Difference between revisions of "Talk:Arveson spectrum"
From Encyclopedia of Mathematics
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| − | # | + | # A rapid check shows that $\hat{x}(n)$ does not exactly satisfy the indicated equation but rather with an inverse |
| − | A rapid check shows that $\hat{x}(n)$ does not exactly satisfy the indicated equation but rather with an inverse | ||
$$U_z \hat{x}(n) = (z)^{-1} \hat{x}(n)$$ | $$U_z \hat{x}(n) = (z)^{-1} \hat{x}(n)$$ | ||
| − | |||
Even if there is no ambiguity, it would also be better if we make it clear that $\hat{x}(n)$ is still a function on $T$ (while usually in Fourier transform, the transform is a function on the "Fourier space" which I admit is just a word that doesn't explain anything) | Even if there is no ambiguity, it would also be better if we make it clear that $\hat{x}(n)$ is still a function on $T$ (while usually in Fourier transform, the transform is a function on the "Fourier space" which I admit is just a word that doesn't explain anything) | ||
| − | + | # A question: "vector-valued Riemann integral", is that the same thing as Bochner integrals? | |
| − | # | ||
| − | A question: "vector-valued Riemann integral", is that the same thing as Bochner integrals? | ||
Revision as of 10:54, 15 May 2014
- A rapid check shows that $\hat{x}(n)$ does not exactly satisfy the indicated equation but rather with an inverse
$$U_z \hat{x}(n) = (z)^{-1} \hat{x}(n)$$ Even if there is no ambiguity, it would also be better if we make it clear that $\hat{x}(n)$ is still a function on $T$ (while usually in Fourier transform, the transform is a function on the "Fourier space" which I admit is just a word that doesn't explain anything)
- A question: "vector-valued Riemann integral", is that the same thing as Bochner integrals?
How to Cite This Entry:
Arveson spectrum. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Arveson_spectrum&oldid=32161
Arveson spectrum. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Arveson_spectrum&oldid=32161