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

By N. L. Carothers

This can be a brief path on Banach house idea with certain emphasis on definite features of the classical thought. particularly, the path specializes in 3 significant subject matters: The undemanding conception of Schauder bases, an advent to Lp areas, and an advent to C(K) areas. whereas those issues may be traced again to Banach himself, our fundamental curiosity is within the postwar renaissance of Banach house thought led to via James, Lindenstrauss, Mazur, Namioka, Pelczynski, and others. Their stylish and insightful effects are worthwhile in lots of modern examine endeavors and deserve larger exposure. in terms of necessities, the reader will desire an user-friendly realizing of sensible research and no less than a passing familiarity with summary degree idea. An introductory direction in topology may even be useful, although, the textual content encompasses a short appendix at the topology wanted for the direction.

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Prove that [xn ] is isometric to c0 and complemented in c0 by a norm one projection. 3. Let ( f n ) be a disjointly supported norm one sequence in C[0, 1]. Prove that [ f n ] is isometric to c0 . Is [ f n ] complemented in C[0, 1]? Will these arguments carry over to disjointly supported sequences in L ∞ [0, 1]? 4. Let (xn ) be a basis for a Banach space X , and let (yn ) be a sequence in a Banach space Y . Suppose that n an yn converges in Y whenever n an x n converges in X , where (an ) is a sequence of scalars.

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 . If X is infinite dimensional, use Riesz’s lemma to construct a sequence of norm one vectors (xn ) in X satisfying xn − xm ≥ 1/2 for all n = m. n on a vector space, prove that 7. Given linear functionals f and (gi )i=1 n n ker f ⊃ i=1 ker gi if and only if f = i=1 ai gi for some a1 , . . , an ∈ R. 8. Given T ∈ B(X, Y ), show that ker T = ⊥ (range T ∗ ), the annihilator of range T ∗ in X , and that ker T ∗ = (range T )⊥ , the annihilator of range T in Y ∗ .

We next formulate a “test” for basic sequences; this, too, is due to Banach. 2. A sequence (xn ) of nonzero vectors is a basis for the Banach space X if and only if (i) (xn ) has dense linear span in X , and (ii) there is a constant K such that n m ai xi ≤ K i=1 ai xi i=1 for all scalars (ai ) and all n < m. ) Proof. The forward implication is clear; note that for n < m we have n m ai xi = Pn m ai xi i=1 i=1 ≤ sup P j j ai xi . i=1 Now suppose that (i) and (ii) hold and let S = span{xi : i ≥ 1}.

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