07 09 2023 E TS

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Mathematical Engineering - A.Y. 2022-23 Real and Functional Analysis - Written exam - September 7, 2023 Exercise 1. Let BV([0, 1])bethespaceoffunctions f : [0 , 1] → Rofboundedvariationendowed withthenorm ∥·∥BV definedby ∥f ∥BV := |f (0)|+V 1 0 (f ),where V 1 0 (f )denotesthetotalvariation of f in [0, 1]. Consider the sequences of functions{fn}n∈N and {gn}n∈N, defined by fn(x) = ( 1 − 1 n , 0 ≤ x ≤ 1 n 0, 1 n < x ≤ 1 , g n(x) = ( 1 − 1 n , 0 ≤ x ≤ 1 2 0, 1 2 < x ≤ 1 . (1) Study the convergence pointwisely (everywhere) of{fn}n∈N and of {gn}n∈N. (2) Do {fn}n∈N and {gn}n∈N belong to BV([0, 1]) for any n ∈ N? Justify your answer. (3) Denoted respectively byf and g the pointwise limits of{fn}n∈N and {gn}n∈N, answer to the following questions: Does {fn}n∈N converges in the normed space(BV([0, 1]), ∥ · ∥BV) to f? Justify. Does {gn}n∈N converges in the normed space(BV([0, 1]), ∥ · ∥BV) to g? Justify. Solution. (1) Let x ∈ [0, 1/2], then gn(x) = 1 − 1/n. Therefore it converges, forn → ∞, to g(x) = 1 . While, if x ∈ (1/2, 1], gn(x) = 0 for every n ∈ N. Hence, the sequence of functions {gn}n∈N converges pointwisely to the function g(x) = ( 1, 0 ≤ x ≤ 1 2 0, 1 2 < x ≤ 1. Let us consider now the sequence{fn}n∈N. For everyx ∈ (0, 1], there existsn0 ∈ N large enough such thatx > 1/n for everyn ≥ n0. Thus fn(x) = 0 eventually. Ifx = 0, thenfn(x) = 1 − 1/n, which converges tof (x) = 1 for n → ∞. In particular,fn converges pointwisely to the function f (x) = ( 1, x = 0 0, 0 < x ≤ 1. (2) Observing that, for everyn ∈ N, fn is a bounded and non-increasing function, we get that fn has bounded variation for everyn ∈ N. Similarly we prove that{gn}n∈N ⊂ BV([0, 1]). (3) We first observe that fn(x) − f (x) =    −1/n, x = 0 1 − 1/n, 0 < x ≤ 1/n 0, 1/n < x ≤ 1. Hence we get ∥fn − f ∥BV ≥ V 1…