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I was wondering how I can define a pseudodifferential operator using Cauchy integral formula.

Consider a differential operator $p(T)$ ($p$ is a polynomial for instance). $p(T)$ can be defined as:

$$\begin{equation} p(T) = \oint_{C} p(z)(zI-T)^{-1}dz \end{equation}$$

where contour $C$ encloses all eigenvalues of $T$ and $I$ is the identity operator.

My question: can a similar definition be used to define $p(x,T)$?

My understanding is that $p(x,T) = \oint_{C} p(x,z)(zI-T)^{-1}dz$ cannot be true in general if $T$ and $x$ do not commute.

Update:

I believe a clarification to this question is necessary given the discussion below. The objective of this question is to find a generalization/modification of Cauchy integral formula to make it applicable to pseduodifferential operators. I am convinced that $p(x,T) = \oint_{C} p(x,z)(zI-T)^{-1}dz$ is not valid in general.

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  • $\begingroup$ 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 $xT=\oint_{C} xz(zI-T)^{-1}dz$ $\endgroup$
    – Mirar
    Feb 1, 2023 at 9:29
  • $\begingroup$ If $xT=Tx$, I believe, the extended definition is valid which can be shown, in case of polynomials, by expanding $p(x,T)$. $\endgroup$
    – Mirar
    Feb 1, 2023 at 9:38
  • $\begingroup$ 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 $p(x,T)=1$. $\endgroup$ Feb 1, 2023 at 11:34
  • $\begingroup$ Exactly! In your example, since x and T donot commute, as stated in the question, the extension definition cannot be correct. $\endgroup$
    – Mirar
    Feb 1, 2023 at 11:38
  • $\begingroup$ 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. $\endgroup$
    – Mirar
    Feb 1, 2023 at 11:52

1 Answer 1

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Q1: Can one define the operator $p(x,T)$ (with $T=d/dx$) using the Cauchy integral formula?

That will not work, for example, take $p(x,T)=Tx-xT=1$, but $$p(x,T) = \oint_{C} p(x,z)(zI-T)^{-1}dz= \oint_{C} (zx-xz)(zI-T)^{-1}dz=0\neq 1.$$

Since the Cauchy integral formula is ignorant of commutation relations, it cannot be applied to non-commuting $x$ and $T$.

Q2: Is there some other way to compute the action of $p(x,T)$ on a (smooth) function $f(x)$?

This problem is a common one in physics. A generic way to proceed is to discretize the $x$ variable in $N$ points, and then to replace the differential operator $T$ by finite differences. The function $p(x,T)$ is represented by an $N\times N$ matrix, and the operation $p(x,T)f(x)$ can be carried out with $N^2$ operations.

An alternative, possibly more efficient method, is the socalled split-operator approach, for $p$ of the form $p(x,T)=\exp[A(x)+B(T)]$. One then starts from the Trotter formula $$e^{A+B} = \lim_{n \rightarrow \infty} (e^{A/n}e^{B/n})^n$$ and inserts Fourier transform operations ${\cal F}$, $$e^{A(x)+B(T)}f=\lim_{n \rightarrow \infty} (e^{A/n}{\cal F}^{-1}e^{\tilde{B}/n}{\cal F})^n f,\;\;\tilde{B}={\cal F}B{\cal F}^{-1}.$$ The operators $A$ and $\tilde{B}$ are both diagonal, so they act on a function simply by multiplication. The time consuming step is the Fourier transform, if $f$ is discretized in $N$ points that operation takes a time that scales as $N\log N$. The total computation scales as $nN\log N$, which for $n\ll N$ improves on the $N^2$ scaling of the generic method.

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  • $\begingroup$ Unfortunately I cannot accept this answer. Q1 is not what was asked and Q2 is very specific. However, I really appreciate your time. $\endgroup$
    – Mirar
    Feb 3, 2023 at 6:27
  • $\begingroup$ I added the "generic" method to my answer to Q2. $\endgroup$ Feb 3, 2023 at 6:49

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