# Talk:Schwartz space

## Wrong definition?

The definition seems to be incorrect. Aren't the multi-indices α,β in the first formula supposed to be n-tuples of natural numbers including zero (see this reference)? In its current form, the definition excludes the zeros from the multi-indices by defining them as n-tuples of positive integers.

## Notation

This article provokes cognitive dissonance : the different notations ((x^α D^β)f and x^{\alpha} D{\beta} f) don't seem to agree. Is this x^α \times D^β f or something else? Kromsson (talk) 16:12, 22 December 2008 (UTC)

## Wrong Formula for Schwartz-Space?

If you compare the current description (7/7/09) with the one on planetmath.org (source), it becomes evident, that a "sup" is missing in front of the sumpremum norm, does it not? 84.190.110.27 (talk) 19:33, 7 July 2009 (UTC)

No, I don't think so. Here, it says that for every multi-index, the supremum norm (defined in a different article) of ${\displaystyle \|f\|_{\alpha ,\beta }}$ is finite, and this is correct I believe. Where do you want to have a "sup"? --Bdmy (talk) 19:47, 7 July 2009 (UTC)

## 2 Remakrs

1. I think it is worthy to explain what the D^β means - that it is a differential operator operating β times on f. Also, that α and β are integers.
2. I think it is worthy to note that S is dense in L2, and that that is how the Fourier transform is defined on L2. --Rockyrackoon (talk) 16:00, 20 January 2010 (UTC)

${\displaystyle \alpha ,\beta }$ are not necessariely integers. They're multiindices, i.e. elements of ${\displaystyle \mathbb {N} ^{n}}$. 84.139.145.50 (talk) 18:18, 25 January 2010 (UTC)

## Internal Inconsistency

The second sentance states "the Fourier transform is an endomorphism on this space", the page on endomorphisms states "An invertible endomorphism of X is called an automorphism", implicitly implying the Fourier transform, in this space, is not, in all cases, invertible. However in the properties section lower down the page Property 4 states "The Fourier transform is a linear isomorphism \mathcal{S} \to \mathcal{S}." Implying the FT is always invertible in this space.

(Comment above was by 131.227.74.40 on 18 May 2010)

Saying that something is an endomorphism doesn't rule out the possibility that it's an automorphism (just like saying a function is greater than -1 doesn't rule out that it's strictly positive). But if by your first "implying" you mean the English language sense i.e. "hinting that", then you're right, it's confusing. I've updated the introduction to say automorphism. Quietbritishjim (talk) 13:50, 20 June 2010 (UTC)