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Infinity Norm June 18, 2009

Posted by putnam120 in Math Related.
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Well here is a question that I used to wonder about for a while.

If f is integrable over U, then

\displaystyle\lim_{n\to\infty}\left\{\int_U|f|^nd\mu\right\}^{\frac 1n}=\sup_U|f|.

Well here is a proof of a simplified version.  I only assume that M=\sup|f|<\infty, and \mu(U)<\infty

proof: By “the world’s most obvious integral inequality” \int_U|f|^nd\mu\le M^n\mu(U). So obviously ||f||_\infty\le M.

For the reverse direction let \epsilon>0 then |f|>M-\epsilon on some set K\subset U. Therefore \int_U|f|^n\ge (M-\epsilon)^n\mu(K). To get the result take n\to\infty then \epsilon\to 0. \mathbb{Q.E.D.}

The only part that bothers me is making sure that K is measurable, but that shoudn’t be too difficult to fix.

Comments»

1. Correct “Infinity Norm” « Thoughts on my Mind - February 6, 2010

[...] in Uncategorized. trackback A while back you might have seen that I wrote a post about the “infinity norm“.  Well in that post I assumed that the measure of the entire space was finite, which was a [...]

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