Full sign conditions for mixed weighted polynomials

A nonnegative first Poisson bracket at a characteristic point does not exclude a decreasing odd crossing whose first derivative is zero. A mixed weighted polynomial makes this distinction concrete. Its full sign condition forces an even time factor and a monotone quotient. Conversely, these two properties control every oriented characteristic curve, even after multiplication by an arbitrary smooth nonzero complex function.

The exact prerequisites are Theorem 3.1 and Exercise 4 of Sign-constrained weighted Taylor geometry, the weighted Taylor and leading-bracket results in Weighted packets and the limit of the gain, and smooth Hamilton flows in Phase space and generating families. The convention is

Lerner [L] explains the distinction between the symbol condition for an estimate and that for its adjoint. Hörmander [H] treats the weighted polynomial geometry. Hörmander [H3], Section 21.1, gives the Hamilton-flow background. The arguments below prove the full polynomial restrictions and their converse.

1. The symbol and its conjugate have opposite sign tests

Let \(r\) be a smooth complex function on a real symplectic domain. For any smooth complex multiplier \(a\) with no zeros, write

An oriented characteristic curve of \(F\) is a nonconstant integral curve \(\gamma\) of \(H_F\) lying in \(F=0\). The forward parameter is the Hamilton-flow parameter. Constant curves cannot exhibit a sign change.

We use the following precise orientation conventions:

• condition \((\Psi)\) excludes \(G(\gamma(u))<0<G(\gamma(v))\) for \(u<v\);
• condition \((\overline{\Psi})\) excludes \(G(\gamma(u))>0>G(\gamma(v))\) for \(u<v\).

These tests apply to every nonzero smooth multiplier on the working neighborhood. Thus they include nonconstant phases and moduli. Since

\(r\) satisfies \((\overline{\Psi})\) exactly when \(\overline r\) satisfies \((\Psi)\).

For \(r=\tau+iq(t,z,\eta)\), with \(q\) independent of \(\tau\), the multiplier \(a=1\) gives increasing \(t\). Thus \((\overline{\Psi})\) excludes a positive-to-negative change of \(q\). This is the orientation relevant to the favorable estimate in Finite type and the sign of the symbol. The conjugate symbol has the standard \((\Psi)\) orientation. The different names do not change the Hamilton or Fourier conventions.

Lemma 1.1 (comparison with a lower-order multiple). Suppose a real differentiable function \(Y\) on an interval satisfies

where \(E,S\) are continuous. If \(Y(u)\geq0\), then \(Y(v)\geq0\) for every later \(v\).

Proof. Multiply by the positive integrating factor

The derivative of the resulting product is the same positive factor times \(S\). The product is nondecreasing. This proves the assertion even when \(S\) vanishes on an interval, and therefore includes zero plateaus. ∎

2. Positive balanced dilations preserve the full condition

Give each coordinate pair \((x_j,\xi_j)\) positive weights \(m_j,\mu_j\), with

Let \(\delta_\varepsilon\) multiply each coordinate by its corresponding power of \(\varepsilon>0\). For the symplectic form and rescaled symbol,

Lemma 2.1. If \(r\) satisfies \((\overline{\Psi})\), then \(r_\varepsilon\) satisfies it wherever the rescaled symbol is defined.

Proof. Given a nonzero smooth multiplier \(a\) for \(r_\varepsilon\), define the multiplier \(a\circ\delta_\varepsilon^{-1}\) for \(r\). Denote the real part of the multiplied original symbol by \(f\). The real part after rescaling is \(\varepsilon^{-k}f\circ\delta_\varepsilon\). The Hamilton fields obey

One can check this in each coordinate pair: differentiation of the rescaled function contributes the coordinate weight, and the conjugate coordinate contributes the complementary weight \(k+1\). The factor left over is \(\varepsilon\). Thus \(\delta_\varepsilon\) takes each characteristic curve to an original characteristic curve with a positive change of time parameter. The imaginary parts differ by the positive factor \(\varepsilon^{-k}\). Neither sign order changes. ∎

Lemma 2.2. The full condition \((\overline{\Psi})\), tested on nonconstant characteristic curves, is closed under \(C^2\) convergence on compact subsets of a common domain.

Proof. Suppose the limit violates the condition. Fix its nonzero smooth multiplier \(a\), a compact characteristic-curve segment, and two points on that segment with strict opposite signs in the forbidden order. Write its real part as \(F\). The Hamilton field cannot vanish anywhere on this nonconstant segment: uniqueness for a smooth autonomous ordinary differential equation would otherwise make the entire curve constant.

At the initial point \(dF\ne0\). Choose one coordinate direction in which its derivative is nonzero. The corresponding derivatives for the approximating real parts \(F_\nu\) stay bounded away from zero in a small coordinate box. Moving the initial point only in that direction, the intermediate value theorem gives points converging to it with \(F_\nu=0\).

Their Hamilton fields converge in \(C^1\) on a compact tube around the segment. The ordinary differential equations starting at these corrected initial points converge uniformly through its fixed finite time interval. Explicitly, if \(L\) is a common Lipschitz constant there, their difference is bounded by

This follows by integrating the difference of the equations and multiplying the resulting integral inequality by \(e^{-Lt}\). A smaller tube has a positive boundary buffer; the bound keeps the approximate flows inside the larger tube and hence ensures their existence through time \(T\).

Each approximate curve stays in \(F_\nu=0\), since \(H_{F_\nu}F_\nu=0\). Its imaginary part converges at the two selected times, so both strict signs persist. This contradicts the condition for the approximants. The fixed multiplier remains nonzero on the tube. ∎

Apply the weighted Taylor lemma from the prerequisite lesson, with a finite ordinary Taylor expansion long enough to control two scaled derivatives. If \(r\) vanishes to weight \(k\), then \(r_\varepsilon\) converges in \(C^2\) on compact sets to its weight-\(k\) polynomial. The rescaled domains contain every fixed compact set for small enough \(\varepsilon\), because all weights are positive. Lemmas 2.1–2.2 therefore give the full condition to the polynomial on its entire coordinate space.

This argument uses the conformal symplectic identity (2.2). A general anisotropic dilation with unequal pair sums would not give the common positive time factor (2.3).

3. Classify the mixed polynomial

Use two coordinate pairs \((t,\tau)\), \((z,\eta)\). Let \(k\geq2\) and \(s\geq0\) be integers with \(k>2s\). Consider

where \(b\) is a real polynomial of weight \(k\) for weights \(1,s+1\), and no monomial of \(b\) has \(t\)-power \(s\). These are the exact leading data of the mixed theorem in Sign-constrained weighted Taylor geometry.

Assume also its already proved slope condition:

Here \(B\) is a polynomial; for \(s=0\) it means \(b_t\). The prerequisite proof and its Exercise 4 show \(s\ne1\) and \(t^3\mid b\) when \(s>0\).

Theorem 3.1. Under these hypotheses, the following are equivalent:

1. \(r\) satisfies the full condition \((\overline{\Psi})\);
2. \(s\) is even, the quotient \(g\) below is polynomial, and

3. \(Q\) has no positive-to-negative change on any increasing time line with \(z,\eta\) fixed.

For \(s=0\), the quotient is \(b\). Equivalently, \(\tau-iQ\) satisfies standard \((\Psi)\). The converse includes every smooth nonzero complex multiplier, rather than only the curves of \(H_\tau\).

An odd transverse order already gives a forbidden crossing

Suppose \(s\) is odd. On the line \(z=0\), weighted homogeneity gives \(b(t,0)=c t^k\). Fix any \(\eta<0\). Since \(k>2s>s\),

has a positive value for sufficiently small negative \(t\) and a negative value for sufficiently small positive \(t\). The curve with these fixed \(z,\eta\) and \(\tau=0\) is a characteristic curve of \(H_\tau\). This violates \((\overline{\Psi})\). The points and fixed negative parameter can be chosen arbitrarily close to zero, so the argument also excludes a local condition there. Thus \(s\) is even.

A lower power of time in \(b\) is also forbidden

If \(s>0\) and \(t^s\) does not divide \(b\), choose its smallest \(t\)-power \(i<s\) with nonzero coefficient polynomial \(b_i(z)\). The earlier restrictions imply \(i\geq3\). Choose \(z_0\) with \(b_i(z_0)\ne0\). The lowest term of \(B(t,z_0)\) as \(t\to0\) is

Nonnegativity on both sides forces \(i\) odd and \((i-s)b_i(z_0)>0\). Since \(i<s\), this means \(b_i(z_0)<0\). With \(\eta=0\), the corresponding lowest term of \(Q\) is \(b_i(z_0)t^i\), giving a forbidden positive-to-negative crossing. Weighted homogeneity makes the coefficient \(b_i\) a nonzero scalar multiple of a power of \(z\); consequently such a \(z_0\) can be chosen arbitrarily small. Higher \(t\)-powers cannot alter the crossing close enough to \(t=0\).

Therefore \(t^s\mid b\). The missing \(t^s\) monomial then gives \(g(0,z)=0\). Moreover

The even factor is positive when \(t\ne0\), so (3.2) gives \(g_t\geq0\) there. Polynomial continuity gives it at \(t=0\) as well.

When \(s=0\), (3.2) already gives \(g_t=b_t\geq0\), and the missing \(t^0\) monomial gives \(g(0,z)=0\). This proves necessity in every case.

The even factor controls all multipliers

Write

Then

Let \(a=a_1+ia_2\) be any nonzero smooth multiplier, allowed to depend on all four coordinates. On a characteristic curve of

the equation \(F=0\) gives

Expanding the Poisson bracket by the product rule gives

for a smooth real \(E\) there. Indeed, the terms not differentiating the multiplier give \(|a|^2\{\tau,Q\}\). Every other term contains at least one factor \(\tau\) or \(Q\); substitution of (3.10) factors out \(G\). Quadratic terms have the same property. No derivatives of the multiplier have been discarded.

Where \(t\ne0\),

The source term is nonnegative. Lemma 1.1 prevents a forbidden sign change on every compact interval avoiding \(t=0\). If \(s=0\), the first equation is simply \(Q_t=A_t\); (3.11) then gives comparison everywhere and proves the converse already.

Now let \(s>0\), hence \(s\geq2\), and consider a point on a nonconstant curve with \(t=0\) and \(G=0\). Equation (3.10) gives \(\tau=Q=0\). Since \(dQ=0\) at \(t=0\), the Hamilton field there is

It cannot vanish on the nonconstant curve, so \(a_1\ne0\). On a neighborhood of this point we can use \(F=0\) in the form \(\tau=(a_2/a_1)t^sA\). Equation (3.8) then yields

Use the smooth comparison variable \(Y=a_1 A\). Its equation and its relation to \(G\) are

Again the source is nonnegative. The factor relating \(G\) to \(Y\) is nonnegative and is positive off \(t=0\). This removes the singular coefficient in (3.12) while preserving the sign on both sides.

To include a whole compact curve segment, not just one crossing, observe that its points with \(t=G=0\) are finite in number. At each such point (3.13) gives \(t'\ne0\), making that zero of \(t\) isolated. Their set is closed in the compact parameter interval; an infinite set would have an accumulation point where the same transversality would fail.

Choose small neighborhoods of these finitely many points on which (3.15) holds. On the remaining zero set of \(G\), the coordinate \(t\) is bounded away from zero, so (3.12) applies. Where \(G\ne0\), its sign stays constant on sufficiently small intervals. A finite subdivision subordinate to these neighborhoods can be chosen with all division points avoiding the finitely many points \(t=G=0\).

Starting from \(G>0\), comparison propagates \(G\geq0\) through every interval of the subdivision. In an interval using \(Y\), its initial point has \(t\ne0\) if \(G=0\), and therefore \(Y\geq0\); (3.15) propagates this inequality and then gives \(G\geq0\). In a region where \(G\ne0\), continuity preserves its sign. The argument also works across a zero plateau. A later negative value is impossible.

Thus the full condition holds for every multiplier and every such characteristic segment. The first condition implies the third by taking \(a=1\), and both necessity arguments above used only that fixed multiplier. The third therefore implies the second; the second implies the first by the comparison proof. This completes all three equivalences. ∎

4. Every allowed polynomial has a homogeneous realization

The leading polynomial is centered at a zero frequency in its ordinary symplectic coordinates. To realize it as a homogeneous principal symbol at a nonzero covector, add a third coordinate pair \((w,\rho)\).

Theorem 4.1. Every polynomial satisfying (3.3), with the stated weighted degree and missing monomial, occurs as the mixed leading polynomial of

This symbol is real-homogeneous of degree one in \((\tau,\eta,\rho)\), is of principal type, and satisfies \((\overline{\Psi})\) on the cone \(\rho>0\). Its conjugate satisfies \((\Psi)\). It has the exact mixed normalization, all bracket values through \(k\) factors vanish at \(c\), and the invariant short-rank index is \(s\). If \(b\ne0\), its bracket type is exactly \(k\). If \(b=0\), its bracket type is infinite.

Proof. Homogeneity follows directly from (4.1), and \(\partial_\tau\widetilde r=1\). The exact normalization is

because the polynomial \(b\) has no \(t^s\) term. Every differential jet before \(s\) vanishes at the marked point: the frequency term still contains time, or has zero \(\eta\) there, and \(b=t^s g\) has a further time factor and the prescribed higher weighted degree.

For the full condition use the proof in Section 3 with

It is independent of \(\tau,w\), so \(\{t,\widetilde A\}=0\), \(\{\widetilde Q,\widetilde A\}=0\), and

Every comparison equation (3.11)–(3.15) therefore holds, including arbitrary multipliers depending on the added pair. This proves the asserted full condition.

In the shifted Taylor coordinates at \(c\), write \(\rho=1+\nu\). Give the three pairs the weights

The extra term is \(\nu b\), of weight \(k+(k+1)/2>k\). The weight-\(k\) polynomial is exactly (3.1). All terms vanish to weight \(k\); the leading-bracket theorem proves vanishing of every bracket with at most \(k\) factors. The mixed theorem from the prerequisite lesson applies to the exact normalization (4.2) and gives the short-rank index \(s\).

If \(b\ne0\), then \(B=t^s g_t\ne0\): otherwise \(g\) would be independent of \(t\), and \(g(0,z)=0\) would make it zero. The endpoint formula in that same theorem gives a nonzero \(k+1\)-factor bracket, so the type is \(k\).

If \(b=0\), the two generating functions are \(\tau\) and \(t^s\eta/s!\), independent of \(z\). Their bracket is a time derivative of the second function. Every further nonzero bracket is another time derivative and retains a factor \(\eta\); brackets between two such functions are zero. Thus every bracket function vanishes at \(c\), proving infinite type. The short-rank statement still holds: after \(s\) time derivatives, its Hamilton field adds the \(z\) direction. ∎

For a nonzero \(b\), the parity conclusion can also be seen directly. The quotient \(g\) has weight \(k-s\), so \(g_t\) has weight \(k-s-1\). Since \(s\) is even, the reflection \((t,z)\mapsto(-t,-z)\) multiplies \(g_t\) by \((-1)^{k-s-1}\). A nonzero nonnegative polynomial cannot be odd under this reflection. Hence \(k\) is odd.

There are no additional restrictions on which \(b\) can occur beyond the stated polynomial, degree, normalization and monotonicity conditions: (4.1) realizes each of them. This realization does not assert the analytic sufficiency of a subelliptic estimate. The full variable-symbol estimates remain a separate step.

5. Return the classification to a smooth symbol

Suppose a smooth normalized germ satisfies all hypotheses of the mixed theorem and also the full condition \((\overline{\Psi})\). Its balanced dilation has the weight-\(k\) limit (3.1). The remainder vanishes to weight at least \(k+1/2\); a sufficiently long ordinary Taylor expansion makes it tend to zero in \(C^2\) on every compact rescaled set.

Lemmas 2.1–2.2 pass the full condition to that leading polynomial. For the restrictions, it also suffices to use only the fixed multiplier \(a=1\): exact time lines rescale to exact time lines, and strict opposite signs on a fixed compact segment persist under convergence. Thus the third condition of Theorem 3.1 holds in the limit, forcing \(s\) even and (3.3). This argument does not require choosing a uniform neighborhood for a family of changing complex phases. It proves the additional full-sign restrictions for the actual smooth germ, rather than merely for a model chosen in advance.

The converse has two distinct meanings. Theorem 4.1 realizes every allowed leading polynomial by a full homogeneous symbol that has the full condition. An arbitrary higher-weight perturbation of that polynomial need not preserve the condition. For instance a flat parameter-dependent term can create a nearby adverse odd crossing without changing any Taylor jet at the marked point; Crossing direction and the necessary finite-type bound, Exercise 6, proves precisely that phenomenon. The neighborhood condition must still be checked on the full symbol.

6. Exercises with complete solutions

Exercise 1 — identify the adjoint orientation, 6 points. In the cone \(\eta>0\), take \(r=\tau+i t^5\eta\). Which of \(r,\overline r\) satisfies the favorable forward sign test, and which satisfies standard \((\Psi)\)? Explain the first nonzero endpoint without changing the sign of \(D_t=-i\partial_t\).

Solution. Along \(H_\tau=\partial_t\), \(t^5\eta\) goes from negative to positive. Thus \(r\) satisfies the favorable \((\overline{\Psi})\) test, while \(\overline r=\tau-i t^5\eta\) satisfies standard \((\Psi)\). The full tests for all multipliers follow from the even-factor comparison proof, with \(s=0\) and \(A=t^5\eta\), for which \(A_t=5t^4\eta\geq0\). The first nonzero derivative at \(t=0\) is \(5!\eta>0\). No Fourier-convention reversal is involved. The favorable operator estimate has gain \(1/6\), as proved in the cited model lesson; the conjugate has the adverse orientation for that estimate.

Exercise 2 — a full mixed realization, 10 points. Set \(s=4,k=15\), and

Verify the full sign condition, write its homogeneous realization, and find its type, short-rank index and necessary loss bound.

Solution. Both monomials have weight fifteen for weights \(1,5\). Here

The quotient vanishes at \(t=0\), and \(s\) is even. Theorems 3.1–4.1 give the full condition to

on \(\rho>0\). The marked covector is \((0,0,1)\). The balanced weights are \((1,15),(5,11),(8,8)\). Since \(B\ne0\), its type is fifteen; its short-rank index is four. The weighted packet theorem gives loss at least \(15/16\), hence gain at most \(1/16\). This is a necessary bound; the polynomial realization alone supplies no general operator sufficiency proof.

Exercise 3 — a slope test misses odd transverse order, 8 points. Take \(s=3,k=9\), \(b=t^9\). Check \(Q=0\Rightarrow Q_t\geq0\), but exhibit a forbidden odd crossing arbitrarily near zero.

Solution. Here \(Q=t^9+t^3\eta/6\) and \(B=6t^8\geq0\). If \(t\ne0\), a zero requires \(\eta=-6t^6\), and the slope there is \(6t^8\). At \(t=0\) the slope is zero. Yet any fixed negative \(\eta\), with \(z=0,\tau=0\), gives

This is a positive-to-negative crossing along \(H_\tau\). It violates \((\overline{\Psi})\), although every first slope test holds.

Exercise 4 — even order is not enough, 8 points. Take \(s=4,k=13\) and \(b=t^{13}-2t^3z^2\). Verify the weighted degree and first slope condition, then determine the missing full-sign restriction.

Solution. The weights of \(t,z\) are \(1,5\); \(3+2\cdot5=13\). We obtain

At nonzero \(t\), elimination of \(\eta\) at a zero of \(Q=b+t^4\eta/24\) gives this slope. At \(t=0\), \(Q_t=0\). But \(t^4\) does not divide \(b\). For fixed \(z\ne0,\eta=0\), its leading term is \(-2z^2t^3\), with third derivative \(-12z^2<0\). Hence the full condition fails. The obstruction is the lower odd time power, not the parity of \(s\).

Exercise 5 — a transverse-dependent complex multiplier, 10 points. For the two-pair polynomial of Exercise 2, use \(a=1+iz\). On a characteristic curve of \(F=\operatorname{Re}(ar)\), compute a nonsingular comparison equation for \(A=\eta/24+t^{11}+3tz^2\). Include the multiplier-derivative term.

Solution. We have \(F=\tau-zQ\), \(G=z\tau+Q\). On \(F=0\), \(\tau=zQ\), so

Since \(\{Q,A\}=0\) and \(\{z,A\}=-1/24\),

Also \(H_Ft=1\), so \(t\) itself is the forward parameter up to a constant. The integrating factor gives

The term \(t^4A/24\) comes from differentiating the multiplier and cannot be omitted. The factor relating \(G\) to \(A\) is nonnegative, and positive except at \(t=0\); hence no forbidden sign change occurs even through that vanishing factor.

Exercise 6 — classify two degrees, 12 points. Keep \(s=4\). Classify all allowed \(g\) when \(k=13\), and when \(k=15\). In the second case give the exact coefficient inequalities, including the degenerate boundary, and a nonzero example whose derivative is a square.

Solution. The weights are \(1,5\). For \(k=13\), \(g\) has weight nine, so

Its derivative is \(9a t^8+4c t^3z\). For fixed \(t\ne0\) the free variable \(z\) makes this negative unless \(c=0\); then it is nonnegative exactly when \(a\geq0\).

For \(k=15\), the quotient has weight eleven, and its complete list of monomials is

Thus the necessary and sufficient conditions are

If \(d>0\), complete the square:

If \(d=0\), nonnegativity for every \(z\) forces \(c=0\), and then requires \(a\geq0\). This is exactly the stated degenerate boundary. Taking \((a,c,d)=(9,11,11)\) gives

The zero derivative on a curve is allowed: the quotient remains nondecreasing, and comparison handles the stationary zeros. Multiplication by \(t^4\) gives an allowed nonzero \(b\) of weight fifteen, fully realized by Theorem 4.1.

References

• [L] Nicolas Lerner, Metrics on the Phase Space and Non-Selfadjoint Pseudodifferential Operators, Birkhäuser, 2010. Author's 2009 manuscript. Sections 3.1.3 and 3.2.1 give the estimate/adjoint sign conventions and multiplier invariance.
• [H] Lars Hörmander, The Analysis of Linear Partial Differential Operators IV: Fourier Integral Operators, Springer, 2009 reprint. Publisher's record. Chapter 27 treats mixed weighted Taylor geometry and its full-sign variants.
• [H3] Lars Hörmander, The Analysis of Linear Partial Differential Operators III: Pseudo-Differential Operators, Springer, 2007 reprint. Publisher's record. Section 21.1: Hamilton fields, Poisson brackets, canonical transformations and symplectic coordinates.

Written by GPT-6.1 Sol (OpenAI), at Ultra reasoning effort, September 2026. Self-checked by the writing AI. Public domain (CC0).