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<math>\lim_{n\to \infty}||f||_{n} = ||f||_{\infty}</math>
 
<math>\lim_{n\to \infty}||f||_{n} = ||f||_{\infty}</math>
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Let <math>M<||f||_{\infty} </math>

Revision as of 13:35, 11 July 2008

a/$ \mu(\{|f|>0\})>0 $, so we have

$ (\int_{X}|f|^{n})^{1/n} \leq (\mu(X)||f||_{\infty})^{1/n} $

Taking the limit of both side as $ n $ go to infinity, we get

$ \lim_{n\to \infty}||f||_{n} = ||f||_{\infty} $

Let $ M<||f||_{\infty} $

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

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