real_analysis_III

# real_analysis_III - 3 Spaces L p 1 In this part we fix a...

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Unformatted text preview: 3 Spaces L p 1. In this part we fix a measure space (Ω , A ,μ ) (we may assume that μ is complete), and consider the A-measurable functions on it. 2. For f ∈ L 1 (Ω ,μ ) set k f k 1 = k f k L 1 = k f k L 1 (Ω ,μ ) = Z Ω | f | dμ. It follows from the above inequalities that for c ∈ C 1 k f + g k 1 ≤ k f k 1 + k g k 1 , k cf k 1 = | c |k f k 1 . With more work we derive that k f k 1 = 0 ⇐⇒ f = 0 μ- a . e .. Factorising by the subspace of functions equal μ-a.e., or redefining the symbol = between to functions, we conclude that L 1 (Ω ,μ ) is a normed vector space . 3. Definition of a normed space from B, Ch.2 (B-2). Prove that spaces in B-2, Examples 1, parts 1, 2, 4, 5, 9 (cases l 1 , l ∞ , and c only ), 13, 14, 15–19 are normed. Sets B ( x ,r ) , B ( x ,r ) , and S ( x ,r ) in a normed space V . Prove that B ( x ,r ) is the closure of B ( x ,r ) . 4. Complete (Banach) spaces, B-2. Which spaces from the above examples are complete? Provide proofs. 5. We show that L 1 (Ω ,μ ) is a Banach space. Theorem 1 For any sequence { f j } , f j ∈ L 1 (Ω ,μ ) , satisfying k f j- f k k 1 → , j,k → ∞ , there exists f ∈ L 1 (Ω ,μ ) such that k f j- f k 1 → , j → ∞ . Proof. 1. First we need to construct f , whose existence is claimed by the theorem. Utilising the Cauchy condition find a subsequence { j n } such that k f j n +1- f j n k 1 < 2- n ....
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real_analysis_III - 3 Spaces L p 1 In this part we fix a...

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