ1 ;Q

for 1 ≤ τ1 ≤ τ2 ≤ ∞. Theorem 1. Suppose that an open set G ⊂ Rn satisfies the flexible horn condition, 1 ≤ p ≤ q ≤ ∞; κ ¯ = cκ, where 1c = max1≤j≤n lj κj ; ν = (ν1 , . . . , νn ), νj ≥ 0 are integers (j ∈ en ); 1 ≤ τ1 ≤ τ2 ≤ ∞; l W (G). εj > 0 (j ∈ en ); and f ∈ Sp,a,κ,τ 1 Then l l l1 W (G) → Lq,b,κ,τ2 (G), Dν : Sp,a,κ,τ W (G) → Sq,b,κ,τ W (G) Dν : Sp,a,κ,τ 1 1 2 and the following are valid: Dν f q,G ≤ C1

 e⊆en j∈en

s

Dν f q,b,κ,τ2 ;G ≤ C2 f Sp,a,κ,τ l where

 se,j =

e

hj e,j Dl f p,a,κ,τ1 ;G ,

1

W (G)

εj

 −νj − (1 − κj aj ) p1j −

1 qj

(p ≤ q < ∞),



(32) (33)

for j ∈ e, for j ∈ e .

If εj − lj1 > 0 (j ∈ en ) then Dν f S l1 W (G) ≤ C3 q

Dν f S l1

q,b,κ,τ2

 e⊆en j∈e

W (G)

s

hj e,j

 j∈e

≤ C4 f Sp,a,κ,τ l

se,j −lj1

hj

1

W (G)

e

Dl f p,a,κ,τ1 ;G ,

(34)

(p ≤ q < ∞);

(35)

moreover, 0 < h ≤ min(1, h0 ), C1 –C4 are constants independent of f , and C1 and C3 are also independent of h. In particular, if εj,0 > 0 (j ∈ en ) then Dν f is continuous on G and sup |Dν f | ≤ C1

x∈G

 e⊆en j∈en

s0

e

hj e,j Dl f p,a,κ,τ1 ;G , 513



where s0e,j

=

for j ∈ e,

εj,0 −νj − (1 −

κj aj ) p1j

for j ∈ e .

l ¯ Proof. Since κ ¯ = cκ, c > 0, we can assume that f ∈ Sp,a, κ,τ ¯ 1 W (G), and replace κ with κ everywhere in (32)–(35) and in the expressions for εj and se,j , j ∈ en . These are the inequalities we will prove (the greater κ, the greater ε). Under the conditions of the theorem, the weak derivative Dν f exists. Indeed, let εj > 0 and hence l l ν ¯ ≤ 1, and f ∈ Sp,a, lj − νj > 0 (j ∈ en ). Since p ≤ q, 0 ≤ a ≤ 1, κ κ,τ ¯ 1 W (G) → Sp W (G), D f exists on G and belongs to Lp (G). Then for almost each point x ∈ G the integral identity (9) holds with the same kernels. Hence, using Minkowski’s inequality, we arrive at

  Ae  . Dν f q,G ≤ C h q,G e⊆en

(ν)

e

By (12) for U = G, ϕ = Me , Φ = Dl f , and γ → ∞, we have (32).

(ν+lj1 )

To prove (34), substitute ν + lj1e for ν in (9). Using again (12) for U = G, ϕ = Me and letting γ → ∞, we obtain 1e

Dν+l f q,G ≤ C1

 e⊆en j∈e

consequently, Dν f S l1 W (G) ≤ C2 q

−νj −(1−κj aj )( p1 − q1 )  j

hj

 e⊆en j∈e

j

j∈e

s

hj e,j

 j∈e

se,j −lj1

hj

εj −lj1

hj

e

, and Φ = Dl f

e

Dl f p,a,κ,τ ¯ 1 ;G ;

e

Dl f p,a,κ,τ ¯ 1 ;G .

Using (22) and (23), we similarly establish (33) and (35). Now, let εj,0 > 0 (j ∈ en ). Show that Dν f is continuous on G. By (9), using (32) for q = ∞ and εj = εj,0 > 0 (j ∈ en ), we obtain

 

 εj,0 e Ae  Dν f − Dν fh ∞,G ≤ hj Dl f p,a,κ,τ ¯ 1 ;G , h ∞,G ≤ C e⊆en e=∅

e⊆en j∈e e=∅

lim Dν f − Dν fh ∞,G = 0.

hj →0 j∈e

Since Dν fh is continuous on G, in our case the convergence in L∞ (G) coincides with uniform convergence; consequently, Dν f is continuous on G. The theorem is proven. Theorem 2. Suppose that the domain G, the vectors p, q, ν , and κ and the parameters τ1 and τ2 l satisfy the conditions of Theorem 1 and f ∈ Sp,a,κ,τ W (G). ν If εj > 0 (j ∈ en ) then D f satisfies the H¨older condition with exponent βj1 on G in the metric of Lq ; more exactly  1 Δ(t, G)Dν f q,G ≤ C f Sp,a,κ,τ |tj |βj , (36) l W (G) j∈en

where βj1 (j ∈ en ) is an arbitrary number satisfying the inequalities ⎧ 0 ≤ βj1 ≤ 1 if εj > 1 for j ∈ e, ⎪ ⎨ 0 ≤ βj1 < 1 if εj = 1 for j ∈ e and 0 ≤ βj1 ≤ 1 for j ∈ e , ⎪ ⎩ 0 ≤ βj1 ≤ εj if εj < 1 for j ∈ e. 514

(37)

If εj,0 > 0 (j ∈ en ) then sup |Δ(t, G)Dν f | ≤ C f Sp,a,κ,τ l

x∈G

 W (G) 1

1

|tj |βj,0 ,

(38)

j∈en

1 satisfy the same conditions as β 1 with ε where βj,0 j,0 instead of εj . j

Proof. As in the proof of Theorem 1, below we can replace κ with κ. ¯ Let t be an n-dimensional vector. By Lemma 8.6 of [3, p. 102], there is a domain Gσ ⊂ G (σ = (σ1 , . . . , σn ), σj = ξj r(x), ξj > 0, r(x) = dist(x, ∂G), and x ∈ G). Suppose that |tj | < σj (j ∈ en ). Then, for every x ∈ Gσ , the segment joining the points x and x + t is contained in G. Identity (9) is valid for all points of the segment with the same kernels. Making simple transformations, we obtain

|Δ(t, G)Dν f | ≤ C1

 e⊆en j∈e

e

−1−νj

hj

e

|t1 | |tn | ··· 0

dvj 2+νj −lj j∈e vj

0

       y ρ(v e + he , x)  e e  |Δ(t, G)Dle f (x + y)| dy × Me(ν) e , , ρ (v + h , x)    v + he v e + he Rn

+C2

 e⊆en j∈e

−2−νj

hj



e

e

h1 hn  |tj | · · ·

j∈en

dvj 3+νj −lj j∈e vj |ten |

|te1 |

      (ν+1)  y ρ(v e + he , x)  e e  ×  Me , ρ (v + h , x)  ,  e e e e v +h v +h Rn

1 ×

e

|Dl f (x + y + t1 u1 + · · · + tn un )| dudy = C1



Aet (x, t) + C2

e⊆en

0



Aet,h (x, t),

(39)

e⊆en

where 0 < h ≤ min(1, h0 ). We also assume that |tj | < hj (j ∈ en ) and consequently |tj | < min(σj , hj ) (j ∈ en ). If x ∈ G\Gσ then, by definition, Δ(t, G)Dν f (x) = 0. By (39), Δ(t, G)Dν f q,G = Δ(t, G)Dν f q,Gσ ≤ C1

  Aet (·, t)

q,Gσ

e⊆en

  Ae (·, t) . th q,Gσ

+ C2

(40)

e⊆en

Using (12) for U = G and ηj = |tj | (j ∈ en ) and letting γ → ∞, we obtain   e le At  ≤ C D f |tj |εj , 1 p,a, κ,τ ¯ ;G q,Gσ

(41)

j∈e

and, using (13) for U = G and ηj = |tj | (j ∈ en ) and letting γ → ∞, we find that    e  le A  ≤ C D f |t | |tj |εj −1 2 j p,a, κ,∞;G ¯ th q,Gσ le

≤ C3 D f p,a,κ,∞;G ¯

 j∈e

|tj |



j∈e le

j∈e

|tj |βj = C3 D f p,a,κ,τ ¯ ;G

j∈e

 j∈e

|tj |



1

|tj |βj ,

(42)

j∈e

where βj1 > βj (j ∈ en ) and βj1 is a number satisfying (37). 515

It follows from (40)–(42) that Δ(t, G)Dν f q,G ≤ C f Sp,a, l κ,τ ¯ W (G)



1

|tj |βj .

j∈e

Suppose now that |tj | ≥ min(σj , hj ) for all j ∈ en . Then Δ(t, G)Dν f q,G ≤ 2 Dν f q,G ≤ C(σ, h) Dν f q,G



1

|tj |βj .

j∈en

Estimating Dν f q,G by means of (32), we obtain the sought inequality in this case as well. The theorem is proven. The article was discussed and approved at the meetings of the seminar of the Department of Mathematical Analysis of the Institute of Mathematics and Mechanics of the National Academy of Sciences of Azerbaidzhan which is headed by Academician A. D. Gadzhiev. Also, the author expresses his deep gratitude to Professors A. D. Dzhabrailov and V. S. Guliev for valuable remarks on this article. References 

1. Nikol ski˘ı S. M., “Functions with dominant mixed derivative satisfying a multiple H¨ older condition,” Sibirsk. Mat. Zh., 4, No. 6, 1342–1364 (1963). 2. Dzhabrailov A. D., “Families of spaces of functions whose mixed derivatives satisfy a multiple-integral H¨ older condition,” Trudy Mat. Inst. Steklov., 117, 139–158 (1972). 3. Besov O. V., Il in V. P., and Nikol ski˘ı S. M., Integral Representations of Functions and Embedding Theorems [in Russian], Mir, Moscow (1996). 4. Nadzhafov A. M., “Spaces with function parameters with mixed derivatives,” submitted to AzNIINTI on 1987, No. 680. A. M. Nadzhafov Azerbaidzhan Architecture and Building University Baku, Azerbaidzhan E-mail address: [email protected]

516