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Hi,

Let $v(z)$ be a 2x2 matrix depending on a complex variable z, defined on an oriented contour $\Sigma$ in the complex plane. Can anyone tell me the meaning of the notation:

$$||v||_{L^{2}(\Sigma)}$$

As far as I can tell from the author, this quantity is intended to be a norm of some description and so should be a positive real number. I had a guess that it might mean "add the maximum of row 1 to the maximum of row 2, then integrate the absolute value squared over $\Sigma$ and finally take the square root".

Is there anyone who's seen this notation before who can confirm/deny my guess? The context of this notation is related to harmonic analysis and singular integral operators.

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  • $\begingroup$ I'm not sure whether your question is appropriate for this site, but anyway your guess is unlikely to be correct (unless max of row 1 plus max of row 2 is somehow important for reasons you're not telling us). There are various ways to take the norm of a matrix, probably the most standard is the operator norm, and most likely the rest of your description is on target --- square the norm at each point, integrate, and take a square root. But context is needed to answer this question. $\endgroup$
    – Nik Weaver
    Commented Apr 19, 2013 at 18:16
  • $\begingroup$ Thanks for your reply. It's from the paper www.math.kth.se/~haakanh/publications/deift.pdf, where the notation I refer to is first used in equation (7.52). I have seen it in a few other papers and books too, but I have never seen an actual definition! Quite frustrating. Also the norms in (7.79) and (7.80)? These are presumably a kind of operator norm? Again, there is no definition I can find in the text. $\endgroup$
    – Nigel
    Commented Apr 19, 2013 at 22:22
  • $\begingroup$ Well, for finite dimensional spaces, like $2\times 2$ matrices, all norms are equivalent. Thus, chose your favorite norm on $2\times 2$ matrices (I'd chose the $\sup$ over the coefficients), and then take the $L^2$ norm (with respect to $z$) of it. That's what I'll use for $\|v\|_{L^2(\Sigma)}$. $\endgroup$
    – Adrien Hardy
    Commented Apr 19, 2013 at 22:52
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    $\begingroup$ Thanks I think this (that all matrix norms are equivalent) answers my question and also explains why the authors didn't bother to define it precisely. Basically, we have $$||v||_{L^{2}(\Sigma)} = \left(\int_{\Sigma}|N(z)|^{2}dz\right)^{1/2}$$ where for each fixed z, N(z) is the supremum over the entries of v. $\endgroup$
    – Nigel
    Commented Apr 19, 2013 at 23:53

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