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 $$ T_{a}^{c} \le T_{a}^{b} \ $$.

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

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.

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

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