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Politecnico di Milano , Mathematical Engineering Real and Functional Analysis – Exercises Prof. F. Punzo, G. Verzini, July 10, 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 function f : [0, 1] → R given by f(x) = sin(√x) cos(√x). (1) 1. Prove that f is absolutely continuous in [0 , 1]. 2. Compute the weak derivative of f. Solution. (1) We show that f is absolutely continuous by showing that f satisfies • f is differentiable almost everywhere with f ′ ∈ L1([0, 1]), • f(x) = f(0) + R x 0 f ′(t) dt, for every x ∈ [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) = 1 2√x cos2(√x) − sin2(√x)
. In particular, f ′ ∈ L1([0, 1]), indeed Z 1 0 |f ′(t)| dt ≤ Z 1 0 1√ t dt = 2. Hence, it remains to verify the calculus formula f(x) − f(0) = Z x 0 f ′(t) dt, ∀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) dt, ∀x ∈ [c, 1] We now aim to pass to the limit as c → 0+ on both sides of the equality above. To this end we first observe that, since f is continuous at 0, 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) dt is absolutely continuous (hence continuous) in [0 , 1], so that lim c→0+ Z x c f ′(t) dt = Z x 0 f ′(t) dt. Gathering these facts, we…