Fall 2014 MATH3060 Mathematical Analysis III Solution to Midterm Examination 1. Let x1 and x2 be two distinct points in the metric space (X, d). Find a bounded, continuous function ϕ in X satisfying ϕ(x1 ) = 0 and ϕ(x2 ) = 1. Solution. There are many such functions. The simplest one is f (x) = d(x, x1 ) . d(x, x2 ) + d(x1 , x2 ) Some tried g(x) = d(x, x1 )/d(x2 , x1 ), but that is no good since it may not be bounded. The modified one n d(x, x ) o 1 h(x) = min ,1 d(x2 , x1 ) is good. 2. Consider C[−1, 1] with the metric induced by the sup-norm. Is the set n o f ∈ C[−1, 1] : f is differentiable at 0 closed in C[−1, 1]? Justify your answer. Solution. No, it is not a closed set. A simple example is to consider the function g(x) which is equal to x for x ∈ [0, ∞) and vanishes on (−∞, 0). It is not differentiable at 0 but, as we shift it, gn (x) = g(x − 1/n) is differentiable at 0 and gn converges uniformly to g on [−1, 1] as n → ∞. The sequence of smooth functions (x2 + 1/n)1/2 , which converges uniformly to |x| on [−1, 1], is another example. 3. Let A be a non-empty set in (X, d) and define g(x) = inf{d(x, y) : y ∈ A}. Show that |g(x) − g(y)| ≤ d(x, y), Solution. x, y ∈ X. See Solution to Assignment. 4. Let {ϕj }∞ j=1 be an orthonormal set in R[−π, π]. Show that it does not admit any convergent subsequence in L2 -distance. √ Solution. Let {fn } be such a sequence. We have kfn − fm k2 = 2 for all distinct n, m. On the other hand, if {fnk } is convergent, it is Cauchy and in particular there is some n0 such that kfnk − fnj k2 < 1, for all nk , nj ≥ n0 . Contradiction holds. ∞ to 5. Give a definition of the fractional derivative L ≡ (d/dx)1/3 so that it maps C2π itself and df ∞ L3 f = , ∀f ∈ C2π , dx holds. Solution. As fb0 (n) = infˆ(n) = eπi/2 fˆ(n), it suggests us to define X Lf (x) = eπi/6 n1/3 cn einx , n where cn is the n-th Fourier coefficient of f . First of all, we need to show this is a well-defined smooth function. Indeed, as f is smooth and 2π-periodic, |cn | = ◦(n−k ) for all k. It follows that |eπi/6 n1/3 cn | = ◦(n−k ) for all k. By M-test the RHS in the definition of Lf defines a smooth function. Now, we can easily check P that L3 f = f 0 . You may also define Lf = n (−i)n1/3 cn eix , since the cube root of i is not unique. (I prefer the root in the first quadrant.) (a) (10 marks) Find the Fourier series of h(x) = |x| on [−π, π]. Solution. By direction computation, ∞ π 4 X cos(2k + 1) h(x) ∼ − . 2 π k=0 (2k + 1)2 (b) (5 marks) Prove ∞ 1 π2 X = . 8 (2k + 1)2 k=0 Quote a theorem to justify your derivation. Solution. A theorem states that the Fourier series of a Lipschitz continuous function converges to the function uniformly. (You should state this result. The key words Lipschitz continuous or piecewise C 1 should come up.) Now h is Lipschitz continuous. By taking x = 0 in (a), we get ∞ 4X 1 π , 0= − 2 π k=0 (2k + 1)2 and the identity follows. (c) (5 marks) Find the sum ∞ X k=0 1 . (2k + 1)4 Quote a theorem to justify your derivation. Solution. As h is continuous, it is integrable on [−π, π]. (Should point out integrable first in order to apply Parseval. This identity does not come for free. Sadly most of you did not. I did not subtract any points this time though.) A theorem states that Parseval’s identity holds in this case. Therefore, we have ∞ X 2 2 khk2 = 2πa0 + π a2k , k=1 which yields ∞ X k=0 1 π4 , = (2k + 1)4 96 after some manipulations. 7. (a) (10 marks) Let f ∈ R2π , that is, it is 2π-periodic and integrable on [−π, π], and Sn f the n-th partial sum of its Fourier series. Derive the formula Z 1 π sin n + 12 z f (x + z) dz . (Sn f )(x) = π −π 2 sin 12 z Solution. (a). See Notes. (b) (10 marks) Let g ∈ R2π which is equal to f on (−ρ, ρ) for some small ρ > 0. Show that limn→∞ Sn g(0) = g(0) if limn→∞ Sn f (0) = f (0). Solution. Let I = [−δ, δ] and Dn be the Dirichlet kernel. We write Z Z Sn g(0) = Dn (z)g(z)dz + Dn (z)g(z)dz I [−π,π]\I and similarly Z Sf (0) = Z Dn (z)f (z)dz + I Dn (z)f (z)dz. [−π,π]\I Using f = g on I, we have Z Dn (z)(g(z) − f (z))dz|, |Sn g(0) − Sn f (0)| = | [−π,π]\I whose RHS tends to 0 as n → ∞ as a result of Riemann-Lebesgue lemma. Note: This is Riemann’s localization principle. It tells us that although the Fourier series of a function depends on its values on [−π, π], its convergence to the function at a particular point is only determined by its values in a small neighborhood of this point. 8. (20 marks) For f ∈ R2π and a ∈ R, define fa (x) = f (x + a) ∈ R2π . Show that Z π |fa (x) − f (x)| dx → 0, as a → 0. −π Solution. I will be sketchy. Let I = [−π, π] and J = [−2π, 2π]. First approximate f by a step function on J (see the proof of Riemann-Lebesgue lemma in notes) and then approximate the step function by a continuous function g on J so that their L1 -distance is very small. (There are finitely many discontinuity for a step function, so approximation is possible.) Therefore, for ε > 0, there is a continuous function g such that Z ε |f (x) − g(x)|dx < . 8π J On the other hand, g is continuous on J and hence uniformly continuous, we can find a small ρ > 0 less than π such that |g(y) − g(x)| < ε/4 for x, y ∈ [−2π, 2π], |x − y| < ρ. Therefore, for a, |a| < ρ, Z Z |fa (x) − f (x)|dx = |f (x + a) − f (x)|dx I Z ZI |f (x + a) − g(x + a)|dx + |g(x + a) − g(x)|dx ≤ I IZ + |g(x) − f (x)|dx I < 2π ε ε ε + + 2π < ε. 8π 4 8π Note: We need to choose a slightly bigger J because x + a may not stay in I for x ∈ I. The proof is based on the same idea by which Riemann-Lebesgue lemma was proved in notes. You may use the cut-off function if you like. 9. (20 marks) Show that every open set in (−∞, ∞) can be written as a disjoint union of open intervals. Solution. Let G be open. Since G is open, for every x ∈ G, there exists some open interval containing x that belongs to G. Denote all these open intervals by S {Ix }. Let Jx = Ix . Then Jx is an open interval containing x that belongs to G. It is the largest one in the sense that when I is an open interval containing x that belongs to G, I ⊂ Jx . It follows that for x, y ∈ G, either Jx = Jy or Jx ∩ Jy = φ. If one introduces a relation on G × G by x ∼ y if y ∈ Jx . It is readily checked that S ∼ is an equivalence relation. So G = [x] Jx is a disjoint union. Saying more, the union is a countable one. For, if there are uncountably many inequivalence classes [x], picking one rational number in each Jx will produce uncountably many rational numbers, which is absurd. Therefore, the union is countable.