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24 01 2024 E TS

Esame completo di Real and functional analysis per il corso di Mathematical Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

Real and functional analysisEsame completo

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Esame completo di Real and functional analysis per il corso di Mathematical Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Mathematical Engineering - A.Y. 2023-24 Real and Functional Analysis - Written exam - January 24, 2024 Exercise 1. Consider the measure space ([0, 1], λ([0, 1])) with the Lebesgue measure. De ne the sequence of functions {fn}n∈N by fn(x) = n  1 − e−n2x sin2(x)  1 + n + x e− 1 n , x ∈ [0, 1], n ∈ N. (1) Study the convergence a.e. of the sequence {fn}n∈N. (2) Study the convergence in measure of {fn}n∈N. (3) Prove that fn ∈ L1([0, 1]) for any n ∈ N. (4) Study the convergence in L1([0, 1]) of the sequence {fn}n∈N. Solution. (1) Let x ∈ (0, 1], then e−n2x → 0 as n → +∞, while for x = 0 we have e−n2x sin2(x) = 0 . On the other side, e− 1 n → 1 as n → +∞. Hence, fn converges pointwisely everywhere (therefore also a.e.) as n → +∞ to the function f de ned by f (x) = 1 + x, x ∈ [0, 1]. (2) By the theory, since λ([0, 1]) < +∞, we have that convergence a.e. implies convergence in measure. Thus, fn converges in measure to f as n → +∞. (3) We need to show that, for any n ∈ N, Z 1 0 |fn(x)| dx < +∞. This is straightforward since fn(·) a continuous function on the whole [0, 1], for any xed n ∈ N. Hence the integral above is a proper integral (thus nite). (4) On account of point (1), we already know that {fn}n converges pointwisely a.e. to the function f ; hence, since convergence in L1 implies a.e. convergence up to subsequences, then the only possible candidate limit for the convergence of {fn}n∈N in L1 is f . To study the convergence of {fn}n to f in L1([0, 1]), we need to check whether ∥fn − f ∥L1([0,1] → 0 as n → +∞ with f as in item (1). We have ∥fn − f ∥L1 = Z 1 0 |fn(x) − f (x)| dx. We have that fn are continuous hence measurable in [0, 1] and |fn(x)| = n  1 − e−n2x sin2(x)  1 + n + x e− 1 n ≤ n  1 − e−n2x sin2(x)  1 + n + |x e− 1 n | ≤ 1 + |x| ≤ 2 =: g(x), for any n ∈ N…

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Prima pagina: 24 01 2024 E TS