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.