Timeline for Is it possible to define pseudodifferential operator $p(x,T)$ using Cauchy integral formula?
Current License: CC BY-SA 4.0
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| Feb 2, 2023 at 21:58 | history | edited | Mirar | CC BY-SA 4.0 |
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| Feb 2, 2023 at 21:35 | history | edited | Mirar | CC BY-SA 4.0 |
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| Feb 2, 2023 at 20:52 | answer | added | Carlo Beenakker | timeline score: 2 | |
| Feb 2, 2023 at 7:25 | comment | added | Mirar | Let us continue this discussion in chat. | |
| Feb 1, 2023 at 15:56 | history | edited | Mirar | CC BY-SA 4.0 |
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| Feb 1, 2023 at 11:55 | comment | added | Mirar | Valid point. Do you believe we cannot define $p(x,T)$ in terms of Cauchy integral in any way? | |
| Feb 1, 2023 at 11:52 | comment | added | Mirar | Anyway, I am convinced that the extended definition is not true in general and I was wondering whether it would be possible to find a similar definition. | |
| Feb 1, 2023 at 11:38 | comment | added | Mirar | Exactly! In your example, since x and T donot commute, as stated in the question, the extension definition cannot be correct. | |
| Feb 1, 2023 at 11:34 | comment | added | Carlo Beenakker | for $T=d/dx$ consider $p(x,T)=Tx-xT$; if you insert that in your proposed formula, you would get $$p(x,T) = \oint_{C} p(x,z)(zI-T)^{-1}dz= \oint_{C} (zx-xz)(zI-T)^{-1}dz=0,$$ because $zx-xz=0$, but the correct answer is $p(x,T)=1$. | |
| Feb 1, 2023 at 9:38 | comment | added | Mirar | If $xT=Tx$, I believe, the extended definition is valid which can be shown, in case of polynomials, by expanding $p(x,T)$. | |
| Feb 1, 2023 at 9:29 | comment | added | Mirar | In my understanding, if it were to be symmetrical, $T$ and $x$ would commute, isn't it? Consider, for instance, $p(x,T)=(x+T)^2$. By expanding, $p(x,T)=x^2+xT+Tx+T^2$ consists of four operators and each of them can be defined by the integral above. E.g., $Tx=\oint_{C} z(zI-T)^{-1}xdz$ and $xT=\oint_{C} xz(zI-T)^{-1}dz$ | |
| Feb 1, 2023 at 6:46 | history | edited | Mirar | CC BY-SA 4.0 |
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| Jan 31, 2023 at 21:21 | history | asked | Mirar | CC BY-SA 4.0 |