15 02 2023 E TS

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Mathematical Engineering - A.Y. 2022-23 Real and Functional Analysis - Written exam - February 15, 2023 Exercise 1. Consider the functionsf, g : [0, 1] → R given by f (x) = ( x2 sin 1 x
, 0 < x ≤ 1 0, x = 0 , g (x) = ( 0 0 < x ≤ 1 1, x = 0 . (1) Is f absolutely continuous in [0, 1]? Is f of bounded variation in [0, 1]? Justify the answers. (2) Is g absolutely continuous in [0, 1]? Is g of bounded variation in [0, 1]? Justify the answers. (3) Consider the function h : [0 , 1] → R, h(x) := f (x) + g(x). Determine whether h is absolutely continuous in[0, 1]. Justify the answer. Solution. (1) We show thatf is absolutely continuous by showing thatf satisfies • f is differentiable almost everywhere withf ′ ∈ L1([0, 1]), • f (x) = f (0) + R x 0 f ′(t) dt, for everyx ∈ [0, 1]. Since f is written as product and composition of elementary functions, it is differentiable in (0, 1], hence f is differentiable a.e. in[0, 1] (λ({0}) = 0). For x ∈ (0, 1], we have f ′(x) = 2 x sin  1 x  − cos  1 x  . In particular, f ′ ∈ L1([0, 1]), indeed f ′ is bounded on a bounded interval. Hence, it remains to verify the calculus formula f (x) − f (0) = Z x 0 f ′(t) d t, ∀x ∈ [0, 1]. Let then 0 < c < 1 be arbitrarily fixed. Since f ∈ C 1([c, 1]), by the standard fundamental theorem of calculus in[c, 1] we have f (x) − f (c) = Z x c f ′(t) d t, ∀x ∈ [c, 1] We now aim to pass to the limit asc → 0+ on both sides of the equality above. To this end we first observe that, sincef is continuous at0, we have f (c) → f (0) as c → 0+. Moreover, since f ′ ∈ L1([0, 1]) then the map c 7→ Z x c f ′(t) d t is absolutely continuous (hence continuous) in[0, 1], so that lim c→0+ Z x c f ′(t) d t = Z x 0 f ′(t) d t. Gathering these facts, we obtain f (x) − f (0) = f (x) − lim c→0+ f (c) = lim c→0+ Z x c f ′(t) d…