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

Revision as of 13:30, 11 July 2008

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

Alumni Liaison

Meet a recent graduate heading to Sweden for a Postdoctorate.

Christine Berkesch