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18.103 MIDTERM IIWednesday, May 2, 2007Name:Numeric Student ID:Instructor’s Name:I agree to abide by the terms of the honor code:Signature:Instructions: Print your name, student ID number and instructor’s name in the spaceprovided. During the test you may not use notes, books or calculators. Read eachquestion carefully and show all your work; full credit cannot be obtained withoutsufficient justification for your answer unless explicitly stated otherwise. Underline yourfinal answer to each question. There are 5 questions. You have 55 minutes to do all theproblems.Question Score Maximum1 102 103 104 105 (Bonus) 5Total 40Question 1 of 5, Page 2 of 7 Solutions1. State and carefully prove the “Strong Law of Large Numbers” expressed in termsof random variables and expectation values.Solution:This is theorem 1 of Section 2.7 in your book. See page 111 for a solution. Manypeople forgot to assert that the random variables fiare bounded, in the assump-tions of the theorem. This is not, in general, necessary, but the proof in the book(which everyone used) requires it to conclude that the integrals, e.g.Zf2idµare indeed finite.Question 2 of 5, Page 3 of 7 Solutions2. Define Lp(X, µ) and prove that for 1 ≤ p < ∞, Lpis a Banach space.Solution:This is proved on p. 187 of your book, in the appendix “On LpMatters,” withthe assumption that the space X is σ-finite. In class, we did a more general proofwithout this assumption, so you can refer to class notes as well. Either proof wascompletely acceptable on the test.Question 3 of 5, Page 4 of 7 Solutions3. Given a constant c > 0, evaluate the integralZ∞−∞dyc2+ y2using the Fourier transform of the functionfc(x) =(e−cx, x ≥ 00, x < 0.Solution:The Fourier transform of fc(x) isˆfc(x) =Z∞−∞fc(x) e−ixydx=Z∞0e−(c+iy)xdx=1c + iySo in particular,|ˆfc(x)|2=1c2+ y2Now, using Plancherel’s theorem, which states that||ˆf||22= 2π||f ||22for functions f in the Schwartz class, we haveZ∞−∞dyc2+ y2= 2πZ∞0|e−cx|2dx =πc.Many people forgot to justify why this could be used for fc, which is not strictlyin the Schwartz class, as there is a discontinuity at the origin. But owing to timeconstraints, this was not counted against you. How would you show this?Question 4 of 5, Page 5 of 7 Solutions4. Let X = Y = [0, 1], the unit interval, and M = N = B[0,1], the Borel sets inthe unit interval. Consider the measure spaces (X, M, µ) with µ = µL, Lebesguemeasure, and (Y, N , ν) with ν the counting measure (i.e., the number of elementsin the set). SetD = {(x, y) ∈ X × Y | x = y}with 1Dthe characteristic function of D.Which, if any, of the following integrals are equal?Z10Z101Ddµ dν,Z10Z101Ddν dµ,ZX×Y1Dd(µ × ν).Here, µ × ν is the product measure, defined as usual by extending the definitionon “rectangles” via outer measure. Be sure to carefully explain your answer.Solution:First, note that Fubini’s theorem does not apply because the counting measure νis not σ-finite on the unit interval. So no two of these integrals are necessarilyequal, and that’s exactly what we can show by direct computation. Recall thatZ101Ddµ = µ(Ey) = 0since the slice at fixed y, Ey, is a single point, which has Lebesgue measure 0. Thisimmediately impliesZ10Z101Ddµ dν =Z100 dν = 0.Similarly,Z101Ddν = ν(Ex) = 1since the slice at fixed x, Ex, also contains a single point, which has countingmeasure 1. Hence,Z10Z101Ddν dµ =Z101 dµ = µL([0, 1]) = 1.Finally, to compute the value for the product measure, recall that µ × ν is anextension of the measure defined byµ × ν(A × B) = µ(A)ν(B)on sets A × B, with A ∈ M, B ∈ N , so-called “rectangles.” SoZX×Y1Dd(µ × ν) = µ × ν(D)Question 4 of 5, Page 6 of 7 Solutionswhich can be computed by finding a sequence of rectangles converging in measureto D. There are many ways to do this. My first thought was to imagine a sequencewhere the nth term contains 2nsquares of width 1/2ncovering the diagonal. Themeasure of each of these squares, with respect to µ×ν, is µ(I)ν(I) = ∞·1/2n= ∞,so the product measure of the diagonal, which is the limit of the measures of thisconverging sequence of squares, is also ∞.Question 5 of 5, Page 7 of 7 Solutions5. (Bonus) Define the convolution operator ∗ for functions f, g ∈ L1(R) byf ∗ g(x) =ZRf(x − y)g(y) dyProve f ∗ g is in L1(R) and that||f ∗ g||1≤ ||f ||1||g||1Solution:This exercise is handled on page 193 of the book. Many people tried to use Fubini’stheorem, which is the right idea, but first you need to prove that the integrand towhich you are applying Fubini’s theorem is inde ed

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