N. L. Carothers's A Short Course on Banach Space Theory PDF

By N. L. Carothers

ISBN-10: 0521842832

ISBN-13: 9780521842839

It is a brief path on Banach house conception with distinct emphasis on sure features of the classical idea. particularly, the path makes a speciality of 3 significant themes: The basic conception of Schauder bases, an creation to Lp areas, and an advent to C(K) areas. whereas those themes may be traced again to Banach himself, our basic curiosity is within the postwar renaissance of Banach house concept led to by way of James, Lindenstrauss, Mazur, Namioka, Pelczynski, and others. Their stylish and insightful effects are worthy in lots of modern examine endeavors and deserve larger exposure. in terms of necessities, the reader will want an hassle-free figuring out of sensible research and at the least a passing familiarity with summary degree thought. An introductory direction in topology might even be useful, notwithstanding, the textual content encompasses a short appendix at the topology wanted for the path.

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Let M = {(x, 0) : x ∈ R} ⊂ R2 . Show that there are uncountably many subspaces N of R2 such that R2 = M ⊕ N . 22 Preliminaries 13. Let M be a finite-dimensional subspace of a normed linear space X . Show that there is a closed subspace N of X with X = M ⊕ N . In fact, if M is nontrivial, then there are infinitely many distinct choices for N . [Hint: Given a basis x1 , . . , xn for M, find f 1 , . . ] 14. Let M and N be closed subspaces of a Banach space X with M ∩ N = {0}. Prove that M + N is closed in X if and only if there is a constant C < ∞ such that x ≤ C x + y for every x ∈ M, y ∈ N .

Prove that every proper subspace M of a normed space X has empty interior. If M is a finite dimensional subspace of an infinite dimensional normed space X , conclude that M is nowhere dense in X . 4. Prove that B(X, Y ) is complete whenever Y is complete. 5. If Y is a dense linear subspace of a normed space X , show that Y ∗ = X ∗ , isometrically. 6. Prove Riesz’s lemma: Given a closed subspace Y of a normed space X and an ε > 0, there is a norm one vector x ∈ X such that x − y > 1 − ε for all y ∈ Y .

Then, 1 = | x j , e |2 + | xk , x j |2 and, hence, | x j , e | < 1. Finally, notice that 1 = |x ∗j (x j )| = | x j , e | · |x ∗j (e)|, which implies that |x ∗j (e)| = 1/| x j , e | > 1. This contradicts the fact that x ∗j = 1. Exercises 33 Exercises 1. Let X be a Banach space with basis (xn ). We know that the expression |||x||| = supn Pn x defines a norm on X that is equivalent to · . Show that under ||| · |||, each Pn has norm one. That is, we can always renorm X so that (xn ) has basis constant K = 1.

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A Short Course on Banach Space Theory by N. L. Carothers

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