User:Kcbetancourt/AnalysisHW7

5.1

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

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

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

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

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

5.3a

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

Clearly, $$ D_{-}f(c)\le D^{-}f(c)$$ 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 $$ h > 0 $$. Then $$ 0 \le \lim_{h\to 0^{+}}\inf \frac{f(c+h)-f(c)}{h} = D_{+}f(c) $$. Furthermore, $$ f(c) < f(c-h) \implies f(c) - f(c-h) < 0 \implies \frac{f(c) - f(c-h)}{h} < 0 \implies D^{-}f(c) = \lim_{h\to 0^{+}}\sup \frac{f(c)-f(c-h)}{h} \le 0 $$. Thus, $$ D_{-}f(c)\le D^{-}f(c)\le 0\le D_{+}f(c)\le D^{+}f(c) $$.

5.8a

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

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 $$ T_{a}^{b}(cf) = |c|T_{a}^{b}(f) $$. $$ T_{a}^{b}(f+g) = \sum_{i=1}^{n} |(f+g)(x_{i}) - (f+g)(x_{i-1})| \le \sum_{i=1}^{n} |f(x_{i}) - f(x_{i-1})| + \sum_{i=1}^{n} |g(x_{i}) - g(x_{i-1})| \le T_{a}^{b}(f) + T_{a}^{b}(g) $$. Thus, $$ T_{a}^{b}(f+g)\le T_{a}^{b}(f) + T_{a}^{b}(g) $$

Let $$ c\in \R\ $$ Suppose $$ c = 0 $$. Then, $$ T_{a}^{b}(cf) = T_{a}^{b}(0) = 0 = |0|T_{a}^{b}(f) =|c|T_{a}^{b}(f) $$.

Now suppose $$ c\ne 0 $$. Then $$ T_{a}^{b}(cf) = \sum_{i=1}^{n} |cf(x_{i}) - cf(x_{i-1})| = |c|\sum_{i=1}^{n} |f(x_{i}) - f(x_{i-1})| \le |c|T_{a}^{b}(f) $$. So, $$ T_{a}^{b}(cf) \le |c|T_{a}^{b}(f) $$.

$$ T_{a}^{b}(f) = \sum_{i=1}^{n} |f(x_{i}) - f(x_{i-1})| = |c^{-1}||c|\sum_{i=1}^{n} |f(x_{i}) - f(x_{i-1})| = |c^{-1}|\sum_{i=1}^{n} |cf(x_{i}) - cf(x_{i-1})| \le |c^{-1}|T_{a}^{b}(cf) $$ $$ \implies T_{a}^{b}(f) \le |c^{-1}|T_{a}^{b}(cf) \implies |c|T_{a}^{b}(f) \le |T_{a}^{b}(cf) $$.

Therefore, $$ T_{a}^{b}(cf) = |c|T_{a}^{b}(f) $$.

5.11

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

$$ f\in BV[a,b] \implies f'(x) $$ 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 $$ \int_{a}^{b}g'\le g(b) - g(a) $$ and $$ \int_{a}^{b}h'\le h(b) - h(a) $$ and $$ g', h' \ge 0 $$. Then $$ \int_{a}^{b}|f'|\le \int_{a}^{b}g' + \int_{a}^{b}h'\le g(b) - g(a) + h(b) - h(a) = T_{a}^{b}(g) + T_{a}^{b}(h) = T_{a}^{b}(f) $$. Therefore, $$ \int_{a}^{b} |f'|\le T_{a}^{b}(f) $$.

5.14a

Show that the sum and difference of two absolutely continuous functions are absolutely continuous.

Let $$ f $$ and $$ g $$ be two absolutely continuous functions on the interval $$ [a,b] $$. Then, given $$ \varepsilon > 0, \exists \delta > 0 $$ such that $$ \sum_{i=1}^{n}|f(x_{i}') - f(x_{i})|<\frac {\varepsilon}{2} $$ and $$ \sum_{i=1}^{n}|g(x_{i}') - g(x_{i})|<\frac {\varepsilon}{2} $$ with $$ \sum_{i=1}^{n}|x_{i}' - x_{i}|<\delta $$ for every finite collection of non-overlapping intervals $$ {(x_{i}, x_{i}')} $$. Then $$ \sum_{i=1}^{n}|(f+g)(x_{i}') - (f+g)(x_{i})| \le \sum_{i=1}^{n}|f(x_{i}') - f(x_{i})| + \sum_{i=1}^{n}|g(x_{i}') - g(x_{i})|< \frac {\varepsilon}{2} + \frac {\varepsilon}{2} = \varepsilon $$. And $$ \sum_{i=1}^{n}|(f-g)(x_{i}') - (f-g)(x_{i})| \le \sum_{i=1}^{n}|f(x_{i}') - f(x_{i})| + \sum_{i=1}^{n}|g(x_{i}') - g(x_{i})|< \frac {\varepsilon}{2} + \frac {\varepsilon}{2} = \varepsilon $$. Therefore, the sum and difference of $$ f $$ and $$ g $$ are absolutely continuous.

5.14b

Show that the product of two absolutely continuous functions are absolutely continuous. Hint: Make use of the fact that they are bounded.

Since $$ f $$ and $$ g $$ are bounded, there exists $$ M $$ such that $$ |f|\le M $$ and $$ |g|\le M $$ almost everywhere on $$ [a,b] $$. Since $$ f $$ and $$ g $$ are absolutely continuous, we can redefine them so that $$ \sum_{i=1}^{n}|f(x_{i}') - f(x_{i})|<\frac {\varepsilon}{2M} $$ and $$ \sum_{i=1}^{n}|g(x_{i}') - g(x_{i})|<\frac {\varepsilon}{2M} $$. Then, $$ \sum_{i=1}^{n}|fg(x_{i}') - fg(x_{i})| \le \sum_{i=1}^{n}|f(x_{i}')||g(x_{i}') - g(x_{i})| + \sum_{i=1}^{n}|g(x_{i})||f(x_{i}') - f(x_{i})| \le M\frac{\varepsilon}{2M} + M\frac{\varepsilon}{2M} = \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$. Therefore, the product of $$ f $$ and $$ g $$ is absolutely continuous.

5.15

The Cantor ternary function (Problem 2.48) is continuous and monotone but not absolutely continuous.

Problem 2.48 tells us that the Cantor ternary function, $$ f $$, is continuous and monotone, and constant on each interval in the complement of C, which implies that $$ f' = 0 $$ almost everywhere on $$ [0,1] $$. Also, $$ \text{m}C = 0 $$. Assume to the contrary that $$ f $$ is absolutely continuous. Then $$ \int_{0}^{1}f' = f(1) - f(0) $$ However, $$ \int_{0}^{1}f' = 0 \ne 1 = f(1) - f(0) $$. Thus we have reached a contradiction, therefore the Cantor function is not absolutely continuous.

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.

Let $$ \varepsilon > 0 $$. Then there exists an open set $$ O $$ such that $$ E\subset O $$ and $$ \text{m}O = \text{m}O - \text{m}E = \text{m}(O\setminus E) < \delta $$. Since $$ O $$ is open, it can be written as a countable union of disjoint open intervals, $$ I_{n} $$, $$ O = \bigcup I_{n} $$. Then, $$ \sum l(I_{n}) = \sum \text{m}O < \delta $$. Since $$ g $$ is continuous on $$ [0,1] $$, $$ \sum l(I_{n}) < \delta \implies \sum l(g(I_{n}\cap [0,1])) < \varepsilon $$. Since $$ g[E] \subset \bigcup g(I_{n}\cap [0,1]) $$, $$ \text{m}g[E] < \varepsilon $$. Let $$ \varepsilon $$ go to zero. Thus, $$ \text{m}g[E] = 0 $$