12 07 2023 E TS

Page 1

Mathematical Engineering - A.Y. 2022-23 Real and Functional Analysis - Written exam - July 12, 2023 Exercise 1. Consider the measure space ([0, +∞), L([0, +∞))) with the Lebesgue measure. Define the sequence of functions{fn}n∈N by fn(x) = e−nx + n2e−x n + n2 + (1 + n2)x2 , x ∈ [0, +∞), n ∈ N. (1) Study the convergence a.e. of the sequence{fn}n∈N. (2) Study the convergence inL1([0, +∞)) of the sequence{fn}n∈N. (3) Consider the sequencegn(x) = fn(x) + χ[n,n+1]. Study the convergence inL1([0, +∞)) of the sequence{gn}n∈N. Solution. (1) Let x ∈ (0, +∞) fn(x) = e−nx + n2e−x n + n2 + (1 + n2)x2 = e−nx n2 + e−x 1 + 1 n + (1+n2) n2 x2 − → e−x 1 + x2 for n → +∞ Therefore the sequence {fn}n∈N converges to the functionf (x) = e−x 1+x2 almost everywhere in [0, ∞). (2) In order to show that the sequence{fn}n∈N converges inL1([0, +∞)) to the functionf, we are going to use the dominated convergence theorem. We first observe that for eachn ∈ N, we have n + n2 + (1 + n2)x2 ≥ n2 + (1 + n2)x2 ≥ n2(1 + x2) and e−nx + n2e−x = n2  e−nx n2 + e−x  ≤ n2(1 + e−x) Therefore we have |fn(x)| = e−nx + n2e−x n + n2 + (1 + n2)x2 ≤ n2(1 + e−x) n2(1 + x2) = 1 + e−x 1 + x2 ≤ 2 1 + x2 Letting g(x) = 2 1+x2 and observing that g ∈ L1([0, +∞)), we get that the sequence{fn}n∈N converges in L1([0, +∞)) to f, thanks to the dominated convergence theorem. (3) Let hn(x) = χ[n,n+1](x). The sequence {hn}n∈N converges almost everywhere toh(x) ≡ 0. Indeed, for every x ∈ [0, +∞), there exists n0 ∈ N such that n > x for every n ≥ n0, which implies hn(x) = 0 for every n ≥ n0. Notice that the sequence {hn}n∈N does not converge in L1([0, +∞)). Indeed, the candidate limit ish, which has norm zero, and ∥hn∥1 = Z n+1 n 1dx = 1 for every n ∈ N. Hence the sequence {hn}n∈N does not converge in L1([0, +∞)). Finally, suppose by…