Jim Arvo wrote:
Hi Outermeasure. If I'm understanding your notation correctly, you are saying that this norm is the finite-dimensional analog of the operator norm on the space of linear functions from
. (Actually, that should be
, right?) But all operator norms are automatically submultiplicative, just as all induced matrix norms are, right? If so, then the norm I defined is not an operator norm, and that's exactly my concern. Here's another way to see the difference. Let's call my norm
denotes the ith column of the identity matrix. So my norm is clearly bounded above by the operator norm that you suggested, but the equality cannot hold in general because the latter is submultiplicative and the former is not (except when p=1). Am I missing or misinterpreting something?
No! It is the operator norm for
. I'll leave you to prove it is the case.
Also, it is not submulticative, and it makes no sense to multiply because they m and n are not equal (which is why you say M is an mxn matrix). Even if they are equal, you still have the problem of
. Of course, if p=1 and m=n, then you do indeed have the usual Banach algebra norm on