Difference between revisions of "Exam3.2 Old Kiwi" - Rhea

Revision as of 13:07, 8 July 2008

2a.

$ ( \Rightarrow ) $ Say $ f $ is A.C. Then $ f $ is of bounded variation, and since $ f $ is clearly nondecreasing, $ f $ must be bounded.

In particular, $ \infty > f(1)=\sum_{n=1}^\infty m([0,1]\cap G_n) \Rightarrow \sum_{n=1}^\infty m(G_n)<\infty $.

$ ( \Leftarrow ) $

Now say $ \sum_{n=1}^\infty m(G_n)<\infty $.

I claim $ f'(x) = \sum_{n=1}^\infty \chi_{G_n}(x) $ a.e. as follows:

For $ h>0 $, $ \frac{f(x+h)-f(x-h)}{2h}=\frac{1}{2h}\sum_{n=1}^\infty m([x-h,x+h]\cap G_n)=\frac{1}{2h}\sum_{n=1}^\infty \int_{x-h}^{x+h} \chi_{G_n}=\frac{1}{2h} \int_{x-h}^{x+h} \sum_{n=1}^\infty \chi_{G_n}(x) $ (MCT)

Taking $ lim_{h\rightarrow 0} $ of both sides we get $ f'(x) $ on the left and $ \sum_{n=1}^\infty \chi_{G_n}(x) $ a.e. on the right (Lebesgue's Differentiation Theorem).

Now, since $ \sum_{n=1}^\infty m(G_n)<\infty $, we have $ \int_0^1 f' = \int_0^1 \sum_{n=1}^\infty \chi_{G_n}= \sum_{n=1}^\infty \int_0^1 \chi_{G_n}=\sum_{n=1}^\infty m(G_n)<\infty $ $ \Rightarrow f' $ is integrable on $ [0,1] $.

Finally, $ f(x)-f(0)=\int_0^x f' $.

So $ f $ is A.C.

b) $ ( \Rightarrow ) $

Say $ f $ is Lipschitz. Then $ \exists $ an integer $ M $ such that $ |f'(x)| < M \ \forall \ x \in [0,1] $. So $ \sum_{n=1}^\infty \chi_{G_n}(x)<M \ a.e. $(by part a)

$ \Rightarrow $ For almost every $ x\in [0,1], x\not\in G_m \forall \ m>M \Rightarrow m(G_m)=0 \ \forall \ m>M $, but $ G_m $ is open $ \Rightarrow G_m= \O \ \forall \ m>M $.

$ (\Leftarrow ) $

Now say $ \exists \ N $ such that $ n>N \Rightarrow G_n $ is empty.

Then for any $ y<x \in \ [0,1] $,

$ \frac{|f(x)-f(y)|}{|x-y|}=\frac{\sum_{n=1}^N m[y,x]\cap G_n}{|x-y|} \leq \frac{\sum_{n=1}^N m[y,x]}{|x-y|} \leq \frac{N |x-y|}{|x-y|} = N $.

So $ f $ is Lipschitz.