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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)dt</math> and <math>f_{n}^{'}</math> are nonnegative almost everywhere. | 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)= \sum_{k=1}^{n}f_{k}(x)=\sum_{1}^{n}\int_{0}^{x}f_{k}^{'}(t)dt=\int_{0}^{x}\sum_{k=0}^{n} </math> | + | Let <math>g_{n}(x)= \sum_{k=1}^{n}f_{k}(x)=\sum_{1}^{n}\int_{0}^{x}f_{k}^{'}(t)dt=\int_{0}^{x}(\sum_{k=0}^{n}f_{k}^{'}(t))dt </math> |
Revision as of 09:25, 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)= \sum_{k=1}^{n}f_{k}(x)=\sum_{1}^{n}\int_{0}^{x}f_{k}^{'}(t)dt=\int_{0}^{x}(\sum_{k=0}^{n}f_{k}^{'}(t))dt $
