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

Degenerate energy and sharp gains of regularity

Ellipticity controls a derivative in every direction. A degenerate equation may control a missing direction only after motion in another direction. This lesson makes that mechanism quantitative for a family of two-dimensional operators. It proves the estimates, checks their sharpness, and explains why the gain supplied by an energy norm differs from the gain supplied by the equation norm.

The prerequisites are Plancherel's theorem, integration by parts, and the Fourier definition of Sobolev spaces. The conventions and general operator mapping results are developed in Symbols, operators and Sobolev scales. No pseudodifferential construction is needed for the estimates below. Basic references are Hörmander's Hypoelliptic second order differential equations [H], Rothschild and Stein's Hypoelliptic differential operators and nilpotent groups [RS], and Laurent and Léautaud's Unique continuation and applications [LL]. The present elementary model isolates one part of the geometry considered in those works.

1. A direction that appears after taking brackets

Fix an integer \(k\geq1\). On \(\mathbb R^2\), with coordinates \((x,y)\), set \[ X=\partial_x,\qquad Y=x^k\partial_y,\qquad L_k=X^*X+Y^*Y=-\partial_x^2-x^{2k}\partial_y^2. \] The adjoints here are for Lebesgue measure. The coefficients of \(Y\) have no \(y\)-dependence, so \(Y^*=-Y\). For \(u\in C_c^\infty(\mathbb R^2)\), define \[ E_k(u)=\|Xu\|_2^2+\|Yu\|_2^2=(L_ku,u)_{L^2}. \] We take the inner product to be linear in its first argument. At \(x=0\), \(Y\) is zero. Nevertheless, \[ [X,Y]=kx^{k-1}\partial_y,\qquad \underbrace{[X,[X,\ldots,[X,Y]]]}_{k\text{ occurrences of }X} =k!\partial_y. \] Thus repeated commutation recovers the missing direction. The length of the last bracket is \(k+1\), since \(Y\) also counts as a factor.

Write \[ \varepsilon=\frac1{k+1},\qquad \|u\|_{H^s}^2=\int_{\mathbb R^2}(1+\xi^2+\eta^2)^s |\widehat u(\xi,\eta)|^2\,d\xi\,d\eta. \] All Fourier transforms in this lesson are unitary. Our two main estimates are \[ \|u\|_{H^\varepsilon}^2\leq13\big(E_k(u)+\|u\|_2^2\big), \tag{1.1} \] and, with a constant \(C_k\) depending only on \(k\), \[ \|u\|_{H^{2\varepsilon}} \leq C_k\big(\|L_ku\|_2+\|u\|_2\big). \tag{1.2} \] Both exponents are optimal in these respective formulations.

The terminology needs care. Formula (1.1) is an energy subelliptic estimate with gain \(\varepsilon\). Formula (1.2) is an estimate in the equation norm with gain \(2\varepsilon\). Relative to a second-order elliptic estimate, the latter loses \(2-2\varepsilon\) derivatives. These two gains should never be identified merely because both estimates concern the same operator.

2. An elementary uncertainty inequality

The missing derivative is recovered by balancing concentration against differentiation.

Lemma 2.1. For every integer \(k\geq1\) and every \(f\in C_c^\infty(\mathbb R)\), \[ \|f\|_2^2\leq12\left(\|f'\|_2^2+\|t^kf\|_2^2\right). \tag{2.1} \]

Proof. If \(t\in[-1,1]\) and \(z\in[1,2]\), the fundamental theorem of calculus and Cauchy–Schwarz give \[ |f(t)|^2\leq2|f(z)|^2+2|t-z|\int_{-1}^2|f'(r)|^2\,dr \leq2|f(z)|^2+6\|f'\|_2^2. \] Integrate first over \(z\in[1,2]\), then over \(t\in[-1,1]\). The result is \[ \int_{-1}^1|f|^2\leq4\int_1^2|f|^2+12\|f'\|_2^2. \] On \(\{|t|\geq1\}\), \(|t|^{2k}\geq1\). Consequently \[ \|f\|_2^2\leq5\|t^kf\|_2^2+12\|f'\|_2^2, \] which proves (2.1). The same argument applies to Schwartz functions by a cutoff limit. \(\square\)

Corollary 2.2. For \(\eta\neq0\), \[ \|f'\|_2^2+\eta^2\|x^kf\|_2^2 \geq\frac1{12}|\eta|^{2\varepsilon}\|f\|_2^2. \tag{2.2} \]

Proof. Put \(a=|\eta|^\varepsilon\) and write \(f(x)=a^{1/2}F(ax)\). This transformation preserves the \(L^2\) norm. It gives \[ \|f'\|_2^2=a^2\|F'\|_2^2, \qquad \eta^2\|x^kf\|_2^2=a^2\|t^kF\|_2^2, \] because \(\eta^2a^{-2k}=a^2\). Apply Lemma 2.1. \(\square\)

The balancing scale is therefore \(|x|\asymp|\eta|^{-1/(k+1)}\). A packet concentrated more tightly pays for its \(x\)-derivative; a broader packet pays for multiplication by \(\eta x^k\).

3. The energy controls a fractional derivative

Theorem 3.1. Estimate (1.1) holds for every \(u\in C_c^\infty(\mathbb R^2)\).

Proof. Take the Fourier transform only in \(y\), and denote the resulting function by \(v(x,\eta)\). Applying (2.2) for each \(\eta\neq0\) and integrating yields \[ \big\||D_y|^\varepsilon u\big\|_2^2\leq12E_k(u). \tag{3.1} \] The value \(\eta=0\) has no effect on this integral. Also \[ \big\||D_x|^\varepsilon u\big\|_2^2 \leq\|u\|_2^2+\|\partial_xu\|_2^2, \] since \(|\xi|^{2\varepsilon}\leq1+\xi^2\). For \(0<\varepsilon\leq1\), subadditivity of \(t^\varepsilon\) on nonnegative numbers gives \[ (1+\xi^2+\eta^2)^\varepsilon \leq1+|\xi|^{2\varepsilon}+|\eta|^{2\varepsilon}. \] Plancherel's theorem now shows \[ \|u\|_{H^\varepsilon}^2 \leq2\|u\|_2^2+\|\partial_xu\|_2^2+12E_k(u) \leq13\big(E_k(u)+\|u\|_2^2\big). \] This is the stated constant. \(\square\)

For \(k=2\), the energy controls one third of a derivative in the ordinary isotropic Sobolev scale. The coefficient \(x^4\) in the equation is not an obstacle that can be removed by treating it as uniformly positive. Its vanishing is precisely what creates this fractional exponent.

4. The equation controls twice as much

To prove (1.2), we need to control a second \(x\)-derivative without losing the frequency scale.

Lemma 4.1. Set \(A=-d^2/dt^2+t^{2k}\). There is \(B_k<\infty\) such that \[ \|f''\|_2\leq B_k\|Af\|_2, \qquad \|f\|_2\leq12\|Af\|_2 \tag{4.1} \] for every \(f\in C_c^\infty(\mathbb R)\).

Proof. Lemma 2.1 and Cauchy–Schwarz imply \[ \frac1{12}\|f\|_2^2\leq(Af,f)\leq\|Af\|_2\|f\|_2. \] If \(f\neq0\), division proves the second assertion. The assertion is also true for \(f=0\).

Put \(V(t)=t^{2k}\). Two integrations by parts give the exact identity \[ \|Af\|_2^2 =\|f''\|_2^2+\|Vf\|_2^2 +2\int V|f'|^2-\int V''|f|^2. \tag{4.2} \] There is a finite number \(M_k\geq0\) with \[ V''(t)=2k(2k-1)t^{2k-2}\leq\tfrac12t^{4k}+M_k \] for all real \(t\). Indeed, the difference of the left side and \(t^{4k}/2\) is bounded above, because its highest-degree term is negative. For \(k=1\) the left side is the constant \(2\). Since \(V\geq0\), (4.2) therefore implies \[ \|f''\|_2^2+\tfrac12\|Vf\|_2^2 \leq\|Af\|_2^2+M_k\|f\|_2^2 \leq(1+144M_k)\|Af\|_2^2. \] Take \(B_k=(1+144M_k)^{1/2}\). \(\square\)

Theorem 4.2. Estimate (1.2) holds for every \(u\in C_c^\infty(\mathbb R^2)\).

Proof. At frequency \(\eta\), the operator is \[ A_\eta=-\partial_x^2+\eta^2x^{2k}. \] For \(\eta\neq0\), the unitary dilation in Corollary 2.2 conjugates \(A_\eta\) to \(|\eta|^{2\varepsilon}A\). Thus (4.1) gives \[ \|\partial_x^2v(\cdot,\eta)\|_2\leq B_k\|A_\eta v(\cdot,\eta)\|_2, \] and \[ |\eta|^{2\varepsilon}\|v(\cdot,\eta)\|_2 \leq12\|A_\eta v(\cdot,\eta)\|_2. \] At \(\eta=0\) the first inequality holds with constant \(1\), and the second has zero left side. Integration and Plancherel give \[ \|\partial_x^2u\|_2\leq B_k\|L_ku\|_2, \qquad \big\||D_y|^{2\varepsilon}u\big\|_2\leq12\|L_ku\|_2. \tag{4.3} \] Here \(2\varepsilon\leq1\), since \(k\geq1\). Consequently \[ (1+\xi^2+\eta^2)^{2\varepsilon} \leq1+|\xi|^{4\varepsilon}+|\eta|^{4\varepsilon} \leq2+\xi^4+|\eta|^{4\varepsilon}. \] It follows that \[ \|u\|_{H^{2\varepsilon}}^2 \leq2\|u\|_2^2+(B_k^2+144)\|L_ku\|_2^2. \] Taking square roots proves the theorem. \(\square\)

For \(k=1\), the equation norm controls one derivative. For \(k=2\), it controls two thirds of a derivative. The energy gains are respectively one half and one third. The theorem is an a priori estimate for the displayed test-function domain; a proof of local regularity for arbitrary distributional solutions must also justify localization and regularization.

5. Wave packets prove that the exponents cannot improve

Theorem 5.1. Suppose an estimate \[ \|u\|_{H^s}^2\leq C\big(E_k(u)+\|u\|_2^2\big) \tag{5.1} \] holds for all test functions supported in a fixed open rectangle containing a point of \(x=0\). Then \(s\leq\varepsilon\). If instead \[ \|u\|_{H^s}\leq C\big(\|L_ku\|_2+\|u\|_2\big), \tag{5.2} \] holds on that domain, then \(s\leq2\varepsilon\).

Proof. Choose nonzero smooth compactly supported functions \(f\) and \(g\), with \(g\) supported in the \(y\)-interval of the rectangle. For \(\lambda\geq1\), set \[ u_\lambda(x,y)=\lambda^{\varepsilon/2} f(\lambda^\varepsilon x)g(y)e^{i\lambda y}. \] For all sufficiently large \(\lambda\), its support lies in the rectangle. Its \(L^2\) norm is the constant \(\|f\|_2\|g\|_2\). Direct differentiation gives \[ \|\partial_xu_\lambda\|_2=O(\lambda^\varepsilon), \qquad \|x^k\partial_yu_\lambda\|_2=O(\lambda^\varepsilon). \] For the second formula, use \(\lambda\lambda^{-k\varepsilon}=\lambda^\varepsilon\). Differentiating twice shows similarly \[ \|L_ku_\lambda\|_2=O(\lambda^{2\varepsilon}). \tag{5.3} \] The \(g'\) and \(g''\) terms have smaller powers of \(\lambda\).

If \(s>0\), choose \(R<\infty\) such that \(\int_{|\zeta|\leq R}|\widehat g(\zeta)|^2\,d\zeta>0\). The Fourier transform in \(y\) is centered at \(\eta=\lambda\). For \(\lambda\geq2R\), the portion \(|\eta-\lambda|\leq R\) lies in \(|\eta|\geq\lambda/2\), so \[ \|u_\lambda\|_{H^s}\geq c_s\lambda^s. \] Substitution in (5.1) gives \(\lambda^{2s}\leq C'\lambda^{2\varepsilon}\), and substitution in (5.2) gives \(\lambda^s\leq C''\lambda^{2\varepsilon}\). Let \(\lambda\to\infty\). Only the asserted exponent ranges are possible. Nonpositive \(s\) already satisfies both bounds. \(\square\)

For an example, choose \(k=3\). A packet of \(y\)-frequency \(\lambda\) occupies an \(x\)-width of order \(\lambda^{-1/4}\). Its energy is of order \(\lambda^{1/2}\), while the equation norm is at most of order \(\lambda^{1/2}\). The Sobolev norm is of order \(\lambda^s\): squaring that norm in the energy estimate accounts for the different optimal exponents.

6. Perturbations in the controlled directions

An arbitrary first-order derivative is not harmless here. A first-order derivative in a controlled direction is.

Proposition 6.1. Let \(K\subset\mathbb R^2\) be compact. Let \(b_0,b_1,b_2\) be bounded measurable complex functions on \(K\), and define on \(C_c^\infty(K^\circ)\) \[ P=L_k+b_1X+b_2Y+b_0. \] Then \[ \|u\|_{H^{2\varepsilon}} \leq C\big(\|Pu\|_2+\|u\|_2\big), \tag{6.1} \] where \(C\) depends on \(k\) and the three coefficient bounds. No derivatives of those coefficients are required for this test-function estimate.

Proof. Write \(B=b_1X+b_2Y+b_0\). For some \(M<\infty\), \[ \|Bu\|_2\leq M\big(E_k(u)^{1/2}+\|u\|_2\big). \] Also \(E_k(u)\leq\|L_ku\|_2\|u\|_2\). The elementary inequality \(\sqrt{ab}\leq\delta a+(4\delta)^{-1}b\), for \(a,b\geq0\), gives \[ \|Bu\|_2\leq M\delta\|L_ku\|_2 +M\big(1+(4\delta)^{-1}\big)\|u\|_2. \] If \(M>0\), choose \(\delta=(2M)^{-1}\). Since \(\|L_ku\|_2\leq\|Pu\|_2+\|Bu\|_2\), absorption yields \[ \|L_ku\|_2\leq2\|Pu\|_2+C_M\|u\|_2. \] The case \(M=0\) is immediate. Apply Theorem 4.2. \(\square\)

For example, \(L_2+(1+i\sin y)\partial_x+\cos x\,x^2\partial_y\) retains the two-thirds estimate. The same argument does not cover replacing \(x^2\partial_y\) by \(\partial_y\): near \(x=0\), the energy gives no uniform bound for that derivative.

7. Exercises

Exercise 7.1 — scaling, 6 points. Find the exponent \(a\) for which the two contributions to the energy of \(\lambda^{a/2}f(\lambda^a x)g(y)e^{i\lambda y}\) have the same leading power of \(\lambda\). Explain why the value is \(1/(k+1)\), rather than \(1/k\).

Solution. The squared \(x\)-derivative norm is proportional to \(\lambda^{2a}\). The leading squared weighted \(y\)-derivative norm is proportional to \(\lambda^{2-2ka}\). Equating the powers gives \(2a=2-2ka\), hence \(a=1/(k+1)\). The additional \(1\) comes from the cost of the \(x\)-derivative. Balancing the coefficient alone would omit that cost.

Exercise 7.2 — an uncontrolled direction, 8 points. Let \(L=-\partial_x^2\) on \(\mathbb R^2\). Prove that no positive isotropic Sobolev gain follows from its energy or its equation norm on a fixed rectangle.

Solution. Take fixed nonzero compactly supported \(f(x),g(y)\) inside the rectangle and let \(u_\lambda=f(x)g(y)e^{i\lambda y}\). The \(L^2\) norm, energy, and \(\|Lu_\lambda\|_2\) are all independent of \(\lambda\). The frequency-window argument in Theorem 5.1 gives \(\|u_\lambda\|_{H^s}\geq c_s\lambda^s\) for every \(s>0\). Either proposed estimate would give a bounded left side, a contradiction. No bracket involving \(\partial_x\) creates \(\partial_y\).

Exercise 7.3 — interpolation, 8 points. Deduce from Theorem 4.2 that for \(0\leq\theta\leq1\), \[ \|u\|_{H^{2\theta\varepsilon}} \leq C_k^\theta\|u\|_2^{1-\theta} (\|L_ku\|_2+\|u\|_2)^\theta. \]

Solution. For \(0<\theta<1\), write the integrand defining the squared norm as \[ \big(|\widehat u|^2\big)^{1-\theta} \big((1+\xi^2+\eta^2)^{2\varepsilon}|\widehat u|^2\big)^\theta. \] Hölder's inequality with exponents \(1/(1-\theta)\) and \(1/\theta\) gives \(\|u\|_{H^{2\theta\varepsilon}}\leq\|u\|_2^{1-\theta}\|u\|_{H^{2\varepsilon}}^\theta\). Use Theorem 4.2. The endpoint cases are equality at \(\theta=0\) and the theorem at \(\theta=1\).

Exercise 7.4 — multiplication order, 10 points. Suppose \(b\) is bounded and measurable. Compare \(bYu\) with \(Y(bu)\). Explain which expression Proposition 6.1 controls, and give an example showing why the other may not even belong to \(L^2\).

Solution. Boundedness of \(b\) gives \(\|bYu\|_2\leq\|b\|_\infty\|Yu\|_2\). This is the expression in Proposition 6.1. Distributionally, \(Y(bu)=bYu+x^k(\partial_yb)u\). Take \(b(y)=\mathbf1_{\{y>0\}}\), and choose \(u\) with \(u(x,0)\neq0\) for some \(x\neq0\). Then \(x^k(\partial_yb)u=x^ku(x,0)\delta_{y=0}\), a nonzero surface measure, which is not an \(L^2\) function. The order of multiplication in the stated operator is essential.

References

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