Saturday, 20 September 2014

real analysis - limprightarrowinfty||x||p=||x||infty given ||x||infty=max(|x1|,|x2|)

I have seen the proof done different ways, but none using the norm definitions provided.



Given:
||x||p=(|x1|p+|x2|p)1/p and ||x||=max(|x1|,|x2|)




Prove:
limp



I have looked at the similar questions:
The l^{\infty} -norm is equal to the limit of the l^{p} -norms. and Limit of \|x\|_p as p\rightarrow\infty but they both seem to use quite different approaches (we have not covered homogeneity so that is out of the question, and the other uses a different definition for the infity norm).

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