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Mathematical Engineering - A.Y. 2022-23 Real and Functional Analysis - Written exam - June 20, 2023 Exercise 1. Consider the functionsf, fn : [0, 1] → R, n ∈ N, defined respectively by f (x) = ( x cos 1 x
, 0 < x ≤ 1 0, x = 0 , f n(x) = ( x cos 1 x
1 2πn < x ≤ 1 1 2πn , 0 ≤ x ≤ 1 2πn . (1) Is f of bounded variation in [0, 1]? Is f absolutely continuous in [0, 1]? Justify the answers. (2) Prove thatfn converges pointwisely a.e. tof as n → +∞. Does fn converges to f in L∞ as n → +∞? Justify the answers. (3) Considering that fn ∈ AC([0, 1]) for any n ∈ N (not to be proven), use items (1),(2) to answer to the following question: is the normed space(AC([0, 1]), ∥ · ∥∞) complete? Justify the answer. Solution. (1) We show thatf is not of bounded variation in[0, 1], hence it is not absolutely continuous in [0, 1] as well (recall thatAC([0, 1]) ⊂ BV([0, 1])). Indeed, for n ∈ N, n ≥ 2, consider for instance the following partitionPn of [0, 1]: Pn = {xk}n k=0, x k =    1, k = 0 1 kπ , k = 1, . . . , n− 1 0, k = n . Then f (xk) =    cos 1, k = 0 (−1)k πk , k = 1, . . . , n− 1 0, k = n , so that we get nX k=1 |f (xk) − f (xk−1)| ≥ n−1X k=2 |f (xk) − f (xk−1)| = n−1X k=2 | (−1)k πk − (−1)k−1 π(k − 1) | = n−1X k=2 2k − 1 πk(k − 1) In conclusion, V 1 0 (f ) ≥ sup n nX k=1 |f (xk) − f (xk−1)| ≥ ∞X k=2 2k − 1 πk(k − 1) = +∞. Hence, f does not belong toBV([0, 1]). (2) The pointwise convergence offn to f follows immediately by noting that for anyx ∈ [0, 1], we have 1 2πn → 0 as n → +∞. Concerning the convergence inL∞([0, 1]), we notice that both fn and f are continuous in [0, 1] and they coincide in  1 2πn , 1  . So that fn − f is continuous 2 and, for anyn ∈ N, we have ∥fn − f ∥∞ = esssup x∈[0,1]|fn(x) − f (x)| = sup x∈[0,1] |fn(x) − f (x)| = sup x∈[0, 1 2πn ] 1 2πn − x…