Difference between revisions of "Problem Set 7 7.6 Old Kiwi" - Rhea

Latest revision as of 14:59, 13 July 2008

6.

First we show $ f_n\rightarrow f $ in $ L^p $.

Let $ \delta > 0 $. Let $ \epsilon = \frac {\delta }{1+2^{p+1}} $ Then by Egorov $ \exists E \subset X $ such that

$ m(E) > m(X)-\epsilon $ and $ |f_n-f|<\frac{\epsilon}{m(X)} $ on $ E \ \forall \ n>N $, some N.

Also, by Fatou, $ \int_X |f^p| \leq lim_n \int |f_n^p| \leq 1 $ (by hypothesis)

So $ \int_X |f-f_n|^p=\int_E |f-f_n|^p+\int_{X-E} |f-f_n|^p\leq m(X)\frac{\epsilon}{m(x)}+\int_{X-E} (|f|+|f_n|)^p\leq \epsilon + 2^p(\int_{X-E} (|f^p|+|f_n^p|) \leq \epsilon + 2^p(2m(X-E) $) (Holder)

So $ \int_X |f-f_n|^p \leq \epsilon (1+2^{p+1})=\delta $ if $ n>N \Rightarrow f_n\rightarrow f in L^p $.

Now,

$ |\int_X fg-f_ng|=|\int_X g(f-f_n)| \leq \|{g}\|_q \|{f-f_n}\|_p \rightarrow 0 $ Thus $ \int_X f_ng \rightarrow \int_X fg $.

--Wardbc 16:56, 10 July 2008 (EDT)

This is wrong due to an incorrect use of Holder at the end of the 5th line. It's not necessarily true that $ f_n \rightarrow f in L_p $

[Slaps forehead with palm]

--Wardbc 15:59, 13 July 2008 (EDT)