Subellipticity and unique continuation · Self-checked by the writing AI

A derivative scale for signed one-dimensional equations

An equation can control a function even where its coefficient vanishes. The useful scale is then determined by the first coefficient derivative that is large. This lesson proves a uniform estimate with the complete derivative scale, rather than choosing one derivative in advance. The direction of a sign change is essential: the coefficient may go from negative to positive, but a change in the opposite direction creates a decaying null solution.

The prerequisites are elementary real analysis, Taylor's formula, Lagrange interpolation, and the kernel and sign conventions of Finite type and the sign of the symbol. Full sign conditions for mixed weighted polynomials explains why the favorable estimate uses the conjugate sign condition. All estimates used here are proved below. Basic references are Lerner [L], Evans–Zworski [EZ] and Hörmander [H].

We first state the derivative-scale estimate. Its proof combines a polynomial small-set bound, a one-sided integral kernel and a partition into intervals whose lengths adapt to the coefficient. This also gives sharp estimates from the derivatives at a single point and handles intervals cut by the domain boundary.

1. The scale that the equation controls

Let \(I\) be an open interval, let \(G\) be real and smooth on \(I\), and use

\[ D=-i\frac d{dt},\qquad L_G=D+iG. \tag{1.1} \]

Fix an integer \(k\geq0\). Assume

\[ |G^{(k+1)}(t)|\leq1\quad(t\in I), \tag{1.2} \]

and the favorable orientation

\[ u<v,\quad G(u)>0\quad\Longrightarrow\quad G(v)\geq0. \tag{1.3} \]

Thus zeros and zero intervals are allowed. Define the continuous scale

\[ M(t)=\max_{0\leq j\leq k} |G^{(j)}(t)|^{1/(j+1)}. \tag{1.4} \]

Theorem 1.1. There are constants \(M_k,C_k>0\), depending only on \(k\), such that if \(M(t)\geq M_k\) everywhere on \(I\), then

\[ \|Mu\|_{L^2(I)} \leq C_k\|L_Gu\|_{L^2(I)} \qquad(u\in C_c^\infty(I)). \tag{1.5} \]

The constants are independent of \(G\), the interval and the location of the support. In particular the statement applies to intervals of length at least one, and also to shorter intervals. We prove it in Section 5 without extending \(G\) beyond \(I\).

The \(j=0\) term is \(|G|\). At a zero where \(G'\) is large, the next term is \(|G'|^{1/2}\). A \(k\)-th derivative instead supplies the scale \(|G^{(k)}|^{1/(k+1)}\). The maximum combines these possibilities and can change its determining derivative as \(t\) moves.

2. A polynomial cannot be small on too much of an interval

All measures below are ordinary lengths. For \(k\geq1\), let \(P\) be a real polynomial of degree at most \(k\), and put

\[ N(P)=\max_{0\leq j\leq k}|P^{(j)}(0)|. \tag{2.1} \]

Lemma 2.1. For a constant \(A_k\) depending only on \(k\),

\[ \left|\{t\in(-1,1):|P(t)|\leq a\}\right| \leq A_k\left(\frac a{N(P)}\right)^{1/k} \tag{2.2} \]

when \(P\ne0\) and \(a>0\). No assumption on its individual roots is needed.

Proof. First fix the nodes \(x_j=-1+2j/k\), \(0\leq j\leq k\), and their Lagrange polynomials \(\ell_j\). Interpolation and its derivatives give

\[ N(P)\leq B_k\|P\|_{L^\infty([-1,1])}, \qquad B_k=\max_{0\leq d\leq k} \sum_{j=0}^k|\ell_j^{(d)}(0)|<\infty. \tag{2.3} \]

Now let \(E=\{|P|\leq a\}\cap(-1,1)\) have measure \(m>0\). We can choose \(k+1\) points \(t_0,\ldots,t_k\) in \(E\) with mutual distances greater than \(m/(2(k+1))\). Choose them successively: after at most \(k\) choices, removing the intervals of that radius about the chosen points removes length at most \(km/(k+1)<m\). A point of \(E\) still remains.

Interpolate again, now at those points. For \(|t|\leq1\), each numerator factor has size at most two, and each denominator factor has size greater than \(m/(2(k+1))\). Therefore

\[ \|P\|_\infty \leq(k+1)a \left(\frac{4(k+1)}m\right)^k. \tag{2.4} \]

Combine this with (2.3) and solve for \(m\). For example,

\[ A_k=4(k+1)\big((k+1)B_k\big)^{1/k} \]

works. If \(m=0\), the assertion is immediate. This proof includes coalescing roots and polynomials of lower degree. ∎

Lemma 2.2. Let \(J\) be any open interval containing zero and contained in \((-1,1)\). Suppose \(G\) is real and smooth on \(J\), satisfies (1.2), and

\[ \rho=\max_{0\leq j\leq k}|G^{(j)}(0)|\geq2. \tag{2.5} \]

For \(k\geq1\), set \(R=\rho^{1/(k+1)}\). Then

\[ E=\{t\in J:|G(t)|\leq R\} \quad\hbox{satisfies}\quad |E|\leq A'_k/R. \tag{2.6} \]

The constant is independent of \(J\) and \(G\).

Proof. Let \(P\) be the degree-\(k\) Taylor polynomial of \(G/\rho\) at zero. Its derivative norm (2.1) is one. Taylor's integral remainder, on the segment from zero to each \(t\in J\), gives

\[ |G(t)-\rho P(t)| \leq\frac{|t|^{k+1}}{(k+1)!}\leq1. \tag{2.7} \]

Hence \(E\) is contained in the set where \(|P|\leq(R+1)/\rho\leq2R/\rho\). Lemma 2.1 applies to that polynomial on the full interval \((-1,1)\), even though \(G\) is defined only on \(J\). Its bound is

\[ A_k(2R/\rho)^{1/k}=2^{1/k}A_k/R. \tag{2.8} \]

This proves the claim. Neither the Taylor remainder nor the coefficient is evaluated outside \(J\). ∎

If \(k=0\), the same role is played by the simpler bound

\[ |G(t)|>\rho-1\geq\rho/2, \quad t\in J, \tag{2.9} \]

since \(|G'|\leq1\). We will use \(R=\rho/2\) and an empty small set in that case.

3. A sharp central estimate without a sign assumption

The small set where the coefficient fails to be large cannot contain much of a function unless that function has a large derivative.

Lemma 3.1. Suppose \(E=\{|G|\leq R\}\subset J\) has measure at most \(A/R\). Then every \(u\in C_c^\infty(J)\) satisfies

\[ R^2\|u\|_2^2 \leq2\|Gu\|_2^2+4A^2\|u'\|_2^2. \tag{3.1} \]

Proof. Extend \(u\) by zero to the real line. It is still smooth. The fundamental theorem of calculus and Cauchy–Schwarz give

\[ \|u\|_\infty^2\leq2\|u\|_2\|u'\|_2. \tag{3.2} \]

Split its squared norm between \(E\) and its complement:

\[ \|u\|_2^2 \leq R^{-2}\|Gu\|_2^2 +\frac{2A}{R}\|u\|_2\|u'\|_2. \]

After multiplication by \(R^2\), bound the last term by \(\tfrac12R^2\|u\|_2^2+2A^2\|u'\|_2^2\), and move the first part to the left. ∎

Theorem 3.2. Under the hypotheses of Lemma 2.2, including its arbitrary truncated interval \(J\), but without any sign assumption,

\[ \rho^{1/(k+1)}\|u\|_2 \leq C_k\big(\|Du\|_2+\|Gu\|_2\big), \qquad u\in C_c^\infty(J). \tag{3.3} \]

Proof. For \(k\geq1\), apply Lemmas 2.2 and 3.1 and take square roots. For \(k=0\), use (2.9) directly. ∎

The sharp power is larger than or equal to \(2^{-k}\), since \(2^k\geq k+1\), as follows by induction. Thus for \(\rho\geq2\) the theorem also proves the estimate with \(\rho^{2^{-k}}\) on the left. The distinction between the two powers matters when seeking the best scale.

This estimate has two separate norms on the right. Replacing them by \(\|(D+iG)u\|_2\) requires the sign hypothesis. Cancellation between the derivative and the potential can otherwise defeat every positive gain.

4. The favorable sign gives a one-sided inverse

Lemma 4.1. Let \(G\) be real on an interval \(J\), satisfy (1.3), and let \(E=\{|G|\leq R\}\) have measure \(m\). Then

\[ \|u\|_2\leq(m+R^{-1})\|L_Gu\|_2 \qquad(u\in C_c^\infty(J)). \tag{4.1} \]

Here \(R>0\), and no derivative hypothesis is needed.

Proof. The negative values of \(G\) all precede its positive values. Hence there is a dividing point \(b\), possibly an endpoint, such that \(G\leq0\) to its left and \(G\geq0\) to its right. If both signs occur, choose \(b\) between the supremum of the negative set and the infimum of the positive set; these are in the stated order by (1.3). If one sign is absent, use the appropriate endpoint. A zero interval between the signs causes no difficulty.

Write \(f=u'-Gu=iL_Gu\). Since \(u\) is zero near the endpoints of \(J=(\alpha,\beta)\), integration of the equation gives

\[ \begin{aligned} u(t)&=\int_\alpha^t \exp\left(\int_s^tG(r)\,dr\right)f(s)\,ds, &&t\leq b,\\ u(t)&=-\int_t^\beta \exp\left(-\int_t^sG(r)\,dr\right)f(s)\,ds, &&t\geq b. \end{aligned} \tag{4.2} \]

At an infinite endpoint the formula starts beyond the compact support. Each integral uses only the corresponding sign side. On both sides the nonnegative absolute kernel is

\[ K(t,s)=\exp\left(-\int_{\min(t,s)}^{\max(t,s)} |G(r)|\,dr\right) \tag{4.3} \]

with the respective ordering and same-side indicators. On any such segment of length \(h\),

\[ \int|G|\geq R(h-m)_+,\qquad K(t,s)\leq e^{-R(h-m)_+}. \tag{4.4} \]

Its row and column integrals are consequently at most

\[ \int_0^\infty e^{-R(h-m)_+}\,dh=m+R^{-1}. \tag{4.5} \]

The two sign blocks are disjoint, so these bounds hold for the whole kernel. For clarity, if both bounds are \(B\), Cauchy–Schwarz in the measure \(K(t,s)\,ds\) gives

\[ \left(\int K(t,s)|f(s)|\,ds\right)^2 \leq B\int K(t,s)|f(s)|^2\,ds. \]

Integration in \(t\) gives an \(L^2\) operator bound \(B\). Apply it to (4.2), and use \(\|f\|_2=\|L_Gu\|_2\). ∎

Theorem 4.2. Under Lemma 2.2 and the favorable sign condition,

\[ \rho^{1/(k+1)}\|u\|_2 \leq C_k\|L_Gu\|_2, \qquad u\in C_c^\infty(J). \tag{4.6} \]

The constant is uniform over all such \(J,G\).

Proof. For \(k\geq1\), (2.6) and (4.1) give the bound with \(R=\rho^{1/(k+1)}\). For \(k=0\), (2.9) makes \(E\) empty with \(R=\rho/2\), and (4.1) applies again. ∎

Corollary 4.3 (no boundary condition on the function). If \(J=(-1,1)\) and \(v\) is any smooth function there, then

\[ \begin{gathered} \rho^{1/(k+1)} \|v\|_{L^2(-1/2,1/2)} \\ \leq C_k\|L_Gv\|_{L^2(-1,1)} \\ +C_k\|v\|_{L^2(-1,1)}. \end{gathered} \tag{4.7} \]

If a norm on the right is infinite, the inequality is understood in that sense. In particular, with \(\varepsilon=2^{-k}\) and \(\rho\geq2\),

\[ \begin{gathered} \rho^{2\varepsilon}\int_{-1/2}^{1/2}|v|^2 \\ \leq C_k\rho^2\int_{-1}^1|L_Gv|^2 \\ +C_k\int_{-1}^1|v|^2. \end{gathered} \tag{4.8} \]

Proof. Choose a fixed smooth cutoff \(\chi\), supported in \((-1,1)\) and equal to one on \([-1/2,1/2]\). Apply (4.6) to \(\chi v\) and use

\[ L_G(\chi v)=\chi L_Gv-i\chi'v. \tag{4.9} \]

This proves (4.7). Since \(\varepsilon\leq1/(k+1)\) and \(\rho\geq1\), its square implies (4.8), even with the extra factor \(\rho^2\) only on the equation term. No compact support or endpoint value of \(v\) has been assumed. ∎

5. Localize at the complete derivative scale

We now prove Theorem 1.1. Use a large fixed number \(\sigma\geq2\), to be chosen solely in terms of \(k\), and define

\[ m(t)=\max_{0\leq j\leq k} \left(\frac{|G^{(j)}(t)|}{\sigma}\right)^{1/(j+1)}, \qquad \ell(t)=1/m(t). \tag{5.1} \]

The original weight satisfies

\[ m(t)\leq M(t)\leq\sigma m(t). \tag{5.2} \]

If \(M(t)\geq\sigma\), then \(m(t)\geq1\).

The reciprocal length is Lipschitz

On a branch determining the maximum in (5.1), \(|G^{(j)}|=\sigma m^{j+1}>0\). Branches near zero lie below the positive maximum and can be ignored locally. Thus \(m\) is locally Lipschitz and its derivative almost everywhere is the derivative of an active branch. If \(j<k\),

\[ |m'| \leq \frac{|G^{(j+1)}|} {(j+1)\sigma m^j} \leq\frac{m^2}{j+1}. \tag{5.3} \]

If \(j=k\), (1.2) instead gives

\[ |m'|\leq\frac1{(k+1)\sigma m^k}\leq m^2. \tag{5.4} \]

The last inequality uses \(m\geq1\) and \(\sigma\geq1\). Therefore \(|\ell'|\leq1\) almost everywhere. Absolute continuity on compact subintervals gives

\[ |\ell(t)-\ell(s)|\leq|t-s|. \tag{5.5} \]

This estimate includes points where the active derivative changes.

Build a partition with a uniformly bounded derivative cost

Fix \(\delta=1/32\). Enumerate a countable dense set of centers in \(I\), retaining a center whenever its small interval of radius \(\delta\ell(t)/5\) is disjoint from all previously retained intervals. Write the retained centers as \(t_j\), and put

\[ r_j=\delta\ell(t_j),\qquad B_j=(t_j-r_j,t_j+r_j),\qquad m_j=m(t_j). \tag{5.6} \]

These intervals may extend beyond \(I\); all functions below are evaluated only on their intersections with \(I\).

Here are the covering details. If a candidate small interval meets a selected one, (5.5) gives

\[ |\ell(t)-\ell(t_j)| \leq\frac{\delta}{5}\big(\ell(t)+\ell(t_j)\big). \tag{5.7} \]

Their lengths are therefore comparable with ratio less than two. The candidate center lies within \(3r_j/5\) of the selected center. Hence the inner intervals of radii \(3r_j/4\) cover the dense set. They cover all of \(I\): near any remaining center its positive continuous \(\ell\) bounds the meeting selected radii below; only finitely many disjoint small intervals of that minimum length fit nearby. Taking a sequence of dense centers then retains the inner covering inequality for at least one of these finitely many intervals.

The same argument proves local finiteness on compact subsets of \(I\). If two \(B_j\) meet, their radii have ratio at most \((1+\delta)/(1-\delta)<2\). At any point, let \(r_*\) be the supremum of the meeting radii; it is finite because \(r_j\leq\delta\). All meeting centers lie within \(r_*\), and their disjoint small intervals lie within an interval of length \(12r_*/5\). Each small interval has length greater than \(r_*/5\). Thus at most sixteen \(B_j\) meet any point; this deliberately loose bound is uniform.

Choose a fixed smooth bump \(0\leq\beta\leq1\), supported in \((-1,1)\) and equal to one on \([-3/4,3/4]\). Set

\[ \beta_j(t)=\beta((t-t_j)/r_j),\qquad W(t)=\left(\sum_j\beta_j(t)^2\right)^{1/2}, \qquad \theta_j(t)=\beta_j(t)/W(t). \tag{5.8} \]

On \(I\), the sums are locally finite and \(1\leq W\leq4\). Hence

\[ \sum_j\theta_j^2=1. \tag{5.9} \]

For \(t\in B_j\cap I\), (5.5) gives

\[ (1-\delta)m(t)\leq m_j\leq(1+\delta)m(t). \tag{5.10} \]

The vector \((\theta_j)\) is the normalization of \((\beta_j)\). Its derivative is the orthogonal projection of \((\beta_j')/W\) onto the perpendicular space of that unit vector. Consequently

\[ \sum_j|\theta_j'(t)|^2 \leq\sum_j|\beta_j'(t)|^2 \leq C_{\mathrm{cov}}m(t)^2. \tag{5.11} \]

The last bound uses the overlap bound, \(r_j=\delta/m_j\) and (5.10). Its constant depends only on the fixed bump and \(\delta\). It does not depend on \(\sigma,G\) or \(I\).

Apply the normalized estimate on each actual interval

For each selected center use

\[ y=m_j(t-t_j),\qquad G_j(y)=G(t_j+y/m_j)/m_j. \tag{5.12} \]

The domain \(J_j=(-1,1)\cap m_j(I-t_j)\) is an open interval containing zero. It may be truncated at the original boundary. The normalized coefficient has

\[ \begin{gathered} \max_{0\leq d\leq k}|G_j^{(d)}(0)|=\sigma, \\ |G_j^{(k+1)}|\leq m_j^{-(k+2)}\leq1. \end{gathered} \tag{5.13} \]

Positive rescaling preserves the favorable sign. Moreover \((\theta_j u)(t_j+y/m_j)\) is compactly supported in \(J_j\): its interval cutoff vanishes for \(|y|\geq\delta\), and \(u\) vanishes near the original endpoints. Theorem 4.2 therefore applies without any exterior coefficient.

Changing variables back gives

\[ \begin{gathered} \sigma^{2/(k+1)}m_j^2\|\theta_j u\|_2^2\\ \leq C_k^2\|L_G(\theta_j u)\|_2^2. \end{gathered} \tag{5.14} \]

For example, the squared norm on the rescaled function is multiplied by \(m_j\), while that on its normalized equation is multiplied by \(m_j^{-1}\); this is the origin of \(m_j^2\).

Sum (5.14), use (5.9)–(5.11), and expand \(L_G(\theta_j u)=\theta_jL_Gu-i\theta_j'u\). We obtain

\[ \begin{gathered} \sigma^{2/(k+1)}(1-\delta)^2\|mu\|_2^2\\ \leq2C_k^2\|L_Gu\|_2^2\\ +2C_k^2 C_{\mathrm{cov}}\|mu\|_2^2. \end{gathered} \tag{5.15} \]

Choose \(\sigma\) large enough that the coefficient on the left is at least four times \(C_k^2C_{\mathrm{cov}}\). All constants on the right are independent of that choice. Absorb the weight term, and use \(M\leq\sigma m\) from (5.2). This proves (1.5) when \(M\geq M_k=\sigma\), with a constant depending only on \(k\).

Only finitely many selected intervals meet the compact support of \(u\), so each summed identity on that support is an ordinary finite identity. The construction works for every interval length. This completes the proof of Theorem 1.1. ∎

6. A uniform estimate with parameters

The derivative-scale theorem also treats coefficients whose highest derivative grows with a large parameter.

Corollary 6.1. Let \(q(t,\vartheta)\) be real and smooth in \(t\), on an open interval that may depend on the parameter \(\vartheta\). Let \(\Lambda\geq1\) be a parameter size. Suppose, uniformly in all parameters and all \(t\),

\[ |\partial_t^{k+1}q|\leq C_0\Lambda,\qquad \sum_{j=0}^k|\partial_t^j q|\geq c_0\Lambda, \tag{6.1} \]

and \(q\) has the favorable sign orientation. For sufficiently large \(\Lambda\), depending only on \(k,C_0,c_0\),

\[ \begin{gathered} \|qv\|_2+\Lambda^{1/(k+1)}\|v\|_2 \\ \leq C\|(D+iq)v\|_2, \qquad v\in C_c^\infty(I). \end{gathered} \tag{6.2} \]

The constant is independent of the interval and the parameters. Nonnegative \(q\) satisfies the required sign condition.

Proof. Define

\[ a=((1+C_0)\Lambda)^{-1/(k+2)},\qquad t=as,\qquad H(s)=a q(as,\vartheta). \tag{6.3} \]

Its highest derivative is bounded by one. Its derivative scale is

\[ \begin{gathered} M_H(s)=a M_q(as),\\ M_q(t)=\max_{0\leq j\leq k} |\partial_t^j q(t,\vartheta)|^{1/(j+1)}. \end{gathered} \tag{6.4} \]

When \(c_0\Lambda/(k+1)\geq1\), at least one term of the sum in (6.1) has that size, so

\[ M_q(t)\geq \left(\frac{c_0\Lambda}{k+1}\right)^{1/(k+1)}. \tag{6.5} \]

It follows that \(M_H\) is bounded below by a positive fixed constant times

\[ \Lambda^{\,1/(k+1)-1/(k+2)} =\Lambda^{\,1/((k+1)(k+2))}. \tag{6.6} \]

This tends to infinity. Theorem 1.1 applies on the rescaled interval for large enough \(\Lambda\). Both the weight and the equation acquire the same factor \(a\), and the same change of \(L^2\) measure, so its conclusion is exactly \(\|M_qv\|_2\leq C_k\|(D+iq)v\|_2\). Now \(M_q\geq|q|\) and (6.5) give (6.2). ∎

In a symbol \(q(t,x',\xi')\), this corollary applies with the transverse phase variables frozen as parameters and \(\Lambda=|\xi'|\), provided the displayed bounds hold uniformly. It proves the full scalar estimate at each parameter value. Passing to the operator \(q(t,x',D')\) requires phase-space localization and quantization estimates; integration over independent scalar parameters alone does not perform that step.

7. Exercises with complete solutions

Exercise 1 — a coefficient with a positive floor, 8 points. Let \(G(t)=\lambda(t^4+2t^2+3)\) on \((-1,1)\), with \(\lambda\geq1\), and take \(k=4\). Compute \(\rho\). Compare the general central power with the stronger estimate coming from the positive floor.

Solution. At zero the derivatives through order four are \(3\lambda,0,4\lambda,0,24\lambda\), so \(\rho=24\lambda\); the fifth derivative is zero. Theorem 3.2 gives the power \((24\lambda)^{1/5}\). However \(G\geq3\lambda\) everywhere and satisfies the favorable orientation. In Lemma 4.1 take \(R=3\lambda\) and an empty small set, disregarding equality on the single point if necessary. The kernel bound is \(1/(3\lambda)\), so

\[ 3\lambda\|u\|_2\leq\|L_Gu\|_2. \]

The first conclusion is uniform for all coefficients with the same finite-derivative information; it does not claim that each particular coefficient has exactly that weakest scale.

Exercise 2 — coalescing roots and a forbidden orientation, 12 points. For \(0\leq d\leq1\), consider \(P_d(t)=t^3-dt\). Explain why its small-set bound is uniform in \(d\), and why the exponent \(1/3\) cannot be improved uniformly. For fixed \(d>0\), show that \(G=\lambda P_d\) can fail every positive equation gain despite its large third derivative.

Solution. The third derivative of \(P_d\) is six, independent of \(d\), so Lemma 2.1 gives a uniform bound \(C a^{1/3}\). At \(d=0\) the set where \(|P_0|\leq a\), for \(0<a<1\), has length exactly \(2a^{1/3}\). A larger power of \(a\) would fail as \(a\to0\).

For \(d>0\), the zero at zero is a decreasing crossing, with \(G'(0)=-\lambda d<0\). Choose \(0<r<R<\sqrt d\) and a smooth cutoff \(\chi\) supported in \((-R,R)\), equal to one on \([-r,r]\). The exact uncut solution is

\[ h_\lambda(t)= \exp\left(\lambda(t^4/4-dt^2/2)\right), \qquad h_\lambda'-G h_\lambda=0. \]

On the support of \(\chi'\), its exponent is at most \(-c\lambda\), while \(\|\chi h_\lambda\|_2\asymp\lambda^{-1/4}\). Thus the normalized compact function \(u_\lambda=\chi h_\lambda/\|\chi h_\lambda\|_2\) has norm one and

\[ \|L_Gu_\lambda\|_2 \leq C\lambda^{1/4}e^{-c\lambda}. \]

No positive power of \(\lambda\) can be bounded by this equation norm. Yet \(\rho=6\lambda\), and the sign-free estimate (3.3) remains valid for the separate derivative and potential norms. This is cancellation, not a failure of the polynomial small-set bound.

Exercise 3 — why unrestricted functions need the error term, 6 points. For \(G=\lambda\) on \((-1,1)\), exhibit a nonzero smooth function with \(L_Gv=0\). Explain why Corollary 4.3 cannot discard its full-interval norm of \(v\).

Solution. The function \(v(t)=e^{\lambda t}\) obeys \(v'-\lambda v=0\), hence \(L_Gv=0\). Its norm on \((-1/2,1/2)\) is positive. A bound for that norm by the equation norm alone is therefore impossible for arbitrary smooth \(v\). The cutoff term in (4.9) accounts for this boundary information. Compact test functions instead have the endpoint vanishing used by the one-sided inverse.

Exercise 4 — sharpness of the whole weight, 10 points. Take \(G_\lambda(t)=\lambda t^5/120\), \(k=5\), and let \(a=1/6\). Compute \(M\) at \(t=\lambda^{-a}y\). Show that no uniform version of (1.5) can replace \(M\) by \(M^\theta\) with \(\theta>1\).

Solution. The derivatives are \(\lambda t^{5-j}/(5-j)!\), \(0\leq j\leq5\). Therefore

\[ M(\lambda^{-a}y)=\lambda^a \max_{0\leq j\leq5} \left(\frac{|y|^{5-j}}{(5-j)!}\right)^{1/(j+1)}. \]

The maximum is at least one because of \(j=5\). The coefficient has the favorable orientation and its sixth derivative is zero. For nonzero \(f\in C_c^\infty(\mathbb R)\), use

\[ u_\lambda(t)=\lambda^{a/2}f(\lambda^a t). \]

For large \(\lambda\) its support lies in \((-1,1)\). Its equation norm is exactly

\[ \lambda^a\|(D_y+i y^5/120)f\|_2, \]

whereas \(\|M^\theta u_\lambda\|_2\) is a fixed positive constant times \(\lambda^{a\theta}\). If \(\theta>1\), the proposed estimate is impossible. This also gives sharpness of the central exponent \(1/(k+1)\); the same scaling works with \(\lambda t^k/k!\), including \(k=0\).

Exercise 5 — a lower derivative gives a better scale, 10 points. Let \(G_\lambda(t)=\lambda(t^3+2t)\), with \(k=3\) and \(\lambda\geq2\). Compare the central estimate based on \(\rho\) with the derivative-scale estimate. Prove that the better power is sharp uniformly in \(\lambda\).

Solution. The fourth derivative is zero. At zero the derivative maximum is \(\rho=6\lambda\), so the general central power is \((6\lambda)^{1/4}\). But

\[ G_\lambda'(t)=\lambda(3t^2+2)\geq2\lambda, \]

and \(G_\lambda\) is increasing through zero. Thus \(M(t)\geq\sqrt{2\lambda}\) everywhere. Theorem 1.1 gives \(\sqrt{\lambda}\|u\|_2\leq C\|L_Gu\|_2\) for large \(\lambda\).

With \(u_\lambda(t)=\lambda^{1/4}f(\sqrt\lambda t)\), its norm is fixed and the conjugated equation is

\[ \sqrt\lambda(D_y+2iy)f +i\lambda^{-1/2}y^3f. \]

Its norm is \(O(\sqrt\lambda)\). Consequently no larger power of \(\lambda\) can hold uniformly. The maximum in (1.4) detects this first-derivative scale even though we allowed a third-order zero in the general hypothesis.

Exercise 6 — a truncated domain is not an exterior coefficient, 8 points. Let \(J=(-h,2h)\subset(-1,1)\), with \(h>0\) possibly much smaller than \(\rho^{-1/(k+1)}\). Explain why Theorems 3.2 and 4.2 retain their constants. For a boundary interval in Section 5, derive the factor \(m_j^2\) in (5.14) and identify exactly where endpoint vanishing is used.

Solution. The Taylor polynomial is defined on all of \((-1,1)\), but its remainder is evaluated only on segments from zero to points of \(J\). The small set in \(J\) is a subset of the polynomial small set on \((-1,1)\), so its bound does not depend on \(h\). A compact test function can be extended by zero for the derivative estimate. The signed kernel integrates only inside \(J\); no value of \(G\) outside \(J\) enters.

For \(v_j(y)=(\theta_j u)(t_j+y/m_j)\), the change of variables gives

\[ \|v_j\|_2^2=m_j\|\theta_j u\|_2^2,\qquad \|(D_y+iG_j)v_j\|_2^2 =m_j^{-1}\|L_G(\theta_j u)\|_2^2. \]

Multiplication by \(m_j\) in the normalized inequality gives (5.14). Compact support of \(u\) makes \(v_j\) zero near the original endpoint of \(J_j\); the interval cutoff makes it zero near the other cutoff endpoints. These are precisely the endpoint values required in (4.2). A short interval can only make the derivative cost more severe; it does not weaken the uniform estimate.

References

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