Difference between revisions of "Solution to number 5 Old Kiwi" - Rhea

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-Fix <math>\epsilon > 0</math>.
-Then <math>\exists \ N</math> such that <math>n>N \Rightarrow \int_X|f_n-f|^p <\epsilon/3</math>.
-Define <math>f_0=f</math>.
-Then for <math>i=0,1,...,N \ \exists \ \delta_i>0 </math> such that <math> m(A) <\delta_i \Rightarrow \int_A|f_i|^p<\epsilon/3</math>, since <math>f_i \in L^p</math>.
-Define <math>\delta=min\{\delta_0, \delta_1,...,\delta_N, \epsilon/3\}</math>.
-Fix A with <math>m(A)<\delta</math>.
-Note: <math>\int_A|f_n| = \int_{\{x\in A:|f_n|\leq1\}}|f_n|+\int_{\{x\in A:|f_n|>1\}}|f_n| \leq m(A) + \int_A|f_n|^p \leq \epsilon/3 + \int_A|f_n|^p \ \ \forall \ n=0,1,...</math>.
-Then if n = 1,...,N
-<math>\int_A|f_n|\leq 2\epsilon/3 < \epsilon</math> (by our choice of <math>\delta</math>).
-and if
-n > N
-then
-<math>\int_A|f_n|\leq \epsilon/3 + \int_A|f_n|^p\leq \epsilon/3 + \int_A|f|^p+\int_A|f_n-f|^p\leq \epsilon/3 + \int_A|f|^p+\int_X|f_n-f|^p\leq \epsilon</math>.
-So <math>m(A)<\delta \Rightarrow \int_A|f_n| <\epsilon \ \forall \ n</math>.

Latest revision as of 20:10, 1 July 2008