User:MCPOliseno /Math710 CHAPTER 5

1 Let $$ f \ $$ be the function defined by f(0)=0 and f(x)=xsin(1/x) for x $$ \ne \ $$ 0. Find $$ D^+ \ $$ f(0), $$ D_+ \ $$ f(0), $$ D^- \ $$ f(0), $$ D_- \ $$ f(0).

$$ D^+ \ $$ f(0) = $$ \overline{lim}_{h \to 0^+} \frac{f(0+h)-f(0)}{h} \ $$ =$$ \overline{lim}_{h \to 0^+} \frac{f(h)}{h} \ $$ = $$ \overline{lim}_{h \to 0^+} \frac{hsin(1/h))}{h} = \overline{lim}_{h \to 0^+} sin (1/h) \ $$ = 1

$$ D_+ \ $$ f(0) = $$ \underline{lim}_{h \to 0^+} \frac{f(h)}{h} \ $$ = $$ \underline{lim}_{h \to 0^+} \ $$ sin (1/h) = -1

$$ D^- \ $$ f(0) = $$ \overline{lim}_{h \to 0^+} \frac{f(x)-f(x-h)}{h} \ $$ = $$ \overline{lim}_{h \to 0^+} \frac{-f(-h)}{h} \ $$ = $$  \overline{lim}_{h \to 0^+} \frac{-(-hsin(1/-h))}{h} \ $$ = $$ \overline{lim}_{h \to 0^+} \ $$ -sin(1/-h) = 1

$$ D_- \ $$ f(0) = $$ \underline{lim}_{h \ to 0^+} \frac{-f(-h)}{h} \ $$ = $$ \underline{lim}_{h \to 0^+} \ $$ -sin(1/-h) = -1

3(a) If $$ f \ $$ is continuous on [a, b] and assumes a local minimum at c $$ \in \ $$ (a, b), then $$ D_- f(c) \le D^- f(c) \le 0 \le D_+ f(c) \le D^+ f(c) \ $$.

Since c is a local minimum $$ \in \ $$ (a, b), then F(c) $$ \le \ $$ f(c+h) and F(c) $$ \le \ $$ f(c-h). Since lim inf $$ \le \ $$ lim sup, $$ D_+ f(c) \le D^+ f(c) \ $$. Now, $$ D_+ f(c) = \underline{lim}_{h \to 0^+} \frac{f(c+h)-f(c)}{h} \ $$, which is positive, sinceF(c) $$ \le \ $$ f(c+h), then 0 $$ \le D_+ f(c) \le D^+ f(c) \ $$. Similarly, since lim inf $$ \le \ $$ lim sup, $$ D_- f(c) \le D^- f(c) \ $$. And $$ D^- f(c) = \overline{lim}_{h \to 0^+} \frac{f(c)-f(c-h)}{h} \ $$ is negative since, F(c) $$ \le \ $$ f(c-h), thus it follows $$ D_- f(c) \le D^- f(c) \le 0 \ $$. Therefore $$ D_- f(c) \le D^- f(c) \le 0 \le D_+ f(c) \le D^+ f(c) \ $$.

8 (a) Show that if $$ a \le c \le b \ $$, then $$ T_{a}^{b} \ $$ = $$ T_{a}^{c} + T_{c}^{b} \ $$ and that hence $$ T_{a}^{c} \le T_{a}^{b} \ $$.

Construct partition P = {($$ x_{i-1}, x_i \ $$) | a = $$ x_0 \le x_1 \le \dots \le x_k \ $$ = b}. Let $$ P_1 \in \ $$ [a, c] and $$ P_2 \in \ $$ [c, b]. Then P = $$ P_1 \cup P_2 \ $$. Then $$ \sum_{P_1} |f(x_{i}) - f(x_{i-1})| \ $$ + $$ \sum_{P_e} |f(x_{i}) - f(x_{i-1})| \ $$ = $$ \sum_{i=1}^{k} |f(x_{i}) - f(x_{i-1})| \ $$ $$ \le T_{a}^{b} \ $$ (f). Suppose c $$ \in (x_{i-1}, x_i) \ $$. Then $$ P_1 \ $$ = {$$ (x_{i-1}, x_i) \ $$ | i = 1 $$ \dots \ $$ j-1} $$ \cup (x_{j-1}, c) \ $$ and $$ P_2 \ $$ = { $$(x_{i-1}, x_i) \ $$ | i = j $$ \dots \ $$ k} $$ \cup (c, x_j) \ $$. Then $$ P_1 + P_2 \ $$ =

$$ \sum_{i=1}^{j-1} \ $$ |f($$ x_{i-1} \ $$)- f($$ x_i \ $$)| + |f(c)- f($$x_{j-1} \ $$)| + $$ \sum_{i=j}^{k} \ $$ |f($$ x_{i-1} \ $$)- f($$ x_i \ $$)| + |f($$ x_j \ $$) - f(c)| $$ \le T_{a}^{b} \ $$ (f). Thus $$ T_{a}^{b} \ $$ = $$ T_{a}^{c} + T_{c}^{b} \ $$ and therefore it follows that $$ T_{a}^{c} \le T_{a}^{b} \ $$.

(b) Show that $$ T_{a}^{b} (f+g) \ $$ $$ \le T_{a}^{b} (f) \ $$ + $$ T_{a}^{b} (g) \ $$.

$$ T_{a}^{b} (f+g) \ $$ =

11 Let $$ f \ $$ be of bounded variation on [a, b]. Show that $$ \int_{a}^{b} |f'| \le T_{a}^{b}(f) \ $$.

Let f $$ \in \ $$ BV[a, b]. We can write f = g - h, with g, h monotone increasing. By Theorem 5.6, g' and h' exist almost everywhere with $$ \int_{a}^{b} \ $$ g' $$ \le \ $$ g(b) - g(a) and $$ \int_{a}^{b} \ $$ h' $$ \le \ $$ h(b) - h(a). Also, g', h' $$ \ge \ $$ 0. So, f' = g' - h', almost everywhere with |f'(x)| $$ \le \ $$ |g'(x)| + |h'(x)| = g'(x) + h'(x) almost everywhere. Then $$ \int_{a}^{b} \ $$ |f'(x)| = $$ \int_{a}^{b} \ $$ g' + $$ \int_{a}^{b} \ $$ h' $$ \le \ $$ g(b) - g(a) + h(b) - h(a) = $$ T_{a}^{b} \ $$ g + $$ T_{a}^{b} \ $$ h = $$ T_{a}^{b} \ $$ f.

14 (a) Show that the sum and difference of two absolutely continuous functions are also absolutely continuous.

Let f, g be absolutely continuous. Then f, g $$ \in \ $$ BV. Then $$ \sum_{i=1}^{n} \ $$ |f(y) - f(x)| < $$ \epsilon \implies \ $$ $$ \sum_{i=1}^{n} \ $$ |y-x| $$ \le \delta \ $$ and $$ \sum_{i=1}^{n} \ $$ |g(y) - g(x)| $$ \le \epsilon \implies \ $$ $$ \sum_{i=1}^{n} \ $$ |y-x| $$ \le \delta \ $$. Then $$ \sum_{i=1}^{n} \ $$ |f+g(y) - f+g(x)| $$ \le \sum_{i=1}^{n} \ $$ |f(y)-f(x) + g(y) -g(x)| $$ \le \ $$ 2 $$ \epsilon \ $$, which implies that $$ \sum_{i=1}^{n} \ $$ |y-x| $$ \le \ $$ 2 $$ \delta \ $$.

(b) Show that the product of two absolutely continuous functions is absolutely continuous. [Hint: Make use of the fact that they are bounded.]

Let f, g be two absolutely continuous functions. Then $$ \exists M \ $$ such that f(x),g(x)< M for $$ x \in \bigcup [x_k,y_k]=S \ $$, and that $$\forall \epsilon \ $$ and pairwise disjoint finite sets of intervals $$[x_k, y_k] \ $$, there is a $$ \delta \ $$ such that

$$\sum_{k=1}^N \left| y_k - x_k \right| < \delta\implies \sum_{k=1}^N | f(y_k) - f(x_k) |, \ \sum_{k=1}^N | g(y_k) - g(x_k) | < \epsilon/(2M)$$,

Then $$\sum_{k=1}^N |(fg)(y_k)-(fg)(x_k)| = \sum_{k=1}^N |f(y_k)g(y_k)-f(x_k)g(x_k)| \ $$

$$ = \sum_{k=1}^N |f(y_k)g(y_k)-f(x_k)g(x_k)-f(y_k)g(x_k)+f(y_k)g(x_k)| = \sum_{k=1}^N |f(y_k)(g(y_k)-g(x_k)) +g(x_k)(f(y_k)-f(x_k))| \ $$

$$ \leq \sum_{k=1}^N |f(y_k)|\cdot|g(y_k)-g(x_k)| +|g(x_k)|\cdot|f(y_k)-f(x_k)|=\sum_{k=1}^N |f(y_k)|\cdot|g(y_k)-g(x_k)| +\sum_{k=1}^N |g(x_k)|\cdot|f(y_k)-f(x_k)| \leq 2M\epsilon/(2M) = \epsilon \ $$

15 The Cantor ternary function is continuous and monotone but not absolutely continuous.

Note a function that is monotone and absolutely continuous takes sets of measure zero to sets of measure zero, and f is monotone and mC=0, mf(C) = 1. Thus f must not be absolutely continuous.

18 Let g be an absolutely continuous monotone function on [0, 1] and $$ E \ $$ a set of measure zero. Then $$ g[E] \ $$ has measure zero.

Note that since g is monotone, $$ x\in(x',y') \ $$ implies $$ g(x)\in(g(x'),g(y')) \ $$. Since g is absolutely continuous, for all $$\epsilon \ $$ and pairwise disjoint finite sets of intervals $$[x_k, y_k] \ $$, $$ \exists \delta \ $$ such that

$$\sum_{k=1}^N \left| y_k - x_k \right| < \delta\implies \sum_{k=1}^N | g(y_k) - g(x_k) | < \epsilon$$. This is true when $$(x'_k, y'_k)\subset [x_k, y_k] \ $$, where E $$\subset \bigcup_{k=1}^\infty (x'_k,y'_k) \ $$ and $$\Sigma (y'_k-x'_k) \leq \epsilon' \ $$.

Then by taking the limit as $$N\to\infty \ $$, then $$\lim_{N\to\infty} \sum_{k=1}^n |y_k-x_k|<\delta \implies \sum_{k=1}^\infty |g(y_k)-g(x_k)|<\epsilon \ $$, but g[E]$$\subset \bigcup (g(x_k),g(y_k)) \ $$, so it follows mg[E]$$ < \epsilon \ $$ and therefore mg[E]=0