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Politecnico di Milano , Mathematical Engineering Real and Functional Analysis – Exercises Prof. F. Punzo, G. Verzini, June 19, 2025 E1 E2 E3 Surname/Name: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Id. No. . . . . . . . . . . . . . . . . . . . . . . . . . . . . [Solutions must be written ONL Y on these sheets, under the exercise and on the back.] [Solutions can be written in English or in Italian.] Exercise 1. [5 points] Consider the sequence of positive functions fn(x) = √n + 1 √n(x2 + 2x) + p n + √n(2x + 3) , x ∈ (−1, +∞). 1. Compute the pointwise limit f of the sequence {fn}+∞ n=1; 2. Does the sequence {fn}+∞ n=1 ⊂ L1(−1, +∞) converge in L1(−1, +∞)? 3. Does the sequence {fn}+∞ n=1 ⊂ L1(1, +∞) converge in L1(1, +∞)? Solution. (1) Let x ∈ R. We get fn(x) = √n + 1 √n(x2 + 2x) + p n + √n(2x + 3) → 1 x2 + 4x + 3 =: f(x), as n → +∞. (2) We notice that f(x) = 1 x2+4x+3 = 1 (x+1)(x+3) is not an L1(−1, +∞) function. Since the L1 convergence implies the existence of a subsequence which converges almost everywhere, so that f is the unique candidate limit for L1-convergence of {fn}n∈N. However, since ( L1(−1, +∞), ∥ · ∥ 1) is a closed space, then if {fn}n∈N converges to f in L1, the limit f would be in L1 which is not the case. Hence {fn}n∈N doesn’t converge in L1. Alternatively, observing that the function fn is positive in ( −1, +∞) for every n ∈ N, we can apply Fatou’s Lemma and get lim inf n→+∞ Z +∞ −1 fn(x)dx ≥ Z +∞ −1 f(x)dx = +∞. Therefore, limn→+∞ R +∞ −1 fn(x)dx = +∞. (3) Since L1 convergence implies the existence of a subsequence which converges almost everywhere and the sequence {fn}n∈N converges to f pointwisely, then f is the unique candidate limit for L1-convergence. We proceed by Dominated…