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Some notes on the inclusion between Morrey spaces
Article in Journal of Mathematical Inequalities · March 2022
DOI: 10.7153/jmi-2022-16-26
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Mathematical Inequalities [3959] Second Galley Proofs
SOME NOTES ON THE INCLUSION BETWEEN MORREY SPACES
PHILOTHEUSE. A. TUERAH ANDNICKYK. TUMALUN∗
(Communicated by L. Liu)
Abstract. In this paper, we show that the Morrey spaces Mqp
1(Rn)cannot be contained in the weak Morrey spaceswMqp
2(Rn)forq16=q2. We also show that the vanishing Morrey spaces V Mqp(Rn)are not empty and properly contained in the Morrey spacesMqp(Rn).
1. Introduction
Let 16p6q<∞ and n>2 . The Morrey space Mqp(Rn) is the set of all functions f ∈Llocp (Rn) for which
kfkMqp= sup
x∈Rn,r>0
|B(x,r)|1q−1pkfkLp(B(x,r))<∞, where
kfkLp(B(x,r))= Z
B(x,r)|f(y)|pdy 1p
.
Here B(x,r) is the open ball in Euclidean space Rn with center x and radius r, and
|B(x,r)| denotes its Lebesgue measure. Meanwhile, theweak Morrey space wMqp(Rn) is defined to be the set of all functions f ∈wLlocp (Rn)for which
kfkwMqp= sup
x∈Rn,r>0
|B(x,r)|1q−1pkfkwLp(B(x,r))<∞, where
kfkwLp(B(x,r))=sup
t>0
t|{y∈B(x,r):|f(y)|>t}|1p,
and|{y∈B(x,r):|f(y)|>t}|also denotes the Lebesgue measure of the set{y∈B(x,r):
|f(y)|>t}. Now, we define V Mp
q(Rn) =
f ∈Mp
q(Rn): lim
r→0
Mf(r) =0
,
Mathematics subject classification(2020): 46E30, 42B35.
Keywords and phrases: Morrey spaces, vanishing Morrey spaces.
∗Corresponding author.
c D l , Zagreb
Paper JMI-3959 1
2 P. E. A. TUERAH ANDN. K. TUMALUN
where
Mf(r) =sup
x∈Rn
|B(x,r)|1q−1pkfkLp(B(x,r)). The setV Mp
q(Rn)is called thevanishing Morrey space. It is clear thatV Mp
q(Rn)is a subset ofMqp(Rn).
The Morrey spaces were introduced by C. B. Morrey [1] and the vanishing Morrey spaces were introduced in [2]. Recently, many authors are attracted in studying the in- clusion properties between Morrey spaces [3, 4, 5, 6, 7, 8]. One interesting result stated in [5, Remark 4.5], that is, the weak Morrey spaces wMqp
1(Rn) cannot be contained in the weak Morrey spaces wMqp
2(Rn) and vice versa, for distinct values q1 and q2. This statement was deduced by a characterization of inclusion between weak Morrey spaces and its parameters, which is proved by using Closed Graph Theorem and Mor- rey norm estimate for the characteristic functions of balls [5, Theorem 4.4]. Regarding to the inclusion between vanishing Morrey spaces and Morrey spaces over a bounded domain, it was stated in [9] that the vanishing Morrey spaces are properly contained in the Morrey spaces without giving an explicit counter example.
In this paper, we will prove that the Morrey spacesMqp
1(Rn)cannot be contained in the weak Morrey spaces wMqp
2(Rn) and the Morrey spaces Mqp
2(Rn) cannot be contained in the weak Morrey spaceswMqp
1(Rn), for different valuesq1 andq2. This result is more general and sharp than the previous result in [5] since we can recover that previous result and the fact that Mqp(Rn) is a proper subset of wMqp(Rn) [6, Theorem 1.2]. We also note that our method here is different than in [5] because we give a function which belongs toMqp
1(Rn)\wMqp
2(Rn)and a function which belongs to Mqp
2(Rn)\wMqp
1(Rn). Furthermore, by using the idea in [8], we also show that the van- ishing Morrey spaces V Mp
q(Rn) are non empty and properly contained in Mqp(Rn) by providing some examples.
The positive constantCthat appears in the proofs of all theorems may vary from line to line and the notationC=C(n,p,q) indicates thatC depends only on n,p and q.
2. A note on the inclusion between weak Morrey spaces
Let 16p<q<∞andγ=nq<n. Define a function f:Rn−→Rby the formula
f(y) =
( |y|−γ, y6=0,
0, y=0. (1)
It is clear that 1−pγn <0 andn−pγ>0 by observing to the given assumptions.
The function f, that appears in Lemma 1 and 3, is defined by (1).
LEMMA1. If x∈Rn and r>0, then kfkLp(B(x,r))6Crnp−γ=Crnp−nq, where C=C(n,p,q).
Proof. Note that Z
B(x,r)|f(y)|pdy= Z
{|y|6|x−y|<r}|y|−pγdy+ Z
{|x−y|<|y|}∩{|x−y|<r}|y|−pγdy
=I+II.
Since n−pγ>0 , we have I6
Z
{|y|<r}|y|−pγdy=C Z r
0
tn−pγ−1dt=Crn−pγ and
II6 Z
{|x−y|<r}|x−y|−pγdy=C Z r
0
tn−pγ−1dt=Crn−pγ, by using polar coordinate for radial function. Therefore
kfkLp(B(x,r))= Z
B(x,r)|f(y)|pdy 1p
6C
Crnp−γ1p
=Crnp−nq, which proves the lemma.
The following lemma is not hard to prove. We leave its proof to the reader.
LEMMA2. Let s>0, M>0, and ϕ:(0,∞)−→[0,∞). If sup
0<t6s
ϕ(t) =M= sup
s<t<∞ϕ(t), then
sup
t>0
ϕ(t) =M.
Using the above lemma, we can compute the weak Lebesgue norm of f on the ballB(0,r)with arbitrary radiusr.
LEMMA3. If r>0, thenkfkwLp(B(0,r))=Cr−γ+np =Cr−nq+np, where C=C(n,p,q). Proof. Let r be an arbitrary positive real number. Note that, for everyt>0 , we have
|{y∈B(0,r):|f(y)|>t}|= n
y∈B(0,r):|y|<t−1γo
=
B(0,r)∩B(0,t−1γ)
. (2) We now defineϕ:(0,∞)−→[0,∞)by the formula
ϕ(t) =t|{y∈B(0,r):|f(y)|>t}|1p. (3)
4 P. E. A. TUERAH ANDN. K. TUMALUN
For everyt>r−γ, we obtaint−1γ <r. Then
|{y∈B(0,r):|f(y)|>t}|=
B(0,t−1γ)
=Ct−nγ, by using (2). This gives us
t|{y∈B(0,r):|f(y)|>t}|1p =Ct t−nγ1p
=Ct1−pnγ, ∀t∈(r−γ,∞). (4) On the other hand, for everyt6r−γ, we have t−1γ >r. Hence
|{y∈B(0,r):|f(y)|>t}|=|B(0,r)|=Crn, which comes from (2). Therefore,
t|{y∈B(0,r):|f(y)|>t}|1p =Ct(rn)1p =Ctrnp, ∀t∈ 0,r−γ
. (5)
We obtain
ϕ(t) =
( Ct1−pnγ, ∀t∈(r−γ,∞)
Ct(rn)1p =Ctrnp, ∀t∈(0,r−γ].
by virtue of (4) and (5). Observing ϕ non increasing on (r−γ,∞) and non decreasing on(0,r−γ], since 1−pγn <0 and np>0 respectively, then
sup
r−γ<t<∞
ϕ(t) =Cr−γ+np = sup
0<t6r−γ
ϕ(t). (6)
Thus
kfkwLp(B(0,r))=sup
t>0
ϕ(t) =Cr−γ+np =Cr−nq+np, that is concluded from Lemma 2.
By taking Lemma 2 and Lemma 3 as the tools, we are ready to state and prove the first main result of this paper.
THEOREM1. Let16p<q1<∞and16p<q2<∞. If q16=q2, thenMqp
1(Rn)* wMqp
2(Rn)andMqp
2(Rn)*wMqp
1(Rn).
Proof. We will only prove that Mqp
1(Rn) is not contained by wMqp
2(Rn). The proof thatMqp
2(Rn)is not contained by wMqp
1(Rn)can be done by similar method.
Letγ1=n/q1 and f1:Rn−→R, defined by the formula f1(y) =
( |y|−γ1, y6=0, 0, y=0.
We will show that f1∈Mqp
1(Rn)\wMqp
2(Rn). Let x∈Rn and r>0 be arbitrarily given. According to Lemma 1, by replacingγ withγ1, we obtain
|B(x,r)|q11−1pkf1kLp(B(x,r))6Cr
n q1−np
r
n p−qn
1 =C<∞.
This gives us
kf1kMqp
1 = sup
x∈Rn,r>0
|B(x,r)|q11−1pkf1kLp(B(x,r))<∞, sincexandrare arbitrary. Whence f1∈Mqp
1(Rn). By virtue to Lemma 3, we have kf1kwMqp
2
>|B(0,r)|q12−1pkf1kwLp(B(0,r))=Crqn2−nprnp−qn1 =Crn
1 q2−q1
1
. Hence kf1kwMqp
2 =∞. This is due to arbitrary r and q16=q2. We conclude that f1∈/wMqp
2(Rn). Thus, we have already proved that Mqp
1(Rn)*wMqp
2(Rn).
As an immediate consequence of Theorem 1, we recover the result from [5] which is stated in the following corollary.
COROLLARY1. Let 16 p<q1<∞ and 16p<q2<∞. If q16=q2, then wMqp
1(Rn)*wMqp
2(Rn)and wMqp
2(Rn)*wMqp
1(Rn).
3. A note on the inclusion between Morrey spaces and vanishing Morrey spaces Let 16p<q<∞ and δ =exp(−2qnp ). Define a function g:Rn−→R by the formula
g(y) =
χB(y)
|y|npq(ln|y|)2
1p
, y6=0,
0, y=0,
(7)
whereχB is a characteristic function defined onB=B(0,δ).
The functiong in the following lemma is defined by (7). This following lemma shows that the vanishing Morrey spaces in a non empty set.
LEMMA4. g∈V Mp
q(Rn).
Proof. Letx∈Rn andr>0 be arbitrarily given. Note that
|B(x,r)|1q−1pkgkLp(B(x,r))6C Z
|y|6|x−y|<r
χB(y)
|x−y|n−npq|y|npq (ln|y|)2 dy
!1p
+C Z
{|x−y|<|y|}
∩{|x−y|<r}
χB(y)
|x−y|n−npq|y|npq(ln|y|)2 dy
!1p
=I+II. (8)
Now we have two cases, that is,δ 6rorr<δ. Assumeδ 6r, then we have I6
Z
|y|<r
χB(y)
|y|n(ln|y|)2dy= Z
|y|<δ
1
|y|n(ln|y|)2dy=C −1
ln(δ)
, (9)
6 P. E. A. TUERAH ANDN. K. TUMALUN
and II=
Z
{|x−y|<|y|<δ}
∩{|x−y|<r}
1
|x−y|n−npq |y|npq(ln|y|)2 dy6
Z
|x−y|<δ
1
|x−y|n(ln|x−y|)2dy
=C −1
ln(δ)
, (10)
since 1/tnp/q(ln(t))2 decreasing on interval(0,δ). Assumer<δ. We have I6
Z
|y|<r
1
|y|n(ln|y|)2dy=C −1
ln(r)
6C −1
ln(δ)
, (11)
and II=
Z
{|x−y|<|y|<r}
∩{|x−y|<r}
1
|x−y|n−npq |y|npq(ln|y|)2 dy6
Z
|x−y|<r
1
|x−y|n(ln|x−y|)2dy
=C −1
ln(r)
6C −1
ln(δ)
, (12)
since 1/tnp/q(ln(t))2 decreasing on interval(0,r)⊆(0,δ). By virtue of (8), (9), (10), (11), and (12), we conclude that
|B(x,r)|1q−1pkgkLp(B(x,r))6I+II6C −1
ln(δ) 1p
,
whereC=C(n,p,q). This meansg∈Mqp(Rn). We remaind to prove
r→0lim
Mf(r) =0. (13)
For every 0<r<δ, we have shown that Mf(r)6C
−1 ln(r)
1p . This means (13) holds and the proof is done.
Now we define a function that will play as an element of Morrey spaces but not in the vanishing Morrey spaces. Let 16p<q<∞. For everyk∈N, withk>3 , we set xk= (2−k, . . . ,0)∈Rn and
uk(y) = (
8
np qk
, y∈B(xk,8−k),
0, y∈/B(xk,8−k).
Define a functionu:Rn−→Rby the formula u(y) =
∞ k=3
∑
uk(y)
!1p
. (14)
We first claim thatubelongs to the Morrey spaces Mqp(Rn).
LEMMA5. u∈Mqp(Rn).
Proof. Let x∈Rn andr>0 be arbitrarily given. There are two casses: (i) x∈/ B(xk,2(4−k)) for every k>3 , or, (ii) x∈B(xj,2(4−j))for some j>3 . Assume (i) holds. Then
2(4−k)6|x−xk|6|x−y|+|y−xk|<r+4−k, for everyy∈B(x,r)∩B(xk,8−k). This meansr
np q−n
64(n−npq)k and r
np q−nZ
B(x,r)|u(y)|pdy6
∞ k=3
∑
4(n−
np q)kZ
B(x,r)∩B(xk,8−k)8
np qk
dy 6C
∞ k=3
∑
2(
np q−n)k
<∞, (15)
whereC depends onn. Assume (ii) holds. Since {B(xk,2(4−k))}k>3 is a disjoint col- lection, then there is only one j>3 such thatx∈B(xj,2(4−j))andx∈/B(xk,2(4−k)) for everyk>3 with k6=j. Note that
rnpq−n Z
B(x,r)∩B(xj,8−j)
uj(y)dy=rnpq−n Z
B(x,r)∩B(xj,8−j)8npq jdy6C<∞, (16) whereC depends on n,p, and q. By virtue of (16) and the computation of (15), we have
rnpq−n Z
B(x,r)|u(y)|pdy=rnpq−n
∞ k=3
∑
Z
B(x,r)∩B(xk,8−k)
uk(y)dy
=r
np q−nZ
B(x,r)∩B(xj,8−j)
uj(y)dy +rnpq−n
∞ k=3
∑
k6=j Z
B(x,r)∩B(xk,8−k)
uk(y)dy
6C+C
∞ k=3
∑
k6=j
2(npq−n)k<∞, (17)
whereC depends onn,p, andq. Combining (16) and (17), whence
|B(x,r)|1q−1pkukLp(B(x,r))=C
rnpq−n Z
B(x,r)|u(y)|pdy 1p
6C<∞, whereC depends onn,p, andq. Thereforeu∈Mqp(Rn).
The following theorem states that the vanishing Morrey spaces is a non empty proper subset of the Morrey spaces. This theorem is the second main result in this paper.
8 P. E. A. TUERAH ANDN. K. TUMALUN
THEOREM2. Let16p<q<∞. ThenV Mp
q(Rn)is a non empty proper subset ofMqp(Rn).
Proof. According to Lemma 4,V Mp
q(Rn)is non empty, and according to Lemma 5, the function u belongs to Mqp(Rn). Therefore, we need only to show thatu does not belong to V Mp
q(Rn). Let 0<r<1 . By the Archimedan property, there is an integerk>3 such that 8−k<r. Then
Mf(r)p
>Cr
np q−nZ
B(xk,r)|u(y)|pdy>C Z
B(xk,8−k)uk(y)dy
=C Z
B(xk,8−k)8npqkdy>C8−nk Z
B(xk,8−k)1dy=C>0,
whereC depends onn. This means Mf(r) is bounded away from zero as r tends to zero. Thusu∈/V Mqp(Rn).
R E F E R E N C E S
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Math. Soc.,43, (1938), 126–166.
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[3] D. I. HAKIM ANDY. SAWANO,Complex interpolation of various subspaces of Morrey spaces, Sci.
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(Received March 22, 2021) Philotheus E. A. Tuerah
Mathematics Department, Faculty of Mathematics and Natural Sciences Universitas Negeri Manado, Tondano Selatan District, Minahasa Regency Sulawesi Utara Province, Indonesia, Indonesia e-mail:[email protected] Nicky K. Tumalun Mathematics Department, Faculty of Mathematics and Natural Sciences Universitas Negeri Manado, Tondano Selatan District, Minahasa Regency Sulawesi Utara Province, Indonesia e-mail:[email protected]
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