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 $