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