Let denote the space of sequences such that , equipped with the norm . Let denote the usual norm on .

Show that , and hence deduce that .

Any help would be appreciated. Thanks in advance

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- Apr 13th 2010, 06:25 AMejgmathNormed sequence space
Let denote the space of sequences such that , equipped with the norm . Let denote the usual norm on .

Show that , and hence deduce that .

Any help would be appreciated. Thanks in advance - Apr 13th 2010, 07:05 AMJose27
- Apr 13th 2010, 07:14 AMLaurent
This will prove the inclusion, but not the comparison of norms, I think.

For that one, you can write for all , which comes from the expansion of the left-hand side (all terms of the expansion are nonnegative, and the right-hand side consists in a few of those). And let . - Apr 13th 2010, 07:19 AMejgmath
Hi, thanks for the response, I am really sorry but I made a small mistake in the original post. I missed the squared out of the 2-norm, does this change much. Also, how do you deduce the second part of the question? I have included the revised version below. Thanks

- Apr 13th 2010, 07:27 AMLaurent
- Apr 14th 2010, 05:14 AMejgmath
Could someone help explain the logic here?

I don't understand how follows from .

If the vectors in have a greater length than the ones in how can be a subset of ?

Many thanks to anyone who can help - Apr 14th 2010, 05:43 AMEnrique2
Just think that means that is finite, and the same for with . You can look your geometric argument by the point of view that there are vectors with "infinity length" with the measure , but with finite length with the smaller masure , as for example . The contrary is not possible because this inequality