TVS II c Gabriel Nagy Topological Vector Spaces II: Constructing New Spaces Out of Old Ones Notes from the Functional Analysis Course (Fall 07 - Spring 08) Convention. Throughout this note K will be one of the fields R or C. All vector spaces mentioned here are over K. In this section we outline several important general methods for constructing topological vector spaces. Notations. Suppose X is a vector space, equipped with a linear topology T, and Y ⊂ X is a linear subspace. Consider the quotient space X /Y of equivalence classes modulo Y. (So x ∼ x0 means x − x0 ∈ Y.) Let π : X → X /Y denote the quotient map. Proposition-Definition 1. With the above notations, the collection −1 π∗T = {A ⊂ X /Y : π (A) ∈ T} has the following properties. (i) π∗T is a linear topology on the quotient space X /Y. (ii) The topology π∗T is the strongest among all topologies that make π continuous. (iii) Given a topological space Z, a map φ : X /Y → Z is continuous relative to π∗T, if and only if the composition φ ◦ π : X → Z is continuous, relative to T. (iv) The topology π∗T is the unique topology which makes the map π both continuous and open1. The topology π∗T is called the quotient topology. Proof. The fact that π∗T is a topology with properties (ii)-(iii) is well-known. (Nothing about linearity is assumed.) Next, let us prove that π is open. Start with some open set A ∈ T, and let us show that π(A) ∈ π∗T. Of course, what we have to show then is the fact that π−1φ(A) is open (in T). But this is quite clear, since [ π−1φ(A) = A + Y = (A + y). y∈Y We now prove the continuity of the addition and scalar multiplication maps: (X /Y) × (X /Y) 3 (v1, v2) 7−→ v1 + v2 ∈ X /Y, (1) K × (X /Y) 3 (α, v) 7−→ αv ∈ X /Y, (2) 1 This means that if A ⊂ X is open, then π(A) is open. 1 For the continuity of (1) start with a pair (v1, v2) ∈ X /Y, and some open set A ∈ π∗T containing v1 + v2, and we need to construct two open sets A1, A2 ∈ π∗T, such that v1 ∈ A1, v2 ∈ A2, and A1+A2 ⊂ A. To get these, we pick two vectors x1, x2 ∈ X , such that v1 = π(x1) −1 and v2 = π(x2), so that v1 + v2 = π(x1 + x2), and we consider the set B = π (A), which is open in X and contains x1 + x2. Using the linearity of T, there exists open sets B1, B2 ∈ T, such that x1 ∈ B1, x2 ∈ B2, and B1 + B2 ⊂ B. Now we can (use the fact that π is open) simply take A1 = π(B1) and A2 = π(B2). The continuity of the multiplication is proved the exact same way. The last point to be proven is the uniqueness statement in part (iv). Assume S is another linear topology on X /Y which makes π both continuous and open. On the one hand, if we −1 start with some A ∈ π∗T, then π (A) is open in X , and then the openness assumption on π (relative to S) forces ππ−1(A) to belong to S. Since π is surjective, we have the equality −1 π π (A) = A. This way we have shown the inclusion π∗T ⊂ S. On the other hand, by property (ii) we know that the continuity of π relative to S forces the other inclusion S ⊂ π∗T, so we must have equality π∗T = S. Remark 1. Using the notations as above, the quotient topology on X /Y is Hausdorff, if and only if the singleton set {0} is closed in X /Y (see Exercise 2 from the previous section), which in turn is equivalent to the condition that Y is closed in X . Based on this observation, one has a simple method of “Hausdorffization” of a linear topology on X : replace X with its quotient modulo Y = {0} – the closure of the singleton set {0} (see TVS I, Exercise 5). Example 1. Consider the spaces Lp (X, A, µ) introduced in the previous section, equipped K with the semi-metric topology. It is not hard to show that the closure of {0} is simply the space N = {f : X → K f measurable, f = 0, µ-a.e.}. The corresponding quotient space – now denoted by Lp (X, A, µ) is now a topological vector K space. The quotient topology on Lp constructed this way is referred to as the standard metric topology. Proposition-Definition 2. Suppose X and Y are vector spaces, T is a linear topology on Y and ψ : X → Y is a linear map (i) The collection ψ∗T = ψ−1(A : A ∈ T is a linear topology on X . (ii) ψ∗T is the weakest among all topologies on X , that makes the map ψ : X → Y contin- uous. (iii) Given a topological space Z, a map φ : Z → X is continuous relative to ψ∗T, if and only if ψ ◦ φ : Z → Y is continuous. (iv) If T is Hausdorff, then so is ψ∗T. The topology ψ∗T is called the ψ-pull-back of T. 2 Proof. Statements (ii)-(iv) are well known from General Topology. To prove the linearity of addition and scalar multiplication, we use (iii), which reduces the proof to continuity of the maps A : X × X 3 (x, y) 7−→ ψ(x + y) ∈ Y, M : K × X 3 (α, x) 7−→ ψ(αx) ∈ Y. But this is clear, becase we can present these maps as compositions ψ×ψ A : X × X −−→Y × Y −−−−−−−−→Yaddition in Y Id×ψ multiplication in Y M : K × X −−−→ K × Y −−−−−−−−−−−→Y involving continuous functions. The above construction can be generalized as follows. Proposition-Definition 3. Suppose (Yi)i∈I is a family of vector spaces, with each Yi equipped a linear topology Ti. Suppose also a vector space X is given, together with a system ψi S ∗ Ψ = (X −→Yi)i∈I of linear maps. Let T be the topology on X , which has S = i∈I ψi Ti as a sub-base2. (i) T is a linear topology on X . (ii) T is the weakest among all topologies that make all maps ψi : X → Yi continuous. (iii) Given a topological space Z, a map φ : Z → X is continuous relative to T, if and only if all the compositions ψi ◦ φ : Z → Yi are continuous, relative to Ti. (iv) If all Ti are Hausdorff, then so is T. The topology T is referred to as the joint Ψ-pull-back. Proof. Statements (ii)-(iv) are well-known from Topology. To prove tha continuity of the addition and scalar multiplication, we argue as before using (iii), which reduces the problem to the continuity of the maps Ai : X × X 3 (x, y) 7−→ ψi(x + y) ∈ Yi Mi : K × X 3 (α, x) 7−→ ψi(αx) ∈ Yi. But the continuity of these maps is again clear, since we can write them as compositions ψi×ψi addition in Yi Ai : X × X −−−→Yi × Yi −−−−−−−−→Yi Id×ψi multiplication in Yi Mi : K × X −−−→ K × Yi −−−−−−−−−−−→Yi involving continuous functions. 2 Equivalently, T consists of arbitrary unions of finite intersections of sets in S. 3 Below are two important applications of (joint) pull-backs. Example 2. Suppose T is a linear topology on a vector space Y and X ⊂ Y is a linear subspace. Consider then the inclusion map ι : X ,→ Y. The pull-back ι∗T, hereafter denoted by T , is simply given as X T X = {A ∩ X : A ∈ T}, and is referred to as the induced topology. By Proposition-Definition 2 (iv), it Y is a topo- logical vector space, then so is Y (with its induced topology). Example 3. Suppose (Xi)i∈I is a family of vector spaces, each Xi equipped with a linear 3 Q topology Ti. If we consider the product space X = i∈I Xi, together with the coordinate πi map system Π = (X −→Xi)i∈I , and we let T be the joint Π-pull-back, then T is the product topology. In this special case, property (iv) from Proposition-Definition 3 has a slightly improved version: (iv) T is Hausdorff, if and only if all Ti are Hausdorff. In other words, X is a topological vector space (with the product topology), if and only if all the Xi are topological vector space. Exercise 1. Give an example of non-Hausdorff linear topology T on Y = K2 and of a linear subspace {0} 6= X ⊂ Y, such that the induced topology T X is Hausdorff. Exercise 2. Suppose T is a linear topology on a vector space Y and X ⊂ Y is a closed linear subspace, such that the induced topology T X is Hausdorff. Show that T is also Hausdorff. Exercise 3. Consider the product space Y = KN equipped with the natural product topology T. For p > 0 regard `p as a linear subspace of Y. Prove that the induced topology K T `p is strictly weaker than the standard metric topology TDp , that is, one has a strict K inclusion T `p ( TDp . K Comment. It is a nice coincidence that the product space construction from Topology matches with the linear structure. Unfortunately, the other important construction from Topology – the topological sum – does not have a suitable linear counterpart. The correct notion of a linear sum of a family (Xi)i∈I of vector space is defined as follows.