17 06 2022 E TS

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Real and Functional Analysis Master Degree Program in Mathematical Engineering, a.y. 2021/22 17/6/2022 Exercises Exercise 1. (7 points) Let (R, L(R), λ) be the Lebesgue measure space. Consider the sequence of functions fn(x) = 1√n e−x2/n, x ∈ R, n ∈ N. 1. Study the convergence a.e. of the sequence{fn}n∈N in R and prove that, for anyn ∈ N, the function fn is integrable inR. 2. For every fixed 1 ≤ p ≤ +∞, study the convergences in Lp(R) of the sequence {fn}n∈N. Moreover, discuss the convergence in measure of{fn}n∈N. 3. Prove that {fn}n∈N does not converge weakly tof0 ≡ 0 in L1(R). Solution. (1) First of all we observe that, since0 < e −t ≤ 1 for every t ≥ 0, we have (⋆) 0 < f n(x) ≤ 1√n for every x ∈ R and n ∈ N. This, together with the fact that1/√n → 0 as n → +∞, immediately implies thatfn(x) → 0 for every x ∈ R as n → +∞, that is, the sequence{fn}n converges pointwise on R to the function f0(x) ≡ 0. To prove the integrability onR of the functionfn, we explicitly compute itsL1-norm: since we have already pointed out thatfn(x) > 0 for every x ∈ R, by the change of variablesx = √ny we get Z R |fn(x)| dx = 1√n Z R e−x2/n dx = 1√n Z R e−y2 · √n dy = Z R e−y2 dy = √π < +∞, and this proves thatfn ∈ L1(R) for every n ∈ N. (2) Since the convergence in measure is implied by the convergence inLp (for some 1 ≤ p ≤ ∞), it is convenient to start by studying the convergence of the sequence{fn}n in Lp(R). To this end we observe that, since we know from point (1) that the sequence{fn}n converges pointwise on R to the function f0 ≡ 0, the unique candidate limit for {fn}n in Lp(R) is f0 (which clearly belongs toLp(R) for every1 ≤ p ≤ ∞); as a consequence, we need to check whether ∥fn − f0∥Lp(R) = ∥fn∥Lp(Rn) → 0 as n → +∞. Due to the different nature of theLp-norm when p < ∞ and p = ∞,…