14 02 2024 E TS

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Mathematical Engineering - A.Y. 2023-24 Real and Functional Analysis - Written exam - February 14, 2024 Exercise 1. Consider the measure space ([0, +∞), L([0, +∞))) with the Lebesgue measure. De ne the sequence of functions {fn}n∈N by fn(x) = ( n n+1 if x ∈ [0, 1], n√x(n+x) if x > 1. (1) Compute the pointwise a.e. limit f of the sequence {fn}n∈N. (2) Determine the exponents p ≥ 1 such that the pointwise a.e. limit f belongs to Lp([0, +∞)). (3) Determine the exponents p ≥ 1 such that {fn}n∈N converges to f in Lp([0, +∞)). (4) Let gn(x) = χ[n,n+1](x) · (x − n), for x ∈ [0, +∞) and n ∈ N. De ne hn = fn + gn for each n ∈ N. Does the sequence {hn}n∈N converge in L4([0, ∞))? Solution. We rst notice that fn ∈ Lp([0, +∞)) for all n ∈ N and any p ≥ 1. (1) Let x ∈ [0, 1] and n ∈ N. It holds that fn(x) = n n+1 . Taking the limit for n → +∞ we get fn(x) → 1. Now, if x > 1 and n ∈ N we get fn(x) = n√x(x+n) , observing that fn(x) = n√x(x + n) = 1√x(1 + x n) , we get fn(x) → 1√x , as n → ∞. Therefore the sequence {fn}n∈N converges pointwisely to f in [0, +∞), where f(x) = ( 1 if x ∈ [0, 1], 1√x if x > 1. (2) We observe that for p ∈ [1, 2] the function f does not belong to Lp([0, +∞)). Indeed, ∥f ∥p p = Z ∞ 0 |f |p dx ≥ Z ∞ 1 1 x p 2 dx. Observing that p 2 ≤ 1 for p ∈ [1, 2], we get that f /∈ Lp([0, +∞)). However, if p > 2, we have ∥f ∥p p = Z ∞ 0 |f |p dx = 1 + Z ∞ 1 1 x p 2 dx < +∞, being p 2 > 1 for p > 2, so we get that f ∈ Lp([0, +∞)). (3) Since Lp convergence implies the existence of a subsequence which converges almost every- where and the sequence {fn}n∈N converges to f pointwisely, then f is the unique candidate limit for Lp-convergence. Moreover, recalling that convergence in Lp implies the convergence of the Lp-norm of fn to the Lp-norm of f , by item (2) we deduce that…