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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
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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
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There is another meaning of "Schwartz space". Namely, a
Schwartz space is a locally convex space satisfying some property (which I don't remember) slightly weaker than that of a nuclear space. The article should at least mention that other meaning.
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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?
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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.
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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?
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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.
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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
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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. --
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is finite, and this is correct I believe. Where do you want to have a "sup"? --
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right, it's confusing. I've updated the introduction to say automorphism.
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are not necessariely integers. They're multiindices, i.e. elements of
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