(New page: 3a) Suppose that there is such an <math>f</math>. Then we may choose <math>N</math> large enough such that <math>m(\{x:f^*<f\}\cap[-N,N])>0</math>. Call this set <math>A</math>. Let <m...) |
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| + | Say <math>f \neq 0 \in L^1</math>. Then <math>\exists \ \epsilon >0</math>, an integer <math>N</math> and a measurable set <math>E</math> such that, <math>m(E)>0, |f(x)|>\epsilon \ \forall \ x \in E</math> and <math>E \subset [-N,N]</math>. | ||
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| + | Then for any <math>x > N</math>, | ||
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| + | <math>f^*(x) \geq \frac{\int_{x-(x+N)}^{x+(x+N)} \epsilon \chi_E}{2(x+N)} \geq \frac{\epsilon m(E)}{2(x+N)}</math> | ||
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| + | Thus, <math>\int_{\mathbb{R}} f^* \geq \int_N^\infty \frac{\epsilon m(E)}{2(x+N)} dx = \infty</math>. | ||
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| + | So <math>f^* \not \in L^1</math>. | ||
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| + | --[[User:Wardbc|Wardbc]] 15:04, 8 July 2008 (EDT) | ||
Latest revision as of 14:04, 8 July 2008
3a)
Suppose that there is such an $ f $. Then we may choose $ N $ large enough such that $ m(\{x:f^*<f\}\cap[-N,N])>0 $. Call this set $ A $.
Let $ E_n=\{ x\in A:|f(x)|\in [n,n+1)\} $. Then $ \exists \ k $ such that $ m(E_k)>0 $.
Now, $ f $ is bounded on $ E_k $, and $ E_k $ has finite measure, so $ f(x) \chi_{E_k}(x) $ is integrable over $ \mathbb{R} $.
So by Lebesgue's Differentiation Theorem, $ f(x)\chi_{E_k}(x)\leq (f(x)\chi_{E_k}(x))^* $ for a.e. $ x\in E_k $
But, for every $ x \in E_k, x \in A \Rightarrow, f^*(x)<f(x)=f(x)\chi_{E_k}(x) $.
Thus, $ f^*(x)<(f(x)\chi_{E_k}(x))^* a.e $.
But, for any $ x $, and any $ Q $ centered at $ x $, $ \frac{\int_Q |f(x)|}{m(Q)}\geq \frac{\int_Q |f(x)|\chi_{E_k}(x)}{m(Q)}. \Rightarrow f^*(x)\geq (f(x)\chi_{E_k}(x))^* $ (taking the sup of both sides over all such $ Q $).
But $ m(E_k)>0 $, so this is a contradiction.
So $ \not\exist $ such an $ f $.
b)
Say $ f \neq 0 \in L^1 $. Then $ \exists \ \epsilon >0 $, an integer $ N $ and a measurable set $ E $ such that, $ m(E)>0, |f(x)|>\epsilon \ \forall \ x \in E $ and $ E \subset [-N,N] $.
Then for any $ x > N $,
$ f^*(x) \geq \frac{\int_{x-(x+N)}^{x+(x+N)} \epsilon \chi_E}{2(x+N)} \geq \frac{\epsilon m(E)}{2(x+N)} $
Thus, $ \int_{\mathbb{R}} f^* \geq \int_N^\infty \frac{\epsilon m(E)}{2(x+N)} dx = \infty $.
So $ f^* \not \in L^1 $.
--Wardbc 15:04, 8 July 2008 (EDT)
