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| − | Since all the <math>f_{n}</math> are AC, there exists <math>f_{n}^{'}</math> such that <math>f_{n}(x)=f_{n}(x)-f_{n}(0)=\int_{0}^{x}f_{n}^{'}(t) | + | Since all the <math>f_{n}</math> are AC, there exists <math>f_{n}^{'}</math> such that <math>f_{n}(x)=f_{n}(x)-f_{n}(0)=\int_{0}^{x}f_{n}^{'}(t)dt<\math> and <math>f_{n}^{'}<\math> are nonnegative almost everywhere. |
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| + | Let <math>g_{n}(x)= \sigma_{1}^{n}f_{n}(x)<\math>S | ||
Revision as of 09:17, 10 July 2008
Since all the $ f_{n} $ are AC, there exists $ f_{n}^{'} $ such that $ f_{n}(x)=f_{n}(x)-f_{n}(0)=\int_{0}^{x}f_{n}^{'}(t)dt<\math> and <math>f_{n}^{'}<\math> are nonnegative almost everywhere. Let <math>g_{n}(x)= \sigma_{1}^{n}f_{n}(x)<\math>S $
