5. Banach Spaces

Definitions and Elementary Properties

Definition. A normed linear space is a vector space N over R or C in which to every vector x there corresponds a real number denoted ||x|| called the norm of x, with the properties

1. ||x|| ≥ 0, and ||x|| = 0 iff x = 0
2. ||x + y|| ≤ ||x|| + ||y||
3. ||αx|| = |α| ||x||   for every scalar α

It is easy to verify that N is a metric space with respect to d defined by d(x,y) = ||x - y||. A Banach space is a complete normed linear space.

Examples. Let p be a real number such 1 ≤ p < ∞. Denote by l_p(n) the space of all n-tuples x = (x₁, ... , x_n) of scalars, with the norm defined by

||x||_p = (∑_{i = 1, ... , n} |x_i|^p)^1/p

We could also take n = ∞, that is, l_p = l_p(∞) is the space of all sequences x = {x₁, x₂, ... , x_n, ...} such that ∑_{i = 1, ... , ∞} < ∞, with the norm defined by

||x||_p = (∑_{i = 1, ... , ∞} |x_n|^p)^1/p

More generally, if X is a measurable space with measure m, we could consider the space of all measurable f such that |f(x)|^p is integrable, with

||f||_p = (∫_X |f(x)|^p dm(x))^1/p

Definition. A Hilbert space is a complex Banach space whose norm arises from an inner product, that is, a complex function (x, y) of vectors such that

1. (αx + βy, z) = α (x, z) + β (y, z)
2. (x, y)^- = (y, x),    where ^- denotes complex conjugation
3. (x, x) = ||x||²

Clearly the above properties imply

(x, αy + βz) = α^- (x, y) + β^- (x, z)

Example. The space l₂, with the inner product of x = {x₁, x₂, ... , x_n, ...} and y = {y₁, y₂, ... , y_n, ... } defined by

(x, y) = ∑_{i = 1, ... , ∞} x_iy_i^-

The Schwarz Inequality. If x and y are any two vectors in a Hilbert space then |(x, y)| ≤ ||x|| ||y||.

Proof. when y = 0, the result is clear, for both sides vanish. When y ≠ 0, the inequality is equivalent to |(x, y/||y||)| ≤ ||x||. Thus we have to prove that if ||y|| = 1 then |(x, y)| ≤ ||x|| for all x. But

0 ≤ ||x - (x, y) y||² = (x - (x, y) y, x - (x, y) y)

= (x, x) - (x, y) (x, y)^- - (x, y) (x, y)^- + (x, y) (x, y)^- (y, y)

= ||x||² - |(x, y)|².

Remark.This inequality in fact is needed to establish that ||x + y|| ≤ ||x|| + ||y||.

Definition. if X and Y are normed linear spaces and T: X → Y is a linear transformation, we say that T is bounded if there is an M such that ||Tx|| ≤ M ||x|| for all x ∈ X. In that case

||T|| = inf {M: ||Tx|| ≤ M ||x||, x ∈ X}

is called the norm of T.

Proposition. The following conditions are equivalent:

1. T is continuous at 0.
2. T is bounded.
3. T is continuous everywhere.

Proof. (1) ⇒ (2): If T is continuous at 0, then for every ε > 0, there is a δ > 0 such that ||x|| < δ implies ||Tx|| < ε. But then for arbitrary x, ||Tx|| = ||x|| ||T(x/||x||)|| = δ^-1 ||x|| ||T(δx/||x||)||. thus T is bounded with ||T|| ≤ εδ^-1.

(2) ⇒ (3). we have ||T(x - y)|| ≤ ||T|| ||x - y|| from the definition. Thus Tx → Ty as x → y and T is continuous at any arbitrary point y.

(3) ⇒ (1) is obvious.

Linear Transformations, Dual Space, and Banach Algebras.

Let B (X, Y) be the set of bounded linear transformations on X into Y. We shall show that B (X, Y) is a normed linear space which is a Banach space if Y is a Banach space.

Let T₁, T₂ ∈ B (X, Y). Define T₁ + T₂ by

(T₁ + T₂)(x) = T₁(x) + T₂(x)

and aT by

(aT)(x) = aT(x)

for every x ∈ X. Then

||T₁ + T₂|| ≤ ||T₁|| + ||T₂||

and

||aT|| = |a| ||T||

Moreover, ||T|| = 0 implies T = 0, so B (X, Y) is a normed linear space.

Suppose now that Y is complete. Let {T_n} be a Cauchy sequence in B (X, Y). Then, for any x ∈ X,

||T_m x - T_n x|| ≤ ||T_m - T_n|| ||x||

so that {T_n x} is a Cauchy sequence in Y. Since Y is complete we may define

Tx = lim_n T_n x   for x ∈ X.

It is obvious that T is linear. Since {T_n}, {||T_n||} is bounded, i.e., there there is an M such that ||T_n|| ≤ M for all n. For every x, ||T_n x|| ≤ M ||x||. So, ||Tx|| ≤ M ||x||. Hence T ∈ B (X, Y).

Let ε > 0. Choose n₀ so that ||T_m - T_n|| < ε if m, n > n₀. Then

||T_m x - T_n x|| < ε ||x|| for m, n > n₀, x ∈ X.

Letting n tend to infinity, we get

||T_m x - Tx|| ≤ ε ||x|| for m > n₀, x ∈ X.

This implies ||T_m - T|| ≤ ε for m > n₀, so that {T_n} converges to T in B (X, Y).

In particular, if y is one-dimensional, we write X' instead of B (X, Y). If X is a normed vector space, the space X' of continuous linear functionals on X, with norm

||x'|| = sup {|x'(x)|: ||x|| = 1}

= inf {M: |x'(x)| ≤ M ||x||, x ∈X}

is called the dual of X. We then have the

Corollary. The dual of a normed linear space is a Banach space.

Definition. Let X be a Banach space with an additional operation of multiplication (x, y) |→ xy defined so that X is a ring. If

||xy|| ≤ ||x|| ||y||

then X is said to be a Banach algebra.

It is easy to see that for any Banach space X, B (X, X) is a Banach algebra if the multiplication operation is taken to be composition of operators.

Two Important Theorems

(A) The Hahn-Banach Theorem. Let M be a linear subspace of a normed linear space, and let f be a functional defined on M. Then f can be extended to a functional f₀ defined on the whole space N, such that ||F₀|| = ||f||.

Lemma. Let M be a linear subspace of a normed linear space N and let f be a functional defined on M. If x₀ is a vector not in M, and if

M₀ = M + [x₀]

is the linear space spanned by M and x₀, then f can be extended to a functional f₀ such that ||f₀|| = ||f||.

Proof. Assume first that N is a real normed space. Without loss of generality, we can take ||f|| = 1 (f can be replaced by f/||f||). Each y ∈ M₀ can be uniquely be written y = x + αx₀, x ∈ M. we can define a functional f₀ on M₀ by f₀ (x + αx₀) = f(x) + αr₀ for any real number r₀. We have now to choose r₀so that |f₀ (x + αx₀)| ≤ ||x + αx₀||, that is

- ||x + αx₀|| ≤ f(x) + αr₀ ≤ ||x + αx₀||

or

- f(x) - ||x + αx₀|| ≤ αr₀ ≤ - f(x) + ||x + αx₀||

which is equivalent to

- f(x/α) - ||x/α + x₀ ≤ r₀ ≤ - f(x/α) + ||x/α + x₀||

(1)

Next, for any vectors x₁, x₂ ∈ M

f(x₂) - f(x₁) ≤ ||f|| ||x₂ - x₁|| = ||x₂ - x₁||

≤ ||x₂ - x₀|| + ||x₁ - x₀||

so

- f(x₁) - ||x₁ + x₀|| ≤ - f(₂) + ||x₂ + x₀||

(2)

Setting

a = sup {- f(x) - ||x + x₀||: x ∈ M}

b = inf {-f(x) + ||x + x₀||: x ∈ M}

Then a ≤ b by (2). Choosing r₀ such that a ≤ r₀ ≤ b, inequality (1) is satisfied, completing the proof.

Let now N be complex, and f be a complex valued functional on M with ||f|| = 1. Then f(x) = g(x) + ih(x) where g and h are real valued functionals on the real space M. Since ||f|| = 1, we have ||g|| ≤ 1. Furthermore, it is easy to see that h(x) = - g(ix) (since f(ix) = if(x)), so f(x) = g(x) - ig(ix). We can extend g to a real valued functional g₀ on the real linear space M₀ so that ||g₀|| = ||g|| and define f₀(x) = g₀(x) - ig₀(x). then f₀ is an extension of f. It remains to prove that ||f₀|| = 1, that is, if x ∈ M₀ and ||x|| = 1, then |f₀(x)| ≤ 1. If f₀(x) is real, f₀(x) = g₀(x) and ||g₀|| ≤ 1. If f₀(x) is complex, write f₀(x) = re^iθ, r > 0. Then

|f₀(x)| = r = e^-iθ f₀(x) = f₀(e^-iθ x)

and our conclusion follows from ||e^iθ x|| = ||x|| = 1 and the fact that f₀(e^-iθ x) is real.

Proof of Hahn-Banach Theorem. The set of all extensions of f to functionals g with the same norm defined on subspaces which contain M is clearly a partially ordered set with respect to the following relation: g₁ ≤ g₂ means that the domain of g₁ is contained in the domain of g₂, and g₂(x) = g₁(x) for all x in the domain of g₁. It is easy to see that the union of any chain of extensions is also an extension and is therefore an upper bound for the chain. Zorn's lemma now implies that there exists a maximal extension f₀. The domain of f₀ must be the entire space N, for otherwise it could be extended further by our lemma and would not be maximal.

Corollary. Let N be a normed linear space and N^* its dual. Then, for x ∈ N

||x|| = sup {|f(x)|: f ∈ N^*, ||f|| = 1}

Hence, for fixed x ∈ N the mapping f |→ f(x) is a bounded linear functional on N^* of norm ||x||, that is, N is isometrically imbedded in N^**, the dual of its dual.

Proof. By the Hahn-Banach theorem, we can always find an f ∈ N^* of norm 1 such that f(x) = ||x||. On the other hand, |f(x)| ≤ ||x|| if ||f|| = 1.

(B) The Uniform Boundedness Theorem. Let B be a Banach space and N a normed linear space. If [T_i} is a non-empty set of continuous linear transformations of B into N with the property that {T_i(x)} is a bounded subset of N for each vector x in B, then {||T_i||} is a bounded set of numbers; that is, {T_i} is bounded as a subset of B (B, N).

Proof. The proof makes use of Baire's Category Theorem.

For each positive integer n, the set

F_n = {x ∈ B: ||T_i(x)|| ≤ n for all i}

is clearly a closed subset of B, and by our assumption we have

B = ∪_{i = 1, ... , ∞}F_n

Since B is complete, Baire's Theorem shows that one of the F_n's, say F_n0 has non-empty interior, and thus contains a closed sphere S₀ with center x₀ and radius r₀ > 0. Then ||T_i(S₀)|| ≤ for all i. It is clear that the translate S₀ - x₀ = {y - x₀: y ∈ S₀} is the closed sphere with radius r₀ centered at the origin, so (S₀ - x₀)/r₀ is the closed unit sphere. Since x₀ ∈ S₀, evidently ||T_i (S₀ - x₀)|| ≤ 2n₀. Thus ||T_i(x)|| ≤ 2n₀/r₀ for ||x|| ≤ 1 and ||T_i|| ≤ 2n₀/r₀ for every i, completing the proof.

Note. The theory of normed linear spaces was first developed by Stefan Banach in a 1931 monograph in Polish which was translated into French in the following year as Théorie des opérations linéaires. Banach also discovered the Contraction Mapping Theorem.

Link. Lawrence Baggett"s online textbook "Functional Analysis" found at http://spot.colorado.edu/~baggett/functional.html gives a comprehensive exposition of the subject.

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