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Timeline for Showing non-attainment of supremum

Current License: CC BY-SA 4.0

14 events
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Sep 23, 2018 at 15:28 history edited Pietro Majer CC BY-SA 4.0
TeX fixed
Mar 11, 2011 at 6:10 vote accept Ashok
Oct 17, 2010 at 7:13 vote accept Ashok
Oct 20, 2010 at 7:20
Oct 17, 2010 at 6:04 comment added Pietro Majer 1. that's a general fact about $L^p$ spaces: if $fj$ is a sequence in $L^p(X,\mu)$ converging to $f$, then $|fj|^p$ converges to $|f|^p$ in $L^1(X,\mu).$ (Sketch of Proof: up to extracting a subsequence we may assume $fj$ converges a.e. and is dominated in $L^p$; then apply dom.conv.thm). 2. Yes, they are different: the latter set is just the set of all $tu$ with $0\le t\le 1$ and $u$ in the former set. So what has been added does not affect the value of the supremum, for the reason that $q(x)\ge 0, \forall x$. The reason for introducing $\hat E$ is just convexity.
Oct 17, 2010 at 5:17 comment added Ashok Thank you very much Pietro Majer. I have two questions on your solution. 1. How does $p_j/\|pj\|_{\alpha}$ converges to $\frac{u}{\|u\|_{\alpha}}inl^{\alpha}$ imply $(p_j/\|p_j\|_{\alpha})^{\alpha}$ converges to $(u/\|u\|_{\alpha})^{\alpha}$ in $l^1$? 2.In the last line,how arethosetwo supremums equal? Wehave $\{\frac{p}{\|p\|_{\alpha}}:p\in E\}\subset \hat{E}$ But these two sets are not equal right?
Oct 16, 2010 at 17:29 comment added Pietro Majer Oh, Piero! :-)
Oct 16, 2010 at 17:27 history edited Pietro Majer CC BY-SA 2.5
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Oct 16, 2010 at 10:57 comment added Piero D'Ancona Try inserting some spaces. Especially before undrescores
Oct 16, 2010 at 10:26 comment added Pietro Majer I met TeX troubles :-( I hope you can read it.
Oct 16, 2010 at 9:49 history edited Pietro Majer CC BY-SA 2.5
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Oct 16, 2010 at 9:31 history edited Pietro Majer CC BY-SA 2.5
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Oct 16, 2010 at 9:23 history edited Pietro Majer CC BY-SA 2.5
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Oct 16, 2010 at 9:16 history edited Pietro Majer CC BY-SA 2.5
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Oct 16, 2010 at 9:10 history answered Pietro Majer CC BY-SA 2.5