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4 Hausdorff Topological Vector Spaces. Quotient Topological Vector Spaces. Continuous Linear Mappings
4 HausdorfT Topological Vector Spaces. Quotient Topological Vector Spaces. Continuous Linear Mappings Throughout this chapter, we denote by E a T VS over the field of complex numbers.
Hausdorff Topological Vector Spaces A topological space X is said to be Hausdorff if, given any two distinct points x and y of X, there is a neighborhood U of x and a neighborhood V of y which do not intersect, i.e., such that U n V = 0. A very important property of HausdorfT topological spaces is the so-called uniqueness of the limit: A filter on a H a u s d o g topological space X converges to at most one point. Indeed,suppose that a filter 9 on X would converge to twodistinct points x and y . Let U (resp. V) be a neighborhood of x (resp. y ) such that U n V = 0. But both U and V must belong to 9, which demands that their intersection be nonempty! I n a Hausdorff space, any set consisting of a single point is closed (there are topological spaces with the same property which are not Hausdorff; but such spaces are not TVS, as will be seen). A T V S E is Hausdorff if, given any two distinct points x and y , there is a neighborhood Uof x which does not containy. As a matter of fact, we have the following result:
PROPOSITION 4.1. A TVS E is Hausdorfl if and only if to every point x # 0 there is a neighborhood U of 0 such that x $ U. T h e necessity of the condition is trivial. Suppose it is satisfied. Let x, y be two distinct points of E, which means that x - y 0. T h e n there is a neighborhood U of 0 such that x - y 6 U. Choose a balanced
+
31
32
[Part I
TOPOLOGICAL VECTOR SPACES
+
neighborhood of 0, V, such that V V C U (Theorem 3.1, p. 21). Since V is balanced, we have - V = V, hence V - V C U. Suppose that the intersection
is nonempty, and let z be one of its points: z = x with x’, y’ E V. We have x
-y
= y’
- x‘
E
+
XI,
z
=y
+ y’,
V - V € u,
which contradicts our choice of U. Thus (4.1) must be empty. PROPOSITION 4.2. In a TVS E, the intersection of all neighborhoods of the origin is a vector subspace of E, which is the closure of the set (0). Let us first prove that the intersection of all the neighborhoods of the origin, which we denote temporarily by N, is a vector subspace of E. Let U be an arbitrary neighborhood of the origin, x, y two elements of N, and a, /3 two complex numbers, which are not both equal to zero. V C U,assume furthermore Let V be a neighborhood of 0 such that V that V is balanced (Definition 3.2, Theorem 3.1, p. 21). As x, y EN, we have x, y € ( 2 + p y v ;
+
+
+
+
+
+
hence, ax /3y E .(a2 p2)-l V /3(a2 p2)-l V C V V C U. This implies that ax /3y E N since U is arbitrary. Let x belong to N. Then every neighborhood U of 0 contains x, which can also be written
+
O€(-U) fx.
But (- U ) + x is an arbitrary neighborhood of x (since multiplication of vectors by -1 is a homeomorphism, that is to say a bicontinuous one-to-one mapping onto). Thus every neighborhood of x contains the origin, which means that XE{O) (see p. 8). Conversely, suppose that an arbitrary neighborhood of x contains 0; such an arbitrary neighborhood can be written - U x, where U is an arbitrary neighborhood of 0; and 0 E - U x is equivalent to x E U , which means that x E N.
+
+
Q.E.D.
For a TVS E to be Hausdor-, it is necessary and su#cient COROLLARY. that the set (0) be closed in E , or that the complement of the origin be open in E.
Chap. 4-31
QUOTIENT TOPOLOGICAL VECTOR SPACES
33
Indeed, to say that {0} is closed in E is equivalent to saying that N = {0}or that no point x # 0 may belong to all the neighborhoods of 0. An important consequence of the corollary of Proposition 4.2 is the next result:
4.3. Let f,g, be two continuous mappings of a topological PROPOSITION space X into a HausdorfJ TVS E. The set A in which f and g coincide, Indeed, A is the preimage of the closed set {0} C E under the continuous mapping x f ( x ) - g(x). -+
PROPOSITION 4.4. Let X, E , f,g be as in Proposition 4.3.Iff and g are equal on a dense subset Y of X , they are equal everywhere in X . Indeed, f = g on a closed subset of X (Proposition 4.3)containing Y.
Quotient Topological Vector Spaces Let M be a vector subspace of E, and let us consider the quotient vector space (p. 15) EIM and the canonical map : E -+ E / M which assigns to every x E E its class $(x) modulo M. We know that the mapping 4 is linear. On E we have a topology (since E is a TVS). We may define then, in a canonical way, a topology on EIM which is called the quotient topology on EIM. As always, we say what the filter of neighborhoods of the origin in EIM is going to be: it is simply the image under C# of the filter of neighborhoods of the origin in E. This is the same as saying the following: a subset 0 of EIM is a neighborhood of zero for the quotient topology if and only if there is a neighborhood U of zero in E whose image i.e., 0= $(U). The neighborhoods of zero in under C# is equal to 0, EIM are the direct images under 4 of the neighborhoods of 0 in E. Note that C#J transforms neighborhoods of a point into neighborhoods of a point. This is not true in general about continuous functions: the preimage of a neighborhood under a continuous function is a neighborhood, but nothing is said about the image. On the other hand, we do not know a priori if 4 is continuous. But it is easy to see that this is indeed so: let 0 be a neighborhood of the origin in EIM; there is a neighborhood U of zero in E such that 4(U) = 0, hence U C + - l ( 0 ) , which proves that +-l( 0) is a neighborhood of zero in E. Going to open sets, we see that 4 transforms open sets into open sets and the preimages of open sets under 4 are open sets.
34
[Part I
TOPOLOGICAL VECTOR SPACES
I t is not true, in general, that the direct images of closed sets, under 4, are closed sets. A familiar counterexample is the following one. Consider in the plane R2 the hyperbola {(xl , x2) E R2; xlx2 = l}. Take for M one of the coordinate axes. Then EIM can be identified with the other coordinate axis and rp with the orthogonal projection on it; all these identifications are also valid for the topologies. The hyperbola above is closed in R2 but its image under 4 is the complement of the origin on a straight line, which is open. The student may easily verify the following point: the quotient topology on EIM is thejnest topology on EIM such that 4 is continuous. From our definition it follows immediately that the quotient topology on E / M is compatible with the linear structure of E / M (see p. 20). PROPOSITION 4.5. Let E be a TVS, and M a vector subspace of E. The two following properties are equivalent: (a) M is closed; (b) EIM is Hausdo$ In view of the corollary of Proposition 4.2, (b) can be restated as saying that the complement of the origin is open in E / M . But the complement of the origin is exactly the image under 4 of the complement of M, and 4 maps open sets into open sets, and is continuous, whence the equivalence of (a) and (b). COROLLARY.The TVS El@) is Hausdor-. The TVS El{@ is said to be the HausdorjJ topological vector space associated with the TVS E. When E itself is Hausdod, 4 : E + (canonical mapping) is one-to-one onto, since then = {0},and E/{@ is identified with E.
{q
Continuous Linear Mappings Let E, F be two TVS, and f a linear map of E into F. We suppose that F is HausdorfT and that f is continuous, in the usual sense (see p. 11 et seq.). Then the kernel (p. 16) of f is closed. Indeed, Kerf is the preimage of the set {0}C F, which is closed when F is Hausdod. Of course, Kerf might be closed also when F is not HausdorfT (Example 1, f = 0; Example 2, f is one-to-one and E is Hausdofi; in this case, Kerf = {0} is closed in E).
Chap. 4-51
CONTINUOUS LINEAR MAPPINGS
35
*
Let us consider the usual diagram (p. 16): E F-m ’If* a I
where i is the natural injection, 4 the canonical map, and f the unique linear map which makes the diagram commutative.
if and only if the map f is continuous. Suppose f continuous and let SZ be an open subset in F or in Im f (an open set in Im f is the intersection of an open set in F with Imf). The preimage of D underfis equal to the image, under 4, of the preimage of SZ under f. By hypothesis, f-’(SZ) is open, and 4 transforms open sets into open sets; therefore the preimage of SZ under 3 is open. Q.E.D. When both 4 andf are continuous, so is f = f o 4.
PROPOSITION 4.6. The map f is continuous
In general, the inverse of 3, which is well defined on Im f , since 3 is one-to-one, will not be continuous; in other words, f will not be bicontinuous.
Dejinition 4.1. I f f is continuous and if the inverse o f f , dejined on Im f (this subspace of F being equipped with the topology induced by F ) , is also continuous, we say that f is a homomorphism. If furthermore f is one-to-one, we say that f is an isomorphism of E into F or onto Im f . The set of continuous linear maps of a TVS E into another TVS F will be denoted by L(E;F). Of course, it is a subset of Y ( E ;F), the vector space of linear maps, continuous or not, from E into F. It is evident that L(E; F) is a vector subspace of Y ( E ;F), hence is a vector space, for the natural addition and multiplication by scalars, of functions. When E = C,one denotes usually L(E;F) by E’ and calls this vector the dual of E (sometimes, the topological dual of E, in order to underline the difference between E’ and E*, the algebraic dual of E; see p. 17). Naturally, E’ is a vector subspace of E*; E’ is the vector space of the continuous linear functionals, or continuous linear forms, on F. Elements of E’ will usually be denoted by x’, y’, etc. The vector spaces E’ and L(E; F) will play an important role in the forthcoming and will be equipped with various topologies. We conclude this section with a property of continuous linear mappings which is well known, and reflects the “homogeneity” of the topology in a TVS:
36
TOPOLOGICAL VECTOR SPACES
PROPOSITION 4.7. Let E, F be two TVS, u a linear map of E into F.
The mapping u is continuous i f (and only if!) u is continuous at the origin.
V
Indeed, an arbitrary neighborhood of u ( x ) (x E E ) in F is of the form u(x), where V is a neighborhood of 0 in F. Since u is linear, we have
+
u-1(
v + u(x)) 3 u-I( V ) + x.
If u is continuous at the origin, rl( V) is a neighborhood of zero in E. Exercises 4.1. Consider the topological vector space T(X;E) defined in Exercise 3.1. Prove that it is Hausdoff if and only if the spaces Emare Hausdoff for all x.
4.2. Prove that the product of a family of TVS E, if every E, is Hausdorff.
(a E A) is
Hausdoff if and only
-
4.3. Let M be a linear subspace of a TVS E. Another linear subspace, N, of E is called an algebraic supplementary of M in E if the mapping (x, y ) x y of M x N into E is an isomorphism onto E for the vector space structure; N is called a topological supplementary of M if (x, y ) x y is an isomorphism of M x N onto E for the TVS structure. One says then that E is the topological direct sum of M and N. Prove the equivalence of the following two properties:
..+ +
+
(a) N is a topological supplementary of M; (b) the restriction to N of the canonical mapping of E onto E/M is an isomorphism (for the TVS structure) of N onto E/M. Prove that M has at least one topological supplementary in E if there is a continuous linear map p of E onto M such that p o p = p (then p(x) = x is equivalent with x E M). 4.4. Let f be a continuous linear map of a TVS E onto another one, F. Prove the equivalence of the following properties: (a) Kerf has a topological supplementary in E (cf. Exercise 4.3); (b) there is a continuous linear map g o f F into E such thatf o g = identity ofF. 4.5. Let E be a TVS, and M a linear subspace of E. For every TVS G, the restriction to M of the continuous linear mappings f : E 4 G defines a linear mapping of L(E, G ) into L(M, G). Prove that this mapping is onto for every TVS G if and only if M has a topological supplementary in E.
Hausdorff Topological Vector Spaces A topological space X is said to be Hausdorff if, given any two distinct points x and y of X, there is a neighborhood U of x and a neighborhood V of y which do not intersect, i.e., such that U n V = 0. A very important property of HausdorfT topological spaces is the so-called uniqueness of the limit: A filter on a H a u s d o g topological space X converges to at most one point. Indeed,suppose that a filter 9 on X would converge to twodistinct points x and y . Let U (resp. V) be a neighborhood of x (resp. y ) such that U n V = 0. But both U and V must belong to 9, which demands that their intersection be nonempty! I n a Hausdorff space, any set consisting of a single point is closed (there are topological spaces with the same property which are not Hausdorff; but such spaces are not TVS, as will be seen). A T V S E is Hausdorff if, given any two distinct points x and y , there is a neighborhood Uof x which does not containy. As a matter of fact, we have the following result:
PROPOSITION 4.1. A TVS E is Hausdorfl if and only if to every point x # 0 there is a neighborhood U of 0 such that x $ U. T h e necessity of the condition is trivial. Suppose it is satisfied. Let x, y be two distinct points of E, which means that x - y 0. T h e n there is a neighborhood U of 0 such that x - y 6 U. Choose a balanced
+
31
32
[Part I
TOPOLOGICAL VECTOR SPACES
+
neighborhood of 0, V, such that V V C U (Theorem 3.1, p. 21). Since V is balanced, we have - V = V, hence V - V C U. Suppose that the intersection
is nonempty, and let z be one of its points: z = x with x’, y’ E V. We have x
-y
= y’
- x‘
E
+
XI,
z
=y
+ y’,
V - V € u,
which contradicts our choice of U. Thus (4.1) must be empty. PROPOSITION 4.2. In a TVS E, the intersection of all neighborhoods of the origin is a vector subspace of E, which is the closure of the set (0). Let us first prove that the intersection of all the neighborhoods of the origin, which we denote temporarily by N, is a vector subspace of E. Let U be an arbitrary neighborhood of the origin, x, y two elements of N, and a, /3 two complex numbers, which are not both equal to zero. V C U,assume furthermore Let V be a neighborhood of 0 such that V that V is balanced (Definition 3.2, Theorem 3.1, p. 21). As x, y EN, we have x, y € ( 2 + p y v ;
+
+
+
+
+
+
hence, ax /3y E .(a2 p2)-l V /3(a2 p2)-l V C V V C U. This implies that ax /3y E N since U is arbitrary. Let x belong to N. Then every neighborhood U of 0 contains x, which can also be written
+
O€(-U) fx.
But (- U ) + x is an arbitrary neighborhood of x (since multiplication of vectors by -1 is a homeomorphism, that is to say a bicontinuous one-to-one mapping onto). Thus every neighborhood of x contains the origin, which means that XE{O) (see p. 8). Conversely, suppose that an arbitrary neighborhood of x contains 0; such an arbitrary neighborhood can be written - U x, where U is an arbitrary neighborhood of 0; and 0 E - U x is equivalent to x E U , which means that x E N.
+
+
Q.E.D.
For a TVS E to be Hausdor-, it is necessary and su#cient COROLLARY. that the set (0) be closed in E , or that the complement of the origin be open in E.
Chap. 4-31
QUOTIENT TOPOLOGICAL VECTOR SPACES
33
Indeed, to say that {0} is closed in E is equivalent to saying that N = {0}or that no point x # 0 may belong to all the neighborhoods of 0. An important consequence of the corollary of Proposition 4.2 is the next result:
4.3. Let f,g, be two continuous mappings of a topological PROPOSITION space X into a HausdorfJ TVS E. The set A in which f and g coincide, Indeed, A is the preimage of the closed set {0} C E under the continuous mapping x f ( x ) - g(x). -+
PROPOSITION 4.4. Let X, E , f,g be as in Proposition 4.3.Iff and g are equal on a dense subset Y of X , they are equal everywhere in X . Indeed, f = g on a closed subset of X (Proposition 4.3)containing Y.
Quotient Topological Vector Spaces Let M be a vector subspace of E, and let us consider the quotient vector space (p. 15) EIM and the canonical map : E -+ E / M which assigns to every x E E its class $(x) modulo M. We know that the mapping 4 is linear. On E we have a topology (since E is a TVS). We may define then, in a canonical way, a topology on EIM which is called the quotient topology on EIM. As always, we say what the filter of neighborhoods of the origin in EIM is going to be: it is simply the image under C# of the filter of neighborhoods of the origin in E. This is the same as saying the following: a subset 0 of EIM is a neighborhood of zero for the quotient topology if and only if there is a neighborhood U of zero in E whose image i.e., 0= $(U). The neighborhoods of zero in under C# is equal to 0, EIM are the direct images under 4 of the neighborhoods of 0 in E. Note that C#J transforms neighborhoods of a point into neighborhoods of a point. This is not true in general about continuous functions: the preimage of a neighborhood under a continuous function is a neighborhood, but nothing is said about the image. On the other hand, we do not know a priori if 4 is continuous. But it is easy to see that this is indeed so: let 0 be a neighborhood of the origin in EIM; there is a neighborhood U of zero in E such that 4(U) = 0, hence U C + - l ( 0 ) , which proves that +-l( 0) is a neighborhood of zero in E. Going to open sets, we see that 4 transforms open sets into open sets and the preimages of open sets under 4 are open sets.
34
[Part I
TOPOLOGICAL VECTOR SPACES
I t is not true, in general, that the direct images of closed sets, under 4, are closed sets. A familiar counterexample is the following one. Consider in the plane R2 the hyperbola {(xl , x2) E R2; xlx2 = l}. Take for M one of the coordinate axes. Then EIM can be identified with the other coordinate axis and rp with the orthogonal projection on it; all these identifications are also valid for the topologies. The hyperbola above is closed in R2 but its image under 4 is the complement of the origin on a straight line, which is open. The student may easily verify the following point: the quotient topology on EIM is thejnest topology on EIM such that 4 is continuous. From our definition it follows immediately that the quotient topology on E / M is compatible with the linear structure of E / M (see p. 20). PROPOSITION 4.5. Let E be a TVS, and M a vector subspace of E. The two following properties are equivalent: (a) M is closed; (b) EIM is Hausdo$ In view of the corollary of Proposition 4.2, (b) can be restated as saying that the complement of the origin is open in E / M . But the complement of the origin is exactly the image under 4 of the complement of M, and 4 maps open sets into open sets, and is continuous, whence the equivalence of (a) and (b). COROLLARY.The TVS El@) is Hausdor-. The TVS El{@ is said to be the HausdorjJ topological vector space associated with the TVS E. When E itself is Hausdod, 4 : E + (canonical mapping) is one-to-one onto, since then = {0},and E/{@ is identified with E.
{q
Continuous Linear Mappings Let E, F be two TVS, and f a linear map of E into F. We suppose that F is HausdorfT and that f is continuous, in the usual sense (see p. 11 et seq.). Then the kernel (p. 16) of f is closed. Indeed, Kerf is the preimage of the set {0}C F, which is closed when F is Hausdod. Of course, Kerf might be closed also when F is not HausdorfT (Example 1, f = 0; Example 2, f is one-to-one and E is Hausdofi; in this case, Kerf = {0} is closed in E).
Chap. 4-51
CONTINUOUS LINEAR MAPPINGS
35
*
Let us consider the usual diagram (p. 16): E F-m ’If* a I
where i is the natural injection, 4 the canonical map, and f the unique linear map which makes the diagram commutative.
if and only if the map f is continuous. Suppose f continuous and let SZ be an open subset in F or in Im f (an open set in Im f is the intersection of an open set in F with Imf). The preimage of D underfis equal to the image, under 4, of the preimage of SZ under f. By hypothesis, f-’(SZ) is open, and 4 transforms open sets into open sets; therefore the preimage of SZ under 3 is open. Q.E.D. When both 4 andf are continuous, so is f = f o 4.
PROPOSITION 4.6. The map f is continuous
In general, the inverse of 3, which is well defined on Im f , since 3 is one-to-one, will not be continuous; in other words, f will not be bicontinuous.
Dejinition 4.1. I f f is continuous and if the inverse o f f , dejined on Im f (this subspace of F being equipped with the topology induced by F ) , is also continuous, we say that f is a homomorphism. If furthermore f is one-to-one, we say that f is an isomorphism of E into F or onto Im f . The set of continuous linear maps of a TVS E into another TVS F will be denoted by L(E;F). Of course, it is a subset of Y ( E ;F), the vector space of linear maps, continuous or not, from E into F. It is evident that L(E; F) is a vector subspace of Y ( E ;F), hence is a vector space, for the natural addition and multiplication by scalars, of functions. When E = C,one denotes usually L(E;F) by E’ and calls this vector the dual of E (sometimes, the topological dual of E, in order to underline the difference between E’ and E*, the algebraic dual of E; see p. 17). Naturally, E’ is a vector subspace of E*; E’ is the vector space of the continuous linear functionals, or continuous linear forms, on F. Elements of E’ will usually be denoted by x’, y’, etc. The vector spaces E’ and L(E; F) will play an important role in the forthcoming and will be equipped with various topologies. We conclude this section with a property of continuous linear mappings which is well known, and reflects the “homogeneity” of the topology in a TVS:
36
TOPOLOGICAL VECTOR SPACES
PROPOSITION 4.7. Let E, F be two TVS, u a linear map of E into F.
The mapping u is continuous i f (and only if!) u is continuous at the origin.
V
Indeed, an arbitrary neighborhood of u ( x ) (x E E ) in F is of the form u(x), where V is a neighborhood of 0 in F. Since u is linear, we have
+
u-1(
v + u(x)) 3 u-I( V ) + x.
If u is continuous at the origin, rl( V) is a neighborhood of zero in E. Exercises 4.1. Consider the topological vector space T(X;E) defined in Exercise 3.1. Prove that it is Hausdoff if and only if the spaces Emare Hausdoff for all x.
4.2. Prove that the product of a family of TVS E, if every E, is Hausdorff.
(a E A) is
Hausdoff if and only
-
4.3. Let M be a linear subspace of a TVS E. Another linear subspace, N, of E is called an algebraic supplementary of M in E if the mapping (x, y ) x y of M x N into E is an isomorphism onto E for the vector space structure; N is called a topological supplementary of M if (x, y ) x y is an isomorphism of M x N onto E for the TVS structure. One says then that E is the topological direct sum of M and N. Prove the equivalence of the following two properties:
..+ +
+
(a) N is a topological supplementary of M; (b) the restriction to N of the canonical mapping of E onto E/M is an isomorphism (for the TVS structure) of N onto E/M. Prove that M has at least one topological supplementary in E if there is a continuous linear map p of E onto M such that p o p = p (then p(x) = x is equivalent with x E M). 4.4. Let f be a continuous linear map of a TVS E onto another one, F. Prove the equivalence of the following properties: (a) Kerf has a topological supplementary in E (cf. Exercise 4.3); (b) there is a continuous linear map g o f F into E such thatf o g = identity ofF. 4.5. Let E be a TVS, and M a linear subspace of E. For every TVS G, the restriction to M of the continuous linear mappings f : E 4 G defines a linear mapping of L(E, G ) into L(M, G). Prove that this mapping is onto for every TVS G if and only if M has a topological supplementary in E.