Revision as of 13:30, 11 July 2008 by Dvtran (Talk)

a/$ \mu({|f|>0})>0 $, so we have $ (\int_{X}|f|^{n})^{1/n} \leq (\mu(X)||f||_{\infty})^{1/n} $

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett