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Oct 1, 2016 at 3:03 comment added fedja "What do you think about the simple proof?" It is even better, of course :-) Anyway, this way, or that, the nice thing is that it works and another nice thing is that you know this stuff now.
Sep 30, 2016 at 22:23 comment added Fedor Petrov If you want to attract someone's attention, write their name after at sign, like @fedja
Sep 30, 2016 at 21:33 comment added user98553 What do you think about the simple proof?
Sep 30, 2016 at 21:32 comment added user98553 I successfully laid out the proof. Estimate derivation is very bulky, it is 'one dimensional' in essence. When I finished the text I suddenly realized that there is a simple way to derive it from the one dimesional Wiener theorem: let us consider $f(z)=\det \varphi(z)$, it belongs to $W^1$ since $W^1$ is an algebra. $\varphi$ is point-wise invertible by assumption hence $f(z)\ne 0$ for all $z$. Therefore by one dimensional Wiener theorem $\frac{1}{\det \varphi} \in W^1$. Hence $\varphi^{-1}\in W$ (its matrix elements are products of $\frac{1}{\det \varphi}$ with matrix elements of $\varphi$).
Sep 30, 2016 at 21:06 history edited user98553 CC BY-SA 3.0
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Sep 18, 2016 at 20:30 comment added fedja You shouldn't be that literal when reading mathematics. Of course, the series should be the one you get from $f^{-1}=(g-h)^{-1}=(I-g^{-1}h)^{-1}g^{-1}=[I+g^{-1}h+g^{-1}hg^{-1}h+g^{-1}hg^{-1}hg^{-1}h+\dots]g^{-1}$. However, if you honestly write the matrix elements, you can put the entries of $g$ together in the products, after which the proof works. Will you figure it out yourself, or you want me to spell it out for you?
Sep 16, 2016 at 16:33 review First posts
Sep 16, 2016 at 16:59
Sep 16, 2016 at 16:20 history answered user98553 CC BY-SA 3.0