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 $$[x_{i-1}, x_i] \ $$.)