# Hilbert Spaces with Applications by Lokenath Debnath-

By Lokenath Debnath-

Overseas reprint version. third variation. Paperback. released in China. Contents in ENGLISH and completely comparable as US version. (Follower dealers, comparable to "gofrl" or "WhiteCoffee", may well promote you different affordable variation books)

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Additional info for Hilbert Spaces with Applications

Example text

Consider the space C ([a, b]) with the norm defined as f = Is this a Banach space ? 45. Show that L(f )(x) = C ([0, 1]) into itself. x 0 f (t)dt b a |f (t)|dt. defines a continuous linear mapping from 46. Give an example of a linear mapping from a normed space into a normed space which is not continuous. 47. Prove that a linear mapping from a normed space into a normed space is continuous if and only if it maps bounded sets to bounded sets. 38 Chapter 1 Normed Vector Spaces 48. 4. 49. 11. 50. Show that B (E1 , E2 ) is a vector space.

If F is continuous at every x ∈ E1 , then we simply say that F is continuous. 2). Here we only make the following simple but useful observation. 3. The norm on a normed space E is a continuous mapping from E into R. Indeed, if xn − x → 0, then | xn − x | ≤ xn − x → 0. Continuity can be described in many different ways. The conditions in the following theorem characterize continuity in terms of open and closed sets. The proof of the theorem is left as an exercise. 4. Let F : E1 → E2 . The following conditions are equivalent: (a) F is continuous; (b) The inverse image F −1 (U) of any open subset U of E2 is open in E1 ; (c) The inverse image F −1 (S) of any closed subset S of E2 is closed in E1 .

It turns out that finite dimensional normed spaces can be characterized by the property that the unit ball is compact. The proof of this fact is based on a theorem that is usually referred to as Riesz’s lemma. 33. (Riesz’s lemma) Let X be a closed proper subspace of a normed space E. For every ε ∈ (0, 1) there exists an xε ∈ E such that xε = 1 and xε − x ≥ ε for all x ∈ X. Proof: Let z ∈ E\X and let d = infx∈X z − x . Since X is closed, d > 0. If ε ∈ (0, 1), then d ≤ z − x0 ≤ d ε for some x0 ∈ X.