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Mathematical Engineering Real and Functional Analysis - Written exam - February 17, 2022 Theory Question 1. Lecture Notes of 20, 22, 28 September. Question 2. 1. For anyn∈ N, letEn :={x∈ X :|f (x)|≤ n},fn := fχEn in X. We have thatfn∈L∞(X),fn→f a.e. inX asn→ +∞. Moreover,|fn|≤| f| a.e. inX for anyn∈ N. Since f∈ L1(X), we can apply the Dominated Convergence Theorem. Hence the conclusion follows. 2. Lecture Notes of 12 October. Question 3. Lecture Notes of1st, 11 and 15 December. Question 4. Lecture Notes of 14 and 15 December. Exercises Exercise 1. Let α∈ R be fixed, and letfa : [0, 1]→ R be the function defined as follows: fα(x) = { sin(xα), if 0<x ≤ 1, 0, if x = 0. (1) Prove that the functionfα is differentiable a.e.in [0, 1] for every α ∈ R; moreover, assuming α> 0, prove thatf′ α∈L1([0, 1]). (2) Assuming thatα> 0, verify whetherfα∈ AC([0, 1]). (3) Isf−1∈ BV([0, 1])? Justify the answer. Solution. (1) Certainly, for everyα∈ R,fα is differentiable in(0, 1], hencefα is differentiable a.e. in [0, 1]. Forx∈ (0, 1], we have f′ α(x) = αxα−1 cos(xα). In particular, ifα> 0, f′ α∈L1([0, 1]), indeed ∫ 1 0 |f′ α(x)| dx =α ∫ 1 0 xα−1| cos(xα)| dx≤α ∫ 1 0 xα−1 dx< +∞, indeed xα−1 is integrable in[0, 1] under the assumptionα> 0. (2) Letα> 0. By the Second Fundamental Theorem of Calculus,fα is absolutely continuous in the interval[0, 1]if andonly iffα is differentiablea.e. in[0, 1],f′ α∈L1([0, 1])and theCalculus formula forfα holds. Hence, it remains to verify that fα(x)−fα(0) = ∫ x 0 f′ α(t)dt, ∀x∈ [0, 1]. Let then 0 < c <1 be arbitrarily fixed. Since fα∈ C 1([c, 1]), by the standard fundamental theorem of calculus in[c, 1] we have fα(x)−fα(c) = ∫ x c f′ α(t)dt, ∀x∈ [c, 1] 2 We now aim to pass to the limit asc→ 0+ on both sides of the equality above. To this end we first observe that,…