06 07 2022 E TS

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Real and Functional Analysis Master Degree Program in Mathematical Engineering, a.y. 2021/22 6/7/2022 Exercises Exercise 1. (7 points) Consider the measurable space ([0, +∞),L([0, +∞))) with the Lebesgue measure. Given the sequence of functions fn(x) = xe−x2[1+ x n ] −1 χ[0,n](x), x ∈ [0, +∞), n ∈ N0, 1. study the convergence a.e. of{fn}n∈N0 in [0, +∞) and prove that, for anyn∈ N0, the function fn is integrable in[0, +∞). 2. Study the convergences inL1([0, +∞)) and in measure of{fn}n∈N0 in [0, +∞). 3. Compute, if it exists finite, the following limit lim n→+∞ ∫ +∞ 0 fn(x) dx. Solution. (1) We have that1+ x n→ 1 andχ[0,n](x)→ 1 asn→ +∞, for allx∈ [0, +∞). Thus,fn(x) converges tof (x) = xe−x2 for anyx∈ [0, +∞), asn→ +∞. Hencef is the limit a.e. of{fn}n∈N in [0, +∞). -Integrability. Note that, for anyn∈ N0, ∫ ∞ 0 |fn(x)|dx = ∫ n 0 xe−x2[1+ x n ] −1 dx. Sincex↦→xe−x2[1+ x n ] −1 is continuous in the compact set[0,n ], it follows that it is Lebesgue integrable over [0,n ]. Thusfn∈L1([0, +∞)). (2) We can proceed for instance in two ways denoted by (a) or (b), both leads to the conclusion that fn are integrable for anyn∈ N0. (a) We notice thatχ[0,n](x)≤ 1 for anyn∈ N0,x∈ [0, +∞); moreover, for any fixedx∈ [0, +∞), we have thatx2[ 1 + x n ]−1 increases asn increases. Hence, for any fixedx∈ [0, +∞), we have xe−x2[1+ x n ] −1 decreases asn increases; in particular xe−x2[1+ x n ] −1 ≤xe−x2 (1+x)−1 , ∀x∈ [0, +∞), n∈ N0. 1 We thus have that |fn(x)| =fn(x)≤xe−x2 (1+x)−1 =:g(x) ∀x∈ [0, +∞), n∈ N0. Sinceg(x)∼xe−x asx→ +∞and ∫ +∞ 0 xe−x dx = 1 < +∞(byparts)thenalso ∫ +∞ 0 g(x) dx converges by asymptotic comparison test, henceg is integrable in[0, +∞) and so isfn for any n∈ N0. (b) We notice thatfn(x) = 0 if x∈ (n, +∞); while, forx∈ [0,n ], n∈ N0, we have 1 + x n≤ 1 + n n = 2, hence x2 [ 1…