How sign orientation controls approximate roots

A large positive transverse derivative makes a zero of a smooth function close to the zero of its linear part. The oriented sign condition controls the true zero. Taylor's remainder then transfers that control to the approximate zero, with an explicit error.

We will prove a bound for the intercept, a two-time inequality for the intercept divided by the slope, and a differential version. We also prove their versions with a positive regularized denominator. The second transverse derivative controls the position error; its time derivative controls the differential error. These are distinct uses of the hypotheses.

The geometric prerequisite is the aligned-gradient theorem in How sign orientation aligns transverse gradients, specialized to one transverse dimension. We give the polynomial interpolation argument and the root construction explicitly. The argument uses Taylor's formula, the intermediate value theorem and ordinary differentiation. Its application to the canonical symbol is a subsequent step.

1. The quantitative quotient theorem

Let $F$ be a real smooth function on

U=\{(t,y):|t|<1,\ |y|<1\}. \tag{1.1}

Put $f(t)=F(t,0)$ and $b(t)=\partial_yF(t,0)$. Fix an integer $k\geq1$, and assume

\begin{gathered} |\partial_y^2F|\leq1,\qquad |\partial_t\partial_y^2F|\leq1\quad\hbox{on }U,\\ |f(t)|\leq1,\qquad |b^{(k)}(t)|\leq1 \quad(|t|<1),\\ F(s,y)>0\ \Longrightarrow\ F(t,y)\geq0 \quad(-1<s<t<1, |y|<1). \end{gathered} \tag{1.2}

The last implication forbids a positive value from becoming a later negative value at a fixed $y$. It does not assert that $F$ is increasing in time.

The positive supremum

N=\sup_{|t|<1}b(t) \tag{1.3}

is finite: the degree-$(k-1)$ Taylor polynomial of $b$ at zero has finite coefficients, and its remainder is bounded by $1/k!$ throughout this interval. We require $N$ to exceed a constant depending only on $k$.

Here are permissible explicit constants. Set

\begin{gathered} K_k=\frac{[8(k+1)]^{k-1}}{(k-1)!},\qquad N_0=32K_k,\\ C_1=6K_k,\qquad C_0=4C_1,\\ C_2=8C_1^2,\qquad C_3=24C_1^2. \end{gathered} \tag{1.4}

All depend only on $k$. They are not claimed to be optimal.

Theorem 1.1 (intercept and approximate-root control). If $N\geq N_0$, then for $|s|<1/2$,

|f(s)|\leq \frac{C_1b(s)}N+\frac{C_1^2}{2N^2}. \tag{1.5}

For $-1/2<t<s<1/2$, if $N\min(b(s),b(t))>C_0$, then

\begin{aligned} &\frac{f(t)}{b(t)}-\frac{f(s)}{b(s)}\\ &\quad\leq\frac{C_2}{N^2b(s)} +\frac{C_2}{N^2b(t)}. \end{aligned} \tag{1.6}

For $|s|<1/2$, if $Nb(s)>C_0$, then

\begin{aligned} &b(s)\frac{d}{ds}\left(\frac{f(s)}{b(s)}\right)\\ &\qquad\geq-\frac{C_2}{N^2} \left(1+\frac{|b'(s)|}{b(s)}\right). \end{aligned} \tag{1.7}

Let

G(s)=b(s)+C_1/N. \tag{1.8}

Then $G(s)\geq C_1/(2N)>0$ on $|s|<1/2$. The regularized inequalities are

\begin{aligned} &\frac{f(t)}{G(t)}-\frac{f(s)}{G(s)}\\ &\quad\leq\frac{C_3}{N^2G(s)} +\frac{C_3}{N^2G(t)} \end{aligned} \tag{1.9}

when $-1/2<t<s<1/2$ and $N\min(G(s),G(t))>C_0+2C_1$, and

\begin{aligned} &G(s)\frac{d}{ds}\left(\frac{f(s)}{G(s)}\right)\\ &\qquad\geq-\frac{C_3}{N^2} \left(1+\frac{|G'(s)|}{G(s)}\right) \end{aligned} \tag{1.10}

when $|s|<1/2$ and $NG(s)>C_0+2C_1$.

The denominator is positive everywhere on the inner interval, but the quantitative quotient inequalities still have the stated lower thresholds. Positivity alone does not discharge those thresholds.

2. Large slopes occur on both sides of the inner interval

The one-dimensional aligned-gradient theorem gives a direction $e\in\{-1,1\}$ with $eb(t)\geq-3$ for all $|t|<1$. Since $N>3$, some slope is greater than three. Consequently $e=-1$ is impossible, and

b(t)\geq-3\quad(|t|<1). \tag{2.1}

Let $T$ be the Taylor polynomial of $b$ at zero of degree at most $k-1$. Taylor's integral remainder gives

|T(t)-b(t)|\leq1/k!\quad(|t|<1). \tag{2.2}

For $j=0,\ldots,k-1$, choose the actual interpolation nodes

s_j^-=-1+\frac{j+1}{2(k+1)},\qquad s_j^+=-s_j^-. \tag{2.3}

Every negative node lies in $(-1,-1/2)$, and every positive node lies in $(1/2,1)$. Their spacing in absolute value is $h=1/[2(k+1)]$.

For either set of nodes, Lagrange interpolation expresses the entire polynomial as

T(t)=\sum_{j=0}^{k-1}T(s_j^\pm) \prod_{\substack{0\leq l<k\\l\ne j}} \frac{t-s_l^\pm}{s_j^\pm-s_l^\pm}. \tag{2.4}

For $|t|\leq1$, each numerator factor has absolute value at most two. The absolute denominator product is $h^{k-1}j!(k-1-j)!$. Hence

\begin{aligned} \sup_{|t|\leq1}|T(t)| &\leq \max_j|T(s_j^\pm)|\, \left(\frac2h\right)^{k-1} \sum_{j=0}^{k-1}\frac1{j!(k-1-j)!}\\ &=K_k\max_j|T(s_j^\pm)|. \end{aligned} \tag{2.5}

For $k=1$, the products are empty and equal one; this is interpolation of a constant. The formula therefore includes that endpoint.

Equations (2.2) and (1.3) imply $\sup|T|\geq N-1/k!\geq N/2$. Thus each node set has a node with $|T(s_j^\pm)|\geq N/(2K_k)$. At such a node $T$ cannot be negative: $N/(2K_k)\geq16$, whereas (2.1) and (2.2) give $T\geq-3-1/k!\geq-4$. It follows that there are times

\begin{gathered} t_-\in(-1,-1/2),\qquad t_+\in(1/2,1),\\ b(t_-)\geq c_kN,\qquad b(t_+)\geq c_kN,\\ c_k=(4K_k)^{-1}. \end{gathered} \tag{2.6}

Indeed $N/(2K_k)-1/k!\geq N/(4K_k)$. No attainment of the supremum in (1.3) was used. The full interpolation inequality, rather than a bound on one selected derivative, produces slopes on both sides.

3. Trap all inner zeros in a narrow strip

Define the exact transverse Taylor remainder

R(t,y)=F(t,y)-f(t)-b(t)y. \tag{3.1}

The two curvature hypotheses give

|R(t,y)|\leq y^2/2,\qquad |\partial_tR(t,y)|\leq y^2/2. \tag{3.2}

For the second estimate, apply Taylor's formula in $y$ to $\partial_tF$. Its value and first transverse derivative at zero are $f'$ and $b'$, respectively. Thus it is exactly the time derivative of the remainder in (3.1).

At either time in (2.6), the term $b(t_\pm)y$ determines the sign of $F$ whenever $c_kN|y|>3/2$, since $|f|+|R|\leq1+1/2$ on $|y|<1$. Put

\delta=C_1/N=3/(2c_kN)<1. \tag{3.3}

For $y>\delta$, the earlier value $F(t_-,y)$ is positive, so the sign implication makes $F(s,y)\geq0$ throughout $|s|<1/2$. For $y<-\delta$, the later value $F(t_+,y)$ is negative. An earlier positive value would contradict the same implication. Thus

\begin{gathered} F(s,y)\geq0\quad(\delta<y<1),\\ F(s,y)\leq0\quad(-1<y<-\delta),\\ |s|<1/2. \end{gathered} \tag{3.4}

The weak inequalities are sufficient. Passing to $y=\pm\delta$ by continuity and using (3.2) gives

-b(s)\delta-\delta^2/2 \leq f(s)\leq b(s)\delta+\delta^2/2. \tag{3.5}

This proves (1.5). The interval in (3.5) must be nonempty, so also

b(s)\geq-\delta/2=-C_1/(2N). \tag{3.6}

Equation (3.6) proves the asserted positive lower bound for $G$.

4. Construct a true zero near the linear zero

Fix $|s|<1/2$ with $Nb(s)>C_0$, and abbreviate $b=b(s)>0$, $f=f(s)$. Since $Nb>C_0=4C_1$, (1.5) implies

|f|\leq2C_1b/N,\qquad h=-f/b,\qquad |h|\leq2C_1/N<1/2. \tag{4.1}

If $h=0$, take $r=0$, which is already a true zero. Suppose $h\ne0$, and set

e=2h^2/b<|h|. \tag{4.2}

The strict inequality follows from $b>4C_1/N\geq2|h|$. At $y=h+e$ and $y=h-e$, all arguments have $|y|<2|h|<1$. Equation (3.2) gives $|R(s,y)|\leq2h^2=be$. Therefore

\begin{gathered} F(s,h+e)=be+R(s,h+e)\geq0,\\ F(s,h-e)=-be+R(s,h-e)\leq0. \end{gathered} \tag{4.3}

The intermediate value theorem supplies a zero $r=h+z$ between those endpoints. In both the zero and nonzero cases,

\begin{gathered} F(s,r)=0,\qquad |r|\leq2|h|\leq4C_1/N,\\ |z|\leq2h^2/b\leq\frac{8C_1^2}{N^2b}. \end{gathered} \tag{4.4}

This construction also gives a positive transverse derivative at the zero. Integrating $|F_{yy}|\leq1$ from zero yields

\partial_yF(s,r)\geq b-|r|>0. \tag{4.5}

When $h\ne0$, use $|r|\leq2|h|<b$; when $h=0$, this derivative is exactly $b>0$. A sufficiently small negative perturbation of $r$ consequently gives a strictly negative value of $F(s,\cdot)$. This last point handles the linear zero $h=0$ as well; no division by $|h|$ is permitted there.

5. Transfer sign orientation to the approximate quotient

Let $-1/2<t<s<1/2$, with both endpoint slopes above $C_0/N$, and use the true zero from Section 4 at the later time $s$. For all sufficiently small $\varepsilon>0$, (4.5) gives $F(s,r-\varepsilon)<0$. The sign implication in (1.2) then gives $F(t,r-\varepsilon)\leq0$: an earlier positive value is forbidden. Letting $\varepsilon\downarrow0$ gives $F(t,r)\leq0$. Expand at the earlier time:

f(t)+b(t)(-f(s)/b(s)+z)+R(t,r)\leq0. \tag{5.1}

Division by $b(t)>0$, followed by (3.2) and (4.4), yields

\begin{aligned} \frac{f(t)}{b(t)}-\frac{f(s)}{b(s)} &\leq |z|+\frac{r^2}{2b(t)}\\ &\leq\frac{8C_1^2}{N^2} \left(\frac1{b(s)}+\frac1{b(t)}\right). \end{aligned} \tag{5.2}

This is (1.6). The proof uses only a true zero at the later time. It does not assume a globally differentiable zero curve.

At any zero $(s,r)$ in the open rectangle, the same sign implication forces

\partial_tF(s,r)\geq0. \tag{5.3}

If that derivative were negative, for small positive $\varepsilon$ the earlier value $F(s-\varepsilon,r)$ would be positive and the later value $F(s+\varepsilon,r)$ negative. This contradicts (1.2). This argument is local in time and does not require that the zero be simple in the time direction.

Using $r=-f/b+z$ in (5.3), we obtain

\begin{aligned} &f'(s)-\frac{b'(s)f(s)}{b(s)}\\ &\quad\geq-|b'(s)|\,|z|-|\partial_tR(s,r)|\\ &\quad\geq-\frac{8C_1^2}{N^2} \left(1+\frac{|b'(s)|}{b(s)}\right). \end{aligned} \tag{5.4}

The expression on the left is exactly $b(s)(f/b)'(s)$. This proves (1.7), including every contribution from the time-dependent remainder. A bound on $R$ alone would not bound that derivative contribution.

6. Regularize the denominator without losing the estimates

Assume $NG(s)>C_0+2C_1$. Because $G=b+C_1/N$, this implies $Nb(s)>C_0+C_1$. In particular the preceding quotient estimates apply. Let

a=\frac{C_0+2C_1}{C_0+C_1}=\frac65. \tag{6.1}

At every such point,

\begin{gathered} 1\leq G/b\leq a,\\ \left|f\left(\frac1G-\frac1b\right)\right| \leq\frac{2C_1^2}{N^2G}. \end{gathered} \tag{6.2}

For the second inequality use $|f|\leq2C_1b/N$ and $G-b=C_1/N$. This retains both denominators rather than estimating them by a common lower constant.

Apply (6.2) at $t$ and $s$. Equation (1.6) and $1/b\leq a/G$ give

\begin{aligned} &\frac{f(t)}{G(t)}-\frac{f(s)}{G(s)}\\ &\quad\leq\frac{aC_2+2C_1^2}{N^2G(s)} +\frac{aC_2+2C_1^2}{N^2G(t)}. \end{aligned} \tag{6.3}

Since $aC_2+2C_1^2\leq C_3$, this proves (1.9).

The additive correction to $b$ is constant in time, so $G'=b'$. Compute the full derivative identity:

\begin{aligned} G(f/G)'&=f'-b'f/G\\ &=b(f/b)'\\ &\quad+b'f\left(\frac1b-\frac1G\right). \end{aligned} \tag{6.4}

The extra term has absolute value at most $2C_1^2N^{-2}|G'|/G$, by (6.2). Also $|b'|/b\leq a|G'|/G$. Thus (1.7) implies

\begin{aligned} &G(f/G)'\\ &\quad\geq-\frac{C_2}{N^2}\\ &\qquad-\frac{aC_2+2C_1^2}{N^2}\frac{|G'|}{G}\\ &\quad\geq-\frac{C_3}{N^2} \left(1+\frac{|G'|}{G}\right). \end{aligned} \tag{6.5}

This proves (1.10) and completes Theorem 1.1. ∎

The proof makes three separate transfers. Polynomial interpolation produces large positive slopes before and after the inner interval. Sign orientation traps its zeros. The true-to-linear zero error then controls both time comparisons and differentiation. Adding a fixed positive denominator correction is a fourth, purely algebraic step.

7. Graded exercises with solutions

Exercise 1 — basic. For $k=1$, compute the constants in (1.4) and the two interpolation nodes. Explain why no derivative of a nonconstant polynomial occurs in Section 2.

Solution. $K_1=1$, $N_0=32$, $C_1=6$, $C_0=24$, $C_2=288$ and $C_3=864$. The negative node is $-3/4$, and the positive node is $3/4$. The Taylor polynomial of $b$ has degree zero. The empty products in (2.4) equal one; (2.2) follows from $|b'|\leq1$. No polynomial derivative bound was silently added.

Exercise 2 — basic. Suppose $F(t,y)=Ny-f_*(t)$, with $|f_*|\leq1$. What condition on $f_*$ is equivalent to the sign implication? Relate its true zero to the quotient in the theorem.

Solution. The implication is equivalent to $f_*$ being nonincreasing. If it increased between $s<t$, choose $Ny$ strictly between those two values; these values lie in $[-1,1]$, so the chosen $|y|<1$ when $N>1$. Then $F(s,y)>0>F(t,y)$, a contradiction. Conversely, a nonincreasing $f_*$ makes $F(t,y)$ nondecreasing for every $y$, so the implication holds. Here $f=-f_*$, $b=N$, and the true and linear zero are both $f_*/N$. Thus $f/b=-f_*/N$ is exactly nondecreasing. The theorem permits a small error for a curved transverse graph.

Exercise 3 — intermediate. Justify the two-time proof when $f(s)=0$. Why is an argument that chooses a nonzero perturbation bounded strictly by $4C_1|h|/(Nb)$ incomplete in that case?

Solution. Here $h=0$, and $r=0$ is a true zero with $F_y(s,0)=b(s)>0$. Hence $F(s,-\varepsilon)<0$ for all sufficiently small positive $\varepsilon$. Orientation gives $F(t,-\varepsilon)\leq0$, and continuity gives $f(t)\leq0$. The left side of (1.6) is then $f(t)/b(t)\leq0$. A strictly positive perturbation bound proportional to $|h|$ becomes zero here and cannot supply the required perturbation. The limiting argument retains this endpoint without division by the linear zero.

Exercise 4 — intermediate. Let $N\geq32$, $r=1/(2N)$, and

F(t,y)=(y-r)\left(N+\frac t2(y+r)\right). \tag{7.1}

Verify the derivative, size and sign hypotheses (1.2) for every $k\geq1$. When applying the theorem, also take $N\geq N_0(k)$. Show that exact monotonicity of $f/b$ is false even though the true zero is constant. Compute the size of its normalized negative derivative.

Solution. The second factor is positive on the full rectangle, since it is at least $N-(1+r)/2>0$. Thus the sign of $F$ is the fixed sign of $y-r$, so orientation holds. Expanding gives

\begin{gathered} F=Ny-Nr+\tfrac t2(y^2-r^2),\\ F_{yy}=t,\qquad F_{tyy}=1,\\ b=N,\qquad b^{(k)}=0,\\ f=-\tfrac12-\frac{t}{8N^2}. \end{gathered} \tag{7.2}

The required absolute bounds hold and the slope supremum is exactly $N$. Yet

(f/b)'=-\frac1{8N^3},\qquad b(f/b)'=-\frac1{8N^2}. \tag{7.3}

The true zero remains $r$; its linear approximation is $r+t/(8N^3)$. Consequently the negative error in the normalized differential inequality can have order $N^{-2}$. Replacing that error by zero is false. The additive regularization of the denominator retains a negative derivative here as well.

Exercise 5 — advanced. In the previous example replace $t$ in (7.1) by $\sin(Lt)$, with $L>1$. Explain why the time derivative of the curvature is a separate hypothesis.

Solution. The positive factor argument, $|F_{yy}|\leq1$, $|f|\leq1$, and all derivatives of the constant slope remain valid. But $F_{tyy}=L\cos(Lt)$, whose supremum is $L$. At zero,

b(f/b)'(0)=-L/(8N^2). \tag{7.4}

Thus no universal constant depending only on $k$ can give (1.7) if that time-curvature bound is removed and $L$ is unrestricted. The position and two-time estimates use the first curvature bound; the differential estimate uses its time derivative too.

Exercise 6 — advanced. For the regularized denominator, derive the threshold on $Nb$ and the comparison $G/b$ directly from $NG>C_0+2C_1$. Verify that both endpoint correction terms and the differentiated correction appear in Section 6.

Solution. Since $Nb=NG-C_1$, the threshold implies $Nb>C_0+C_1$. Hence $\begin{aligned} G/b&=1+C_1/(Nb)\\ &\leq(C_0+2C_1)/(C_0+C_1)=a. \end{aligned} \tag{7.5}$ The reciprocal correction is exactly $-C_1/(NbG)$; multiplying by $|f|\leq2C_1b/N$ gives $2C_1^2/(N^2G)$ at each endpoint. Both are needed in (6.3). Differentiation gives (6.4), including $b'f(1/b-1/G)$. That term costs $2C_1^2N^{-2}|G'|/G$, while converting $|b'|/b$ to $|G'|/G$ costs the factor $a$. These are exactly the two contributions in (6.5).

References

The theorem is the full scalar quotient lemma in Hörmander, The Analysis of Linear Partial Differential Operators IV, Chapter 27, Section 27.4 P, Lemma 27.4.13 and its regularized-denominator conclusions. The proof above supplies the interpolation constants, true-root construction, zero-intercept endpoint and all regularization terms explicitly. The aligned-gradient prerequisite and every bridge used here are proved in this course; the book citation supplies mathematical credit. This lesson does not assume the subsequent two-variable local subelliptic estimate.

P = paywalled source; access may require institutional access or payment. The citation preserves human-source credit; every proof used here is supplied in this lesson or its exact linked programme prerequisites.

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