User:MCPOliseno /Math710 FINAL

MICHELLE POLISENO

FINAL

1) Let $ f \ $ have bounded variation on [0, 1] and define $ g(x) = f(a) + \int_{a}^{x} f'(t)dt \ $. Support the following assertions: (a) $ f'(x) = g'(x) \ $ a. e.; (b) A function of bounded variation can be expressed uniquely (up to additive constant) as the sum of an absolutely continuous function and a singular function (A singular function is a function $ s \ $ for which $ s'(x) \ $ = 0 a.e.)

(a) Let $ f \ $ have bounded variation on [0, 1] and define $ g(x) = f(a) + \int_{a}^{x} f'(t)dt \ $. Then $ g(x) \le f(a) + f(x) - f(a) \ $ = $ f(x) \ $. Note that since $ f \ $ is of bounded variation, $ f'(x) \ $ exists for almost all $ x \in \ $ [0, 1]. Thus $ g'(x) = f'(x) \ $ almost everywhere.

(b) A function $ f \ $ has bounded variation if and only if $ f = g + h \ $ where $ g, h \ $ are bounded monotone increasing functions, and $ f' \ $ exists almost everywhere on [0, 1]. Set $ h= f-g \ $, then since $ g(x) = f(a) + \int_{a}^{x} f'(t)dt \ $, we have that $ g \in AC[0, 1] \ $. (i.e. $ g \ $ is absolutely continuous on [0 ,1]). And by (a), $ g' = f' \ $ almost everywhere. Thus $ h' = 0 \ $ almost everywhere. And therefore $ h \ $ is singular.

To show uniqueness, suppose we also had $ f = g_1 + h_1 \ $ with $ g_1 \ $ absolutely continuous and $ h_1 \ $ singular. Then $ g - g_1 = 0 = h - h_1 \ $ and so $ g - g_1 \ $ and $ h - h_1 \ $ are both absolutely continuous and singular and therefore are constant.

2) Let $ f \in L^1 (\R) \ $ be non-negative. For each measurable set E, define $ \mu (E) = \int_E f \ $. Show that $ \mu \ $ is countable additive on the sigma algebra of measurable sets.

Let $ E \ $ be a measurable set such that $ \mu (E) = \int_E f \ $.

3) Compute $ \int_{0}^{1} f \ $ for the Cantor function $ f \ $.

Let the Cantor function be $ f: [0, 1] \to \R \ $ be Lebesgue measurable. Note that $ 1-f(x) = f(1-x) \ $. Then 1 = $ \int_{0}^{1} (1) = \int_{0}^{1} (f + (1-f)) \ $ = $ \int_{0}^{1} f - \int_{0}^{1} (1-f) \ $ = $ \int_{0}^{1}f + \int_{0}^{1}(f(1-x))dx \ $ = $ \int_{0}^{1}f + \int_{0}^{1} f = 2 \int_{0}^{1} f \ $, thus $ \int_{0}^{1} f \ $ = 1/2.

4) Suppose $ E \ $ has a finite measure. Show that $ L^2 (E) \subseteq L^1 (E) \ $ and that the map $ \psi: L^2(E) \to L^1(E) \ $ defined by $ \phi(f) = f \ $ is continuous.

Suppose $ E \ $ has a finite measure. Let $ \psi: L^2(E) \to L^1(E) \ $ be defined by $ \phi(f) = f \ $.

5) A trigonometric polynomial is a function of the form

$ p(x) = a_0 + \sum_{k=1}^{n} (a_k cos kx + b_k sin kx) \ $.

Let $ P \ $ denote the space of trig polynomials.

Lemma 1 ''Let $ f \ $ be a continuous 2$ \pi \ $ periodic real-valued function on $ \R \ $. For each positive number $ \epsilon \ $ there exists a trigonometric polynomial $ p \ $ such that |$ f(x) - p(x) \ $| < $ \epsilon \ $ for all $ x \ $.

(a) Let $ -\pi \le a < b \le \pi \ $. Use the above lemma to prove there exists $ p \in P \ $ such that $ \int_{-\pi}^{\pi} \ $ $|X_{[a,b]} (x) - p(x)|^2 < \epsilon \ $.

Let $ f \in C[X] $ where $ C[X] \ $ contains all continuous 2$ \pi \ $ periodic real-valued functions on $ \R \ $, and let $ X= [-\pi, \pi]\ $, where $ \pi, -\pi \ $ are identified. Then $ P \subset C[X] \ $. Then by Lemma 1, $ \forall \epsilon > 0 \exists \ $ a trigonometric polynomial $ p \ $ such that |$ f(x) - p(x) \ $| < $ \epsilon \ $ for all $ x \in \ [-\pi, \pi] \ $.

(b) Let $ f \in L^2 [-\pi, \pi] \ $ and $ [a, b] \ $ be as in part (a). Use part (a) and the Cauchy Schwarz inequality to prove there exists a trigonometric polynomial $ p \ $ such that $ | \int_{a}^{b} f - \int_{-\pi}^{\pi} fp | \ $ < $ \epsilon \ $.

(c) Suppose $ f \in L^2 [-\pi, \pi] \ $ has the property that $ \int_{-\pi}^{\pi} f(x) cos mx dx = \int_{-\pi}^{\pi} f(x) sin mx dx = 0 \ $ for $ m = 0, 1, 2, 3, \dots \ $. Prove that $ \int_{a}^{b} f(x) dx = 0 \ $ on every interval $ [a, b] \subseteq [-\pi, \pi ] \ $.

(d) Let $ f \ $ be as in part (c). Prove that $ f = 0 \ $ almost everywhere.

Let $ f \ $ be as in part (c), then $ f \ $ is integrable on $ [-\pi, \pi] \ $ and $ \int_{-\pi}^{\pi} f \ $ = 0. Then, let $ -\pi \le \alpha < \beta \le \pi \ $. Then $ \int_{\alpha}^{\beta} f \ $ = $ \int_{-\pi}^{\alpha} f - \int_{\beta}^{\pi} f \ $ = 0. So, $ \int_{\alpha,\beta} f \ $ = 0 $ \forall \ $ open intervals $ (\alpha, \beta) \ $. Now, let $ G \subset [-\pi, \pi] \ $. Then for all disjoint intervals, $ G = \bigcup_{i} (\alpha_i, \beta_i) \ $. So, $ \int_G f = \sum_{i} \int_{\alpha_i, \beta_i} f \ $ = 0, by the Lebesgue Dominent Convergence Thm. So now, let $ V \subset [-\pi, \pi] \ $ be closed. Then $ \int_V f + \int_{[-\pi,\pi]\ V} = \int_{[-\pi,\pi]} f \ $ = 0. Therefore $ \int_V \ $ = 0.

Then let $ E \ $ = {$ x \in [-\pi,\pi] : f(x) > 0 \ $}. Then suppose that $ mE > 0 \ $. Then $ E = \bigcup_{n=1}^{\infty} \ $ {$x : f(x) > 1/n \ $}. Then $ \exists n \in \N \ $ such that $ mE_n > 0 \ $. By the approximation theorem for measurable sets, $ \exists \ $ a closed set $ V \in E_n \ $ such that $ mV > (1/2)mE_n \ $. Hence, 0 = $ \int_V f \ge (1/n)mV > (1/n)(1/2)mE \ $ but this is a contradiction. Thus, $ f(x) \le 0 \ $ almost everywhere. Similarly, $ f(x) \ge 0 \ $ almost everywhere. And therefore, $ f = 0 \ $ almost everywhere.

(e) Show that the functions $ {1/ \sqrt{2\pi}, 1/\sqrt{\pi} cos nx,1/\sqrt{\pi}  sin nx: n = 1, 2, 3, \dots} \ $ is an orthonormal basis of $ L^2 [-\pi, \pi] \ $.

$ <1/ \sqrt{2\pi}, 1/\sqrt{\pi} cos nx, 1/\sqrt{\pi}  sin n> \ $ = $||1/ \sqrt{2\pi} + 1/\sqrt{\pi}  cos nx + 1/\sqrt{\pi}  sin nx || \ $ = $ ||1/ \sqrt{2\pi}|| + ||1/\sqrt{\pi}  cos nx|| + ||1/\sqrt{\pi}  sin nx || \ $ = $ \int_{-\pi}^{\pi} |1/ \sqrt{2\pi}| + \int_{-\pi}^{\pi} |1/\sqrt{\pi}  cos nx| + \int_{-\pi}^{\pi} + |1/\sqrt{\pi}  sin nx | \ $ = 0. Thus it is an orthonormal basis.

6) Let $ f \ $ be an integrable function on a measurable set $ E \ $. Define its distribution function $ F \ $ as follows: $ F (x) = m \ $ {$t:f(t) \le x \ $}.

Show (a) $ F \ $ is a non-negative, non-decreasing, and continuous from the right.

For all $ x, y \in E \ $ and $ 0 \le \lambda \le 1 \ $, $ F((1- \lambda)x + \lambda y) \le (1- \lambda)y(x) - \lambda F(y) \ $ $ \le x \ $. Thus $ F \ $ is convex, and is therefore non-negative and non-decreasing.

Let $ a \le U < X < Y < V \le b \ $. Then since $ F \ $ is convex, $ Y \ $ lies on or above $ UX \ $ and $ X \ $ lies on or below $ UY \ $. Now, Let $ y \to x^+ \ $. Then $ y(x) \le lim_{y \to x^+} inf F(y) \ $ and $ F(x) \ge lim_{y \to x^+} sup F(y) \ $ Thus $ lim_{y \to x^+} F(x) = y(x) \ $ and therefore $ F \ $ is continuous from the right.

(b) $ lim_{x \to -\infty} F(x) = 0 \ $.

(7) Let $ f: [0, 2] \to \R \ $ be the characteristic function of the interval (1/2, 1]. Find the distribution function for $ f\ $.

Let $ A \ $ = (1/2, 1]. Then let $ f_A (x) = \ $$ \begin{cases} 2,          & x \in A             \\ 0,     & x \notin A   \end{cases} \ $ is the characteristic function $ f: [0, 2] \to \R \ $.

The distribution function is $ F(x) \ $ = $ m \ $ {$t : f(t) \in [0, 2] \ $}, $ \forall x \in A \ $.

(8) Let $ f \ $ be a bounded measurable function fro [0, 1] into [0, $ M \ $] and let $ F \ $ be its distribution function. Show that $ \int_{0}^{1} f = \int_{0}^{M} x dF(x) \ $, where the second integral is the Riemann-Stieltjes integral of $ x \ $. (An approximating sum for $ \int_{0}^{M} g(x)dF(x) \ $ is the Riemann-Stieltjes sum given by $ \sum_{i=1}^{n} g(x_{i}^{*})(F(x_i)-F(x_{i-1})) \ $, where $ {x_0, x_1, \dots, x_n} \ $ is a partition of [0, $ M \ $] and $ x_{i}^{*} \ $ denotes a sample point in $[x_{i-1}, x_i] \ $.)