| Line 1: | Line 1: | ||
| − | 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< | + | 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. |
| − | Let <math>g_{n}(x)= \sigma_{1}^{n}f_{n}(x)< | + | Let <math>g_{n}(x)= \sigma_{1}^{n}f_{n}(x)</math> |
Revision as of 10:17, 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 $ and $ f_{n}^{'} $ are nonnegative almost everywhere.
Let $ g_{n}(x)= \sigma_{1}^{n}f_{n}(x) $
