Namespaces
Variants
Actions

Difference between revisions of "Talk:Arveson spectrum"

From Encyclopedia of Mathematics
Jump to: navigation, search
Line 1: Line 1:
# Numbered list item
+
# 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?
# Numbered list item
 
A question: "vector-valued Riemann integral", is that the same thing as Bochner integrals?
 

Revision as of 10:54, 15 May 2014

  1. 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)

  1. 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=32160