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1. Introduction
It is well-known that Sobolev spaces play a key role in the study of elliptic partial differential equations (PDEs) on domains in . There are many resources for properties of integer order Sobolev spaces of functions and their applications in PDEs (see, e.g., [1–3]). Also, there are variety of resources for properties of real order Sobolev spaces of functions and their applications (see, e.g., classical references such as [4–8] or more recent works such as [9–13]). Likewise, the study of elliptic PDEs on manifolds naturally leads to the study of Sobolev spaces of functions and more generally Sobolev spaces of sections of vector bundles on manifolds. As it turns out, the study of certain differential operators between Sobolev spaces of sections of vector bundles on manifolds equipped with rough metric and the study of low regularity geometry on Riemannian and semi-Riemannian manifolds are closely related to the study of spaces of locally Sobolev functions on domains in the Euclidean space (see, e.g., [14–16]).
In this paper, we focus on certain properties of spaces of locally Sobolev functions that are particularly useful in the study of differential operators on manifolds. Our work can be viewed as a continuation of the excellent work of Antonic and Burazin [17]; their work is mainly concerned with the properties of spaces of locally Sobolev functions with integer smoothness degree. In particular, they study the following fundamental questions for locally Sobolev spaces with integer smoothness degree:
(i) Topology and metrizability
(ii) Density of smooth functions
(iii) Reflexivity and the nature of the dual
(iv) Continuity of differentiation between certain spaces of locally Sobolev functions
Our main goal here is to provide a self-contained manuscript in which the known results are collected and stated in the general setting of Sobolev-Slobodeckij spaces and then develop certain other results that are useful in the study of differential operators on manifolds. In particular, we will discuss the following topics:
(i) General embedding results
(ii) Pointwise multiplication
(iii) Invariance under composition
The results of this type and other related results have been used in the literature—particularly in the study of Einstein constraint equations on manifolds equipped with rough metric—without complete proof. This paper should be viewed as a part of our efforts to fill some of the gaps. Interested readers can find other results in this direction in [13, 15, 16, 18]. Our hope is that the detailed presentation of this manuscript, along with these other four manuscripts, will help in better understanding the structure of the proofs and the properties of Sobolev-Slobodeckij spaces and locally Sobolev functions.
2. Notation and Conventions
Throughout this paper, denotes the set of real numbers, denotes the set of positive integers, and denotes the set of nonnegative integers. For any nonnegative real number , the integer part of is denoted by . The letter is a positive integer and stands for the dimension of the space. For all , is the set of all invertible matrices with real entries.
is a nonempty open set in . The collection of all compact subsets of will be denoted by . If is any function space on and , then denotes the collection of elements in whose support is inside . Also,
If and , we denote the extension by zero of to the entire by , that is,
Lipschitz domain in refers to a nonempty bounded open set in with Lipschitz continuous boundary. We say that a nonempty open set has the interior Lipschitz property provided that for each compact set there exists a bounded open set with Lipschitz continuous boundary such that .
Each element of is called a multi-index. For a multi-index , we let . Also, for sufficiently smooth functions (or for any distribution ), we define the th order partial derivative of as follows:
We use the notation to mean , where is a positive constant that does not depend on the nonfixed parameters appearing in and . We write if and .
If and are two topological spaces, we use the notation to mean , and the inclusion map is continuous.
3. Background Material
In this section, we collect some useful tools and facts we will need from topology and analysis. Statements without proof in this section are mainly taken from Rudin’s functional analysis [19], Grubb’s distributions and operators [20], excellent presentation of Reus [21], Treves’ topological vector spaces [22] and the reference [16], or are direct consequences of statements in the aforementioned references.
3.1. Topological Vector Spaces
Definition 1.
A topological vector space is a vector space together with a topology with the following properties:
(i) For all , the singleton is a closed set
(ii) The maps
are continuous where and are equipped with the product topology
Definition 2.
Suppose is a topological vector space and .
(i) is said to be convex if for all and it is true that
(ii) We say is bounded if for any neighborhood of the origin (i.e., any open set containing the origin), there exits such that
Definition 3.
Let be a topological vector space. is said to be metrizable if there exists a metric whose induced topology is . In this case, we say that the metric is compatible with the topology .
Theorem 4 ([19, 20]).
Let be a topological vector space. The following are equivalent:
(i) is metrizable
(ii) There exists a translation invariant metric on whose collection of open sets is the same as . Translation invariant means
(iii) has a countable local base at the origin
(Recall that a subcollection of is said to be a local base at the origin if for any open set containing the origin there is such that .)
Remark 5.
It can be shown that if and are two translation invariant metrics that induce the same topology on , then the Cauchy sequences of will be exactly the same as the Cauchy sequences of .
Definition 6.
Let be a topological vector space. We say is locally convex if it has a convex local base at the origin.
Definition 7.
Let be a metrizable locally convex topological vector space. Let be any translation invariant metric on that is compatible with . We say that is complete if and only if the metric space is a complete metric space. A complete metrizable locally convex topological vector space is called a Frechet space.
Definition 8.
A seminorm on a vector space is a real-valued function such that
If is a family of seminorms on , then we say is separating provided that for all , there exists at least one such that (that is, if for all , then ). It easily follows from the definition that any seminorm is a nonnegative function.
Theorem 9.
Suppose that is a normed space. Let be a seminorm on . If is continuous, then there exists a constant such that
Proof.
is continuous at so there exists such that if , then . If , then has norm equal to , and so for all , . Hence, for all , we have
Since , clearly the above inequality also holds for .
Definition 10.
Suppose is a separating family of seminorms on a vector space . The natural topology induced by is the smallest topology on that is translation invariant and with respect to which every is a continuous function from to . (Recall that translation invariant means if is open, then is open for every .)
Remark 11.
Suppose that and are two separating family of seminorms on a vector space . Let and be the corresponding natural topologies on . It follows immediately from the definition that if (1) is continuous for each and (2) is continuous for each , then .
The following theorem can be viewed as an extension of Theorem 9.
Theorem 12 ([21], page 157).
Let be a vector space and suppose is a separating family of seminorms on . Equip with the corresponding natural topology. Then, a seminorm is continuous if and only if there exist and such that for all
Theorem 13 ([19, 20]).
Suppose is a separating family of seminorms on a vector space and is the corresponding natural topology on . Then, is a locally convex topological vector space. Moreover, if is countable, then the locally convex topological vector space is metrizable, and the following translation invariant metric on is compatible with :
Corollary 14.
Suppose is a countable separating family of seminorms on a vector space and is the corresponding natural topology on . Then, is a Frechet space if and only if it is complete.
Theorem 15 ([23], Sections 6.4 and 6.5).
Let be a locally convex topological vector space. Then, there exists a separating family of seminorms on whose corresponding natural topology is .
Theorem 16 ([19], page 28).
Suppose is a separating family of seminorms on a vector space and is the corresponding natural topology on . Then, a set is bounded if and only if is a bounded set in for all .
Corollary 17.
Suppose is a separating family of seminorms on a vector space and is the corresponding natural topology on . It follows from Theorem 12 and Theorem 16 that if is bounded, then for any continuous seminorm , is a bounded set in .
Theorem 18 ([20], page 436, [23], Section 6.6).
Let be a topological vector space. Suppose is a separating family of seminorms on a vector space and is the corresponding natural topology on . Then, a linear map is continuous if and only if for each , is continuous on .
Theorem 19 ([20]).
Let be a Frechet space and let be a topological vector space. When is a linear map of into , the following two properties are equivalent:
(1) is continuous
(2) in in
Theorem 20 ([19, 20]).
Let and be two vector spaces and suppose and are two separating families of seminorms on and , respectively. Equip and with the corresponding natural topologies. Then,
(1) A sequence converges to in if and only if for all ,
(2) A linear operator is continuous if and only if
(3) A linear operator is continuous if and only if
Definition 21.
Let be a locally convex topological vector space.
(i) The weak topology on is the natural topology induced by the separating family of seminorms where
It can be shown that this topology is the smallest (weakest) topology with respect to which all the linear maps in are continuous. A sequence converges to in with respect to the weak topology if and only if in for all . In this case, we may write . We denote the weak topology on by . It can be shown that is the same set as
(ii) The weaktopology on is the natural topology induced by the separating family of seminorms where
It can be shown that this topology is the weakest topology with respect to which all the linear maps (from to ) are continuous. A sequence converges to in with respect to the weak topology if and only if in for all . We denote the weak topology on by
(iii) The strong topology on is the natural topology induced by the separating family of seminorms where for any bounded subset of ,
(It can be shown that for any bounded subset of and , is a bounded subset of ; see Theorem 16 and Theorem 28)
Remark 22.
(1) If is a normed space, then the topology induced by the norm
on is the same as the strong topology on ([22], page 198)
(2) In this manuscript, unless otherwise stated, we consider the topological dual of a locally convex topological vector space with the strong topology. Of course, it is worth mentioning that for many of the spaces that we will consider (including or where is an open subset of ), a sequence in converges with respect to the weak topology if and only if it converges with respect to the strong topology (for more details on this, see the definition and properties of Montel spaces in Section 34.4, page 356 of [22])
Theorem 23.
Let be a locally convex topological vector space. Then, the evaluation map
is a well-defined injective linear map. ( is called the bidual of ).
Definition 24.
Let be a locally convex topological vector space. Let denote the strong topology on as the dual of .
(i) If the evaluation map is bijective, then we say that is a semireflexive space
(ii) If the evaluation map is a linear topological isomorphism, then we say that is a reflexive space
Theorem 25 ([24], pages 16 and 17).
(i) Strong dual of a reflexive topological vector space is reflexive
(ii) Every semireflexive space whose topology is defined by the inductive limit of a sequence of Banach spaces is reflexive
(iii) Every semireflexive Frechet space is reflexive
Theorem 26.
Let and be two locally convex topological vector spaces. For all , let be the linear map defined by . Then,
(1) a linear map is continuous if and only if for all , the linear map is continuous
(2) a linear map is continuous if and only if for all , the linear map is continuous
Theorem 27 ([21], page 163, [20], page 46).
Let and be locally convex topological vector spaces and suppose is a continuous linear map. Either equip both and with the strong topology or equip both with the weak topology. Then,
(1) the map
is well-defined, linear, and continuous. ( is called the adjoint of )
(2) If is dense in , then is injective
Theorem 28 ([19], page 70).
Let be a locally convex topological vector space. Then, a set is bounded with respect to if and only if it is bounded with respect to .
Corollary 29.
If is a locally convex topological vector space and (i.e., converges to with respect to ), then is bounded with respect to both and .
Theorem 30.
Let and be two locally convex topological vector spaces. If is continuous, then is continuous. In particular, if (i.e., converges to with respect to ), then in .
Proof.
For all , let be the map . By Theorem 26, is continuous if is continuous for all . Let .
(1) By definition of the weak topology on , we know that the linear map is continuous
(2) By assumption is a continuous linear map
Therefore, belongs to . Since is the weakest topology on that makes all elements of continuous, we can conclude that is continuous.
Theorem 31 ([25], page 13).
Let be a Frechet space. Then, is reflexive if and only if every bounded set in is relatively weakly compact (i.e., the closure of with respect to is compact with respect to ).
Theorem 32 ([26], page 167).
Let be a separable Frechet space. If is relatively weakly compact, then every infinite sequence in has a subsequence that converges in .
The next theorem is an immediate consequence of the previous theorems.
Theorem 33.
Suppose that is a separable reflexive Frechet space. Then, every bounded sequence in has a weakly convergent subsequence, that is, a subsequence that converges with respect to .
Theorem 34 ([27], page 61).
Let and be two Banach spaces. Let be a linear map. Then, is continuous if and only if it is weak-weak continuous; that is, is continuous if and only if is continuous.
Theorem 35.
Let be a Banach space and be a closed subspace of with the induced norm. Suppose that is a sequence in and . If in , then in .
Proof.
This is a direct consequence of the fact that the following two topologies on the space are the same (see [27], page 70):
(1) The topology induced by
(2) The topology
Definition 36.
Let be a vector space and let be a family of vector subspaces of with the property that
(i) for each , is equipped with a topology that makes it a locally convex topological vector space, and
(ii)
The inductive limit topology on with respect to the family is defined to be the largest topology with respect to which
(1) is a locally convex topological vector space, and
(2) all the inclusions are continuous
Theorem 37 ([21]).
Let be a vector space equipped with the inductive limit topology with respect to as described above. If is a locally convex vector space, then a linear map is continuous if and only if is continuous for all .
Definition 38.
Let be a vector space and let be an increasing chain of subspaces of :
Suppose that
(i) each is equipped with a locally convex topology
(ii) for each , the inclusion is a linear topological embedding with closed image
Then, the inductive limit topology on with respect to the family is called a strict inductive limit topology.
Theorem 39 ([21]).
Suppose that is equipped with the strict inductive limit topology with respect to the chain . Then, a subset of is bounded if and only if there exists such that is bounded in .
3.2. Function Spaces and Distributions
Definition 40.
Let be a nonempty open set in and .
Let . A function is called -Holder continuous if there exists a constant such that
Clearly, a -Holder continuous function on is uniformly continuous on . 1-Holder continuous functions are also called Lipschitz continuous functions or simply Lipschitz functions. We define
Theorem 41 [20].
Let be a nonempty open set in and let . There is a function taking values in such that on a neighborhood containing .
Theorem 42 (exhaustion by compact sets) [20].
Let be a nonempty open subset of . There exists a sequence of compact subsets such that and
Moreover, as a direct consequence, if is any compact subset of the open set , then there exists an open set such that .
Theorem 43 [20].
Let be a nonempty open subset of . Let be an exhaustion of by compact sets. Define
Then,
(1) each is an open bounded set and
(2) the cover is locally finite in ; that is, each compact subset of has nonempty intersection with only a finite number of the ’s
(3) there is a family of functions taking values in such that and
Let be a nonempty open set in . For all , , and , we define
For all and , is a seminorm on . We define to be equipped with the natural topology induced by the separating family of seminorms . It can be shown that is a Frechet space.
For all , we define to be equipped with the subspace topology. Since is a closed subset of the Frechet space , is also a Frechet space.
We define equipped with the inductive limit topology with respect to the family of vector subspaces . It can be shown that if is an exhaustion by compacts sets of , then the inductive limit topology on with respect to the family is exactly the same as the inductive limit topology with respect to .
Remark 44.
Suppose is a topological space and the mapping is such that for some . Since , if is continuous, then will be continuous.
Theorem 45 (convergence and continuity for ).
Let be a nonempty open set in . Let be a topological vector space whose topology is induced by a separating family of seminorms .
(1) A sequence converges to in if and only if for all and
(2) Suppose is a linear map. Then, the following are equivalent:
(i) is continuous
(ii) For every , there exist and , and such that
(iii) If in , then in
(3) In particular, a linear map is continuous if and only if there exist and , and such that
(4) A linear map is continuous if and only if
Theorem 46 (convergence and continuity for ).
Let be a nonempty open set in and . Let be a topological vector space whose topology is induced by a separating family of seminorms .
(1) A sequence converges to in if and only if for all
(2) Suppose is a linear map. Then, the following are equivalent:
(i) is continuous
(ii) For every , there exists and such that
(iii) If in , then in
Theorem 47 (convergence and continuity for ).
Let be a nonempty open set in . Let be a topological vector space whose topology is induced by a separating family of seminorms .
(1) A sequence converges to in if and only if there is a such that and in
(2) Suppose is a linear map. Then, the following are equivalent:
(i) is continuous
(ii) For all , is continuous
(iii) For every and , there exists and such that
(iv) If in , then in
(3) In particular, a linear map is continuous if and only if for every , there exists and such that
Remark 48.
Let be a nonempty open set in . Here are two immediate consequences of the previous theorems and remark:
(1) The identity map
is continuous (that is, )
(2) If is a continuous linear map such that for all (i.e., is a local continuous linear map), then restricts to a continuous linear map from to . Indeed, the assumption implies that . Moreover is continuous if and only if for is continuous. Since , this map is continuous if and only if is continuous (see Remark 44). However, since the topology of is the induced topology from , the continuity of the preceding map follows from the continuity of
Theorem 49.
Let be a nonempty open set in . Then is separable.
Definition 50.
Let be a nonempty open set in . The topological dual of , denoted , is called the space of distributions on . Each element of is called a distribution on . The action of a distribution on a function is sometimes denoted by or simply .
Remark 51.
Every function defines a distribution as follows
In particular, every function defines a distribution . It can be shown that the map which sends to is an injective linear continuous map ([21], page 11). Therefore, we can identify with a subspace of ; we sometimes refer to the map as the “identity map.”
Theorem 52 ([20], page 47).
Let be a nonempty open set in . Equip with the weak topology. Then, under the above identification, is dense in .
Theorem 53 ([22], page 302).
Let be a nonempty open set in . Equip with the strong topology. Then, under the identification described in Remark 51, is sequentially dense in .
Remark 54.
(i) Clearly sequential density is a stronger notion than density. So is dense in strong topology)
(ii) Recall that, according to Remark 22, a sequence converges in if and only if it converges in strong topology). This together with the fact that weak topology is weaker than the strong topology implies that convergent sequences in both topologies converge to the same limit. Therefore, it follows from Theorem 53 that is sequentially dense in . Hence, Theorem 52 can be viewed as a corollary of Theorem 53
Theorem 55 ([21], page 9).
is reflexive. So can be identified with the topological vector space .
Definition 56 (restriction of a distribution).
Let be an open subset of and be an open subset of . We define the restriction map as follows:
This is well-defined; indeed, is a continuous linear map as it is the adjoint of the continuous map . Given , we sometimes write instead of .
Definition 57 (support of a distribution).
Let be a nonempty open set in . Let .
(i) We say is equal to zero on some open subset of if
(ii) Let be the collection of all open subsets of such that is equal to zero on . Let . The support of is defined as follows:
Note that supp is closed in but it is not necessarily closed in .
Theorem 58 ([21]).
Let be a nonempty open set in and let . If vanishes on a neighborhood containing , then .
Theorem 59 ([21]).
Let be a sequence in , , and such that in and for all . Then, also .
Theorem 60 ([28], page 38).
Let be a nonempty open set in . Suppose that is a sequence in with the property that for all , exists. Then, there exists such that
Definition 61 (Sobolev-Slobodeckij spaces).
Let be a nonempty open set in . Let and .
(i) If ,
(ii) If
(iii) If ,
(iv) is defined as the closure of in
(v) If ,
(vi) For all compact sets , we define
with . Note that for , can be viewed as a subspace of (see Theorem 68), and the support of is interpreted as the support of a distribution.
(vii) . is equipped with the inductive limit topology with respect to the family of vector subspaces . It can be shown that if is an exhaustion by compacts sets of , then the inductive limit topology on with respect to the family is exactly the same as the inductive limit topology with respect to
Theorem 62.
Let be a nonempty open set in , , and . Then, if and only if , and for all , .
Proof.
We consider two cases:
Case 1: .
Case 2: , , and .
Remark 63.
Let be a nonempty open set in , , and . Clearly for , . For , it is easy to see that for all , the map which sends to belongs to . The map is one-to-one and we can use it to identify with a subspace of ; we sometimes refer to the map that sends to as the “identity map.” So we can talk about the identity map from to for all .
Theorem 64 ([16]).
Let be a nonempty open set in , , and . Then, is a reflexive Banach space.
Corollary 65.
Let be a nonempty open set in , , and . A closed subspace of a reflexive space is reflexive, so is reflexive. Dual of a reflexive Banach space is a reflexive Banach space, so is a reflexive Banach space.
Remark 66.
Let be a nonempty open set in , , and . Since is reflexive, it can be identified with and we may write and talk about the duality pairing . To be more precise, we notice that the identification of and is done by the evaluation map
Therefore, for all and ,
Theorem 67.
Let be a nonempty open set in , , and . Then, is dense in . We may write this as .
Proof.
Our proof will be based on a similar argument given in page 65 of [1]. Let be the mapping introduced in Remark 63. Our goal is to show that the set
is dense in . To this end, it is enough to show that if is such that for all , then . Indeed, let be such an element. By reflexivity of , there exists such that
Thus, for all , we have
So, by the fundamental lemma of the calculus of variations (see [27], page 110), we have (as an element of ) and therefore .
Theorem 68.
Let be a nonempty open set in , , and . Equip with weak topology or strong topology. Then
Proof.
Recall that the convergent sequences in equipped with strong topology are exactly the same as the convergent sequences of equipped with the weak topology (see Remark 22). This together with Theorem 19 implies that in the study of the continuity of the inclusion map from to , it does not matter whether we equip with the strong topology or weak topology. In the proof, as usual, we assume is equipped with the strong topology. We consider two cases:
Case 1: . The continuity of the embedding has been studied in [16]. Also clearly . The former continuous embedding holds by the definition of and the latter embedding is continuous because if in , then for all ,
So, in . This implies the continuity of the inclusion map from to by Theorem 19
Case 2: . Since , it follows from previous case that . Also since is dense in (see Theorem 52, Theorem 53, and Remark 54), it follows that the inclusion map from to is continuous with dense image. Thus, by Theorem 27, . Here, we used the facts that (1) the strong dual of the normed space is and that (2) the dual of is (see Theorem 55). It remains to show that . It follows from Case 1 that and by definition is dense in . So, by Theorem 27,
Remark 69.
Note that for , is the same as . For , is a subspace of which contains . So it follows from the previous theorem that
To be more precise, we should note that for , we identify with the corresponding distribution in . Under this identification, for all the “identity map” is continuous with dense image, and so its adjoint will be an injective continuous map (Theorem 27), and we have
We usually identify with its image under and view as a subspace of . So, under this identification, we can rewrite the above equality as follows:
Finally, noting that for all and , (see Definition 61, Theorem 67, and Corollary 65), we can write
Theorem 70.
Let be a nonempty open set in , , and . Then,
(1) the mapping is an isometric isomorphism between and
(2) suppose . If is continuous, then (more precisely, there is a unique element in whose corresponding distribution is ). Moreover,
Proof.
The first item has been studied in [16]. Here, we will prove the second item. Since is continuous, it can be extended to a continuous linear map . So . However, is reflexive, therefore there exists a unique such that where is the evaluation map defined by . To finish the proof, it is enough to show that as elements of . For all , we have
Also,
Corollary 71.
Let be a nonempty open set in , , and . Suppose that . As a direct consequence of Theorem 70, we have the following:
(i) If , then and
(ii) If , then and
That is, for any and , in order to show that belongs to , it is enough to prove that
and in fact .
Theorem 72.
Let be a nonempty open set in , , and . Suppose that . Then, is a closed subspace of .
Proof.
It is enough to show that if is a sequence of elements in such that in , then , i.e., . By Theorem 68, we have in . Now it follows from Theorem 59 that . Note that for any , we have ; in this proof, we implicitly used the fact that for functions in , the usual definition of support agrees with the distributional definition of support.
Next, we list several embedding theorems for Sobolev-Slobodeckij spaces.
Theorem 73 ([29], Section 2.8.1).
Suppose and satisfy . Then, . In particular, .
Theorem 74 ([30, 18]).
Let be a nonempty bounded open subset of with Lipschitz continuous boundary. Suppose ( does NOT need to be less than or equal to ) and satisfy . If , additionally assume that . Then, . Furthermore, if , then the embedding is compact.
Theorem 75 ([16]).
Let be an arbitrary nonempty open set.
(1) Suppose and satisfy . Then, for all
(2) For all with and ,
(3) If and , then
(4) If are such that and , then
(5) If , , and , then
Theorem 76 ([6]).
Let be a nonempty bounded open subset of with Lipschitz continuous boundary or . If , then and is a Banach algebra.
In the next several theorems, we will list certain multiplication properties of Sobolev spaces. Suppose and . If , then the product has a clear meaning. What if ? In this case, is a distribution and by the product we mean the distribution ; then is in if is continuous. Because then it possesses a unique extension to a continuous linear map from to and so it can be viewed as an element of . See Theorem 70 and Corollary 71. Also see Remark 89.
Theorem 77 (multiplication by smooth functions I, [31], page 203).
Let , , and . Then, the linear map
is well-defined and bounded.
Theorem 78 (multiplication by smooth functions II, [16]).
Let be a nonempty bounded open set in with Lipschitz continuous boundary.
(1) Let and . If , then the linear map defined by is well-defined and bounded
(2) Let and . If , then the linear map defined by is well-defined and bounded
Theorem 79 (multiplication by smooth functions III, [16]).
Let be any nonempty open set in . Let .
(1) If and (that is, and is Lipschitz), then
is a well-defined bounded linear map
(2) If and , then
is a well-defined bounded linear map
(3) If and or and , then
is a well-defined bounded linear map
Theorem 80 (multiplication by smooth functions IV, [16]).
Let be a nonempty open set in , , , and or or . If , then the linear map
is well-defined and bounded.
Theorem 81 (multiplication by smooth functions V, [16]).
Let be a nonempty bounded open set in with Lipschitz continuous boundary. Let . Suppose and . If , then the linear map defined by is well-defined and bounded.
In the next definition, we introduce the notion of smooth multiplication triple which will play a key role in several theorems that will follow.
Definition 82 (smooth multiplication triple).
Let be a nonempty open set in , and .
(i) We say that the triple is a smooth multiplication triple if for all , the map
is well-defined and bounded
(ii) We say that the triple is an interior smooth multiplication triple if for all and , the map
is well-defined and bounded
Remark 83.
(1) Every smooth multiplication triple is also an interior smooth multiplication triple
(2) It is a direct consequence of Theorems 77, 78, and 79 that
(i) if or is bounded with Lipschitz continuous boundary, then for all and , is a smooth multiplication triple
(ii) if is any open set in , , and is not a noninteger with magnitude greater than , then is a smooth multiplication triple
(3) It is a direct consequence of Theorem 80 and Theorem 81 that
(i) if or is bounded with Lipschitz continuous boundary, then for all and , is an interior smooth multiplication triple
(ii) if is any open set in , , and is not a noninteger less than , then is an interior smooth multiplication triple
(4) If is a smooth multiplication triple and , then (see the proof of Theorem 7.31 in [16]). Of course, if , then and so holds for all , and open sets
Theorem 84.
Let be a nonempty open set in , , and . If is a smooth multiplication triple so is .
Proof.
Let . For all and , we have
The last inequality holds because is a smooth multiplication triple. It follows from Corollary 71 that and ; that is, is well-defined and continuous.
Theorem 85.
Let be a nonempty open set in , and . If , further assume that is a smooth multiplication triple. Suppose that and . Then,
(1) for all ,
(2) for all ,
Proof.
The claim follows from the argument presented in the proofs of Corollary 7.39 and Theorem 7.46 in [16].
Theorem 86 (([30], pages 598-605), ([6], Section 1.4)).
Let , , and . Suppose is a nonempty open set in . Then,
(1) the linear operator is well-defined and bounded
(2) for , the linear operator is well-defined and bounded
(3) for and , the linear operator is well-defined and bounded
(4) if is bounded with Lipschitz continuous boundary, and if , (i.e., the fractional part of is not equal to ), then the linear operator for is well-defined and bounded
Theorem 87.
Assumptions:
(i) or is a bounded domain with Lipschitz continuous boundary
(ii) , for
(iii) for
(iv)
(v)
Claim: If and , then and moreover the pointwise multiplication of functions is a continuous bilinear map
Remark 88.
A number of other results concerning the sufficient conditions on the exponents that guarantee the multiplication is well-defined and continuous are discussed in detail in [18].
Remark 89.
Suppose that is a smooth multiplication triple with . is the dual of and is the action of the functional on the function . As it was discussed before, if is a function in , is defined as a product of a smooth function and a distribution. Since is a smooth multiplication triple, will also be a smooth multiplication triple, and that means is continuous (see the note right after Theorem 76). We interpret as an element of to be the unique continuous linear extension of to the entire . It is easy to see that, this unique linear extension is given by
that is, the above map is linear continuous and its restriction to is the same as . (Note that since is a smooth multiplication triple, is indeed an element of .)
Theorem 90 ([32]).
Let and , and let
If is such that and (in particular, note that every with satisfies these conditions), then the map is well-defined and continuous from into .
Corollary 91.
Let , , and be as in the previous theorem. Moreover, suppose . Then, the map is well-defined and continuous from into . The reason is that when , we have .
In the remaining of this section, we will state certain useful properties of the topological vector space . The properties we will discuss here echo the ones stated in [24] for spaces .
Theorem 92.
Let be a nonempty open set in , , and . Then, is continuously embedded in .
Proof.
For all , we have
This together with the fact that the image of under the identity map is inside implies that
Also, by the definition of the inductive limit topology on , we have
It follows from (72) and (73) that for all ,
which, by Theorem 37, implies that .
Theorem 93.
Let be a smooth multiplication triple. Then, is dense in .
Proof.
We will follow the proof given in [24] for spaces . Let . It is enough to show that there exists a sequence in that converges to in (this proves sequential density which implies density). By Meyers-Serrin theorem, there exists a sequence such that in . Let be such that on a neighborhood containing (see Theorem 41). Let . Since is a smooth multiplication triple, multiplication by is a linear continuous map on and so in . Now, we note that and for all , are in with support inside . Consequently, in . Now, since , we may conclude that is a sequence in that converges to in .
Remark 94.
As a consequence, if is a smooth multiplication triple, then (equipped with the strong topology) is continuously embedded in . More precisely, the identity map is continuous with dense image, and therefore, by Theorem 27, the adjoint is a continuous injective map. We have
We usually identify with its image under and view as a subspace of . So, under this identification, we can rewrite the above equality as follows:
Next, we will prove that if is a smooth multiplication triple, then is separable. To this end, we need the following lemma.
Lemma 95.
Let and be two topological spaces. Suppose that
(1) is dense in
(2) is continuous
(3) is dense in
Then, is dense in .
Proof.
It is enough to show that intersects every nonempty open set in . So let be nonempty. Since is dense in , we have and so is nonempty. Also, since is continuous, . is dense in , so . Therefore,
Theorem 96.
Let be a smooth multiplication triple. Then, is separable.
Proof.
According to Theorems 92 and 93, is continuously embedded in and it is dense in . Since is separable, it follows from Lemma 95 that is separable.
Theorem 97.
Let be an interior smooth multiplication triple. Let be the partition of unity introduced in Theorem 43. Let be the collection of all sequences whose terms are nonnegative integers. For all sequences define by
Then, is a separating family of seminorms on and the natural topology induced by this family on is the same as the inductive limit topology on .
Proof.
Note that support of every is compact, so for each , only finitely many of ’s are nonzero. Thus, the sum in the definition of is a finite sum. Now it is not hard to show that each is a seminorm and is separating. Here, we will show that the topologies are the same. Let us denote the inductive limit topology on by and the natural topology induced by the given family of seminorms .
In what follows, we implicitly use the fact that both topologies are locally convex and translation invariant.
Step 1 (). We will prove that for each , . This together with the definition of (the biggest topology with this property) implies that . Let . By Theorem 18, it is enough to show that for all , is continuous. Since is compact, there are only finitely may ’s such that ; let us call them . So, for all ,
By assumption, is an interior smooth multiplication triple, so for each , the mapping from is continuous. Hence, must be continuous.
Step 2 (). Since is a locally convex topological vector space, there exists a separating family of seminorms whose corresponding natural topology is (see Theorem 15). We will prove that for all , is continuous. This together with the fact that is the smallest topology with this property shows that . Let . By Theorem 12, it is enough to prove that there exists such that
For all , we have . Since has compact support, only finitely many terms in the sum are nonzero, and so by the finite subadditivity of a seminorm, we get
Now, note that belongs to the normed space . Since is continuous and , we can conclude that is continuous. Thus, by Theorem 9, there exists a positive integer such that
It follows that for all ,
where
4. Spaces of Locally Sobolev Functions
Let , . Let be a nonempty open set in . We define
We equip with the natural topology induced by the separating family of seminorms (see Definition 10) where
When and are clear from the context, we may just write or instead of . It is easy to show that for all , is a seminorm on . The fact that the family of seminorms is separating will be proved in Theorem 103.
Remark 98.
Note that, by item (1) of Theorem 20, in if and only if in for all .
Remark 99.
Clearly if is a smooth multiplication triple, then .
An equivalent description of locally Sobolev functions is described in the following theorem.
Theorem 100.
Suppose that is a smooth multiplication triple. Then, is in if and only if for every precompact open set with there is such that .
Proof.
: suppose and let be a precompact open set such that . Let be such that on a neighborhood containing . Let . is a locally Sobolev function, so ; also clearly .
: suppose has the property that for every precompact open set with there is such that . Let . We need to show that . Note that is compact, so there exists a bounded open set such that
By assumption, there exists such that . It follows from the hypothesis of the theorem that . Clearly on . Therefore, .
5. Overview of the Basic Properties
Material of this section is mainly an adaptation of the material presented in the excellent work of Antonic and Burazin [17], which is restricted to integer order Sobolev spaces, and Peterson [24], which is restricted to Hilbert spaces . We have added certain details to the statements of the theorems and their proofs to ensure all the arguments are valid for both integer and noninteger order Sobolev-Slobodeckij spaces.
Definition 101.
If is a subset of with the property that,
then we say is an admissible family of functions.
Remark 102.
Note that if is an admissible family of functions, then for all , the set is also an admissible family of functions.
Theorem 103.
Let be an interior smooth multiplication triple. If is an admissible family of functions then
(1)
(2) The collection is a separating family of seminorms on
(3) The natural topology induced by the separating family of seminorms is the same as the topology of
Proof.
(1) Let be such that for all . We need to show that if , then . By the definition of , for all , there exists such that . Define
Clearly, , and since is continuous, is an open set. is an open cover of the compact set . So there exist points such that . If , then there exists such that and so . So the smooth function is nonzero on . Thus, on we have
Indeed, if we define
then is smooth with compact support in and
on the entire . Now, note that for each , is in (because by assumption for all ). So . Since and have compact support and is an interior smooth multiplication triple, it follows that
(2) Now, we prove that is a separating family of seminorms. We need to show that if has the property that for all , then . By definition of locally Sobolev functions, is an element of . So, in order to show that , it is enough to prove that for all , . We consider two cases:
Case 1. . Let be such that on a neighborhood containing . By assumption in and so it is zero in . Now we have
which is exactly what we wanted to prove
Case 2. . We claim that if for all , then for any , , and so this case reduces to the previous case. Indeed, if is an arbitrary element of , then by what was proved in item (1),
where by assumption, for each , is zero as an element of . Hence, in
(3) Finally, we show that the natural topology induced by is the same as the natural topology induced by . Obviously is a subset of , so it follows from the definition of natural topology induced by a family of seminorms (see Definition 10) that . In order to show that , it is enough to prove that for all , the map is continuous. By what was shown in item (1), we can write
where the implicit constant does not depend on . In the last inequality, we used the assumption that is an interior smooth multiplication triple. Now, it follows from Theorem 20 that is continuous
Lemma 104.
There exists an admissible family that has only countably many elements.
Proof.
Let be an exhaustion by compact sets for . For each , let be a nonnegative function such that on and outside . Clearly, is a countable admissible family of functions.
Corollary 105.
Let be an interior smooth multiplication triple. Considering Theorem 13, it follows from the previous lemma and Theorem 103 that is metrizable. Indeed, if is a countable admissible family, then
is a compatible translation invariant metric on .
Theorem 106.
Let be an interior smooth multiplication triple. Then, is a Frechet space.
Proof.
By Corollary 14, it is enough to show that equipped with the metric in (95) is complete. Note that all admissible families result in equivalent topologies in . So we can choose the functions ’s in the definition of to be the partition of unity introduced in Theorem 43. Now, suppose is a Cauchy sequence with respect to . In what follows, we will prove that converges to a distribution in . For now, let us assume this is true. We need to show that is an element of ; that is, we need to show that for all , .
It follows from the definition of that for each , is a Cauchy sequence in . Since is a Banach space, there exists in such that in . Note that , so in , and thus, for all we have
Hence, in . Since we can conclude that .
It remains to show that converges in . To this end, it is enough to show that for all , the sequence converges in (see Theorem 60). Let . Since is compact, there are only finitely many of ’s that are nonzero on the support of (see Theorem 43) which we denote by . So for each , . We have
all exist (since is Cauchy in , it is convergent in , and so it is convergent in ). Therefore, exists.
Theorem 107.
Let be a smooth multiplication triple (so we know that and is metrizable). Then, .
Proof.
Since both spaces are metrizable, it suffices to show that if in , then in . To this end, let be an arbitrary element of . We need to show that if in , then in . But this is a consequence of the fact that is a smooth multiplication triple.
Theorem 108.
Let be a nonempty open set in , , and . Then, is continuously embedded in ; i.e., the “identity map” from to is continuous.
Proof.
By Theorem 45, it is enough to show that if in , then in ; that is, for all , in .
Let and let denote multiplication by . Multiplication by smooth functions is a continuous linear operator on ([21]). So is continuous. The range of this map is in the subspace . So is continuous. However, . Hence, is continuous. As a consequence, since in , in . Finally, since , we can conclude that in .
Corollary 109.
Since , it follows that under the hypotheses of Theorem 108, is continuously embedded in .
Theorem 110.
Let be a smooth multiplication triple. Then, is dense in .
Proof.
Let . It is enough to show that there exists a sequence in such that in , i.e.,
First, note that, since is a smooth multiplication triple, for all , there exists a constant such that
Let be the admissible family introduced in the proof of Lemma 104. For each , there exists a number such that for all , on . So,
Clearly, by definition of , for each , , also has compact support, so (see Remark 83). Hence, for each , there exists such that . We claim that in . Indeed, given and , let be such that . Then, for , we have
Remark 111.
As a consequence, if is a smooth multiplication triple, then (equipped with the strong topology) is continuously embedded in . More precisely, the identity map is continuous with dense image, and therefore, by Theorem 27, the adjoint is a continuous injective map. We have
We usually identify with its image under and view as a subspace of . So, under this identification, we can rewrite the above equality as follows:
Theorem 112.
Let be a smooth multiplication triple. Then, is separable.
Proof.
is continuously embedded in and it is dense in . Since is separable, it follows from Lemma 95 that is separable.
As a direct consequence of the definitions, locally Sobolev functions and Sobolev functions with compact support are both subsets of the space of distributions. The next two theorems establish a duality connection between the two spaces. But first, we need to state a simple lemma.
Lemma 113.
Let and be two topological spaces. Suppose that is Hausdorff. Let and be two continuous functions that agree on a dense subset of . Then, everywhere. (So, in particular, in order to show that two continuous mappings from to are equal, we just need to show that they agree on some dense subset.)
Proof.
Suppose that there exists such that . Since is Hausdorff, there exist open neighborhoods and of and , respectively, such that . is a nonempty ( is in it) open set in so its intersection with is nonempty. Let be a point in the intersection of and . Clearly, and ; but since , we have . This contradicts the assumption that .
Theorem 114.
Suppose that and are smooth multiplication triples. Define the mapping by
where is any function in that is equal to 1 on a neighborhood containing the support of . Then,
(1) does not depend on the choice of
(2) For all , is indeed an element of
(3) is bijective
(4) Suppose is equipped with the strong topology. Then, the bijective linear map is a topological isomorphism; i.e., it is continuous with continuous inverse. So can be identified with as topological vector spaces
Proof.
(1) For the first item, it is enough to show that if is equal to zero on a neighborhood containing , then . Note that is not necessarily in , so we cannot directly apply the duality pairing identity stated in Remark 69. Let be sequence in such that in . Let be such that on and outside . By assumption, is a smooth multiplication triple and so in . Since elements of dual are continuous, we have
Note that is zero outside and in , so everywhere
(2) In order to show that is an element of , we need to prove that is linear and continuous. Linearity is obvious. In order to prove continuity, we need to show that for all , is continuous (see Theorem 37). Let and fix a function which is equal to 1 on a neighborhood containing . For all we have
which proves the continuity of the linear map
(3) In order to prove that is bijective, we give an explicit formula for the inverse. Recall that by definition, is a subspace of and by Remark 94, can also be viewed as a subspace of . More precisely, if we let be the “identity map” and be the adjoint of , then is a continuous injective linear map and
Moreover, if , then , and therefore, if , then and
Now, we claim that the image of is in and in fact is the inverse of . Let us first prove that the image of is in . Let . We need to show that for all , . To this end, we make use of Corollary 71. Let and let . For all , we have
which, by Corollary 71, proves that . Now, we prove is the inverse of . Note that for all and ,
Therefore, is identity. Next, we show that for all , . Note that and both are in and so they are continuous functions from to . Since is dense in , according to Lemma 113, it is enough to show that for all , we have .
(4) Let us denote the topology of by and the strong topology on by . Our goal is to show that and are both continuous maps. To this end, we make use of Theorem 20. Recall that is induced by the family of seminorms where . Also is induced by the family of seminorms where varies over all bounded sets in and
Step 1. Let be a bounded subset of . Since is bounded, there exists such that is bounded in (see Theorem 39; note that the topology of can be constructed as the inductive limit of where is an increasing chain of compact subsets of ). So there exists a constant such that for all , . Let be a function in which is equal to 1 on a neighborhood containing . For all , we have
It follows from Theorem 20 that is continuous
Step 2. Let . Let be a compact set whose interior contains . Since is a smooth multiplication triple, there exists a constant such that for all , we have .
We have
So, if we let be the ball of radius centered at in (clearly is a bounded set in ), we get
Corollary 115.
Suppose that and are both smooth multiplication triples. By the previous theorem, can be identified with . Also, by Remark 94, is continuously embedded in . Therefore, is continuously embedded in . Since is a Frechet space, it follows from Theorem 19 and Remark 22 that the preceding statement remains true even if we consider equipped with the weak topology. So,
Theorem 116.
Suppose that and are smooth multiplication triples. Define the mapping by
where is any function in that is equal to on a neighborhood containing the support of . Then,
(1) does not depend on the choice of
(2) For all , is indeed an element of
(3) is bijective
(4) Suppose is equipped with the strong topology. Then, the linear map is bijective and continuous. In particular, and are isomorphic vector spaces
Proof.
(1) Note that since is a smooth multiplication triple, is in . Also by assumption, is a smooth multiplication triple. Therefore, for each , , and hence, . So the pairing in the definition of makes sense. The fact that the output is independent of the choice of follows directly from Theorem 58
(2) Clearly, is linear. Also is continuous (so it is an element of ). The reason is as follows: for all , we have
That is, for all , we have . It follows from Theorem 20 that is continuous
(3) In order to prove that is bijective we give an explicit formula for the inverse. Recall that by Remark 111, can also be viewed as a subspace of . More precisely, if we let be the “identity map” and be the adjoint of , then is a continuous injective linear map and
Now, we claim that the image of is in and in fact is the inverse of . Let us first prove that the image of is in . is continuously and densely embedded in (continuity is proved in Theorem 108 and density is a direct consequence of the density of in ). Therefore, is indeed a subspace of and so elements of can be identified with distributions in that have compact support. It remains to show that if , then . To this end, we make use of Corollary 71. For all , we have
So, by Corollary 71, we can conclude that . In the above, we used the fact that , and so for , we have .
Now, we prove that is the inverse of . For all and , we have
Therefore, for all .
Now, we prove that is also the identity map. Considering Lemma 113, since is dense in , it is enough to show that for all and , . We have
which shows
(4) Let us denote the topology of by and the strong topology on by . Our goal is to show that is continuous. To this end, we make use of Theorem 20. Recall that is induced by the family of seminorms where (here, we are using the notation introduced in Theorem 97). Also is induced by the family of seminorms where varies over all bounded sets in and .
Let be a bounded subset of . Since is bounded, for all , the set is bounded in (see Theorem 16). Thus, for all , there exists a positive integer such that for all , . Recall that in the definition of denotes a fixed partition of unity. For each , let be a function in which is equal to 1 on a neighborhood containing the support of . For all , we have
where . Note that the inequality holds because has compact support and so only finitely many terms in the sum are nonzero, so we can use the subadditivity of the seminorm and linearity of .
It follows from Theorem 20 that is continuous
Remark 117.
According to the previous two theorems, we have the following:
(i) When is viewed as an element of , we have
where is any function in that is equal to 1 on a neighborhood containing
(ii) When is viewed as an element of , we have
where is any function in that is equal to on a neighborhood containing
Corollary 118.
Suppose that and are both smooth multiplication triples. As a direct consequence of the previous theorems, the bidual of is itself. So is semireflexive. It follows from Theorem 25 that is reflexive and subsequently its dual is reflexive.
Now, we put everything together to build general embedding theorems for spaces of locally Sobolev-Slobodeckij functions.
Theorem 119 (embedding theorem I).
Let be a nonempty open set in . If and are such that , then .
Proof.
We have
So, . Now, note that for all ,
So, it follows from Theorem 20 that the inclusion map from to is continuous.
Theorem 120 (embedding theorem II).
Let be a nonempty open set in that has the interior Lipschitz property. Suppose that and are such that for all bounded open sets with Lipschitz continuous boundary. If , further assume that is a smooth multiplication triple. If , further assume that is a smooth multiplication triple. Then, .
Proof.
Suppose and . Let be an open set in that contains and has Lipschitz continuous boundary. We have
Since can be any element of , we can conclude that if , then . In order to prove the continuity of the inclusion map, we can proceed as follows: let and choose as before.
So, it follows from Theorem 20 that the inclusion map from to is continuous.
A version of compact embedding for spaces with integer smoothness degree has been studied in [17]. In what follows, we will state the corresponding theorem and its proof for spaces of locally Sobolev-Slobodeckij functions.
Lemma 121.
Suppose that and are smooth multiplication triples. If converges weakly to in , then
Proof.
The proof is based on the following well-known fact:
Fact 1. Let be a topological space and suppose that is a point in . Let be a sequence in . If every subsequence of contains a subsequence that converges to , then .
Let . By Fact 1, it is enough to show that every subsequence of has a subsequence that converges weakly to in . Let be a subsequence of . We have
Since is reflexive, there exists a subsequence that converges weakly to some . To finish the proof, it is enough to show that . We have
In the first line, we used Theorem 30 and the fact that (see Corollary 115). In the second line, we used the fact that multiplication by smooth functions is a continuous operator on .
Similarly, since , it follows from Theorem 30 that
Consequently, as elements of and subsequently as elements of .
Theorem 122 (compact embedding).
Let be a nonempty open set in that has the interior Lipschitz property. Suppose that and are smooth multiplication triples. If , further assume that is a smooth multiplication triple. Moreover, suppose that , and are such that is compactly embedded in for all bounded open sets with Lipschitz continuous boundary. Then, every bounded sequence in has a convergent subsequence in .
Proof.
The proof makes use of the following well-known fact:
Fact 2. Let and be Banach spaces. Suppose that is a linear compact operator. If the sequence converges weakly (i.e., with respect to the weak topology) to in , then converges to (with respect to the norm of ) in .
Let be a bounded sequence in . By Theorem 33, since is a separable reflexive Frechet space, there exists and a subsequence such that in . We claim that converges to in , that is, for all , in . Suppose that and let . By Lemma 121, we have
So, by Theorem 35,
Let be an open bounded set in with Lipschitz continuous boundary such that . By Theorem 85, the restriction map from to is well-defined and continuous. It follows from Theorem 34 that this restriction map is weak-weak continuous. So in implies that in . By assumption the identity map from to is compact, so it follows from Fact 2 that in which subsequently implies in by Theorem 85.
6. Other Properties
The main results of this section do not appear to be in the literature in the generality appearing here, and they play a fundamental role in the study of the properties of differential operators between Sobolev spaces of sections of vector bundles on manifolds equipped with nonsmooth metric (see [15, 16]).
Theorem 123.
Let be a nonempty open set in , , and . Then, if and only if and for all , .
Proof.
Note that . Since , we have
Therefore,
Theorem 124.
Let be a nonempty open set in , , and . Then, if and only if for all .
Proof.
We prove the claim by induction on . For , we have
Now, suppose the claim holds for . For , we have
Theorem 125.
Let , , and . Suppose is a nonempty bounded open set in with Lipschitz continuous boundary. Then,
(1) the linear operator is well-defined and continuous
(2) for , the linear operator is well-defined and continuous
(3) for and , the linear operator is well-defined and continuous
(4) if , (i.e., the fractional part of is not equal to ), then the linear operator for is well-defined and continuous
Proof.
This is the counterpart of Theorem 86 for locally Sobolev functions. Here, we will prove the first item. The remaining items can be proved using a similar technique.
Step 1. First we prove by induction on that if , then . Let ; we need to show that . If , there is nothing to prove. If , there exists such that . We have
By assumption, , and so it follows from Theorem 86 that the first term on the right hand side is in . Also, since , the second term on the right hand side is in . Hence, . Now, suppose the claim holds for all . Suppose is a multi-index such that . Clearly, there exists such that where is a multi-index with . By the induction hypothesis, and so by the argument that was presented for the base case, we have
Step 2. In this step, we prove the continuity. Again, we use induction on . Let . Choose as in the previous step. For every , we have
On the right hand side, we have sum of two of the seminorms that define the topology of . It follows from item (2) of Theorem 20 that is continuous. Now, suppose the claim holds for all . Suppose is a multi-index such that . Clearly, there exists such that where is a multi-index with . We have
for some and in . It follows from item (2) of Theorem 20 that is continuous
Next, we want to establish a counterpart of Theorem 76 for locally Sobolev-Slobodeckij spaces. To this end, first we state and prove a simple lemma.
Lemma 126.
Let be a nonempty open subset of . Suppose and are such that If is continuous, then .
Proof by Contradiction.
Suppose . Since belongs to the complement of , which is an open set, there exists such that and . Since , there exists such that . is continuous, therefore there exists such that for all . But a.e. on . This contradicts the fact that .
Theorem 127.
Let be a nonempty bounded open set in with Lipschitz continuous boundary or . Suppose where . Then, has a continuous version.
Proof.
Let and be as in Theorem 43. Note that . For all , so by Theorem 76, there exists such that on where is a set of measure zero. Also by Lemma 126. Therefore for any , only finitely many of ’s are nonzero. So we may define by . Clearly, on where (so is a set of measure zero). Consequently . It remains to show that is indeed continuous. To this end, suppose in . We need to prove that . Let be such that . So intersects only finitely many of ’s; let us denote them by . Also since , there exists such that for all , . Hence,
Recall that is a finite sum of continuous functions and so it is continuous. Thus,
Remark 128.
In the above proof, the only place we used the assumption of being Lipschitz was in applying Theorem 76. We can replace this assumption by the weaker assumption that has the interior Lipschitz property. Then, since is compact, there exists with Lipschitz boundary that contains . Then, by Theorem 85, and so it has a continuous version . Since almost everywhere on and outside of the compact set , we can conclude that is in and it is almost everywhere equal to . We set . The rest of the proof will be exactly the same as before.
Theorem 129.
Let or be a bounded open set in with Lipschitz continuous boundary. Suppose and are such that
Then,
(1)
(2) for all , . In particular, if , then the mapping is a well-defined continuous linear map from to
Remark 130.
In the above theorem, since the locally Sobolev spaces on are metrizable, the continuity of the mapping
can be interpreted as follows: if in and in , then in . Also since is considered as a normed subspace of , we have a similar interpretation of the continuity of the mapping in item (2).
Proof.
(1) Suppose and . First, we show that is in . Clearly, the set is an admissible family of test functions. So in order to show that , it is enough to show that for all , is in . This is clearly true because , , and by assumption
In order to prove the continuity of the map , suppose in and in . We need to show that in . That is, we need to prove that for all ,
We have
By assumption, , so
(2) Suppose and . First, we show that is in . To this end, let be such that on a neighborhood containing . We have
Now, we prove the continuity. Suppose in and in . Let be as before. We have
This together with the assumption that implies in . Since and , we conclude that in
Remark 131.
In the above theorem, the assumption that is Lipschitz or was used only to ensure that we can apply Theorem 103 and to make sure that the locally Sobolev spaces involved are metrizable. For item (1), we can use the weaker assumption that , , and are interior smooth multiplication triples. For item (2), we just need to assume that is an interior smooth multiplication triple.
Corollary 132.
Let be the same as the previous theorem. If , then is closed under multiplication. Moreover, if
then
The next theorem plays a key role in the study of differential operators on manifolds equipped with nonsmooth metrics (see [15]).
Theorem 133.
Let or let be a nonempty bounded open set in with Lipschitz continuous boundary. Let and be such that . Let . Suppose for all and , . Then,
(1)
(2) moreover if for each and for all in , then in
Proof.
(1) By the Leibniz formula, we have
By assumption, for all , is in . Since , it follows from Corollary 132 that
(2) Since in , it again follows from Corollary 132 that for all ,
Thus, in
Theorem 134.
Let or let be a nonempty bounded open set in with the Lipschitz continuous boundary. Let and be such that .
(1) Suppose that and that for all where is some interval in . If is a smooth function, then
(2) Suppose that in and that for all and , where is some open interval in . If is a smooth function, then in
(3) If is a smooth function, then the map taking to is continuous from to
Proof.
The proof of part (1) generalizes the argument given in [33]. Let . First, we show that . To this end, we fix a multi-index and we show that (see Theorem 124).
It follows from the chain rule (and induction) that is a sum of the terms of the form
where and . It is a consequence of Theorem 129 that if and , then . As a consequence,
Considering this and the fact that , we have
for all . In particular, . Also, since is smooth and is continuous, . Therefore,
So, where . Now, for noninteger , we use a bootstrapping argument to show that in fact belongs to .
is smooth; therefore, . Also (note that ). By Theorem 129, we have
provided that
Therefore, for all such that . Consequently, for all such that (see Theorem 123). Now, we can repeat this argument by starting with “ is smooth, therefore for all such that .” This results in for all such that . Repeating this, a finite number of times shows that .
Our next goal is to prove items (2) and (3). First, we note that if then without loss of generality (WLOG) we may assume that . Indeed, the constant function is an element of . So,
where . Thus WLOG we may assume that .
Let , , and be as in Theorem 43. Clearly, is an admissible family of functions. Therefore, in order to show that in , it is enough to prove that
Let be those admissible test functions whose support intersects the support of . So,
Consequently,
Since in , for all , we have
and so,
Since , we have
Consequently, for the continuous representatives of and , we have uniform convergence. From this point, we work with these continuous versions. The continuous function attains its max and min on the compact set which we denote by and , respectively. Note that
So, and are in (that is, ). Let be such that . By (169) there exists such that
Let be such that on and outside of . Define by
Clearly, is a smooth function and . Moreover, on . Also, for all and , we have
Indeed, if , then both sides are equal to zero. If , then
and so,
is a smooth function and its value at is . Also, by assumption, . Therefore, the mapping from to is continuous. Hence,
That is,
So, we proved item (2). Finally, we note that is metrizable. So continuity of the mapping is equivalent to sequential continuity which was proved in item (2).
7. Conclusion
Sobolev-Slobodeckij spaces play a key role in the study of elliptic differential operators in nonsmooth setting. The study of certain differential operators between Sobolev spaces of sections of vector bundles on compact manifolds equipped with rough metric is closely related to the study of locally Sobolev functions on domains in the Euclidean space. In the present paper, we provided a self-contained rigorous study of certain fundamental properties of locally Sobolev-Slobodeckij spaces. In particular, by introducing notions such as “smooth multiplication triple” and “interior smooth multiplication triple,” we rigorously studied completeness, separability, nature of the dual space, general embedding results, continuity of differentiation, and invariance under composition by smooth functions for locally Sobolev-Slobodeckij spaces.
Acknowledgments
An earlier draft of this manuscript was posted to arXiv: https://arxiv.org/abs/1806.02188. MH was supported in part by NSF (Awards 1262982, 1620366, and 2012857). AB was supported by NSF (Award 1262982).
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