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- Banach spaces
1.1. Direct sums and complemented subspaces. As in the Linear Algebra courses you have done, we shall use the term “direct sum”, and the notation X ⊕ Y for two related, but distinct, notions. I hope this will not lead to confusion. Let us start by recalling the algebraic version. When U and W are vector spaces (over the same field) we may define U ⊕ W to be the cartesian product U × W , equipped with coordinate-wise vector space operations. (i.e. (u, w) + (u′, w′) = (u + u′, w + w′) etc). This is sometimes called an “exterior” direct sum. When U and V are subspaces of a vector space V , there is always a linear mapping U ⊕W → U +W ≤ V , given by (u, w) 7 → u+w. If this mapping is an isomorphism, we allow ourselves to say that the subspace U + W of V is a direct sum, and even to write U + W = U ⊕ W. As you will remember, this occurs exactly when X ∩ Y = { 0 }.
When X and Y are normed spaces, we introduce an analogous idea, but the addition of little topology. To start with, there are many obvious ways to equip the (external) direct sum X ⊕ Y with a norm. For instance,
‖(x, y)‖∞ = ‖x‖ ∨ ‖y‖;
‖(x, y)‖p = (‖x‖p^ + ‖y‖p)^1 /p^.
All these norms are easily seen to be equivalent. We write (X ⊕ Y )p to denote X ⊕ Y equipped with ‖ · ‖p. The ∞ norm is often the easiest to do calculations with.
Definition 1.1. Let X and Y be subspaces of a normed space Z such that X ∩ Y = { 0 }. If the linear mapping (x, y) 7 → x + y : (X ⊕ Y )∞ → X + Y ⊆ Z is a (topological) isomorphism, we shall say that X + Y is the topological direct sum of X and Y , and write X + Y = X ⊕ Y.
Theorem 1.2. Let X and Y be subspaces of a normed space Z such that X ∩ Y = { 0 } and let W be the sum X + Y ⊆ Z. The following are equivalent:
(1) W is a topological direct sum; (2) the linear projection P : W → X defined by P (x + y) = x, when x ∈ X and y ∈ Y , is continuous; (3) there exists a positive real number δ such that ‖x − y‖ ≥ δ whenever x ∈ X, y ∈ Y and ‖x‖ = 1; (4) the linear mapping R : W → W defined by R(x + y) = x − y (x ∈ X, y ∈ Y ) is bounded.
Proof. We note first that the operator S : (X ⊕ Y )∞; (x, y) 7 → x + y is always continuous, with ‖S‖ ≤ 2. At issue is whether S−^1 is continuous. Indeed, W is a topological direct sum if and only if this is so. Now we can write S−^1 (w) = (P (w), (I − P )(w)) and this enables us to see easily that
S−^1 continuous =⇒ P continuous with‖P ‖ ≤ ‖S−^1 ‖ P continuous =⇒ S−^1 continuous with‖S−^1 ‖ ≤ ‖P ‖ + 1.
This establishes (1) ⇐⇒ (2).
It is easy to see that (2) and (3) are equivalent (with δ = ‖P ‖−^1 ), and (4) is equivalent to (2) because R = 2P − I.
It is an important consequence of Baire’s “Category” Theorem that if a Banach space is the algebraic direct sum of two closed subspaces then it is automatically the topological direct sum of those subspaces. We shall prove this later in the course when we review Baire’s theorem and its applications.
Definition 1.3. Let X be a subspace of a normed space Z. We say that X is complemented in Z if there exists a subspace Y such that Z = X ⊕ Y. By Theorem 1.2 an equivalent condition is that here exist a continuous linear projection P : Z → X.
Notice that any complemented subspace is closed: indeed if P is a continuous projection from Z onto its subspace X then X = ker(I − P ), which is closed by continuity of P. It is an important result from Section B that any closed subspace of a Hilbert space is complemented.
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1.2. Quotient spaces.
Proposition 1.4. Let X be a normed space and let Z be a closed linear subspace. We may define a norm on the quotient space X/Z by setting ν(C) = inf{‖y‖ : y ∈ C} = inf{‖y‖ : Z + y = C},
when C is a coset of Z in X. Equivalently, writing C = Z + x,
ν(Z + x) = inf{‖x − z‖ : z ∈ Z}.
If π : X → X/Z is the canonical quotient mapping x 7 → Z + x then π is a bounded operator which maps the open unit ball of X onto the open unit ball of X/Y.
If X is a Banach space, so is X/Z.
Proof. The norm axioms N2 and N3 are easily verified. For instance, for cosets C, D we have
ν(C + D) ≤ ‖x + y‖ ≤ ‖x‖ + ‖y‖,
whenever x ∈ C and y ∈ D. Taking the infimum over all such x, y yields N2 for the function ν. For axiom N1, we consider C = Z + x such that ν(C) = 0 and are required to show that C equals Z, the zero element of X/Z. By the second version of the definition of ν, there exist zn ∈ Z such that ‖x − zn‖ ≤ 1 /n (n ∈ N). This implies that x ∈ Z. Since we are assuming the Z is closed, we have x ∈ Z and so Z + x = Z.
The assertions about π follow immediately from the definitions of the norm on X/Z.
We now assume that X is a Banach space. We shall use Proposition 0.21 to show that X/Z is complete. Suppose then that xn ∈ X are such that
j=1 ‖π(xn)‖^ converges. For each^ n^ we may choose^ x ′ n ∈^ X^ such that^ π(x ′ n) =^ π(xn) and ‖x′ n‖ ≤ 2 ‖π(xn)‖. The series
x′ n is thus absolutely convergent, and hence convergent in X. Continuity of the linear operator π now assures us that
π(x′ n) is convergent in X/Z.
Definition 1.5. We call the norm defined in Proposition 1.4 the quotient norm on X/Z. Notice that the canonical quotient mapping π has norm 1 and maps the open unit ball of X onto the open unit ball of X/Z.
By analogy with algebraic theories, such as those of groups and rings, it is natural to hope for some sort of “First Isomorphism Theorem” stating that the image T [X] of an arbitrary operator T ∈ L(X; Y ) should be isomorphic to the quotient X/ ker T. In our linear topological framework, this need not be true (though some important cases where it is true will appear in Chapter ??). The point is that, according to Definition 0.26, the sort of isomorphism we are supposed to be interested in is a linear homeomorphism, and in general a continuous linear operator from a normed space X onto a normed space Y need not be an isomorphism in this sense.
Proposition 1.6. Let X and Y be normed spaces and let T : X → Y be a bounded linear surjection. Then the linear bijection T˜ : X/ ker T → Y defined by T˜ (ker T + x) = T (x) is continuous and ‖ T˜ ‖ = ‖T ‖.
Proof. Let C be a coset of ker T in X, then T˜ (C) = T (x)
for all x ∈ C and so ‖ T˜ (C)‖ ≤ ‖T ‖ ‖x‖
for all x ∈ C. Taking the infimum, we see that T˜ is bounded with ‖ T˜ ‖ ≤ ‖T ‖. On the other hand, for any x we have
‖T (x)‖ = ‖ T˜ (ker T + x)‖ ≤ ‖T ‖ ‖ ker T + x‖ ≤ ‖ T˜ ‖ ‖x‖,
yielding ‖T ‖ ≤ ‖ T˜ ‖.
Definition 1.7. We shall say that a linear surjection T : X → Y is a quotient operator if the operator T˜ of the above proposition is an isomorphism. We shall say that T is an isometric quotient operator if T˜ is an isometry.
Not every linear surjection is a quotient operator.
Exercise 1.8. Let ‖ · ‖ and ‖ · ‖′^ be norms on a space X such that ‖ · ‖ dominates, but is not equivalent to, ‖ · ‖′. (For a suitable example look at Example 0.18.) Let X′^ denote X equipped with ‖ · ‖′^ and let T : X → X′^ be the identity mapping. Then T is a continuous surjection and ker T = { 0 }. But the induced bijection T˜ : X/ ker T → X′^ (which is essentially the same thing as T because ker T is trivial) is not an isomorphism because its inverse is not continuous.
Corollary 1.11. If X and Y are as in the above lemma and T : X → Y is a bounded linear operator such that T [BX (0; M ) contains a dense subset of BY (0; 1) then T [BX (0; M ) ⊇ BY (0; 1).
Theorem 1.12. Let X and Y be normed spaces and let T : X → Y be a bounded linear operator, with dual operator T ∗^ : Y ∗^ → X∗. Then
(1) T is an isomorphic embedding if and only if T ∗^ is a quotient operator; (2) T is a quotient operator if and only if T ∗^ is an isomorphic embedding.
Let Z be a closed linear subspace of a normed space X. Then there are natural isometric isomorphisms (X/Z)∗^ ∼= Z⊥ and Z∗^ ∼= X∗/Z⊥.
Proof. (1) Assume first that T is an isomorphic embedding, with ‖x‖ ≤ M ‖T (x)‖ for all x ∈ X. Let Z be the image T [X] ⊆ Y. For any f ∈ X∗^ may well-define h ∈ Z∗^ by h(T (x)) = f (x); we note that |h(T (x))| = |f (x)| ≤ ‖f ‖ ‖x‖ ≤ M ‖f ‖ ‖T (x)‖ so that ‖h‖ ≤ M ‖f ‖. Now by the Hahn–Banach theorem we mat extend h to g ∈ Y ∗^ with g Z = h and ‖g‖ = ‖h‖. This yields g ∈ Y ∗^ with T ∗(g) = f and ‖g‖ ≤ M ‖f ‖ and shows that T ∗^ is a quotient operator.
Conversely, if T ∗^ is a quotient operator, and x ∈ X we use Corollary 0.32 to obtain f ∈ X∗^ with ‖f ‖ = 1 and f (x) = ‖x‖. The quotient condition tells us that there exists g ∈ Y ∗^ with T ∗(g) = f and ‖g‖ ≤ M. Thus
‖x‖ = f (x) = g(T (x)) ≤ ‖g‖ ‖T (x)‖ ≤ M ‖T (x)‖,
showing that T is an isomorphic embedding.
(2) Now suppose that T is a quotient operator and that y ∈ Y ∗; Let π : X → X/Z be the canonical quotient mapping. For each h ∈ (X/Z)∗, the composite h ◦ π is in X∗^ with
‖h ◦ π‖ ≤ ‖h‖‖π‖ = ‖h‖.
Moreover, if z ∈ Z we have h(π(x)) = h(0) = 0, so that h ◦ π ∈ Z⊥. So we have a linear map π∗^ : (X/Z)∗^ → Z⊥^ satisfying π∗‖ ≤ 1. We shall see that this mapping is surjective and isometric. Suppose that g ∈ Z⊥; since g(x) = g(x′) whenever x − x′^ ∈ Z, we may well-define a linear mapping h : X/Z → R by h(π(x)) = g(x). If we show that h is bounded, with ‖h‖ ≤ ‖g‖ we shall have finished the first part. For any x ∈ X, we have
h(π(x)) = g(x) = g(x′) ≤ ‖g‖‖x′‖,
whenever x − x′^ ∈ Z. Taking the infimum, we get
h(π(x)) ≤ ‖g‖‖π(x)‖,
which is what we want. The proof of the converse assertion (where we assume that T ∗^ is an embedding) relies on the Hahn-Banach Separation Theorem and is deferred until later in the course.
For the proof of the second part, consider the restriction operator R : X∗^ → Z∗^ defined by R(f ) = f Z. Clearly, R is linear and ‖R‖ ≤ 1. By the Hahn-Banach extension theorem, R is surjective. So (“by algebra”) there is a linear bijection R˜ : X∗/ ker R → Z∗^ given by R˜(Z⊥^ + f ) = R(f ). We want to show that R˜ is isometric.
We finish with a sketch proof of a useful theorem. Providing the details is left as an exercise to the reader.
Theorem 1.13. Every separable Banach space is isometrically isomorphic to a quotient of ` 1.
Proof. Let Y be a separable Banach space and let D = {yn : n ∈ N} be a countable dense subset of the unit sphere SY. There is a bounded linear operator T : 1 → Y such that T (en) = yn for all n. This maps the open unit ball of 1 onto the open unit ball of Y and hence is an isometric quotient operator.
1.3. Biduals, reflexivity and completions. For any normed space X, the dual space X∗^ is a Banach space and does itself also have a dual space X∗∗. We call this the bidual of X. In the case of a finite-dimensional space we know from Linear Algebra that X∗∗^ can be identified with X. This is sometimes, but not always, true for infinite dimensional spaces.
Proposition 1.14. For any normed space X there is a linear isometric embedding JX : X → X∗∗^ defined by
(JX (x))(f ) = f (x) (f ∈ X∗, x ∈ X).
Proof. Trivial verifications (c.f. Section A Algebra) show that JX (x) is a linear mapping X∗^ → R and that the map x 7 → JX (x) is linear. We need to check that each JX (x) is continuous and that ‖JX (x)‖ = ‖x‖. First, we have
|(JX (x))(f )| = |f (x)| ≤ ‖f ‖ ‖x‖,
so that JX (x) is indeed continuous with ‖JX (x) ≤ ‖x‖. To get the other inequality, we consider any x ∈ X and apply Corollary 0.32 obtaining f ∈ X∗^ with ‖f ‖ = 1 and f (x) = ‖x‖. We now have
‖JX (x)‖ ≥ |(JX (x))(f )| = ‖x‖.
Definition 1.15. Let X be a normed space. A completion of X is a pair (J, Y ) where Y is a Banach space and J : X → Y is an isometric embedding with J[X] = Y.
Theorem 1.16. Every normed space has a completion.
Proof. It is usual to prove this result by taking a quotient space of the vector space of all Cauchy sequences in X. This is an important technique, but an easier approach is to take J = JX and Y = JX [X] ⊆ X∗∗.
Definition 1.17. We say that a normed space X is reflexive if JX : X → X∗∗^ is surjective.
Examples 1.18.
(1) From 0.25 we see that p is reflexive for 1 < p < ∞ but that c 0 is not reflexive (because c∗∗ 0 identifies with∞). (2) Every Hilbert space is reflexive, and so, in particular, L 2 (μ) is reflexive for any measure μ. (3) We shall see in due course that Lp(μ) is reflexive whenever 1 < p < ∞.
Since and dual space (and hence any bidual space) is always complete, we see that a reflexive normed space must be a Banach space.
Theorem 1.19. Let X be a normed space. The following are equivalent:
(1) X is reflexive; (2) X∗^ is reflexive and X is complete.
Proof. First suppose that X is reflexive and let φ ∈ X∗∗∗. We have to find f ∈ X∗^ with JX∗ (f ) = φ. An obvious candidate is to put f = φ ◦ JX. If ξ ∈ X∗∗^ we have ξ = JX (x) for a suitably chosen x ∈ X and so
(JX∗^ (f ))(ξ) = ξ(f ) = (JX (x))(f ) = f (x) = φ(JX (x)) = φ(ξ).
Thus φ = JX∗^ (f ) as required.
Now assume that X is complete and that X∗^ is reflexive. Since JX is an isometry, the image JX [X] is complete and hence closed in X∗∗. To show that JX [X] is the whole of X∗^ it is enough to show that if φ ∈ X∗∗∗^ and φ JX [X]= 0 then φ = 0. (This is Proposition 0.34.)
By the assumed reflexivity of X∗, φ can be written JX∗^ (f ) for some f ∈ X∗. If φ JX [X]= 0 then we have
f (x) = (JX (x))(f ) = φ(JX (x)) = 0
for all x ∈ X. Thus f = 0 and so φ = JX∗ (f ) = 0.
Corollary 1.20. The space ` 1 is not reflexive.
