(Undo revision 3588 by Dvtran (Talk))
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<\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)dtS</math>
 
+
Let <math>g_{n}(x)= \sigma_{1}^{n}f_{n}(x)<\math>S
+

Revision as of 09:16, 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)dtS $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin