Book 6B

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Course 6Book 6B: Linear PDE: Characteristics, Waves and RegularityChapter 8

Lᵖ Theory and Strong Solutions

Calderón–Zygmund estimates, strong solutions, and the Alexandrov–Bakelman–Pucci maximum principle.

42 min read · Updated Oct 20, 2026

Read with Jost, Partial Differential Equations, sections 12.1–12.2 (strong solutions and a survey of the LpL^p theory) and section 3.2 (the maximum principle of Alexandrov and Bakelman); Taylor III, §13.5 ("Singular integral operators on LpL^p"); Gilbarg and Trudinger, chapter 9, for the full theory. Stein's Singular Integrals and Differentiability Properties of Functions, chapter I, is the classic source for the Calderón–Zygmund method. This chapter proves Marcinkiewicz interpolation, the Calderón–Zygmund decomposition, the LpL^p bound for second derivatives, and the Alexandrov–Bakelman–Pucci estimate.

In this chapter · 8 sections
  1. 8.1From vorticity to velocity
  2. 8.2Distribution functions and weak LpL^pLp
  3. 8.3The Calderón–Zygmund decomposition
  4. 8.4Singular integrals are bounded on LpL^pLp
  5. 8.5Strong solutions and the W2,pW^{2,p}W2,p estimates
  6. 8.6The Alexandrov–Bakelman–Pucci maximum principle
  7. 8.7History
  8. 8.8Exercises
  9. ↗Free resources

6B.7 Schauder Theory and the Continuity Method measured the gain of two derivatives in Hölder norms. This chapter measures it in LpL^p norms: if Δu=f\Delta u = f with f∈Lpf \in L^p, then every second derivative of uu is in LpL^p, for every 1<p<∞1 < p < \infty. The proof is a different use of the same cancellation. In 6B.7 Schauder Theory and the Continuity Method Hölder continuity of ff near the singularity made the singular kernel integrable. Here nothing is assumed about ff beyond integrability, and the cancellation is exploited on average: the operator f↦D2(Φ∗f)f \mapsto D^2(\Phi * f) is bounded on L2L^2 by the Fourier transform, and almost bounded on L1L^1 (it maps L1L^1 into "weak L1L^1") by a geometric decomposition of ff into a good part and bad parts with mean zero, the Calderón–Zygmund decomposition. Interpolation between the two endpoints, and duality, give every LpL^p in between.

The LpL^p estimates hold with merely continuous leading coefficients, which makes them the natural tool for nonlinear equations whose coefficients depend on the solution; with Sobolev embedding they bootstrap to Hölder estimates. They lead to strong solutions: functions with second derivatives in LpL^p that satisfy the equation almost everywhere. For strong solutions the pointwise maximum principle needs replacing, and the replacement, the Alexandrov–Bakelman–Pucci estimate, is a beautiful piece of convex geometry: the maximum of a function controls how much the gradient map spreads out, and that spread is bounded by the equation. Its sharpest application, at the end of the chapter, is a proof of the isoperimetric inequality.

By the end of this chapter you will be able to:

  • use distribution functions and weak LpL^p, and prove the Marcinkiewicz interpolation theorem;
  • prove the Calderón–Zygmund decomposition, and use it to show that singular integrals of Calderón–Zygmund type are weak (1,1)(1,1);
  • prove ∥D2u∥Lp≤Cp∥Δu∥Lp\|D^2u\|_{L^p} \leq C_p\|\Delta u\|_{L^p} for 1<p<∞1 < p < \infty, and explain why it fails for p=1p = 1 and p=∞p = \infty;
  • state the interior W2,pW^{2,p} estimate and the existence theory for strong solutions;
  • prove the Alexandrov–Bakelman–Pucci maximum principle, and use its method to prove the isoperimetric inequality.

From vorticity to velocity

In the world Model The Biot–Savart law and the limits of L∞L^\infty

In a two-dimensional incompressible fluid, the velocity vv is recovered from the vorticity ω=∂1v2−∂2v1\omega = \partial_1v_2 - \partial_2v_1 (how fast fluid elements spin) by the Biot–Savart law: v=∇⊥ψv = \nabla^\perp\psi, where the stream function solves Δψ=ω\Delta\psi = \omega. So the velocity gradient ∇v\nabla v is made of second derivatives of ψ\psi, that is, of the singular integrals of this chapter applied to ω\omega. The LpL^p theory gives

∥∇v∥Lp≤Cp∥ω∥Lp(1<p<∞),\|\nabla v\|_{L^p} \leq C_p\|\omega\|_{L^p} \qquad (1 < p < \infty),

with a constant CpC_p that grows like pp as p→∞p \to \infty, and the estimate is false for p=∞p = \infty: bounded vorticity, such as a patch of uniformly spinning fluid with a sharp edge, does not give a Lipschitz velocity field.

This is not a technicality. The Euler equations move vorticity along the flow, so the natural information is a bound on ω\omega in L∞L^\infty, and to prove that the flow is uniquely determined one needs control of ∇v\nabla v. Victor Yudovich showed in 1963 that the growth Cp∼pC_p \sim p is just slow enough: it implies that the velocity is log-Lipschitz, ∣v(x)−v(y)∣≤C∣x−y∣log⁡1∣x−y∣|v(x) - v(y)| \leq C|x - y|\log\frac{1}{|x - y|}, and that is enough for the particle paths, and the solution, to be unique. The precise rate at which the LpL^p constant blows up decided a basic question in fluid mechanics.

Distribution functions and weak LpL^p

For a measurable function ff on Rn\mathbb{R}^n, its distribution function is μf(t)=∣{x:∣f(x)∣>t}∣\mu_f(t) = |\{x : |f(x)| > t\}| for t>0t > 0. Every LpL^p norm can be computed from it:

∫∣f∣p dx=p∫0∞tp−1μf(t) dt\int|f|^p\,dx = p\int_0^\infty t^{p-1}\mu_f(t)\,dt

(Exercise 8.9, the layer-cake formula of 3A.5 Product Measures and Change of Variables). Chebyshev's inequality says μf(t)≤∥f∥pptp\mu_f(t) \leq \frac{\|f\|_p^p}{t^p}. A function is in weak LpL^p if μf(t)≤Cptp\mu_f(t) \leq \frac{C^p}{t^p} for some CC and all tt; every LpL^p function is, but not conversely: ∣x∣−n|x|^{-n} is in weak L1L^1 (its distribution function is α(n)/t\alpha(n)/t) but not in L1L^1.

An operator TT is strong (p,p)(p, p) if ∥Tf∥p≤A∥f∥p\|Tf\|_p \leq A\|f\|_p, and weak (p,p)(p, p) if μTf(t)≤(A∥f∥pt)p\mu_{Tf}(t) \leq \big(\frac{A\|f\|_p}{t}\big)^p. Strong implies weak, by Chebyshev. It is sublinear if ∣T(f+g)∣≤∣Tf∣+∣Tg∣|T(f + g)| \leq |Tf| + |Tg| (linear operators are).

Theorem 8.1 Marcinkiewicz interpolation

Let TT be sublinear, weak (1,1)(1, 1) with constant A1A_1 and strong (2,2)(2, 2) with constant A2A_2. Then for 1<p<21 < p < 2, TT is strong (p,p)(p, p):

∥Tf∥p≤Ap∥f∥p,App=p 2p(A1p−1+A222−p).\|Tf\|_p \leq A_p\|f\|_p, \qquad A_p^p = p\,2^p\Big(\frac{A_1}{p - 1} + \frac{A_2^2}{2 - p}\Big).

Proof. Fix t>0t > 0 and split f=f1+f2f = f_1 + f_2 with f1=f 1{∣f∣>t}f_1 = f\,\mathbf 1_{\{|f|>t\}} (the large values) and f2=f 1{∣f∣≤t}f_2 = f\,\mathbf 1_{\{|f|\leq t\}} (the small values). Then f1∈L1f_1 \in L^1 and f2∈L2f_2 \in L^2 when f∈Lpf \in L^p, and ∣Tf∣≤∣Tf1∣+∣Tf2∣|Tf| \leq |Tf_1| + |Tf_2|, so

μTf(2t)≤μTf1(t)+μTf2(t)≤A1t∫∣f∣>t∣f∣+A22t2∫∣f∣≤t∣f∣2,\mu_{Tf}(2t) \leq \mu_{Tf_1}(t) + \mu_{Tf_2}(t) \leq \frac{A_1}{t}\int_{|f|>t}|f| + \frac{A_2^2}{t^2}\int_{|f|\leq t}|f|^2,

using the weak (1,1)(1, 1) bound on f1f_1 and Chebyshev with the strong (2,2)(2, 2) bound on f2f_2. Now integrate with the layer-cake formula, substituting s=2ts = 2t:

∥Tf∥pp=p 2p∫0∞tp−1μTf(2t) dt≤p 2p[A1∫0∞tp−2 ⁣ ⁣∫∣f∣>t∣f∣ dx dt+A22∫0∞tp−3 ⁣ ⁣∫∣f∣≤t∣f∣2 dx dt].\|Tf\|_p^p = p\,2^p\int_0^\infty t^{p-1}\mu_{Tf}(2t)\,dt \leq p\,2^p\Big[A_1\int_0^\infty t^{p-2}\!\!\int_{|f|>t}|f|\,dx\,dt + A_2^2\int_0^\infty t^{p-3}\!\!\int_{|f|\leq t}|f|^2\,dx\,dt\Big].

Exchange the order of integration (Tonelli, 3A.5 Product Measures and Change of Variables). In the first, for each xx the tt-integral runs over 0<t<∣f(x)∣0 < t < |f(x)| and gives ∣f(x)∣p−1p−1\frac{|f(x)|^{p-1}}{p-1}; in the second it runs over t≥∣f(x)∣t \geq |f(x)| and gives ∣f(x)∣p−22−p\frac{|f(x)|^{p-2}}{2 - p}. Both produce ∫∣f∣p\int|f|^p, with the stated constants.

The idea, cutting ff at a height depending on the level tt being measured, is that large values are handled by the L1L^1 estimate (which is cheap on them) and small values by the L2L^2 estimate. The constant blows up as p→1p \to 1, as it must: the theorem fails at p=1p = 1.

The Calderón–Zygmund decomposition

A dyadic cube is a cube of the form 2−k([0,1)n+m)2^{-k}\big([0, 1)^n + m\big) with k∈Zk \in \mathbb{Z} and m∈Znm \in \mathbb{Z}^n. Two dyadic cubes are either disjoint or nested, and each is the union of 2n2^n dyadic children of half its side. Write fQ=1∣Q∣∫Qff_Q = \frac{1}{|Q|}\int_Qf for the average over QQ.

Lemma 8.2 The Calderón–Zygmund decomposition

Let f∈L1(Rn)f \in L^1(\mathbb{R}^n) and t>0t > 0. There is a family of disjoint dyadic cubes QkQ_k such that

  1. t<1∣Qk∣∫Qk∣f∣≤2ntt < \frac{1}{|Q_k|}\int_{Q_k}|f| \leq 2^nt for each kk;
  2. ∑k∣Qk∣≤1t∥f∥1\sum_k|Q_k| \leq \frac1t\|f\|_1;
  3. ∣f∣≤t|f| \leq t almost everywhere outside ⋃kQk\bigcup_kQ_k.

Proof. Choose k0k_0 so large and negative that every dyadic cube of side 2−k02^{-k_0} has 1∣Q∣∫Q∣f∣≤∥f∥1∣Q∣≤t\frac{1}{|Q|}\int_Q|f| \leq \frac{\|f\|_1}{|Q|} \leq t. Start with these cubes and subdivide each into its 2n2^n children. Select a child if its average of ∣f∣|f| exceeds tt, and stop subdividing it; otherwise subdivide it again, and repeat. The selected cubes are disjoint (nested dyadic cubes are never both selected, because subdivision stops at selection). Each selected cube QQ has average greater than tt, and its parent was not selected, so the parent's average is at most tt; since ∣parent∣=2n∣Q∣|{\rm parent}| = 2^n|Q|, 1∣Q∣∫Q∣f∣≤2n∣parent∣∫parent∣f∣≤2nt\frac{1}{|Q|}\int_Q|f| \leq \frac{2^n}{|{\rm parent}|}\int_{\rm parent}|f| \leq 2^nt. That is (1). For (2), ∣Qk∣<1t∫Qk∣f∣|Q_k| < \frac1t\int_{Q_k}|f|, and summing over the disjoint cubes gives at most 1t∥f∥1\frac1t\|f\|_1. For (3), a point xx outside every selected cube lies in dyadic cubes of arbitrarily small side, all with average at most tt; by the Lebesgue differentiation theorem (3A.6 Modes of Convergence and Differentiation) the averages over these shrinking cubes tend to ∣f(x)∣|f(x)| for almost every xx, so ∣f(x)∣≤t|f(x)| \leq t.

This is a stopping-time argument: subdivide until something happens, and stop there. Given the cubes, split f=g+bf = g + b with

g={foutside ⋃Qk,fQkon Qk,b=∑kbk,bk=(f−fQk)1Qk.g = \begin{cases}f & \text{outside }\bigcup Q_k,\\ f_{Q_k} & \text{on } Q_k,\end{cases}\qquad b = \sum_kb_k, \quad b_k = (f - f_{Q_k})\mathbf 1_{Q_k}.

The good part gg is bounded, ∣g∣≤2nt|g| \leq 2^nt, and ∥g∥1≤∥f∥1\|g\|_1 \leq \|f\|_1. The bad parts bkb_k are supported in the cubes and have mean zero: ∫bk=0\int b_k = 0. Mean zero is the property that makes a singular integral of bkb_k small far from QkQ_k.

Singular integrals are bounded on LpL^p

For f∈Cc∞(Rn)f \in C_c^\infty(\mathbb{R}^n) let w=Φ∗fw = \Phi * f, so −Δw=f-\Delta w = f (6A.2 Harmonic Functions), and define

Tijf=∂ijw=∂ij(Φ∗f).T_{ij}f = \partial_{ij}w = \partial_{ij}(\Phi * f).

Away from the support of ff, differentiating under the integral is allowed, and Tijf(x)=∫K(x−y)f(y) dyT_{ij}f(x) = \int K(x - y)f(y)\,dy with the kernel K=∂ijΦK = \partial_{ij}\Phi, which satisfies ∣K(z)∣≤C∣z∣−n|K(z)| \leq C|z|^{-n} and ∣∇K(z)∣≤C∣z∣−n−1|\nabla K(z)| \leq C|z|^{-n-1}. (At points of the support, 6B.7 Schauder Theory and the Continuity Method's formula, or the principal value plus a multiple of f(x)f(x), applies; we won't need it.)

Lemma 8.3 The L2L^2 bound

∥Tijf∥L2≤∥f∥L2\|T_{ij}f\|_{L^2} \leq \|f\|_{L^2} for f∈Cc∞(Rn)f \in C_c^\infty(\mathbb{R}^n), and TijT_{ij} extends to a bounded operator on L2L^2.

Proof. On the Fourier side (4A.6 The Fourier Transform), −Δw=f-\Delta w = f reads 4π2∣ξ∣2w^=f^4\pi^2|\xi|^2\hat w = \hat f, and ∂ijw^=−4π2ξiξjw^=−ξiξj∣ξ∣2f^\widehat{\partial_{ij}w} = -4\pi^2\xi_i\xi_j\hat w = -\frac{\xi_i\xi_j}{|\xi|^2}\hat f. The multiplier ξiξj∣ξ∣2\frac{\xi_i\xi_j}{|\xi|^2} has absolute value at most 11, so Plancherel gives the bound. (The function ww itself need not be in L2L^2; but ∂ijw\partial_{ij}w is, and the identity holds in the sense of tempered distributions.) A bounded linear operator on a dense subspace extends uniquely (4A.6 The Fourier Transform).

The multiplier ξiξj∣ξ∣2\frac{\xi_i\xi_j}{|\xi|^2} is homogeneous of degree 00: bounded, but discontinuous at ξ=0\xi = 0, which is the Fourier picture of the kernel's slow decay. The equivalent statement ∫∣D2w∣2=∫∣Δw∣2\int|D^2w|^2 = \int|\Delta w|^2 is the Hessian identity of 6A.6 Weak Solutions and Elliptic Regularity, which holds here because D2wD^2w decays like ∣x∣−n|x|^{-n} and DwDw like ∣x∣1−n|x|^{1-n} at infinity, so the boundary terms vanish.

Theorem 8.4 The weak (1,1)(1, 1) bound

There is A1=A1(n)A_1 = A_1(n) with ∣{∣Tijf∣>t}∣≤A1t∥f∥L1|\{|T_{ij}f| > t\}| \leq \frac{A_1}{t}\|f\|_{L^1} for all f∈L1∩L2f \in L^1\cap L^2 and t>0t > 0.

Proof. Apply Lemma 8.2 at height tt and split f=g+bf = g + b. Since ∣Tijf∣≤∣Tijg∣+∣Tijb∣|T_{ij}f| \leq |T_{ij}g| + |T_{ij}b|, it suffices to bound the measure where each exceeds t2\frac t2.

The good part. ∥g∥22≤∥g∥∞∥g∥1≤2nt∥f∥1\|g\|_2^2 \leq \|g\|_\infty\|g\|_1 \leq 2^nt\|f\|_1, so by Chebyshev and Lemma 8.3, ∣{∣Tijg∣>t2}∣≤4t2∥g∥22≤4⋅2nt∥f∥1|\{|T_{ij}g| > \frac t2\}| \leq \frac{4}{t^2}\|g\|_2^2 \leq \frac{4\cdot2^n}{t}\|f\|_1.

Near the cubes. Let yky_k be the centre of QkQ_k and Bk∗B_k^* the ball about yky_k of radius diam⁡Qk\operatorname{diam}Q_k. Then ∣Bk∗∣≤Cn∣Qk∣|B_k^*| \leq C_n|Q_k|, so ∣⋃Bk∗∣≤Cn∑∣Qk∣≤Cnt∥f∥1\big|\bigcup B_k^*\big| \leq C_n\sum|Q_k| \leq \frac{C_n}{t}\|f\|_1. We give up on this set.

Away from the cubes. For x∉Bk∗x \notin B_k^*, the singularity is not met, and since ∫bk=0\int b_k = 0,

Tijbk(x)=∫QkK(x−y)bk(y) dy=∫Qk(K(x−y)−K(x−yk))bk(y) dy.T_{ij}b_k(x) = \int_{Q_k}K(x - y)b_k(y)\,dy = \int_{Q_k}\big(K(x - y) - K(x - y_k)\big)b_k(y)\,dy.

For y∈Qky \in Q_k and x∉Bk∗x \notin B_k^* we have ∣x−yk∣≥diam⁡Qk≥2∣y−yk∣|x - y_k| \geq \operatorname{diam}Q_k \geq 2|y - y_k|, so by the mean value theorem and the bound on ∇K\nabla K, ∣K(x−y)−K(x−yk)∣≤C∣y−yk∣∣x−yk∣−n−1|K(x - y) - K(x - y_k)| \leq C|y - y_k||x - y_k|^{-n-1} (the segment from x−yx - y to x−ykx - y_k stays at distance at least 12∣x−yk∣\frac12|x - y_k| from 00). Integrating in xx over the complement of Bk∗B_k^*,

∫Rn∖Bk∗∣Tijbk∣ dx≤∫Qk∣bk(y)∣ C∣y−yk∣∫∣z∣≥2∣y−yk∣∣z∣−n−1dz dy=C′∥bk∥1.\int_{\mathbb{R}^n\setminus B_k^*}|T_{ij}b_k|\,dx \leq \int_{Q_k}|b_k(y)|\,C|y - y_k|\int_{|z|\geq2|y-y_k|}|z|^{-n-1}dz\,dy = C'\|b_k\|_1.

This is the Hörmander condition at work: the kernel varies slowly far away, and mean zero converts slow variation into integrability. Summing, and using ∑∥bk∥1≤2∥f∥1\sum\|b_k\|_1 \leq 2\|f\|_1, ∫Rn∖⋃Bk∗∣Tijb∣≤2C′∥f∥1\int_{\mathbb{R}^n\setminus\bigcup B_k^*}|T_{ij}b| \leq 2C'\|f\|_1, and by Chebyshev the set there where ∣Tijb∣>t2|T_{ij}b| > \frac t2 has measure at most 4C′t∥f∥1\frac{4C'}{t}\|f\|_1. (Here Tijb=∑TijbkT_{ij}b = \sum T_{ij}b_k, the sum converging in L2L^2 since b=f−g∈L2b = f - g \in L^2.)

Adding the three contributions gives the bound.

Theorem 8.5 The Calderón–Zygmund inequality

For 1<p<∞1 < p < \infty there is Cp=C(n,p)C_p = C(n, p) with

∥D2u∥Lp(Rn)≤Cp∥Δu∥Lp(Rn)for all u∈Cc2(Rn).\|D^2u\|_{L^p(\mathbb{R}^n)} \leq C_p\|\Delta u\|_{L^p(\mathbb{R}^n)} \qquad\text{for all } u \in C^2_c(\mathbb{R}^n).

Proof. For 1<p≤21 < p \leq 2: Lemma 8.3, Theorem 8.4 and Theorem 8.1 make each TijT_{ij} strong (p,p)(p, p) on L1∩L2L^1\cap L^2, hence on LpL^p by density. For 2<p<∞2 < p < \infty: TijT_{ij} is symmetric, ∫(Tijf)g=∫f(Tijg)\int(T_{ij}f)g = \int f(T_{ij}g) (its multiplier is real and even), so by duality (4A.4 Hahn–Banach and Duality) its norm on LpL^p equals its norm on Lp′L^{p'}, p′=pp−1<2p' = \frac{p}{p-1} < 2. Finally, for u∈Cc2u \in C^2_c, the representation formula (6B.4 Green’s Functions, Separation of Variables and Transforms) gives u=Φ∗(−Δu)u = \Phi * (-\Delta u), so ∂iju=−Tij(Δu)\partial_{ij}u = -T_{ij}(\Delta u).

The constants blow up at both ends: like 1p−1\frac{1}{p-1} as p→1p \to 1 (from Marcinkiewicz) and like pp as p→∞p \to \infty (by duality), which is the growth Yudovich used. At the endpoints the inequality is false. For p=∞p = \infty, 6B.7 Schauder Theory and the Continuity Method's example u=xylog⁡(x2+y2)u = xy\log(x^2 + y^2) (cut off away from the origin) has bounded Δu\Delta u and unbounded D2uD^2u. For p=1p = 1, there are compactly supported uu with Δu∈L1\Delta u \in L^1 and D2u∉L1D^2u \notin L^1 (Exercise 8.12).

Strong solutions and the W2,pW^{2,p} estimates

Localising Theorem 8.5 with a cut-off, exactly as 6B.7 Schauder Theory and the Continuity Method localised the Newtonian estimate, and handling lower-order terms by interpolation and absorption, gives the interior estimate for the Laplacian. Freezing coefficients extends it to variable coefficients, and here the error term is (aij(x0)−aij(x))∂iju(a^{ij}(x_0) - a^{ij}(x))\partial_{ij}u, which is small in LpL^p on a small ball as soon as aija^{ij} is merely continuous (its oscillation on the ball is small), with no Hölder modulus needed. The result (Gilbarg–Trudinger, Theorem 9.11):

Theorem 8.6 Interior W2,pW^{2,p} estimate (statement)

Let Lu=aij∂iju+bi∂iu+cuLu = a^{ij}\partial_{ij}u + b^i\partial_iu + cu be uniformly elliptic on Ω\Omega, with aija^{ij} continuous and bib^i, cc bounded, and 1<p<∞1 < p < \infty. If u∈Wloc2,p(Ω)∩Lp(Ω)u \in W^{2,p}_{\mathrm{loc}}(\Omega)\cap L^p(\Omega) and Lu=f∈Lp(Ω)Lu = f \in L^p(\Omega), then for every Ω′⊂⊂Ω\Omega' \subset\subset \Omega,

∥u∥W2,p(Ω′)≤C(∥u∥Lp(Ω)+∥f∥Lp(Ω)),\|u\|_{W^{2,p}(\Omega')} \leq C\big(\|u\|_{L^p(\Omega)} + \|f\|_{L^p(\Omega)}\big),

with CC depending on nn, pp, Ω′\Omega', Ω\Omega, the ellipticity constants and the moduli of continuity of the aija^{ij}.

A strong solution of Lu=fLu = f is a function u∈Wloc2,pu \in W^{2,p}_{\mathrm{loc}} satisfying the equation almost everywhere. Strong solutions sit between weak solutions (6A.6 Weak Solutions and Elliptic Regularity), which need only one derivative, and classical ones. With a C1,1C^{1,1} boundary and global versions of the estimate, the Dirichlet problem Lu=fLu = f in Ω\Omega, u−φ∈W01,p(Ω)u - \varphi \in W^{1,p}_0(\Omega), has a unique strong solution in W2,p(Ω)W^{2,p}(\Omega) for every f∈Lpf \in L^p when c≤0c \leq 0 (Gilbarg–Trudinger, Theorem 9.15), proved by the method of continuity of 6B.7 Schauder Theory and the Continuity Method with LpL^p estimates in place of Schauder estimates.

The LpL^p theory feeds back into the Hölder theory through Sobolev embedding (4A.11 Sobolev Embeddings and Critical Exponents): if f∈Lpf \in L^p with p>np > n, then u∈W2,pu \in W^{2,p}, so ∇u∈C0,1−n/p\nabla u \in C^{0,1 - n/p} by Morrey's inequality. This is often the first step of a regularity proof for a nonlinear equation: LpL^p estimates with only continuous coefficients give C1,βC^{1,\beta}, after which the coefficients (depending on uu and ∇u\nabla u) are Hölder, and Schauder theory takes over.

The Alexandrov–Bakelman–Pucci maximum principle

The classical maximum principle (6A.5 Maximum Principles) looks at a maximum point and uses the signs of the second derivatives there. For a strong solution the second derivatives exist only almost everywhere, and a single point carries no information. The Alexandrov–Bakelman–Pucci (ABP) estimate replaces the single point by a set of positive measure, the points where the graph of uu lies below a tangent plane, and replaces "the Hessian is negative there" by an inequality integrated over that set. It bounds sup⁡u\sup u by an LnL^n norm of the right-hand side, the natural norm, since LnL^n is invariant under the scaling u(x)↦u(λx)u(x) \mapsto u(\lambda x), f↦λ2f(λx)f \mapsto \lambda^2f(\lambda x) (Exercise 8.13).

Let Ω\Omega be bounded with diameter dd. For u∈C2(Ω)∩C(Ω‾)u \in C^2(\Omega)\cap C(\overline\Omega) define the upper contact set

Γ+={x∈Ω:u(y)≤u(x)+∇u(x)⋅(y−x) for all y∈Ω},\Gamma^+ = \big\{x \in \Omega : u(y) \leq u(x) + \nabla u(x)\cdot(y - x)\ \text{for all } y \in \Omega\big\},

the points where some tangent plane lies above the whole graph. At such points D2u≤0D^2u \leq 0 (the graph is touched from above by a plane).

Lemma 8.7 The gradient image contains a ball

Let M=sup⁡Ωu−sup⁡∂Ωu>0M = \sup_\Omega u - \sup_{\partial\Omega}u > 0. Then every p∈Rnp \in \mathbb{R}^n with ∣p∣<Md|p| < \frac Md equals ∇u(x)\nabla u(x) for some x∈Γ+x \in \Gamma^+.

Proof. Let ∣p∣<Md|p| < \frac Md and consider v(x)=u(x)−p⋅xv(x) = u(x) - p\cdot x on Ω‾\overline\Omega. Since uu is continuous on the compact set Ω‾\overline\Omega, it attains its supremum at some x0x_0, which lies in Ω\Omega because M>0M > 0. For y∈∂Ωy \in \partial\Omega,

v(y)−v(x0)=u(y)−u(x0)−p⋅(y−x0)≤−M+∣p∣d<0.v(y) - v(x_0) = u(y) - u(x_0) - p\cdot(y - x_0) \leq -M + |p|d < 0.

So the maximum of vv over Ω‾\overline\Omega is attained at an interior point x1x_1. There ∇v=0\nabla v = 0, so ∇u(x1)=p\nabla u(x_1) = p, and v(y)≤v(x1)v(y) \leq v(x_1) for all yy, which says u(y)≤u(x1)+p⋅(y−x1)u(y) \leq u(x_1) + p\cdot(y - x_1): x1∈Γ+x_1 \in \Gamma^+.

Theorem 8.8 The Alexandrov–Bakelman–Pucci estimate

Let Lu=aij∂ijuLu = a^{ij}\partial_{ij}u with (aij)(a^{ij}) positive definite at each point, and let D∗=(det⁡aij)1/nD^* = (\det a^{ij})^{1/n}. If u∈C2(Ω)∩C(Ω‾)u \in C^2(\Omega)\cap C(\overline\Omega) satisfies Lu≥fLu \geq f in Ω\Omega, then

sup⁡Ωu≤sup⁡∂Ωu+dn α(n)1/n∥f−D∗∥Ln(Γ+),\sup_\Omega u \leq \sup_{\partial\Omega}u + \frac{d}{n\,\alpha(n)^{1/n}}\Big\|\frac{f^-}{D^*}\Big\|_{L^n(\Gamma^+)},

where f−=max⁡(−f,0)f^- = \max(-f, 0) and α(n)\alpha(n) is the volume of the unit ball.

Proof. Assume M=sup⁡u−sup⁡∂Ωu>0M = \sup u - \sup_{\partial\Omega}u > 0. By Lemma 8.7, the ball B(0,Md)B(0, \frac Md) is contained in ∇u(Γ+)\nabla u(\Gamma^+), the image of Γ+\Gamma^+ under the C1C^1 map ∇u\nabla u, whose Jacobian determinant is det⁡D2u\det D^2u. The measure of the image of a set under a C1C^1 map is at most the integral of the absolute Jacobian over the set (the area formula, an inequality for maps that need not be injective, 3A.5 Product Measures and Change of Variables). So

α(n)(Md)n≤∣∇u(Γ+)∣≤∫Γ+∣det⁡D2u∣ dx.\alpha(n)\Big(\frac Md\Big)^n \leq |\nabla u(\Gamma^+)| \leq \int_{\Gamma^+}|\det D^2u|\,dx.

On Γ+\Gamma^+, −D2u-D^2u is positive semidefinite. For a positive definite AA and a positive semidefinite BB, the eigenvalues of A1/2BA1/2A^{1/2}BA^{1/2} are non-negative, and the arithmetic–geometric mean inequality gives

det⁡A det⁡B=det⁡(A1/2BA1/2)≤(tr⁡(AB)n)n.\det A\,\det B = \det(A^{1/2}BA^{1/2}) \leq \Big(\frac{\operatorname{tr}(AB)}{n}\Big)^n.

With A=(aij)A = (a^{ij}) and B=−D2uB = -D^2u, tr⁡(AB)=−aij∂iju=−Lu≤−f≤f−\operatorname{tr}(AB) = -a^{ij}\partial_{ij}u = -Lu \leq -f \leq f^- (and tr⁡(AB)≥0\operatorname{tr}(AB) \geq 0 on Γ+\Gamma^+). So ∣det⁡D2u∣≤(f−nD∗)n|\det D^2u| \leq \big(\frac{f^-}{nD^*}\big)^n on Γ+\Gamma^+, and

α(n)(Md)n≤∫Γ+(f−nD∗)ndx.\alpha(n)\Big(\frac Md\Big)^n \leq \int_{\Gamma^+}\Big(\frac{f^-}{nD^*}\Big)^ndx.

Take nn-th roots.

The estimate depends only on the determinant of the coefficients and on the LnL^n norm of ff, not on any continuity of the coefficients, and that is its power. Lower-order terms can be included (Gilbarg–Trudinger, Theorem 9.1). It extends to strong solutions in Wloc2,nW^{2,n}_{\mathrm{loc}}, and it is the starting point of the Krylov–Safonov theory (1979–80): Hölder continuity and a Harnack inequality for solutions of non-divergence equations with merely bounded measurable coefficients, the non-divergence counterpart of De Giorgi–Nash–Moser (6B.9 De Giorgi–Nash–Moser), and the foundation of the regularity theory of fully nonlinear equations (6C.10 Fully Nonlinear Equations and Monge–Ampère).

Where this goes Calderón–Zygmund and ABP beyond this chapter

Calderón–Zygmund theory is the analytic core of much of modern PDE: it gives LpL^p estimates for the parabolic operator ∂t−Δ\partial_t - \Delta (6A.7 Parabolic Regularity), for Riesz transforms and for the Hodge decomposition (4A.5 Hilbert Spaces and Lax–Milgram and Taylor III, §13.5–13.7), and it underlies the estimates for the Navier–Stokes and Euler equations. ABP is the basic tool for non-divergence equations, from the Krylov–Safonov Harnack inequality to Caffarelli's regularity theory for fully nonlinear and Monge–Ampère equations (6C.10 Fully Nonlinear Equations and Monge–Ampère). In geometry, Xavier Cabré's ABP-type argument gives a short proof of the isoperimetric inequality (Exercise 8.15), and versions of it on Riemannian manifolds are proved with the Laplacian comparison theorem (9B.1 Laplacian Comparison).

History

Józef Marcinkiewicz proved his interpolation theorem in 1939; he died in the Second World War, and Antoni Zygmund published the result in 1956. Alberto Calderón and Zygmund introduced their decomposition and proved the LpL^p bounds for singular integrals in 1952, and Lars Hörmander formulated the condition on the kernel in 1960. Aleksandr Alexandrov (1960s), Ilya Bakelman (1961) and Carlo Pucci (1966) developed the maximum principle that bears their names. Nikolai Krylov and Mikhail Safonov proved their Harnack inequality in 1979–80. Victor Yudovich's theorem on the Euler equations dates from 1963, and Cabré's proof of the isoperimetric inequality from around 2000.

Recall Where we stand

Weak LpL^p is controlled by the distribution function, and Marcinkiewicz interpolation turns weak (1,1)(1, 1) plus strong (2,2)(2, 2) into strong (p,p)(p, p) for 1<p<21 < p < 2. The Calderón–Zygmund decomposition splits an L1L^1 function at height tt into a bounded good part and mean-zero bad parts on dyadic cubes. The second-derivative operators Tijf=∂ij(Φ∗f)T_{ij}f = \partial_{ij}(\Phi * f) are bounded on L2L^2 by Plancherel and weak (1,1)(1, 1) by the decomposition and the Hörmander condition, so ∥D2u∥p≤Cp∥Δu∥p\|D^2u\|_p \leq C_p\|\Delta u\|_p for 1<p<∞1 < p < \infty, with constants blowing up at both ends, and false at p=1,∞p = 1, \infty. Freezing with merely continuous coefficients gives interior W2,pW^{2,p} estimates and strong solutions. The ABP estimate bounds sup⁡u\sup u by ∥f−/D∗∥Ln\|f^-/D^*\|_{L^n} over the upper contact set, using only that the gradient image of the contact set contains a ball and the AM–GM inequality for determinants. 6B.9 De Giorgi–Nash–Moser takes the last step towards rough coefficients: divergence-form equations with bounded measurable coefficients, and the theorem of De Giorgi, Nash and Moser.

Exercises

Exercise 8.9 The layer-cake formula

Prove ∫∣f∣p=p∫0∞tp−1μf(t) dt\int|f|^p = p\int_0^\infty t^{p-1}\mu_f(t)\,dt by writing ∣f(x)∣p=∫0∣f(x)∣ptp−1dt|f(x)|^p = \int_0^{|f(x)|}pt^{p-1}dt and exchanging the order of integration.

Solution

∫∣f∣pdx=∫∫0∞ptp−11{t<∣f(x)∣} dt dx=∫0∞ptp−1∣{∣f∣>t}∣ dt\int|f|^pdx = \int\int_0^\infty pt^{p-1}\mathbf 1_{\{t<|f(x)|\}}\,dt\,dx = \int_0^\infty pt^{p-1}|\{|f| > t\}|\,dt by Tonelli, since the integrand is non-negative.

Exercise 8.10 Weak L1L^1 is not L1L^1

Show that f(x)=∣x∣−nf(x) = |x|^{-n} on Rn\mathbb{R}^n has μf(t)=α(n)t\mu_f(t) = \frac{\alpha(n)}{t}, so it is in weak L1L^1 with constant α(n)\alpha(n), but ∫f=∞\int f = \infty. Then show that the Hilbert transform Hf(x)=1π p.v.∫f(y)x−ydyHf(x) = \frac1\pi\,\mathrm{p.v.}\int\frac{f(y)}{x - y}dy of f=1[0,1]f = \mathbf 1_{[0,1]} is 1πlog⁡∣x∣∣x−1∣\frac1\pi\log\frac{|x|}{|x - 1|}, which is not integrable on R\mathbb{R} but is in weak L1L^1: the one-dimensional version of TijT_{ij} maps L1L^1 into weak L1L^1 and no better.

Solution

{∣x∣−n>t}={∣x∣<t−1/n}\{|x|^{-n} > t\} = \{|x| < t^{-1/n}\}, a ball of volume α(n)/t\alpha(n)/t. In polar coordinates ∫∣x∣−ndx=nα(n)∫0∞r−1dr=∞\int|x|^{-n}dx = n\alpha(n)\int_0^\infty r^{-1}dr = \infty. For the Hilbert transform, for x∉[0,1]x \notin [0, 1], ∫01dyx−y=log⁡∣x∣∣x−1∣\int_0^1\frac{dy}{x - y} = \log\frac{|x|}{|x - 1|}, and the principal value gives the same formula for x∈(0,1)x \in (0, 1). Near x=0x = 0 and x=1x = 1 it has logarithmic singularities, which are integrable; but for large ∣x∣|x| it behaves like 1πx\frac{1}{\pi x}, which is not integrable. Its distribution function is O(1/t)O(1/t): for small tt the set where ∣Hf∣>t|Hf| > t lies within ∣x∣≲1πt|x| \lesssim \frac{1}{\pi t}, and for large tt it consists of two intervals around 00 and 11 of length about e−πte^{-\pi t}.

Exercise 8.11 A Calderón–Zygmund decomposition by hand

On R\mathbb{R}, let f=4⋅1[0,1/4)f = 4\cdot\mathbf 1_{[0, 1/4)}, so ∥f∥1=1\|f\|_1 = 1, and take t=12t = \frac12. Starting from the dyadic intervals of length 22, carry out the stopping-time selection, find the selected intervals, and check properties 1–3 of Lemma 8.2.

Solution

On [0,2)[0, 2) the average is 12≤t\frac12 \leq t; subdivide. [0,1)[0, 1) has average 1>121 > \frac12: select it, and stop. [1,2)[1, 2) has average 00; its descendants never get selected. Negative intervals have average 00. So the only selected interval is [0,1)[0, 1), with average 1∈(12,2⋅12]1 \in (\frac12, 2\cdot\frac12]; total length 1≤∥f∥1t=21 \leq \frac{\|f\|_1}{t} = 2; outside it f=0≤tf = 0 \leq t. The good part is g=1[0,1)g = \mathbf 1_{[0,1)} (the average) and the bad part b=f−1[0,1)b = f - \mathbf 1_{[0,1)}, which has mean zero.

Exercise 8.12 Failure at p=1p = 1

Show that the operator T12f=∂12(Φ∗f)T_{12}f = \partial_{12}(\Phi * f) is not bounded on L1(R2)L^1(\mathbb{R}^2), by taking fε=1πε21B(0,ε)f_\varepsilon = \frac{1}{\pi\varepsilon^2}\mathbf 1_{B(0,\varepsilon)} (an approximation of δ0\delta_0, ∥fε∥1=1\|f_\varepsilon\|_1 = 1) and showing ∂12(Φ∗fε)(x)=∂12Φ(x)=x1x2π∣x∣4\partial_{12}(\Phi * f_\varepsilon)(x) = \partial_{12}\Phi(x) = \frac{x_1x_2}{\pi|x|^4} for ∣x∣>ε|x| > \varepsilon, whose L1L^1 norm over ε<∣x∣<1\varepsilon < |x| < 1 is about 2πlog⁡1ε→∞\frac{2}{\pi}\log\frac1\varepsilon \to \infty.

Solution

For ∣x∣>ε|x| > \varepsilon, Φ∗fε(x)=Φ(x)\Phi * f_\varepsilon(x) = \Phi(x) by the mean value property (the average of the harmonic function y↦Φ(x−y)y \mapsto \Phi(x - y) over the disc B(0,ε)B(0, \varepsilon), which does not contain xx, is its value at the centre). So ∂12(Φ∗fε)=∂12Φ=2x1x22π∣x∣4\partial_{12}(\Phi * f_\varepsilon) = \partial_{12}\Phi = \frac{2x_1x_2}{2\pi|x|^4} there. In polar coordinates ∣x1x2∣π∣x∣4=∣cos⁡θsin⁡θ∣πr2\frac{|x_1x_2|}{\pi|x|^4} = \frac{|\cos\theta\sin\theta|}{\pi r^2}, and ∫02π∣cos⁡θsin⁡θ∣ dθ=2\int_0^{2\pi}|\cos\theta\sin\theta|\,d\theta = 2, so ∫ε<∣x∣<1=2π∫ε1drr=2πlog⁡1ε\int_{\varepsilon<|x|<1} = \frac2\pi\int_\varepsilon^1\frac{dr}{r} = \frac2\pi\log\frac1\varepsilon. Since ∥fε∥1=1\|f_\varepsilon\|_1 = 1, no bound ∥T12f∥1≤C∥f∥1\|T_{12}f\|_1 \leq C\|f\|_1 can hold.

Exercise 8.13 Why LnL^n

Let uu satisfy Lu≥fLu \geq f on Ω\Omega and set uλ(x)=u(λx)u_\lambda(x) = u(\lambda x) on Ω/λ\Omega/\lambda, which satisfies Lλuλ≥fλL_\lambda u_\lambda \geq f_\lambda with fλ(x)=λ2f(λx)f_\lambda(x) = \lambda^2f(\lambda x) and the coefficients aij(λx)a^{ij}(\lambda x). Show that d ∥f∥Lnd\,\|f\|_{L^n} is unchanged by this scaling, while d ∥f∥Lpd\,\|f\|_{L^p} for p≠np \neq n is not. Explain why the ABP bound must therefore use LnL^n.

Solution

The diameter becomes d/λd/\lambda, and ∥fλ∥Lp(Ω/λ)=λ2λ−n/p∥f∥Lp(Ω)\|f_\lambda\|_{L^p(\Omega/\lambda)} = \lambda^2\lambda^{-n/p}\|f\|_{L^p(\Omega)}. So d∥f∥Lpd\|f\|_{L^p} is multiplied by λ1−n/p\lambda^{1 - n/p}, which is 11 exactly when p=np = n. The left side, sup⁡u−sup⁡∂u\sup u - \sup_{\partial}u, is unchanged by the scaling, so any estimate of this form valid for all λ\lambda must use a scale-invariant quantity.

Exercise 8.14 ABP on a ball

On the ball BRB_R, u=R2−∣x∣22nu = \frac{R^2 - |x|^2}{2n} satisfies Δu=−1\Delta u = -1 and u=0u = 0 on the boundary. Compare sup⁡u\sup u with the ABP bound (with aij=δija^{ij} = \delta_{ij}, D∗=1D^* = 1, f=−1f = -1, d=2Rd = 2R). How far from sharp is it?

Solution

sup⁡u=R22n\sup u = \frac{R^2}{2n}. Here Γ+=BR\Gamma^+ = B_R (uu is concave), so the bound is 2Rnα(n)1/n∣BR∣1/n=2Rnα(n)1/nα(n)1/nR=2R2n\frac{2R}{n\alpha(n)^{1/n}}|B_R|^{1/n} = \frac{2R}{n\alpha(n)^{1/n}}\alpha(n)^{1/n}R = \frac{2R^2}{n}, four times the true value. The estimate is not sharp, but it has the right dependence on RR, which is what matters in applications. (Most of the loss is in Lemma 8.7: the gradient image of uu is the ball of radius Rn\frac Rn, while the lemma only guarantees radius Md=R4n\frac{M}{d} = \frac{R}{4n}.)

Exercise 8.15 Rehearsal: the isoperimetric inequality by the ABP method (Cabré)

Let Ω⊂Rn\Omega \subset \mathbb{R}^n be a bounded domain with smooth boundary, and let uu solve the Neumann problem

Δu=∣∂Ω∣∣Ω∣  in Ω,∂νu=1  on ∂Ω\Delta u = \frac{|\partial\Omega|}{|\Omega|} \ \text{ in } \Omega, \qquad \partial_\nu u = 1 \ \text{ on } \partial\Omega

(solvable, since the compatibility condition ∫ΩΔu=∫∂Ω∂νu\int_\Omega\Delta u = \int_{\partial\Omega}\partial_\nu u holds; 6A.6 Weak Solutions and Elliptic Regularity, and smooth by regularity). Let Γ−={x∈Ω:u(y)≥u(x)+∇u(x)⋅(y−x) ∀y∈Ω‾}\Gamma^- = \{x \in \Omega : u(y) \geq u(x) + \nabla u(x)\cdot(y - x)\ \forall y \in \overline\Omega\}, the lower contact set. (a) Show that for every pp with ∣p∣<1|p| < 1, the minimum of u(y)−p⋅yu(y) - p\cdot y over Ω‾\overline\Omega is not attained on ∂Ω\partial\Omega (its outward normal derivative there is 1−p⋅ν>01 - p\cdot\nu > 0), hence is attained at an interior point x∈Γ−x \in \Gamma^- with ∇u(x)=p\nabla u(x) = p. So the unit ball lies in ∇u(Γ−)\nabla u(\Gamma^-). (b) On Γ−\Gamma^-, D2u≥0D^2u \geq 0, and AM–GM gives det⁡D2u≤(Δun)n\det D^2u \leq \big(\frac{\Delta u}{n}\big)^n. Deduce

α(n)≤∫Γ−det⁡D2u≤∣Ω∣(∣∂Ω∣n∣Ω∣)n,that is,∣∂Ω∣≥n α(n)1/n∣Ω∣(n−1)/n.\alpha(n) \leq \int_{\Gamma^-}\det D^2u \leq |\Omega|\Big(\frac{|\partial\Omega|}{n|\Omega|}\Big)^n, \qquad\text{that is,}\qquad |\partial\Omega| \geq n\,\alpha(n)^{1/n}|\Omega|^{(n-1)/n}.

(c) Check that equality holds for a ball. This is the sharp isoperimetric inequality of 4A.11 Sobolev Embeddings and Critical Exponents, with its best constant; the same argument on a manifold with non-negative Ricci curvature uses the Laplacian comparison theorem (9B.1 Laplacian Comparison) and gives isoperimetric inequalities that are among the comparison results behind Perelman's noncollapsing (12A.4 κ-Noncollapsing).

Solution

(a) If the minimum of v(y)=u(y)−p⋅yv(y) = u(y) - p\cdot y were at y0∈∂Ωy_0 \in \partial\Omega, then moving from y0y_0 a little way inward would not decrease vv, so ∂νv(y0)≤0\partial_\nu v(y_0) \leq 0; but ∂νv=1−p⋅ν≥1−∣p∣>0\partial_\nu v = 1 - p\cdot\nu \geq 1 - |p| > 0. So the minimum is at an interior point xx, where ∇u(x)=p\nabla u(x) = p and v(y)≥v(x)v(y) \geq v(x) for all yy, which says x∈Γ−x \in \Gamma^-. (b) At an interior minimum of vv, D2v=D2u≥0D^2v = D^2u \geq 0. For a positive semidefinite matrix, AM–GM on the eigenvalues gives det⁡≤(tr⁡n)n\det \leq (\frac{\operatorname{tr}}{n})^n. The area formula gives ∣B1∣≤∫Γ−det⁡D2u≤∫Ω(∣∂Ω∣n∣Ω∣)n=∣Ω∣(∣∂Ω∣n∣Ω∣)n|B_1| \leq \int_{\Gamma^-}\det D^2u \leq \int_\Omega\big(\frac{|\partial\Omega|}{n|\Omega|}\big)^n = |\Omega|\big(\frac{|\partial\Omega|}{n|\Omega|}\big)^n; rearrange. (c) For the unit ball, ∣∂B∣=nα(n)|\partial B| = n\alpha(n) and ∣B∣=α(n)|B| = \alpha(n), so the right side is nα(n)1/nα(n)(n−1)/n=nα(n)n\alpha(n)^{1/n}\alpha(n)^{(n-1)/n} = n\alpha(n): equality. (Indeed u=∣x∣22u = \frac{|x|^2}{2} solves the problem, Γ−=B\Gamma^- = B, ∇u(B)=B\nabla u(B) = B exactly, and AM–GM is an equality because D2u=ID^2u = I.)

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