By Rene Erlin Castillo, Humberto Rafeiro

ISBN-10: 3319300326

ISBN-13: 9783319300320

ISBN-10: 3319300342

ISBN-13: 9783319300344

Introduces reader to fresh subject matters in areas of measurable functions

Includes component of difficulties on the finish of every bankruptcy

Content allows use with mixed-level classes

Includes non-standard functionality areas, viz. variable exponent Lebesgue areas and grand Lebesgue spaces

This booklet is dedicated solely to Lebesgue areas and their direct derived areas. particular in its sole commitment, this booklet explores Lebesgue areas, distribution services and nonincreasing rearrangement. furthermore, it additionally bargains with vulnerable, Lorentz and the more moderen variable exponent and grand Lebesgue areas with massive aspect to the proofs. The publication additionally touches on simple harmonic research within the aforementioned areas. An appendix is given on the finish of the booklet giving it a self-contained personality. This paintings is perfect for academics, graduate scholars and researchers.

Topics

Abstract Harmonic Analysis

Functional research

**Read Online or Download An Introductory Course in Lebesgue Spaces PDF**

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**Additional resources for An Introductory Course in Lebesgue Spaces**

**Example text**

Proof. Let us check equality. , f p = B A g p . Moreover, f +g p = B g+g A = p B+A g A p = B g A p+ g p = f p+ g p. When p = ∞ and p = 1 the inequality is immediate, as well as when f p = g p = 0. Suppose that 1 < p < ∞ and f p = α = 0 and g p = β = 0, then there are functions f0 and g0 such that | f | = α f0 and |g| = β g0 with f0 p = g0 p = 1. 4) = (α + β ) (λ f0 (x) + (1 − λ )g0 (x)) ≤ (α + β ) p [λ ( f0 (x)) p + (1 − λ )(g0 (x)) p ]. , f + g ∈ L p (X, A , μ ). Finally, f +g p p ≤( f p+ g p)p ≤ f p+ g p, thus f +g p which ends the proof.

15 (a) The space c0 is the closure of c00 in ∞ . (b) The space c and c0 are Banach spaces. (c) The space c00 is not complete. 21. Show that (s, ρ ) is a complete metric space, where s is the set of all sequences x = (x1 , x2 , . ) and ρ is given by ρ (x, y) = ∞ |xk − yk | 1 ∑ 2k 1 + |xk − yk| . 22. Let p (w), p ≥ 1 be the set of all real sequences x = (x1 , x2 , . ) such that ∞ ∑ |xk | p wk < ∞ k=1 where w = (w1 , w2 , . ) and wk > 0. Does N : N (x) := ∞ p (w) ∑ |xk | −→ R given by 1 p p wk k=1 defines a norm in p (w)?

17) with z = 1p . 18. 17 is a two line proof if we remember that ˆ∞ xα −1 dx = B(α , β ) (1 + x)α +β 0 and the fact that B(1 − α , α ) = π sin πα , 0 < α < 1, see Appendix C. Before stating and proving the Hilbert inequality we need to digress into the concept of double series. Let xk, j j,k∈N be a double sequence, viz. a real-valued function x : N × N → R. We say that a number L is the limit of the double sequence, denoted by lim xk, j = L, k, j→∞ if, for all ε > 0 there exists n = n(ε ) such that |xk, j − L| < ε whenever k > n and j > n.