We know that $\lim_{p\rightarrow\infty}\left\Vert \left(x_{1},\cdots,x_{n}\right)\right\Vert _{p}=\max\left\{ \left|x_{1}\right|,\cdots,\left|x_{n}\right|\right\} =:\left\Vert x\right\Vert _{\infty}$ for any $\left(x_{1},\cdots,x_{n}\right)\in R^{n}$. Now do we have $\lim_{p\rightarrow0}\left\Vert \left(x_{1},\cdots,x_{n}\right)\right\Vert _{p}=\left\Vert \left(x_{1},\cdots,x_{n}\right)\right\Vert _{0}:=\mbox{cardinality}\left\{ x_{i}:x_{i}\neq0\right\} $?
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2$\begingroup$ I tried to make the question a little more sensible. $\endgroup$– Reid BartonCommented Mar 21, 2010 at 6:49
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2$\begingroup$ Depends on what you mean by $\|\cdot\|_p$ when $0<p<1$. This is obviously false if $\|x\|_p=(\sum |x_k|^p)^{1/p}$ (consider $n=1$) and obviously true if $\|x\|_p=\sum |x_k|^p$. $\endgroup$– Jonas MeyerCommented Mar 21, 2010 at 6:56
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3$\begingroup$ The first bullet point at mathoverflow.net/faq#whatnot lists some sites where you can find further help. $\endgroup$– Jonas MeyerCommented Mar 21, 2010 at 7:11
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3$\begingroup$ Should add that I have downvoted this question based on lack of context (is this an exercise? a curiosity? something needed to make progress? etc.) and low level of question. $\endgroup$– Yemon ChoiCommented Mar 21, 2010 at 17:43
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1$\begingroup$ In addition to Jonas' comment, there is a third interesting limit: $\lim_{p \to 0} ((1/n) \sum_{k=1}^n x_k^p)^{1/p}$ has a very pretty value. $\endgroup$– David E SpeyerCommented Mar 22, 2010 at 16:21
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1 Answer
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Hardy-Littlewood-Polya Inequalities.
Contains considerable material on these $(|x_1|^p+\cdots+|x_n|^p)^{1/p}$, including for example $p=-1$ and of course limiting cases at $p=\infty$, $p=0$, $p=-\infty$.
answered Mar 21, 2010 at 12:44