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A formal pseudodifferential operator is a series in integer powers of $\partial =d/dx$, bounded above but possibly infinite towards negative powers. Its coefficients are smooth functions. Multiplication uses the formal pseudodifferential composition rule. Negative powers are formal inverses, not globally defined integration operators on all smooth functions. Every coefficient in a product is a finite sum, so multiplication is defined without analytic convergence.

The formal residue is the coefficient of ${\partial }^{-1}$ after coefficients have been moved to the left by the formal pseudodifferential composition rule. It is not an analytic residue at a point. For scalar coefficients, the residue of a commutator is a total derivative; integrating it with vanishing boundary terms produces the cyclic Adler trace.

Treat $k$ as a formal spectral parameter near infinity. If $P={\sum }_{j\ge 0}{g}_j{\partial }^{\alpha -j}$, the formal pseudodifferential composition rule gives $Pw=k^{\alpha +\beta }{e}^{kx}{\sum }_{s\ge 0}{k}^{-s}{\sum }_{i+j+r=s}{g}_j\left(\genfrac{}{}{0ex}{}{\alpha -j}{r}\right)w_i^{(r)}$. Each coefficient contains finitely many terms. This describes a formal module action, not a claim that the infinite series converges at finite $k$.

Write the two formal pseudodifferential operators in normal order aswhere $p={\alpha }_1$ and $q={\alpha }_2$. The formal pseudodifferential composition rule givesThus $\boxed{\mathrm{ord}(P_1{P}_2)=p+q}$ whenever $ab$ is not identically zero, in particular when both leading coefficients are nowhere zero. In general the rigorous statement over the specified coefficient algebra is $\mathrm{ord}(P_1{P}_2)\le p+q$.

The leading scalar coefficients commute. Subtracting the reversed composition also cancels the terms involving $a_1,b_1$, leavingTherefore the order filtration of formal pseudodifferential operators giveswith equality precisely when the displayed coefficient is not identically zero. The bound can be strict: constant-coefficient powers of $\partial$ commute, giving the zero operator, whose order can be assigned $-\infty$.

There is also a genuine qualification to product-order addition in $C^{\infty }(\mathbb{R})$. Choose nonzero smooth functions $a,b$ with disjoint compact supports and take $P_1=m_a,P_2=m_b$. Each has order zero, but their product is zero. Nonzero coefficients as elements of this algebra need not have a nonzero product. Thus an unqualified assertion of exact product order requires an additional leading-symbol hypothesis.

Use the formal pseudodifferential composition rule on each coefficient $w_i(x)$ and then the exponential eigenfunction convention. For every integer $n$,ConsequentlyCollecting all terms of a fixed power of the spectral parameter gives the useful answerThe sum defining each $v_s$ is finite. For example,andThis is the action on a formal pseudodifferential wave function. It is a formal expansion near $k=\infty$, with $k^{\beta }$ carried as a formal prefactor. It does not assert convergence for finite $k$, and negative powers are not evaluated at $k=0$.

First let $n\ge 0$. The ordinary Leibniz rule, applied to a test function $h$, givesEquivalently, the composition of differential operators isFor completeness, this formula follows by induction: applying $\partial$ differentiates either the coefficient or the remaining derivative of $h$, and the two contributions combine by $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$. The initial case is multiplication by $f$.

For all integers $n$, define $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n(n-1)\cdots (n-r+1)/r!$. The formal pseudodifferential composition rule isFor $n=-s<0$, $\left(\genfrac{}{}{0ex}{}{-s}{r}\right)=(-1{)}^r(\genfrac{}{}{0ex}{}{s+r-1}{r})$, so this is usually an infinite series. In particularApplying $\partial$ on the left gives $m_f$: all later coefficient derivatives cancel in pairs. This uniquely determines the normal-ordered inverse expansion. Repeating the inverse construction, or using the same binomial recurrence downward in $n$, gives every negative power. More explicitly, applying $\partial$ to the proposed $n$-series yields coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ at $f^{(r)}{\partial }^{n+1-r}$, so it is consistent with the already determined $(n+1)$-series. Negative powers here are formal; differentiation on all of $C^{\infty }(\mathbb{R})$ has a nontrivial kernel and has no genuine two-sided inverse there.

Use the canonical representatives of functionals of $u$: $P_f=f{\partial }^{-1}$, $P_g=g{\partial }^{-1}$. The formal pseudodifferential composition rule givesand the reversed product has $f,g$ exchanged. ThereforeMultiplying by $L={\partial }^2+u$ on the left, only the leading displayed commutator term can produce a residue. Differentiation of its coefficient produces powers at most ${\partial }^{-2}$, and multiplication by $u$ produces powers at most ${\partial }^{-3}$. Thusand, with the trace convention of part (i),There is no $u$ dependence in this bracket.

Moreover $\mathrm{res}(P_{f}L)=fu$, so $l_{P_f}=\int fudx$ and $\delta l_{P_f}/\delta u=f$. If $K(x,y)=\{u(x),u(y)\}$ is the distributional field bracket, it is characterized bySince $\int {\partial }_x\delta (x-y)g(y)dy=g^{\prime }(x)$, the answer isThe Dirac delta is periodic in the periodic setting, or the ordinary Dirac delta on the real line with admissible test functions. This is the first Hamiltonian structure of the KdV equation, with Poisson operator $-2{\partial }_x$. Its skewness follows by integration by parts; its coefficients are constant, so the functional Jacobi identity has no coefficient-variation terms.

A qualification is needed if the displayed trace formula is interpreted as a bracket on the reduced space for arbitrary operator representatives. Multiplication by a function $h$ gives $l_{m_h}({\partial }^2+u)=0$, yet direct expansion givesWith $h=g=\cos x$ on a period $[0,2\pi ]$, the latter is $\pi$. Hence the formula does not descend through every possible representative of a functional on this reduced affine space. The requested $f{\partial }^{-1},g{\partial }^{-1}$ representatives, used consistently for variational gradients, give the well-defined constant field bracket just computed; a general restriction from the ambient operator algebra needs an appropriate reduction.

Put every formal pseudodifferential operator in normal order, with all coefficient functions on the left. Its formal pseudodifferential residue is the coefficient of ${\partial }^{-1}$. The Adler trace isHere and below, use periodic coefficients and integrate over a period, or impose convergence and vanishing boundary terms on the real line. Equivalently, the algebraic integration functional must annihilate total derivatives. This convention is needed for the requested trace property; arbitrary elements of $C^{\infty }(\mathbb{R})$ do not supply it automatically.

It suffices to compute with monomials $A=a{\partial }^p$, $B=b{\partial }^q$. Write $r=p+q+1$. If $r<0$, no term in either product has power ${\partial }^{-1}$, so both residues are zero. If $r\ge 0$, the formal pseudodifferential composition rule givesSince $q=r-1-p$, reversal of the factors in the numerator provesAfter $r$ integrations by parts, $\int a{b}^{(r)}dx=(-1{)}^r\int b{a}^{(r)}dx$. Hence the two product traces agree. More explicitly, for $r\ge 1$ their residue difference is the total derivativeFor $r=0$ the residue difference is $ab-ba=0$. In products of operators bounded above in order, only finitely many monomial pairs can contribute to a residue: the required $r$ is nonnegative, so the two downward coefficient indices have a bounded sum. Summing the monomial result therefore proves

The boundary convention is substantive. Without it, take $A={\partial }^2$, $B=(\tanh x){\partial }^{-2}$. Both individual residues vanish, but $\mathrm{res}[A,B]=2(\tanh x{)}^{\prime }=2{\mathrm{sech}}^{2}x$ and its integral over $\mathbb{R}$ is $4$, not zero. The primitive has unequal limits at the two ends. Also a generic smooth residue need not have a convergent integral at all. Thus the proof supplies the intended cyclic trace under its necessary analytic or formal integration convention.