Ph.D. Qualifying Exam â Spring 2004

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Butu k + v k Infpn (u p + v p ) = w nThereforew n < u + ǫ + v kThis holds for every k n, so taking the infimum over k yields∀n NLet n tend to ∞ in order to obtainw n < u + ǫ + V nlim inf (u n + v n ) u + ǫ + lim inf v nn→∞ n→∞Since this holds for every positive ǫ, we getlim inf (u n + v n ) u + lim inf v nn→∞ n→∞But the converse inequality holds in general, as stated earlier. Hencelim infn→∞ (u n + v n ) = u + lim infn→∞ v n □12
Seventh problemSuppose that (f n ) n∈N is a decreasing sequence of continuous functions on [0, 1], convergingpointwise to a continuous function f. Prove that the convergence is uniform.SolutionLet ǫ be a positive real number. Define∀n ∈ N O n = { x ∈ [0, 1] ∣ ∣ fn (x) − f(x) < ǫ }The sequence (f n ) n∈N is decreasing. Thus, if p < n are integers, and x is in O p , wehavef n (x) − f(x) < f p (x) − f(x) < ǫwhich proves that x ∈ O n . So the sequence (O n ) n∈N is increasing.Since the (f n ) n∈N ’s and f are continuous, all the (O n ) n∈N ’s are open. Because (f n ) n∈Nconverges pointwise to f, every x ∈ [0, 1] belongs to at least one of these sets. So (O n ) n∈Nis an open cover of the compact set [0, 1]. Thus there is a finite subcover: there existintegers n 1 · · · n k such that[0, 1] = ( O n1 ∪ · · · ∪ O nk)⊂(Onk ∪ · · · ∪ O nk)= OnkBut using again the fact that the (O n ) n∈N are increasing, we get∀n n k[0, 1] = O nIn other words ∀n n k ∀x ∈ [0, 1] 0 f n (x) − f(x) < ǫConclusion (f n ) n∈N converges uniformly to f.13
Page 1 and 2: Ph.D. Qualifying Exam - Spring 2004
Page 3 and 4: Second problemSuppose (X, A , µ) i
Page 5 and 6: Third problemLet f : (a, b) −→
Page 7 and 8: Define a function on X as follows:
Page 9 and 10: f is indeed the limit of a sequence
Page 11: 2 For every integer n, the function
Page 15 and 16: Ninth problemLet X be a vector spac
Page 17 and 18: Tenth problemSuppose f : R −→ R

Ph.D. Qualifying Exam â Spring 2004

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Ph.D. Qualifying Exam â Spring 2004