Book 6B

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Course 6Book 6B: Linear PDE: Characteristics, Waves and RegularityChapter 9

De Giorgi–Nash–Moser

Moser iteration, the Harnack inequality and Hölder continuity for equations with rough coefficients.

46 min read · Updated Oct 23, 2026

Read with Jost, Partial Differential Equations, chapter 14 ("The Moser iteration method and the regularity theorem of de Giorgi and Nash": the Moser–Harnack inequality, properties of solutions, semilinear equations, regularity of minimisers); Taylor III, §14.9 ("Elliptic regularity III (DeGiorgi-Nash-Moser theory)"); Gilbarg and Trudinger, chapter 8, for the full theory. This chapter proves Caccioppoli's inequality, local boundedness by Moser iteration, the logarithmic estimate, and the passage from Harnack to Hölder continuity; the John–Nirenberg lemma is stated.

In this chapter · 10 sections
  1. 9.1Heat in a composite
  2. 9.2Setting: weak sub- and supersolutions
  3. 9.3Caccioppoli's inequality
  4. 9.4Local boundedness: Moser iteration
  5. 9.5Positive supersolutions: the logarithm and the crossover
  6. 9.6From Harnack to Hölder continuity
  7. 9.7Hilbert's nineteenth problem
  8. 9.8The parabolic theory
  9. 9.9History
  10. 9.10Exercises
  11. ↗Free resources

Every estimate so far needed some regularity of the coefficients: Hölder continuity for Schauder (6B.7 Schauder Theory and the Continuity Method), continuity for LpL^p (6B.8 Lᵖ Theory and Strong Solutions). This chapter proves estimates that need none. Consider a divergence-form equation

div⁡(A(x)∇u)=0,\operatorname{div}\big(A(x)\nabla u\big) = 0,

where the matrix A(x)A(x) is only assumed measurable, bounded and uniformly elliptic: it may jump wildly from point to point, as the conductivity of a composite material does. Ennio De Giorgi (1957) and John Nash (1958), independently and by completely different methods, proved that every weak solution is nevertheless Hölder continuous, with an exponent and a constant that depend only on the dimension and the ellipticity ratio. Jürgen Moser (1960–61) then gave a third proof, by an iteration that is now the standard tool, and with it the Harnack inequality for such equations.

Why does such a theorem matter, when equations with rough coefficients are rare in geometry? Because of nonlinearity. If uu minimises ∫F(∇u)\int F(\nabla u) with FF smooth and uniformly convex, then each derivative ∂ku\partial_ku satisfies div⁡(F′′(∇u)∇∂ku)=0\operatorname{div}(F''(\nabla u)\nabla\partial_ku) = 0: an equation of exactly this form, whose coefficients F′′(∇u)F''(\nabla u) are only as good as ∇u\nabla u, which is what we are trying to control. De Giorgi–Nash makes ∇u\nabla u Hölder, Schauder then bootstraps to C∞C^\infty, and Hilbert's nineteenth problem, whether minimisers of such "regular" variational problems are smooth, is solved. The same circle of ideas runs through the analysis of geometric flows: the Moser iteration is how L2L^2 or LpL^p bounds on curvature are upgraded to pointwise bounds (11B.3 Compactness of Ricci Flows), and the parabolic Harnack inequality is the template for Li–Yau and Hamilton (6A.12 Entropy, Information and Diffusion, 11B.2 Ancient Solutions and the Harnack Inequality).

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

  • define weak sub- and supersolutions of divergence-form equations with measurable coefficients, and prove Caccioppoli's inequality;
  • prove local boundedness of subsolutions by Moser iteration, combining Caccioppoli with the Sobolev inequality;
  • prove the logarithmic estimate for positive supersolutions, and explain how the John–Nirenberg lemma turns it into the weak Harnack inequality;
  • deduce Hölder continuity from the Harnack inequality by oscillation decay;
  • explain how the theorem solves Hilbert's nineteenth problem, what fails for systems, and what the parabolic version says.

Heat in a composite

In the world Model Groundwater and heat in heterogeneous materials

Groundwater flows through an aquifer by Darcy's law: the flux is −K(x)∇h-K(x)\nabla h, where hh is the hydraulic head and KK the hydraulic conductivity. In steady state, conservation gives div⁡(K(x)∇h)=0\operatorname{div}(K(x)\nabla h) = 0. Real aquifers are layered and fractured mixtures of gravel, sand, silt and clay, and their hydraulic conductivities range over many orders of magnitude, from roughly 10−210^{-2} m/s for clean gravel to below 10−910^{-9} m/s for clay. The conductivity jumps across every layer boundary and inclusion. Heat conduction in composites, electrical current in mixtures, and diffusion in biological tissue have the same structure: a divergence-form equation whose coefficient is, for all practical purposes, an arbitrary bounded measurable function between two positive limits.

De Giorgi–Nash–Moser says what can be said about the solution with no knowledge of the coefficient's fine structure. The head (or temperature, or potential) is Hölder continuous, with a modulus depending only on the dimension and the contrast Λ/λ\Lambda/\lambda between the largest and smallest conductivities; it satisfies a Harnack inequality, so a positive head cannot vary by more than a fixed factor across a region; and its gradient, the flux, can be discontinuous, as it must be when the conductivity jumps (6A.6 Weak Solutions and Elliptic Regularity's composite rod). The Hölder exponent can be very small when the contrast is large, and for systems of equations, such as elasticity in a composite, even continuity can fail. Homogenisation theory (4A.7 Weak Convergence and the Direct Method) asks the complementary question: what equation do the large-scale averages satisfy?

Setting: weak sub- and supersolutions

Let Ω⊆Rn\Omega \subseteq \mathbb{R}^n be open and A=(aij)A = (a^{ij}) a measurable matrix function on Ω\Omega, symmetric (for simplicity), with

λ∣ξ∣2≤aij(x)ξiξj≤Λ∣ξ∣2for all ξ∈Rn, x∈Ω,\lambda|\xi|^2 \leq a^{ij}(x)\xi_i\xi_j \leq \Lambda|\xi|^2 \qquad\text{for all } \xi \in \mathbb{R}^n,\ x \in \Omega,

for constants 0<λ≤Λ0 < \lambda \leq \Lambda. A function u∈Hloc1(Ω)u \in H^1_{\mathrm{loc}}(\Omega) is a weak subsolution of div⁡(A∇u)=0\operatorname{div}(A\nabla u) = 0 if

∫Ωaij∂iu ∂jφ dx≤0for all φ∈H01(Ω) with compact support and φ≥0,\int_\Omega a^{ij}\partial_iu\,\partial_j\varphi\,dx \leq 0 \qquad\text{for all } \varphi \in H^1_0(\Omega) \text{ with compact support and } \varphi \geq 0,

a weak supersolution if the inequality is reversed, and a weak solution if it is an equality for all such φ\varphi (of any sign). For a smooth uu and smooth AA, integrating by parts shows that subsolution means div⁡(A∇u)≥0\operatorname{div}(A\nabla u) \geq 0, the analogue of subharmonic (6B.6 Perron’s Method and Barriers): "the equation pushes uu up".

Two facts about Sobolev functions are used without comment (Gilbarg–Trudinger, Lemma 7.6): if u∈H1u \in H^1, then u+=max⁡(u,0)u^+ = \max(u, 0) and (u−k)+(u - k)^+ are in H1H^1, with ∇(u−k)+=∇u\nabla(u - k)^+ = \nabla u on {u>k}\{u > k\} and 00 elsewhere (almost everywhere); and products of H1H^1 functions with bounded Lipschitz functions are in H1H^1. In the proofs we apply the equation to test functions such as ζ2(u+)β\zeta^2(u^+)^\beta; when uu is unbounded these must first be truncated at a height NN and the estimates proved uniformly in NN, a routine step we note once and then suppress.

Caccioppoli's inequality

The basic estimate is a reverse Poincaré inequality: for subsolutions, the gradient is controlled by the function itself on a slightly larger ball.

Lemma 9.1 Caccioppoli's inequality

Let uu be a weak subsolution in Ω\Omega, k∈Rk \in \mathbb{R}, and ζ\zeta a Lipschitz function with compact support in Ω\Omega. Then

∫ζ2∣∇(u−k)+∣2 dx≤4Λ2λ2∫∣∇ζ∣2((u−k)+)2 dx.\int\zeta^2|\nabla(u - k)^+|^2\,dx \leq \frac{4\Lambda^2}{\lambda^2}\int|\nabla\zeta|^2\big((u - k)^+\big)^2\,dx.

In particular, if ζ=1\zeta = 1 on BρB_\rho, ζ=0\zeta = 0 outside BrB_r and ∣∇ζ∣≤1r−ρ|\nabla\zeta| \leq \frac{1}{r - \rho}, then ∫Bρ∣∇(u−k)+∣2≤C(r−ρ)2∫Br((u−k)+)2\int_{B_\rho}|\nabla(u - k)^+|^2 \leq \frac{C}{(r - \rho)^2}\int_{B_r}((u - k)^+)^2.

Proof. Write w=(u−k)+w = (u - k)^+ and test with φ=ζ2w≥0\varphi = \zeta^2w \geq 0. Since ∇w=∇u\nabla w = \nabla u where w>0w > 0 and w=0w = 0 elsewhere, aij∂iu ∂jφ=ζ2aij∂iw ∂jw+2ζw aij∂iw ∂jζa^{ij}\partial_iu\,\partial_j\varphi = \zeta^2a^{ij}\partial_iw\,\partial_jw + 2\zeta w\,a^{ij}\partial_iw\,\partial_j\zeta. The subsolution inequality gives

λ∫ζ2∣∇w∣2≤∫ζ2aij∂iw ∂jw≤−2∫ζw aij∂iw ∂jζ≤2Λ∫ζ∣∇w∣ w∣∇ζ∣.\lambda\int\zeta^2|\nabla w|^2 \leq \int\zeta^2a^{ij}\partial_iw\,\partial_jw \leq -2\int\zeta w\,a^{ij}\partial_iw\,\partial_j\zeta \leq 2\Lambda\int\zeta|\nabla w|\,w|\nabla\zeta|.

By Cauchy's inequality, 2Λζ∣∇w∣w∣∇ζ∣≤λ2ζ2∣∇w∣2+2Λ2λw2∣∇ζ∣22\Lambda\zeta|\nabla w|w|\nabla\zeta| \leq \frac\lambda2\zeta^2|\nabla w|^2 + \frac{2\Lambda^2}{\lambda}w^2|\nabla\zeta|^2. Absorb the first term into the left side: λ2∫ζ2∣∇w∣2≤2Λ2λ∫w2∣∇ζ∣2\frac\lambda2\int\zeta^2|\nabla w|^2 \leq \frac{2\Lambda^2}{\lambda}\int w^2|\nabla\zeta|^2.

For a harmonic function this says that ∫Br/2∣∇u∣2≤Cr2∫Bru2\int_{B_{r/2}}|\nabla u|^2 \leq \frac{C}{r^2}\int_{B_r}u^2: the energy on a ball is bounded by the size of the function on a ball twice as large, with the scaling r−2r^{-2} of 6A.2 Harmonic Functions's derivative estimates. Unlike those, it needs no mean value property, only the equation in weak form and ellipticity, and so it holds for every equation in our class.

Local boundedness: Moser iteration

Caccioppoli trades a derivative for a cut-off. The Sobolev inequality (4A.11 Sobolev Embeddings and Critical Exponents) trades a derivative for integrability: for v∈H01(Rn)v \in H^1_0(\mathbb{R}^n), n≥3n \geq 3,

(∫∣v∣2χ)1/χ≤CS∫∣∇v∣2,χ=nn−2>1.\Big(\int|v|^{2\chi}\Big)^{1/\chi} \leq C_S\int|\nabla v|^2, \qquad \chi = \frac{n}{n - 2} > 1.

(For n=2n = 2, any χ>1\chi > 1 works, with a constant depending on χ\chi and on the size of the support.) Together they give a gain of integrability with no loss: if uu is a subsolution in LqL^q on a ball, then it is in LχqL^{\chi q} on a smaller ball. Iterating infinitely often, with the balls shrinking to a limit ball and the exponents growing to infinity, gives a bound in L∞L^\infty. The art is to keep the constants under control.

Theorem 9.2 Local boundedness of subsolutions

Let uu be a weak subsolution in B2=B(x0,2)B_2 = B(x_0, 2), n≥3n \geq 3. Then u+∈L∞(B1)u^+ \in L^\infty(B_1) and

sup⁡B1u+≤C(∫B2(u+)2 dx)1/2,C=C(n,Λ/λ).\sup_{B_1}u^+ \leq C\Big(\int_{B_2}(u^+)^2\,dx\Big)^{1/2}, \qquad C = C(n, \Lambda/\lambda).

By scaling, sup⁡Bru+≤C(  -∫B2r(u+)2)1/2\sup_{B_r}u^+ \leq C\big(\mathop{\rlap{\,\,-}\int}\nolimits_{B_{2r}}(u^+)^2\big)^{1/2} on any ball.

Proof. The basic step. Let w=u+w = u^+, β≥1\beta \geq 1, and test with φ=ζ2wβ\varphi = \zeta^2w^\beta (truncated, as noted). Since ∇(wβ)=βwβ−1∇w\nabla(w^\beta) = \beta w^{\beta-1}\nabla w and ∇u=∇w\nabla u = \nabla w where w>0w > 0,

β∫ζ2wβ−1aij∂iw ∂jw≤−2∫ζwβaij∂iw ∂jζ.\beta\int\zeta^2w^{\beta-1}a^{ij}\partial_iw\,\partial_jw \leq -2\int\zeta w^\beta a^{ij}\partial_iw\,\partial_j\zeta.

The left side is at least βλ∫ζ2wβ−1∣∇w∣2\beta\lambda\int\zeta^2w^{\beta-1}|\nabla w|^2, and the right side at most 2Λ∫ζwβ∣∇w∣∣∇ζ∣2\Lambda\int\zeta w^\beta|\nabla w||\nabla\zeta|. Splitting ζwβ∣∇w∣=(ζwβ−12∣∇w∣)(wβ+12)\zeta w^\beta|\nabla w| = (\zeta w^{\frac{\beta-1}{2}}|\nabla w|)(w^{\frac{\beta+1}{2}}) and using Cauchy's inequality as in Lemma 9.1,

βλ2∫ζ2wβ−1∣∇w∣2≤2Λ2βλ∫wβ+1∣∇ζ∣2.\frac{\beta\lambda}{2}\int\zeta^2w^{\beta-1}|\nabla w|^2 \leq \frac{2\Lambda^2}{\beta\lambda}\int w^{\beta+1}|\nabla\zeta|^2.

Now let v=wβ+12v = w^{\frac{\beta+1}{2}}, so ∣∇v∣2=(β+12)2wβ−1∣∇w∣2|\nabla v|^2 = \big(\frac{\beta+1}{2}\big)^2w^{\beta-1}|\nabla w|^2. Then

∫ζ2∣∇v∣2≤(β+1)24⋅4Λ2β2λ2∫v2∣∇ζ∣2≤4Λ2λ2∫v2∣∇ζ∣2,\int\zeta^2|\nabla v|^2 \leq \frac{(\beta + 1)^2}{4}\cdot\frac{4\Lambda^2}{\beta^2\lambda^2}\int v^2|\nabla\zeta|^2 \leq \frac{4\Lambda^2}{\lambda^2}\int v^2|\nabla\zeta|^2,

using β+1β≤2\frac{\beta+1}{\beta} \leq 2. So ∫∣∇(ζv)∣2≤2∫ζ2∣∇v∣2+2∫v2∣∇ζ∣2≤C1∫v2∣∇ζ∣2\int|\nabla(\zeta v)|^2 \leq 2\int\zeta^2|\nabla v|^2 + 2\int v^2|\nabla\zeta|^2 \leq C_1\int v^2|\nabla\zeta|^2 with C1=2+8Λ2λ2C_1 = 2 + \frac{8\Lambda^2}{\lambda^2}.

Gain of integrability. Apply the Sobolev inequality to ζv\zeta v, with ζ=1\zeta = 1 on BρB_\rho, supported in BrB_r, ∣∇ζ∣≤2r−ρ|\nabla\zeta| \leq \frac{2}{r - \rho}:

(∫Bρv2χ)1/χ≤4CSC1(r−ρ)2∫Brv2.\Big(\int_{B_\rho}v^{2\chi}\Big)^{1/\chi} \leq \frac{4C_SC_1}{(r - \rho)^2}\int_{B_r}v^2.

In terms of ww and the exponent q=β+1≥2q = \beta + 1 \geq 2, since v2=wqv^2 = w^q and v2χ=wqχv^{2\chi} = w^{q\chi}:

∥w∥Lχq(Bρ)≤(C2(r−ρ)2)1/q∥w∥Lq(Br),C2=4CSC1.\|w\|_{L^{\chi q}(B_\rho)} \leq \Big(\frac{C_2}{(r - \rho)^2}\Big)^{1/q}\|w\|_{L^q(B_r)}, \qquad C_2 = 4C_SC_1.

Iteration. Let qk=2χkq_k = 2\chi^k and rk=1+2−kr_k = 1 + 2^{-k}, so qk→∞q_k \to \infty, rk↓1r_k \downarrow 1, r0=2r_0 = 2, and rk−rk+1=2−k−1r_k - r_{k+1} = 2^{-k-1}. Applying the last inequality with q=qkq = q_k, r=rkr = r_k, ρ=rk+1\rho = r_{k+1}:

∥w∥Lqk+1(Brk+1)≤(C24k+1)12χk∥w∥Lqk(Brk).\|w\|_{L^{q_{k+1}}(B_{r_{k+1}})} \leq \big(C_24^{k+1}\big)^{\frac{1}{2\chi^k}}\|w\|_{L^{q_k}(B_{r_k})}.

Chaining from k=0k = 0,

∥w∥Lqm(B1)≤∏k=0m−1(C24k+1)12χk∥w∥L2(B2)≤C2∑k12χk 4∑kk+12χk∥w∥L2(B2).\|w\|_{L^{q_m}(B_1)} \leq \prod_{k=0}^{m-1}\big(C_24^{k+1}\big)^{\frac{1}{2\chi^k}}\|w\|_{L^2(B_2)} \leq C_2^{\sum_k\frac{1}{2\chi^k}}\,4^{\sum_k\frac{k+1}{2\chi^k}}\|w\|_{L^2(B_2)}.

Both exponents are convergent series because χ>1\chi > 1, so the product is bounded by a constant CC independent of mm. Letting m→∞m \to \infty, ∥w∥L∞(B1)=lim⁡m∥w∥Lqm(B1)≤C∥w∥L2(B2)\|w\|_{L^\infty(B_1)} = \lim_m\|w\|_{L^{q_m}(B_1)} \leq C\|w\|_{L^2(B_2)} (3A.7 Lᵖ Spaces and Jensen’s Inequality).

The iteration is a machine: Caccioppoli + Sobolev = reverse Hölder with a gain χ>1\chi > 1, and an infinite product that converges because the gains compound geometrically while the losses (powers of 4k4^k from the shrinking cut-offs) are only geometric in kk inside exponents that shrink like χ−k\chi^{-k}. The same machine works for subsolutions of parabolic equations, on manifolds that satisfy a Sobolev inequality, and for nonlinear quantities that satisfy a differential inequality (Exercise 9.13). Running it with q0q_0 any positive number (not only 22) and an interpolation argument gives sup⁡B1u+≤Cp(∫B2(u+)p)1/p\sup_{B_1}u^+ \leq C_p\big(\int_{B_2}(u^+)^p\big)^{1/p} for every p>0p > 0 (Gilbarg–Trudinger, Theorem 8.17).

Positive supersolutions: the logarithm and the crossover

For the Harnack inequality we need a lower bound for positive supersolutions. Running the iteration with negative exponents β<0\beta < 0 in the test function ζ2uβ\zeta^2u^\beta gives, by the same computation, a bound

inf⁡B1u≥c(  -∫B2u−p)−1/p\inf_{B_1}u \geq c\Big(\mathop{\rlap{\,\,-}\int}\nolimits_{B_2}u^{-p}\Big)^{-1/p}

for every p>0p > 0 (Gilbarg–Trudinger, Theorem 8.18; Jost, section 14.1). This bounds the infimum by a negative power average. To connect it with a positive power average, which is what the upper bound uses, there is a crossover step: we need, for some small p0>0p_0 > 0,

(  -∫Bup0)(  -∫Bu−p0)≤C.\Big(\mathop{\rlap{\,\,-}\int}\nolimits_Bu^{p_0}\Big)\Big(\mathop{\rlap{\,\,-}\int}\nolimits_Bu^{-p_0}\Big) \leq C.

That says uu is not much larger than 1/(1/u)1/(1/u) on average, a statement about the oscillation of log⁡u\log u. Moser's insight was that the logarithm of a positive supersolution is controlled by the equation.

Lemma 9.3 The logarithmic estimate

Let u>0u > 0 be a weak supersolution in Ω\Omega and v=log⁡uv = \log u. For every ball B2r=B(x,2r)⊂ΩB_{2r} = B(x, 2r) \subset \Omega,

∫Br∣∇v∣2 dx≤C rn−2,C=C(n,Λ/λ).\int_{B_r}|\nabla v|^2\,dx \leq C\,r^{n-2}, \qquad C = C(n, \Lambda/\lambda).

Consequently, by the Poincaré inequality,   -∫Br∣v−vBr∣ dx≤C′\mathop{\rlap{\,\,-}\int}\nolimits_{B_r}|v - v_{B_r}|\,dx \leq C', where vBrv_{B_r} is the average of vv over BrB_r.

Proof. Test the supersolution inequality with φ=ζ2u≥0\varphi = \frac{\zeta^2}{u} \geq 0, where ζ\zeta is a cut-off equal to 11 on BrB_r, supported in B2rB_{2r}, with ∣∇ζ∣≤2r|\nabla\zeta| \leq \frac2r. Since ∇φ=2ζ∇ζu−ζ2∇uu2\nabla\varphi = \frac{2\zeta\nabla\zeta}{u} - \frac{\zeta^2\nabla u}{u^2} and ∇v=∇uu\nabla v = \frac{\nabla u}{u},

0≤∫aij∂iu ∂jφ=2∫ζ aij∂iv ∂jζ−∫ζ2aij∂iv ∂jv.0 \leq \int a^{ij}\partial_iu\,\partial_j\varphi = 2\int\zeta\,a^{ij}\partial_iv\,\partial_j\zeta - \int\zeta^2a^{ij}\partial_iv\,\partial_jv.

So λ∫ζ2∣∇v∣2≤2Λ∫ζ∣∇v∣∣∇ζ∣≤2Λ(∫ζ2∣∇v∣2)1/2(∫∣∇ζ∣2)1/2\lambda\int\zeta^2|\nabla v|^2 \leq 2\Lambda\int\zeta|\nabla v||\nabla\zeta| \leq 2\Lambda\big(\int\zeta^2|\nabla v|^2\big)^{1/2}\big(\int|\nabla\zeta|^2\big)^{1/2}, hence

∫ζ2∣∇v∣2≤4Λ2λ2∫∣∇ζ∣2≤4Λ2λ2⋅4r2∣B2r∣=Crn−2.\int\zeta^2|\nabla v|^2 \leq \frac{4\Lambda^2}{\lambda^2}\int|\nabla\zeta|^2 \leq \frac{4\Lambda^2}{\lambda^2}\cdot\frac{4}{r^2}|B_{2r}| = Cr^{n-2}.

For the second statement, the Poincaré inequality on a ball (4A.10 Sobolev Spaces; for mean-zero functions, scaled to radius rr) gives   -∫Br∣v−vBr∣≤(  -∫Br∣v−vBr∣2)1/2≤Cr(  -∫Br∣∇v∣2)1/2≤Cr(r−nrn−2)1/2=C\mathop{\rlap{\,\,-}\int}\nolimits_{B_r}|v - v_{B_r}| \leq \big(\mathop{\rlap{\,\,-}\int}\nolimits_{B_r}|v - v_{B_r}|^2\big)^{1/2} \leq Cr\big(\mathop{\rlap{\,\,-}\int}\nolimits_{B_r}|\nabla v|^2\big)^{1/2} \leq Cr\big(r^{-n}r^{n-2}\big)^{1/2} = C.

A function whose mean oscillation over every ball is bounded is said to have bounded mean oscillation, written v∈BMOv \in \mathrm{BMO}. Bounded functions are in BMO, but so is log⁡∣x∣\log|x|, which is unbounded. The point of BMO is the following theorem.

Theorem 9.4 The John–Nirenberg lemma (statement)

There are constants p0>0p_0 > 0 and CC, depending only on nn and the BMO bound, such that if   -∫B∣v−vB∣≤K\mathop{\rlap{\,\,-}\int}\nolimits_B|v - v_B| \leq K for every ball B⊂B2rB \subset B_{2r}, then

  -∫Brep0∣v−vBr∣ dx≤C.\mathop{\rlap{\,\,-}\int}\nolimits_{B_r}e^{p_0|v - v_{B_r}|}\,dx \leq C.

Fritz John and Louis Nirenberg proved it in 1961; the proof is a Calderón–Zygmund decomposition (6B.8 Lᵖ Theory and Strong Solutions) of ∣v−vB∣|v - v_B| at height proportional to KK, iterated: the set where ∣v−vB∣|v - v_B| exceeds jKjK has measure decaying geometrically in jj (Gilbarg–Trudinger, Theorem 7.21; Exercise 9.10 does a simple case). Applied to v=log⁡uv = \log u,

  -∫up0⋅  -∫u−p0=  -∫ep0(v−vB)⋅  -∫e−p0(v−vB)≤C2,\mathop{\rlap{\,\,-}\int}u^{p_0}\cdot\mathop{\rlap{\,\,-}\int}u^{-p_0} = \mathop{\rlap{\,\,-}\int}e^{p_0(v - v_B)}\cdot\mathop{\rlap{\,\,-}\int}e^{-p_0(v - v_B)} \leq C^2,

which is the crossover. Combined with the negative-power iteration, this gives the weak Harnack inequality for positive supersolutions,

inf⁡B1u≥c(  -∫B2up0)1/p0,\inf_{B_1}u \geq c\Big(\mathop{\rlap{\,\,-}\int}\nolimits_{B_2}u^{p_0}\Big)^{1/p_0},

and, combined with local boundedness for p=p0p = p_0, the full theorem.

Theorem 9.5 Moser's Harnack inequality

If u≥0u \geq 0 is a weak solution of div⁡(A∇u)=0\operatorname{div}(A\nabla u) = 0 in B4B_4, then

sup⁡B1u≤Cinf⁡B1u,C=C(n,Λ/λ).\sup_{B_1}u \leq C\inf_{B_1}u, \qquad C = C(n, \Lambda/\lambda).

It is the Harnack inequality of 6A.2 Harmonic Functions for harmonic functions, now for every equation in the class, with a constant that does not see the coefficients beyond their ellipticity. The proof for harmonic functions used the mean value property; this one uses nothing but Caccioppoli, Sobolev and John–Nirenberg.

From Harnack to Hölder continuity

Theorem 9.6 De Giorgi–Nash–Moser

Every weak solution of div⁡(A∇u)=0\operatorname{div}(A\nabla u) = 0 in B4B_4 is Hölder continuous in B1B_1: there are α∈(0,1)\alpha \in (0, 1) and CC, depending only on nn and Λ/λ\Lambda/\lambda, with

∣u(x)−u(y)∣≤C ∣x−y∣αsup⁡B4∣u∣(x,y∈B1).|u(x) - u(y)| \leq C\,|x - y|^\alpha\sup_{B_4}|u| \qquad (x, y \in B_1).

Proof. Oscillation decay. For a ball B(x0,4r)⊂B4B(x_0, 4r) \subset B_4, let Ms=sup⁡B(x0,s)uM_s = \sup_{B(x_0,s)}u, ms=inf⁡B(x0,s)um_s = \inf_{B(x_0,s)}u and ω(s)=Ms−ms\omega(s) = M_s - m_s (finite by local boundedness). The functions M4r−uM_{4r} - u and u−m4ru - m_{4r} are non-negative weak solutions in B(x0,4r)B(x_0, 4r), so by Theorem 9.5 (scaled to radius rr),

M4r−mr=sup⁡Br(M4r−u)≤Cinf⁡Br(M4r−u)=C(M4r−Mr),M_{4r} - m_r = \sup_{B_r}(M_{4r} - u) \leq C\inf_{B_r}(M_{4r} - u) = C(M_{4r} - M_r),
Mr−m4r=sup⁡Br(u−m4r)≤Cinf⁡Br(u−m4r)=C(mr−m4r).M_r - m_{4r} = \sup_{B_r}(u - m_{4r}) \leq C\inf_{B_r}(u - m_{4r}) = C(m_r - m_{4r}).

Adding, ω(4r)+ω(r)≤C(ω(4r)−ω(r))\omega(4r) + \omega(r) \leq C\big(\omega(4r) - \omega(r)\big), that is,

ω(r)≤γ ω(4r),γ=C−1C+1<1.\omega(r) \leq \gamma\,\omega(4r), \qquad \gamma = \frac{C - 1}{C + 1} < 1.

Iteration. Let x0∈B1x_0 \in B_1; then B(x0,3)⊂B4B(x_0, 3) \subset B_4, so the decay ω(r)≤γω(4r)\omega(r) \leq \gamma\omega(4r) holds for 4r≤34r \leq 3, and ω(3)≤2sup⁡B4∣u∣\omega(3) \leq 2\sup_{B_4}|u|. Applying it kk times, ω(3⋅4−k)≤γkω(3)\omega(3\cdot4^{-k}) \leq \gamma^k\omega(3). For 0<s≤30 < s \leq 3, choose kk with 3⋅4−k−1<s≤3⋅4−k3\cdot4^{-k-1} < s \leq 3\cdot4^{-k}, and let α=log⁡41γ\alpha = \log_4\frac1\gamma, so γk=4−kα<(4s3)α\gamma^k = 4^{-k\alpha} < \big(\frac{4s}{3}\big)^\alpha. Then

ω(s)≤γkω(3)≤2(4s3)αsup⁡B4∣u∣.\omega(s) \leq \gamma^k\omega(3) \leq 2\Big(\frac{4s}{3}\Big)^\alpha\sup_{B_4}|u|.

For x,y∈B1x, y \in B_1, s=∣x−y∣<2≤3s = |x - y| < 2 \leq 3, and ∣u(x)−u(y)∣≤ω(s)|u(x) - u(y)| \leq \omega(s) for the ball centred at xx, which is the claim.

The proof shows where the exponent comes from: each fourfold shrinking of the ball reduces the oscillation by the factor γ\gamma, which depends on the Harnack constant, which depends only on nn and Λ/λ\Lambda/\lambda. With a large contrast, γ\gamma is close to 11 and α\alpha is small. That is not an artefact of the proof: there are equations in the class whose solutions are Hölder with no better exponent than α→0\alpha \to 0 as Λ/λ→∞\Lambda/\lambda \to \infty (Exercise 9.9).

De Giorgi's method, the original proof, works with level sets instead of powers: Caccioppoli's inequality for (u−k)+(u - k)^+ at every level kk (a function satisfying it for all kk belongs to a De Giorgi class), combined with an isoperimetric-type lemma saying that a function cannot pass from a low level to a high level on a set of small measure without paying in gradient. It proves boundedness by an iteration over levels and Hölder continuity by a "shrinking measure" argument, and it extends to quasi-minimisers of variational problems, where there is no equation at all (Giaquinta; Jost, section 14.4). Nash's method used the fundamental solution and an entropy argument (6A.12 Entropy, Information and Diffusion).

Hilbert's nineteenth problem

In his 1900 list of problems, David Hilbert asked whether the solutions of "regular" variational problems are always analytic. In the simplest form: if u∈H1(Ω)u \in H^1(\Omega) minimises E[u]=∫ΩF(∇u) dxE[u] = \int_\Omega F(\nabla u)\,dx, where FF is smooth (or analytic) and uniformly convex, λ∣ξ∣2≤Fpipj(p)ξiξj≤Λ∣ξ∣2\lambda|\xi|^2 \leq F_{p_ip_j}(p)\xi_i\xi_j \leq \Lambda|\xi|^2, is uu smooth?

The difficulty is circular. The Euler–Lagrange equation div⁡(Fp(∇u))=0\operatorname{div}(F_p(\nabla u)) = 0 (6A.11 Calculus of Variations and Gradient Flows) is nonlinear. Formally differentiating it in xkx_k gives, for w=∂kuw = \partial_ku,

∂i(Fpipj(∇u) ∂jw)=0,\partial_i\big(F_{p_ip_j}(\nabla u)\,\partial_jw\big) = 0,

a linear equation for ww with coefficients aij(x)=Fpipj(∇u(x))a^{ij}(x) = F_{p_ip_j}(\nabla u(x)). Schauder theory would make ww Hölder if the coefficients were Hölder, that is, if ∇u\nabla u were Hölder: what we want to prove. But the coefficients are certainly measurable and uniformly elliptic, because of the convexity of FF, whatever ∇u\nabla u is. So De Giorgi–Nash–Moser applies: w=∂kuw = \partial_ku is Hölder, for each kk. The circle is broken. Now the coefficients F′′(∇u)F''(\nabla u) are Hölder, Schauder gives ∇u∈C1,α\nabla u \in C^{1,\alpha}, and bootstrapping gives C∞C^\infty (and analyticity, by a further argument of Morrey and of Friedman). Making the formal differentiation rigorous uses difference quotients, as in 6A.6 Weak Solutions and Elliptic Regularity: the minimiser is first shown to be in Hloc2H^2_{\mathrm{loc}}, so that ∂ku∈Hloc1\partial_ku \in H^1_{\mathrm{loc}} is a weak solution of the differentiated equation (Jost, section 14.4; Exercise 9.12).

Systems are different. For vector-valued minimisers u:Ω→Rmu : \Omega \to \mathbb{R}^m, the differentiated equations form an elliptic system with measurable coefficients, and there De Giorgi–Nash–Moser fails. De Giorgi himself gave a counterexample in 1968: a linear elliptic system with bounded measurable coefficients in dimension n≥3n \geq 3 with an unbounded solution, u(x)=∣x∣−γxu(x) = |x|^{-\gamma}x for a suitable γ\gamma. Minimisers of variational problems for maps can be singular, and the regularity theory for systems is partial: smooth outside a closed set of small dimension. This is the situation for harmonic maps (6C.7 Harmonic Maps and Their Heat Flow) and for Yang–Mills fields, and it is why their singularities are a subject of their own.

The parabolic theory

Nash's 1958 paper was about parabolic equations ut=div⁡(A(x,t)∇u)u_t = \operatorname{div}(A(x, t)\nabla u) with measurable coefficients, and proved Hölder continuity for them too. Moser proved the parabolic Harnack inequality in 1964: if u≥0u \geq 0 solves such an equation in Q=B2×(0,4)Q = B_2\times(0, 4), then

sup⁡B1×(1,2)u≤Cinf⁡B1×(3,4)u,C=C(n,Λ/λ).\sup_{B_1\times(1,2)}u \leq C\inf_{B_1\times(3,4)}u, \qquad C = C(n, \Lambda/\lambda).

The supremum is taken at earlier times than the infimum, with a gap between. The gap cannot be removed: for the heat kernel Φ(x,t)\Phi(x, t) the ratio Φ(x,t)Φ(y,t)=e(∣y∣2−∣x∣2)/4t\frac{\Phi(x, t)}{\Phi(y, t)} = e^{(|y|^2 - |x|^2)/4t} at a single time is unbounded as t→0t \to 0 (Exercise 9.11). Heat must have time to arrive. The Li–Yau inequality (6A.12 Entropy, Information and Diffusion) is the sharp form of this Harnack inequality for smooth coefficients on manifolds with Ric⁡≥0\operatorname{Ric} \geq 0, and Hamilton's Harnack inequality for the Ricci flow (11B.2 Ancient Solutions and the Harnack Inequality) is its geometric descendant.

Where this goes Moser iteration and Harnack in geometric analysis

Moser iteration needs only two ingredients: a Caccioppoli-type inequality, which comes from the structure of the equation, and a Sobolev inequality, which comes from the geometry. On a Riemannian manifold, a uniform Sobolev inequality on balls holds when the Ricci curvature is bounded below and balls are not collapsed (Saloff-Coste; 9B.2 Volume Comparison), and then the whole theory of this chapter goes through. In the analysis of geometric flows, Moser iteration upgrades integral curvature bounds to pointwise ones, for instance in the theory of Ricci flows with bounded scalar curvature, and the uniform Sobolev inequality along the Ricci flow (Qi S. Zhang, 4A.11 Sobolev Embeddings and Critical Exponents's rehearsal) exists to make such iterations possible. Krylov–Safonov (6B.8 Lᵖ Theory and Strong Solutions) is the non-divergence counterpart, and it is the one used for fully nonlinear equations such as the complex Monge–Ampère equation of Kähler–Ricci flow (6C.10 Fully Nonlinear Equations and Monge–Ampère).

History

David Hilbert posed his nineteenth problem in 1900; Sergei Bernstein (1904) and others solved it under extra smoothness assumptions on the minimiser. Ennio De Giorgi's 1957 paper solved it in general, and John Nash's 1958 paper, written independently and published shortly afterwards, proved the parabolic theorem by another route; Nash learned of De Giorgi's priority after his work was done. Jürgen Moser gave his iteration proof of the elliptic theorem in 1960, the elliptic Harnack inequality in 1961, and the parabolic Harnack inequality in 1964. Fritz John and Louis Nirenberg introduced BMO and proved their lemma in 1961. De Giorgi's counterexample for systems dates from 1968. Krylov and Safonov proved the non-divergence Harnack inequality in 1979–80.

Recall Where we stand

For divergence-form equations with merely measurable, uniformly elliptic coefficients, weak subsolutions satisfy Caccioppoli's inequality, and combining it with the Sobolev inequality gives a gain of integrability; Moser's iteration over shrinking balls and growing exponents turns this into sup⁡B1u+≤C∥u+∥L2(B2)\sup_{B_1}u^+ \leq C\|u^+\|_{L^2(B_2)}. For positive supersolutions, log⁡u\log u has bounded mean oscillation, John–Nirenberg turns that into the crossover   -∫up0  -∫u−p0≤C\mathop{\rlap{\,\,-}\int}u^{p_0}\mathop{\rlap{\,\,-}\int}u^{-p_0} \leq C, and the negative-power iteration gives the weak Harnack inequality; together with local boundedness this is Moser's Harnack inequality. Harnack applied to M−uM - u and u−mu - m gives oscillation decay by a fixed factor, hence Hölder continuity with an exponent depending only on nn and Λ/λ\Lambda/\lambda. Applied to derivatives of minimisers, this solves Hilbert's nineteenth problem; for systems it fails. The parabolic version has a Harnack inequality with a time lag. 6B.10 Harmonic Coordinates and the Regularity of Metrics turns the elliptic theory of 6B.7 Schauder Theory and the Continuity Method–6B.9 De Giorgi–Nash–Moser on the most important unknown of geometric analysis, a Riemannian metric, and finds the coordinates in which the Ricci tensor is elliptic.

Exercises

Exercise 9.7 Caccioppoli for harmonic functions

Let uu be harmonic in B2rB_{2r}. (a) Use Lemma 9.1 with k→−∞k \to -\infty (or prove it directly by testing with ζ2u\zeta^2u) to show ∫Br∣∇u∣2≤Cr2∫B2ru2\int_{B_r}|\nabla u|^2 \leq \frac{C}{r^2}\int_{B_{2r}}u^2. (b) Check it for u=x1u = x_1 on Rn\mathbb{R}^n, and show that the power r−2r^{-2} is forced by scaling.

Solution

(a) Testing ∫∇u⋅∇(ζ2u)=0\int\nabla u\cdot\nabla(\zeta^2u) = 0: ∫ζ2∣∇u∣2=−2∫ζu∇u⋅∇ζ≤12∫ζ2∣∇u∣2+2∫u2∣∇ζ∣2\int\zeta^2|\nabla u|^2 = -2\int\zeta u\nabla u\cdot\nabla\zeta \leq \frac12\int\zeta^2|\nabla u|^2 + 2\int u^2|\nabla\zeta|^2, so ∫ζ2∣∇u∣2≤4∫u2∣∇ζ∣2≤4r2∫B2ru2\int\zeta^2|\nabla u|^2 \leq 4\int u^2|\nabla\zeta|^2 \leq \frac{4}{r^2}\int_{B_{2r}}u^2 with ∣∇ζ∣≤1r|\nabla\zeta| \leq \frac1r. (b) For u=x1u = x_1, the left side is ∣Br∣=α(n)rn|B_r| = \alpha(n)r^n, and ∫B2rx12=1n∫B2r∣x∣2=α(n)(2r)n+2n+2\int_{B_{2r}}x_1^2 = \frac{1}{n}\int_{B_{2r}}|x|^2 = \frac{\alpha(n)(2r)^{n+2}}{n + 2}, so the right side is Cα(n)2n+2rnn+2\frac{C\alpha(n)2^{n+2}r^n}{n + 2}: both are multiples of rnr^n. Under u(x)↦u(λx)u(x) \mapsto u(\lambda x) the left side scales like λ2−n\lambda^{2-n} and ∫u2\int u^2 like λ−n\lambda^{-n}, so the factor must be r−2r^{-2}.

Exercise 9.8 The infinite product

Compute ∑k≥012χk\sum_{k\geq0}\frac{1}{2\chi^k} and ∑k≥0k+12χk\sum_{k\geq0}\frac{k + 1}{2\chi^k} for χ=nn−2\chi = \frac{n}{n-2}, and find the resulting bound for the constant in Theorem 9.2 in terms of C2C_2 when n=3n = 3.

Solution

With θ=1χ<1\theta = \frac1\chi < 1: ∑θk2=12(1−θ)\sum\frac{\theta^k}{2} = \frac{1}{2(1 - \theta)} and ∑(k+1)θk2=12(1−θ)2\sum\frac{(k + 1)\theta^k}{2} = \frac{1}{2(1 - \theta)^2}. For n=3n = 3, χ=3\chi = 3, θ=13\theta = \frac13: the sums are 34\frac34 and 98\frac98, so C≤C23/449/8C \leq C_2^{3/4}4^{9/8}.

Exercise 9.9 The Hölder exponent can be small

In the plane, let u(x)=∣x∣α−1x1u(x) = |x|^{\alpha - 1}x_1 for 0<α<10 < \alpha < 1 (in polar coordinates, u=rαcos⁡θu = r^\alpha\cos\theta). Show that uu is a weak solution of div⁡(A∇u)=0\operatorname{div}(A\nabla u) = 0 where, in the polar frame, AA has eigenvalue 11 in the radial direction and α2\alpha^2 in the angular direction, that is, A(x)=x⊗x∣x∣2+α2(I−x⊗x∣x∣2)A(x) = \frac{x\otimes x}{|x|^2} + \alpha^2\big(I - \frac{x\otimes x}{|x|^2}\big) (measurable and bounded, with Λ/λ=α−2\Lambda/\lambda = \alpha^{-2}). Conclude that the best Hölder exponent can be as small as λ/Λ\sqrt{\lambda/\Lambda}.

Hint

Compute A∇uA\nabla u in polar coordinates: ∇u=αrα−1cos⁡θ er−rα−1sin⁡θ eθ\nabla u = \alpha r^{\alpha-1}\cos\theta\,e_r - r^{\alpha-1}\sin\theta\,e_\theta, so A∇u=αrα−1cos⁡θ er−α2rα−1sin⁡θ eθA\nabla u = \alpha r^{\alpha-1}\cos\theta\,e_r - \alpha^2r^{\alpha-1}\sin\theta\,e_\theta. Then use div⁡(fer+geθ)=1r∂r(rf)+1r∂θg\operatorname{div}(fe_r + ge_\theta) = \frac1r\partial_r(rf) + \frac1r\partial_\theta g.

Solution

div⁡(A∇u)=1r∂r(αrαcos⁡θ)+1r∂θ(−α2rα−1sin⁡θ)=α2rα−2cos⁡θ−α2rα−2cos⁡θ=0\operatorname{div}(A\nabla u) = \frac1r\partial_r(\alpha r^\alpha\cos\theta) + \frac1r\partial_\theta(-\alpha^2r^{\alpha-1}\sin\theta) = \alpha^2r^{\alpha-2}\cos\theta - \alpha^2r^{\alpha-2}\cos\theta = 0 away from the origin. Near the origin ∣∇u∣2∼r2α−2|\nabla u|^2 \sim r^{2\alpha-2}, integrable in the plane, and the boundary terms on small circles vanish, so uu is a weak solution in the whole disc. It is Hölder with exponent exactly α\alpha at the origin, and α=λ/Λ\alpha = \sqrt{\lambda/\Lambda}.

Exercise 9.10 John–Nirenberg for log⁡∣x∣\log|x|

On the interval (−1,1)(-1, 1), let v(x)=log⁡∣x∣v(x) = \log|x|. (a) Show that 1∣I∣∫I∣v−vI∣≤C\frac{1}{|I|}\int_I|v - v_I| \leq C for every subinterval II. (b) Show that ∫−11ep∣v∣dx<∞\int_{-1}^1e^{p|v|}dx < \infty exactly when p<1p < 1. So the exponential integrability given by John–Nirenberg holds only for small exponents, and BMO functions need not be bounded.

Solution

(a) By scaling (v(λx)=v(x)+log⁡λv(\lambda x) = v(x) + \log\lambda, and subtracting the average removes constants), it suffices to treat intervals of length 11 at distance dd from 00: if d≥1d \geq 1, ∣v′∣≤1|v'| \leq 1 on II and the oscillation is at most 11; if d<1d < 1, I⊂(−2,2)I \subset (-2, 2) and ∫−22∣log⁡∣x∣∣+∣vI∣≤C\int_{-2}^2|\log|x|| + |v_I| \leq C. (b) ep∣log⁡∣x∣∣=∣x∣−pe^{p|\log|x||} = |x|^{-p} near 00, integrable iff p<1p < 1.

Exercise 9.11 The time lag is necessary

For the one-dimensional heat kernel Φ(x,t)=(4πt)−1/2e−x2/4t\Phi(x, t) = (4\pi t)^{-1/2}e^{-x^2/4t}, compute Φ(0,t)Φ(1,t)\frac{\Phi(0, t)}{\Phi(1, t)} and show it is unbounded as t→0t \to 0. Explain why a Harnack inequality sup⁡B1u(⋅,t)≤Cinf⁡B1u(⋅,t)\sup_{B_1}u(\cdot, t) \leq C\inf_{B_1}u(\cdot, t) at a single time cannot hold for all positive solutions. Then use the Li–Yau Harnack inequality of 6A.12 Entropy, Information and Diffusion to show that Φ(0,t1)≤CΦ(1,2t1)\Phi(0, t_1) \leq C\Phi(1, 2t_1) with CC independent of t1≥1t_1 \geq 1.

Solution

Φ(0,t)Φ(1,t)=e1/4t→∞\frac{\Phi(0, t)}{\Phi(1, t)} = e^{1/4t} \to \infty. Any constant CC would be violated by the heat kernel (shifted in time so it is positive on the cylinder) at small tt. With a lag, Li–Yau gives Φ(0,t1)≤Φ(1,t2)(t2t1)1/2e1/4(t2−t1)\Phi(0, t_1) \leq \Phi(1, t_2)\big(\frac{t_2}{t_1}\big)^{1/2}e^{1/4(t_2 - t_1)}; for t2=2t1t_2 = 2t_1 this is 2e1/4t1Φ(1,t2)\sqrt2e^{1/4t_1}\Phi(1, t_2), bounded by 2e1/4Φ(1,t2)\sqrt2e^{1/4}\Phi(1, t_2) for t1≥1t_1 \geq 1. The lag must be comparable to the square of the distance for a uniform constant: parabolic scaling.

Exercise 9.12 Differentiating the Euler–Lagrange equation

Let uu be a smooth minimiser of ∫F(∇u)\int F(\nabla u), so ∂i(Fpi(∇u))=0\partial_i(F_{p_i}(\nabla u)) = 0. (a) Differentiate in xkx_k and show that w=∂kuw = \partial_ku satisfies ∂i(aij∂jw)=0\partial_i(a^{ij}\partial_jw) = 0 with aij=Fpipj(∇u)a^{ij} = F_{p_ip_j}(\nabla u). (b) For the area functional F(p)=1+∣p∣2F(p) = \sqrt{1 + |p|^2}, compute FpipjF_{p_ip_j} and show it is uniformly elliptic only as long as ∣∇u∣|\nabla u| is bounded: the ellipticity constants are λ=(1+∣p∣2)−3/2\lambda = (1 + |p|^2)^{-3/2} and Λ=(1+∣p∣2)−1/2\Lambda = (1 + |p|^2)^{-1/2}. Explain why a gradient bound is the first step in the regularity of minimal graphs (6C.6 Quasilinear Elliptic Equations and Minimal Surfaces).

Solution

(a) ∂k∂i(Fpi(∇u))=∂i(Fpipj(∇u)∂j∂ku)=0\partial_k\partial_i(F_{p_i}(\nabla u)) = \partial_i\big(F_{p_ip_j}(\nabla u)\partial_j\partial_ku\big) = 0. (b) Fpi=pi1+∣p∣2F_{p_i} = \frac{p_i}{\sqrt{1 + |p|^2}} and Fpipj=11+∣p∣2(δij−pipj1+∣p∣2)F_{p_ip_j} = \frac{1}{\sqrt{1 + |p|^2}}\big(\delta_{ij} - \frac{p_ip_j}{1 + |p|^2}\big), with eigenvalues (1+∣p∣2)−1/2(1 + |p|^2)^{-1/2} (on vectors orthogonal to pp) and (1+∣p∣2)−3/2(1 + |p|^2)^{-3/2} (on pp). The ratio Λ/λ=1+∣p∣2\Lambda/\lambda = 1 + |p|^2 is bounded only when ∣∇u∣|\nabla u| is. So De Giorgi–Nash–Moser applies to the minimal surface equation once a gradient bound is known, and the regularity theory for minimal graphs reduces to proving one.

Exercise 9.13 Rehearsal: Moser iteration on a manifold

Suppose that on a Riemannian manifold, on a ball B2rB_{2r}, the Sobolev inequality holds in the scale-invariant form (  -∫B2r∣v∣2χ)1/χ≤CSr2  -∫B2r∣∇v∣2\big(\mathop{\rlap{\,\,-}\int}\nolimits_{B_{2r}}|v|^{2\chi}\big)^{1/\chi} \leq C_Sr^2\mathop{\rlap{\,\,-}\int}\nolimits_{B_{2r}}|\nabla v|^2 for all vv supported in B2rB_{2r}, for some χ>1\chi > 1. (a) Show that the proof of Theorem 9.2, for u≥0u \geq 0 with Δu≥0\Delta u \geq 0 (subharmonic), goes through word for word with averages in place of integrals, giving sup⁡Bru≤C(n,CS)(  -∫B2ru2)1/2\sup_{B_r}u \leq C(n, C_S)\big(\mathop{\rlap{\,\,-}\int}\nolimits_{B_{2r}}u^2\big)^{1/2}. (b) Which geometric hypothesis was used, and where? (c) In Ricci flow, Shi's estimates (6A.7 Parabolic Regularity) need a pointwise curvature bound; explain how a uniform Sobolev inequality along the flow, as in Qi S. Zhang's theorem (4A.11 Sobolev Embeddings and Critical Exponents), allows Moser iteration to turn integral bounds on a quantity satisfying a subsolution inequality into pointwise bounds.

Solution

(a) Caccioppoli uses only the weak form of Δu≥0\Delta u \geq 0 and integration by parts with a cut-off, valid on any manifold; the Sobolev step uses the assumed inequality, now with averages, and the constants r2r^2 from ∣∇ζ∣≤2r|\nabla\zeta| \leq \frac{2}{r} cancel the r2r^2 in the Sobolev inequality, so the iteration constants are scale-free. (b) The Sobolev inequality on balls with a uniform constant; on manifolds it follows from a lower Ricci bound plus a volume (noncollapsing) lower bound. It is used once per iteration step. (c) If a quantity Q≥0Q \geq 0 satisfies (∂t−Δ)Q≤CQ(\partial_t - \Delta)Q \leq CQ (a subsolution inequality), the parabolic version of the iteration with a uniform Sobolev constant gives sup⁡Q\sup Q on a parabolic cylinder in terms of an L2L^2 (or LpL^p) average on a larger one. A uniform Sobolev inequality along the flow is exactly the ingredient that makes the constants uniform in time.

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