User:J D Bowen/Math710 HW6

5.1) Let $$ f \ $$ be the function defined by $$ f(0) = 0 \ $$ and $$ f(x) = x\sin (1/x) \ $$ for $$ x\ne 0 \ $$. Observe that

$$ D^{+}f(0) = \lim_{h\to 0^{+}}\sup \sin \frac{1}{h} = 1$$

$$ D_{+}f(0) = \lim_{h\to 0^{+}}\inf \sin \frac{1}{h} = -1$$

$$ D^{-}f(0) = \lim_{h\to 0^{+}}\sup \sin \frac{1}{-h} = 1$$

$$ D_{-}f(0) = \lim_{h\to 0^{+}}\inf \sin \frac{1}{-h} = -1$$

5.3a) Suppose $$ f \ $$ is continuous on $$ [a,b] \ $$ and assumes a local minimum at $$ c\in (a,b) $$.

Then $$ D_{-}f(c)\le D^{-}f(c)$$ and $$ D_{+}f(c)\le D^{+}f(c) $$. Since $$c \ $$ is a local minimum, there exists a $$ \delta > 0 \ $$ such that $$ f(c) < f(c+h) \ $$ and $$ f(c) < f(c-h) \ $$ for $$ 0 < h < \delta \ $$. Then

$$ f(c) < f(c+h) \implies f(c) - f(c+h) < 0 \implies f(c+h) - f(c) > 0 \implies \frac{f(c+h) - f(c)}{h} > 0 \ $$,

since $$ h > 0 \ $$. So $$ 0 \le \lim_{h\to 0^{+}}\inf \frac{f(c+h)-f(c)}{h} = D_{+}f(c) \ $$. Observe that

$$ f(c) < f(c-h) \implies f(c) - f(c-h) < 0 \implies \frac{f(c) - f(c-h)}{h} < 0 \implies D^{-}f(c) = \lim_{h\to 0^{+}}\sup \frac{f(c)-f(c-h)}{h} \le 0 $$.

Hence, $$ D_{-}f(c)\le D^{-}f(c)\le 0\le D_{+}f(c)\le D^{+}f(c) \ $$.

5.8a) We aim to show that given $$ a\le c\le b $$, we must have $$ T_{a}^{b} = T_{a}^{c} + T_{c}^{b} $$ and $$ T_{a}^{c} \le T_{a}^{b} $$.

Partition $$ [a,b] $$ such that $$ a=x_{0}<x_{1}<\cdots <x_{n}=b $$ and let $$ c = x_{k}, 0\le k\le n $$. Then $$ T_{a}^{b} = \sum_{i=1}^{n} |f(x_{i}) - f(x_{i-1})| = \sum_{i=1}^{k} |f(x_{i}) - f(x_{i-1})| + \sum_{i=k+1}^{n} |f(x_{i}) - f(x_{i-1})| \le T_{a}^{c} + T_{c}^{b}$$.

Now let $$ a=x_{0}<x_{1}<\cdots <x_{k}=c $$ be a partition of $$ [a,c] $$, and $$ c=x_{k}<x_{k+1}<\cdots <x_{n}=b $$ be a partition of $$ [c,b] $$.

Then $$ T_{a}^{c} + T_{c}^{b} = \sum_{i=1}^{k}|f(x_{i} - f(x_{i-1})| + \sum _{i=k+1}^{n}|f(x_{i} - f(x_{i-1})| \le T_{a}^{b}(f) $$. Therefore, $$ T_{a}^{b} = T_{a}^{c} + T_{c}^{b} $$. And since $$ T_{c}^{b} $$ is positive it directly follows that $$ T_{a}^{c} \le T_{a}^{b} $$.

5.8b)

Show that $$ T_{a}^{b}(f+g)\le T_{a}^{b}(f) + T_{a}^{b}(g) $$ and $$ T_{a}^{b}(cf) = |c|T_{a}^{b}(f) $$.

5.11) Suppose $$ f $$ is a function of bounded variation on $$ [a,b] \ $$.  We write this $$ f\in BV[a,b] \ $$.  This implies $$f'(x) \ $$ exists almost everywhere in $$ [a,b] \ $$. Let $$ f = g-h \ $$ where $$ g,h \ $$ are monotone and increasing. Then $$ g' \ $$ and $$ h' \ $$ exist almost everywhere with

$$ \int_{a}^{b}g'\le g(b) - g(a) $$,

$$ \int_{a}^{b}h'\le h(b) - h(a) $$,

and $$ g', h' \ge 0 $$.

Then

$$ \int_{a}^{b}|f'|\le \int_{a}^{b}g' + \int_{a}^{b}h'\le g(b) - g(a) + h(b) - h(a) = T_{a}^{b}(g) + T_{a}^{b}(h) = T_{a}^{b}(f) $$.

Hence, $$ \int_{a}^{b} |f'|\le T_{a}^{b}(f) $$.

5.14a) Let $$f, g \ $$ be two absolutely continuous functions.  This means $$\forall \epsilon \ $$ and pairwise disjoint finite sets of intervals $$[x_k, y_k] \ $$, there is a $$\delta \ $$ such that

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

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

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

Since $$\epsilon \ $$ was arbitrary, the sum or difference is absolutely continuous.

5.14b) Let $$f, g \ $$ be two absolutely continuous functions.  This means there is some $$M \ $$ 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 we have

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

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

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

5.18)  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 $$[x_k, y_k] \ $$, there is a $$\delta \ $$ such that

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

In particular, 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 $$\Sigma (y'_k-x'_k) \leq \epsilon' \ $$.

Taking the limit as $$N\to\infty \ $$, we have

$$\lim_{N\to\infty} \sum_{k=1}^n |y_k-x_k|<\delta \implies \sum_{k=1}^\infty |g(y_k)-g(x_k)|<\epsilon \ $$

but

$$g(E)\subset \bigcup (g(x_k),g(y_k)) \ $$

so

$$mg(E) < \epsilon \ $$.

Hence $$mg(E)=0 \ $$.

5.15) Remember that $$f \ $$ is monotone and that $$mC=0, \ mf(C) = 1 \ $$. Since a function which is monotone and absolutely continuous takes sets of measure zero to sets of measure zero, $$f \ $$ must not be absolutely continuous.