Algebra. Homework 7 Solutions

Homework 7. Solutions Problem 1. For each positive integer N , let N 1 X Dn (t) N + 1 n=0 FN (t) = denote the Fejer kernel, defined on [−π, π]. (a) Prove that 1 sin2 ((N + 1)t/2) . N +1 2 sin2 (t/2) FN (t) = (b) The function FN is periodic and non-negative. Rπ (c) The integral −π FN = π. (d) For each a > 0, Z lim N →∞ FN (t) dt = 0. a≤|t|≤π sin(N + 1/2)t 1 + cos t + · · · + cos nt = . 2 2 sin t/2 (b) Part (a) shows that FN ≥ 0. RIt is 2π-periodic because it is a sum of 2π-periodic functions. π (c) We have proven in class that 0 Dn (t) = π/2, so Solution. (a) The Dirichlet kernel Dn (t) = Z π Z FN = 2 −π π FN 0 Thus FN (t) = N Z 1 X π Dn = π. = N + 1 n=0 0 N X 1 sin(2N + 1)(t/2) 2(N + 1) sin(t/2) n=0 and the result follows because for any real number x not a multiple of π, we have sin x + sin 3x + · · · + sin(2n − 1)x = sin2 nx . sin x (d) If 0 < a ≤ |t| ≤ π, then there is a constant Ca > 0 such that sin2 (t/2) ≥ Ca . Hence FN (t) ≤ a ≤ |t| ≤ π, and so Z 0≤ π FN ≤ a π−a → 0, Ca (N + 1) as N → ∞ Problem 2. Let f be integrable on [−π, π]. For each positive integer N , define σN f (x) = N 1 X Sn f (x). N + 1 n=0 1 1 for Ca (N + 1) (a) Prove that 1 σN f (x) = π Z π FN (x − t)f (t) dt. −π (b) Show that if f is continuous on [−π, π] and f (−π) = f (π), then σN f converges to f uniformly on [−π, π] as N → ∞. Rπ Solution. (a) We have proven that Sn f (x) = π1 −π Dn (x − t)f (t) dt. Therefore (sums below are finite) σN f (x) = N 1 X Sn f (x) N + 1 n=0 N = = = 1 1X Dn (x − t)f (t) dt N + 1 π n=0 Z N 1 π 1 X Dn (x − t)f (t) dt π −π N + 1 n=0 Z 1 π FN (x − t)f (t) dt. π −π Problem 3. If f has continuous derivative on [−π, π] then there is a constant M such that |ak | + |bk | ≤ M/k for all k > 0. Solution. The hypothesis imply that the periodic extension of f is piecewise smooth. The Fourier series of f 0 are a0k = kbk and b0k = −kak for k ≥ 1, and thus, since f 0 is continuous on [−π, π], the series ∞ X k=1 1 k (|bk | + |ak | ) ≤ 2π 2 2 2 Z π |f 0 |2 < ∞ −π converges. Thus the√sequence k 2 (|ak |2 + |bk |2 ) is bounded. If M > 0 is such that k 2 (|ak |2 + |bk |2 ) ≤ M for all k ≥ 1, then |ak | + |bk | ≤ 2 M /k for all k ≥ 1. Problem 4. Let f be a continuous function on [−π, π], and suppose that all its Fourier coefficients ak = 0 for k ≥ 0, and bk = 0 for k ≥ 1. Show that f is identically equal to 0. Solution. The periodic extension of f to R is 2π-periodic and piecewise continuous, but it may notR be continuous, π so its Fourier series may not converge to the value of the function for all x. However, because (1/π) −π f = a0 = 0, Rx the function F (x) = 0 f is 2π-periodic and continuous, with piecewise continuous derivative. It thus equals its Fourier series ∞ A0 X F (x) = + (Ak cos kx + Bk sin kx) 2 k=1 for all x. As we know, the Fourier coefficients of F and f are related by Ak = −bk /k and Bk = ak /k for all k ≥ 1. Therefore, Ak = Bk = 0 for k ≥ 1. It follows that F (x) = A0 /2 for all x; thus F is a constant function and its derivative F 0 (x) = f (x) = 0 for all x. Problem 5. Perhaps you have heard about the Dirac delta function. This is the “function” δ on [−π, π] such that Z π f (x)δ(x − a) dx = f (a). −π (i) Although δ is not really a function, you may pretend that it is and try to compute its Fourier series. What is this series? (ii) Suppose that δ was not only just a function, but also a differentaible function with derivative δ 0 . Compute the Fourier series of δ 0 . Solution. (i) At least formally, the Fourier coefficients of δ are Z 1 π ak = δ(x) cos kx dx = cos 0 = 1, π −π for all k, and bk = 1 π Z π δ(x) sin kx dx = sin 0 = 0, −π for all k. The Fourier series for δ is thus ∞ δ(x) ∼ 1 X + cos kx. 2 k=1 0 (ii) The Fourier series for the “derivative” δ is δ 0 (x) ∼ − ∞ X k=1 k sin kx.