How sign orientation aligns transverse gradients

An oriented sign condition restricts how a symbol can tilt in transverse directions as time changes. If an earlier positive value cannot become a later negative value, two large transverse gradients cannot point in substantially different directions. A bound on transverse curvature turns this observation into a uniform estimate.

We will prove the finite-dimensional geometry behind that estimate, obtain a common oriented direction for the gradients, and quantify how its replacement by another direction changes the bound. This supplies the next geometric ingredient after An adaptive scale for repeated brackets. The arguments below use scalar Taylor estimates and Euclidean geometry.

Throughout, \(\langle\cdot,\cdot\rangle\) and \(|\cdot|\) denote the Euclidean inner product and norm. For a unit vector \(e\), write

The projection \(\Pi_e\) depends only on the line through \(e\). The ray \(\mathbb R_+e\) also remembers its orientation. Minimizing \(|v-se|^2\) over \(s\geq0\) gives \(\operatorname{dist}(v,\mathbb R_+e)^2=|v|^2-\max(\langle v,e\rangle,0)^2\). In particular this distance is continuous in \(v\) and in the unit direction \(e\).

1. A convex combination certifies the absence of a joint direction

Let \(Y_1,Y_2\) be nonzero vectors in \(\mathbb R^N\), and let \(a_1,a_2\) be real numbers. The thresholds may be negative.

Lemma 1.1 (two-direction alternative). There is no \(y\) satisfying

if and only if some \(\theta\in[0,1]\) satisfies

Proof. If (1.3) holds, put \(Z=\theta Y_1+(1-\theta)Y_2\) and \(b=\theta a_1+(1-\theta)a_2\). A vector satisfying both strict inequalities in (1.2) would give \(\langle y,Z\rangle>b\). But \(\langle y,Z\rangle\leq |y||Z|\leq |Z|\leq b\). This is a contradiction.

For the converse, minimize the continuous convex function

on \([0,1]\). Suppose its minimum is positive, and let \(\theta_0\) attain that minimum. We will construct a vector in the closed unit ball for which both strict inequalities hold. Shrinking it slightly then gives a vector in the open unit ball.

First suppose \(Z_0=\theta_0Y_1+(1-\theta_0)Y_2\ne0\), and set \(y=Z_0/|Z_0|\). Write

At \(\theta_0\), the value of \(\varphi\) is \(\theta_0r_1+(1-\theta_0)r_2\), and its derivative is \(r_1-r_2\). If \(0<\theta_0<1\), the derivative is zero, so \(r_1=r_2=\varphi(\theta_0)>0\). If \(\theta_0=0\), the right derivative is nonnegative. Thus \(r_1\geq r_2=\varphi(0)>0\). If \(\theta_0=1\), the left derivative is nonpositive. Thus \(r_2\geq r_1=\varphi(1)>0\). Every case gives the required strict inequalities.

It remains to consider \(Z_0=0\). Since the two endpoint vectors are nonzero, \(0<\theta_0<1\). Put \(D=Y_1-Y_2\), which is nonzero. Near \(\theta_0\), the norm in (1.4) is exactly \(|\theta-\theta_0||D|\). The two one-sided minimum conditions give \(|a_1-a_2|\leq |D|\). Choose

The two quantities in (1.5) are now equal, since \(\langle y,D\rangle=a_1-a_2\). Their convex average equals

Thus both are positive here too. In every case, multiplying \(y\) by a number sufficiently close to one and less than one preserves both strict inequalities and puts it in the open unit ball. This contradicts the assumed absence of (1.2). Therefore the minimum of \(\varphi\) is nonpositive, which is (1.3). ∎

This proof includes a vanishing convex combination and minima at either endpoint. Those cases matter when the thresholds have different signs.

2. From the certificate to a distance from a ray

Lemma 2.1 (opposite-ray estimate). Suppose there is no \(y\) satisfying (1.2), and

Then the vectors are nonzero, \(R=\max(a_1,a_2)\geq0\), and

Proof. If \(Y_1=0\), (2.1) gives \(a_1<0\). The strict inequality \(|Y_2|>a_2\) ensures there is a point in the open unit ball with \(\langle y,Y_2\rangle>a_2\): the supremum of this linear functional on that ball is \(|Y_2|\). That point also satisfies \(0>a_1\), contradicting the hypothesis. The same argument excludes \(Y_2=0\).

Lemma 1.1 gives a point \(Z\) on the segment from \(Y_1\) to \(Y_2\) satisfying

In particular \(R\geq0\). Put \(u=|Y_1|\), \(v=|Y_2|\).

If \(\langle Y_1,Y_2\rangle<0\), the orthogonal projection of \(Y_2\) onto the line through \(Y_1\) lies on its negative ray. When the vectors are collinear, the distance in (2.2) is zero. Otherwise let \(\ell\) be the line through \(Y_1,Y_2\). Computing the area of their planar triangle by its two choices of base gives

The middle inequality follows because \(Z\in\ell\) and \(|Z|\leq R\). Dividing by \(u\) proves (2.2). For an algebraic check of the area identity, the square of either area numerator is \(u^2v^2-\langle Y_1,Y_2\rangle^2\).

If \(\langle Y_1,Y_2\rangle\geq0\), the closest point on the negative ray is zero, so the distance is \(v\). For every \(\theta\in[0,1]\),

The second inequality is the minimum of the displayed scalar quadratic. Applying it to the point in (2.3) gives

This proves (2.2) in the remaining case. ∎

The same-side case requires the ray rather than just its supporting line. The estimate (2.5) supplies that case directly.

3. Sign orientation gives a common direction for a whole gradient family

Let \(N\geq1\), and let \(F\) be a real \(C^2\) function on \((-1,1)\times B_1(0)\subset\mathbb R^{1+N}\). Assume

The Hessian norm is its Euclidean operator norm. Put \(g(t)=\nabla_yF(t,0)\).

Theorem 3.1 (aligned transverse gradients). There is a unit vector \(e\in\mathbb R^N\) such that, for every \(|t|<1\),

In fact the proof gives

No bound on the time derivatives or on \(\sup_{|t|<1}|g(t)|\) is assumed.

Proof. Taylor's formula in \(y\) gives

Write \(h=3/2\). If \(s<t\) and \(\langle g(s),y\rangle>h\), with \(|y|<1\), then \(F(s,y)>0\). The sign condition makes \(F(t,y)\geq0\). On the other hand, \(\langle g(t),y\rangle<-h\) would give \(F(t,y)<0\), by (3.4). Consequently there is no \(y\) in the open unit ball with

Fix an anchor time \(s_0\) with \(u=|g(s_0)|>h\), and set \(e_0=g(s_0)/u\). We claim

If \(|g(t)|\leq h\), the distance is at most \(|g(t)|\leq h\), so the claim holds. If \(t>s_0\) and \(|g(t)|>h\), use Lemma 2.1 with \(Y_1=g(s_0)\), \(Y_2=-g(t)\), and both thresholds equal to \(h\). Negating both the vector and the ray in its conclusion gives (3.6).

If \(t<s_0\) and \(|g(t)|>h\), (3.5) forbids a joint direction for \(g(t)\) and \(-g(s_0)\). Interchange their roles in Lemma 2.1, taking \(Y_1=-g(s_0)\), \(Y_2=g(t)\). Its negative ray is now precisely \(\mathbb R_+e_0\), which again proves (3.6). At \(t=s_0\) the distance is zero.

Let \(M=\sup_{|t|<1}|g(t)|\), allowing \(M=+\infty\). If \(M\leq h\), any unit vector gives (3.3), since distance to a ray is at most the norm.

Suppose \(h<M<\infty\). Choose times \(s_j\) with \(|g(s_j)|\to M\), and pass to a subsequence such that \(g(s_j)/|g(s_j)|\to e\). The unit sphere is compact. For each fixed \(t\), (3.6) and continuity of distance to a ray in its unit direction give

The same subsequence works for every \(t\): each of the inequalities (3.6) already holds for all \(t\).

If \(M=+\infty\), choose anchors with norms tending to infinity and again pass to a convergent subsequence of their unit directions. For each fixed \(t\), (3.6) now gives the sharper bound

These cases prove (3.3). To recover (3.2), if \(\langle g(t),e\rangle\geq0\), distance to the ray is its perpendicular component. If that inner product is negative, distance to the ray is \(|g(t)|\), so both the perpendicular component and the size of the negative projection are at most three. ∎

The direction is obtained from the large gradients of the whole family. It need not be the direction of a gradient at any one selected time.

4. Quantify a change of direction

Let \(e,e'\) be unit vectors and define

This measures the angle between their lines and is unchanged when either direction is negated.

Lemma 4.1 (projection difference). The Euclidean operator norm satisfies

Proof. If \(\eta=0\), the vectors span the same line and their projections agree. Otherwise write \(e'=ce+\eta f\), where \(c=\langle e,e'\rangle\), \(f\perp e\), and \(|f|=1\). On the plane with orthonormal basis \((e,f)\), the matrix of \(\Pi_{e'}-\Pi_e\) is

It is symmetric, and its square is \(\eta^2\) times the identity, since \(c^2+\eta^2=1\). Its norm on the plane is therefore \(\eta\). On the orthogonal complement it is zero, proving (4.2). ∎

Corollary 4.2 (stability of the transverse bound). If \(e\) satisfies (3.2), then every unit \(e'\) satisfies

Proof. Decompose \((I-\Pi_{e'})g=(I-\Pi_e)g+(\Pi_e-\Pi_{e'})g\). The first term has norm at most three, and Lemma 4.1 bounds the second by \(\eta|g|\). The second displayed bound follows from \(\eta\geq0\). ∎

The bound with coefficient four is the version commonly used in the subsequent symbol estimates. The exact projection calculation supplies the stronger coefficient one as well.

For example, if \(g(s)\ne0\) and \(e'=g(s)/|g(s)|\), (3.2) gives \(\eta\leq3/|g(s)|\). Thus

This estimate explains why one can use a selected gradient direction when its norm is not small compared with the gradients being estimated. The bound is pointwise in \(t\); an arbitrary selected time supplies no uniform ratio when the whole family is unbounded.

A change from \(e\) to \(-e\) preserves every perpendicular estimate but reverses the ray. Consequently (4.4) alone supplies no lower bound for \(\langle g(t),e'\rangle\).

5. Rescale the curvature and the transverse ball

The normalized constant three has a useful dimensional form.

Corollary 5.1. Let \(F\) be real and \(C^2\) on an open time interval \(I\) times \(B_r(0)\), where \(r>0\). Suppose, for \(A,B\geq0\),

There is a unit \(e\) such that

The same right-hand side bounds the perpendicular component and the size of a negative projection onto \(e\).

Proof. Set \(y=rz\). Taylor's remainder is at most \(Br^2|z|^2/2\). The argument giving (3.5) therefore applies to \(Y(t)=r\nabla_yF(t,0)\) with threshold

All of the anchor and compactness argument in Theorem 3.1 works on any ordered interval and with this threshold. It gives \(\operatorname{dist}(Y(t),\mathbb R_+e)\leq2h\). Dividing by \(r\) proves (5.2).

For completeness, if \(h=0\), Taylor's formula says \(F(t,y)=\langle\nabla_yF(t,0),y\rangle\). If every gradient is zero, any direction works. Otherwise choose a nonzero anchor. For any other nonzero gradient, the sign condition forbids (1.2) for the appropriately ordered pair with thresholds zero. Lemma 2.1 then gives zero distance to the anchor's positive ray. Zero gradients already belong to that ray. This proves the zero-threshold case too. ∎

These estimates turn a scalar sign condition into quantitative control of transverse directions. The next geometric step uses a large transverse gradient to construct explicit canonical coordinates and estimates in those coordinates. The remaining covering and analytic estimates are still required for the general subelliptic theorem.

6. Exercises with complete solutions

Exercise 1 — basic. Derive (3.5) directly from (3.1), retaining the strict open-ball inequality and the Hessian remainder. Is a bound on \(\partial_tF\) needed?

Solution. With \(s<t\) and \(|y|<1\), if \(\langle g(s),y\rangle>3/2\), then

The sign hypothesis makes \(F(t,y)\geq0\). If simultaneously \(\langle-g(t),y\rangle>3/2\), then

which is impossible. Only the two center-height bounds, the transverse Hessian bound and the sign orientation were used. No estimate on \(\partial_tF\) appears.

Exercise 2 — basic. Take \(Y_1=(4,0)\), \(Y_2=(-3,2)\), and \(a_1=a_2=3/2\). Find a certificate (1.3), compute the distance to the negative ray, and verify (2.2).

Solution. For \(\theta=3/7\),

Thus the certificate rules out a joint direction in the open unit ball. Both vector norms exceed their thresholds. The negative ray through \(Y_1\) is the nonpositive horizontal axis. The closest point to \(Y_2\) is \((-3,0)\), so its distance is two. The bound in (2.2) is \(\frac32(1+\sqrt{13}/4)\), which is larger than two: indeed it exceeds \(\frac32(1+3/4)=21/8>2\).

Exercise 3 — intermediate. Take \(Y_1=(4,0)\), \(Y_2=(0,3)\), and \(a_1=a_2=12/5\). Verify a certificate, and explain why distance to the negative ray must be treated using the same-side argument.

Solution. Choose \(\theta=9/25\). Then

The certificate again rules out both strict inequalities. Each vector norm exceeds \(12/5\). Their inner product is zero, so the nearest point on the negative horizontal ray is the origin and the distance is three. Formula (2.2) bounds it by \((12/5)(1+3/4)=21/5\). Here the scalar quadratic in (2.5) has its minimum at \(9/25\), giving the exact lower bound \(R\geq12/5\). This is the case in which a projection onto a full line, by itself, does not identify a point on the negative ray.

Exercise 4 — intermediate. For \(N\geq1\), consider \(F(t,y)=y_1/(1-t)\) on \((-1,1)\times B_1(0)\). Check every hypothesis of Theorem 3.1, identify a direction \(e\), and explain why a finite supremum of the gradients cannot be assumed.

Solution. The function is \(C^2\) on the stated open domain. Its center value and transverse Hessian are both zero. For fixed \(y\), the sign of \(F(t,y)\) is the sign of \(y_1\), since \(1-t>0\). Thus the sign orientation holds. Its gradient is

Choose \(e=(1,0,\ldots,0)\). Every gradient lies exactly on its positive ray, so both errors vanish. Nevertheless \(\sup_{|t|<1}|g(t)|=+\infty\). The domain is open at \(t=1\), and the hypotheses place no bound on time derivatives near that endpoint. The infinite-supremum part of the proof is therefore necessary.

Exercise 5 — advanced. Let \(e=(1,0)\), \(e'=(\cos\theta,\sin\theta)\), and \(g=(L,2)\). Compute the perpendicular component relative to \(e'\), verify Corollary 4.2, and consider \(\theta=\pi\).

Solution. A unit vector perpendicular to \(e'\) is \((-\sin\theta,\cos\theta)\). Hence

Here \(|g-\Pi_eg|=2\leq3\), \(\eta=|\sin\theta|\), and \(|L|\leq|g|\). Thus the stronger bound \(3+\eta|g|\), and consequently the bound \(3+4\eta|g|\), both hold. When \(\theta=\pi\), \(\eta=0\) and the perpendicular component stays equal to two. But \(\langle g,e'\rangle=-L\), which can be arbitrarily negative for positive \(L\). A line-projection estimate cannot preserve the oriented projection after a sign reversal.

Exercise 6 — advanced. Suppose (5.1) holds on a transverse ball of radius three with \(A=2,B=1/2\). Compute the bound obtained with that radius. Then optimize (5.2) over the smaller available radii.

Solution. With \(r=3\), (5.2) gives

For \(A,B>0\), differentiation gives the minimum of \(2A/r+Br\) at \(r=\sqrt{2A/B}\), with value \(2\sqrt{2AB}\). In this example the minimizing radius is \(\sqrt8<3\), so that smaller ball is available and all the same hypotheses remain valid there. The resulting bound is \(2\sqrt2\). If the minimizing radius exceeds the available radius, it cannot be used; the minimum over the permitted radii is then at the available endpoint.

References

The same results are treated in Hörmander, The Analysis of Linear Partial Differential Operators IV, Chapter 27, Section 27.4, Lemmas 27.4.3–27.4.5 and the direction-replacement remark. The scalar minimization proof, explicit same-side case, endpoint analysis, projection calculation and exercises above are independently written. The exact projection norm strengthens the stated coefficient-four bound while retaining that bound for later use.

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