The Two Hyperplane Conjecture David Jerison (MIT) In honor of Steve Hofmann, ICMAT, May 2018 David Jerison The Two Hyperplane Conjecture
Outline ◮ The two hyperplane conjecture ◮ Whence it came (level sets) ◮ Where it may lead ◮ Quantitative connectivity: Isoperimetric, Poincar´ e and Harnack inequalities. David Jerison The Two Hyperplane Conjecture
Isoperimetric set E relative to µ P µ ( E ) ≤ P µ ( F ) for all F , µ ( F ) = µ ( E ) . Perimeter of E relative to a measure µ on R n µ ( E δ ) − µ ( E ) P µ ( E ) := liminf ( E δ = δ -nbd of E ) . δ δ ց 0 Example: µ = 1 Ω dx , Ω open, convex: P µ ( E ) = H n − 1 (Ω ∩ ∂ E ) ( E open) . David Jerison The Two Hyperplane Conjecture
Conjecture 1. There is b ( n ) > 0 such that if Ω ⊂ R n is convex, symmetric ( − Ω = Ω), E ⊂ Ω is isoperimetric, | E | = | Ω | / 2, then there is a half space H such that H ∩ Ω ⊂ E , ( − H ∩ Ω) ⊂ Ω \ E , | H ∩ Ω | ≥ b ( n ) | Ω | . The interface Ω ∩ ∂ E is trapped between hyperplanes. David Jerison The Two Hyperplane Conjecture
Conjecture 1. There is b ( n ) > 0 such that if Ω ⊂ R n is convex, symmetric ( − Ω = Ω), E ⊂ Ω is isoperimetric, | E | = | Ω | / 2, then there is a half space H such that H ∩ Ω ⊂ E , ( − H ∩ Ω) ⊂ Ω \ E , | H ∩ Ω | ≥ b ( n ) | Ω | . The interface Ω ∩ ∂ E is trapped between hyperplanes. Conjecture 1 ∗ (Two hyperplane conjecture) b ( n ) ≥ b ∗ > 0 an absolute constant. David Jerison The Two Hyperplane Conjecture
Conjecture 2 (qualitative form). If Ω ⊂ R n is convex, E ⊂ Ω is isoperimetric, 0 < | E | < | Ω | , then hull( E ) � = Ω Open question, even in R 3 . David Jerison The Two Hyperplane Conjecture
Conjecture 2 (qualitative form). If Ω ⊂ R n is convex, E ⊂ Ω is isoperimetric, 0 < | E | < | Ω | , then hull( E ) � = Ω Open question, even in R 3 . First enemy of both conjectures: E = Ω \ B ( B a ball) . David Jerison The Two Hyperplane Conjecture
First serious enemy: the Simons cone S = { ( x , y ) ∈ R n × R n : | x | = | y |} ⊂ R 2 n , 2 n ≥ 8 . S is area-miniminizing for fixed boundary conditions in any Ω. S 1 = S ∩ B 1 , Ω := hull( S 1 ) , E := { ( x , y ) ∈ Ω : | x | > | y |} , | E | = | Ω | / 2 . David Jerison The Two Hyperplane Conjecture
First serious enemy: the Simons cone S = { ( x , y ) ∈ R n × R n : | x | = | y |} ⊂ R 2 n , 2 n ≥ 8 . S is area-miniminizing for fixed boundary conditions in any Ω. S 1 = S ∩ B 1 , Ω := hull( S 1 ) , E := { ( x , y ) ∈ Ω : | x | > | y |} , | E | = | Ω | / 2 . E is not stable for the isoperimetric problem. This is a slightly sharpened version of a theorem of Sternberg and Zumbrun from 1990s. David Jerison The Two Hyperplane Conjecture
Log-concave measures on R n µ = e − V dx , V is convex . Ω convex is achieved in the limit: V ( x ) = ∞ , x ∈ R n \ Ω . µ = 1 Ω dx ; V ( x ) = 0 , x ∈ Ω , David Jerison The Two Hyperplane Conjecture
Log-concave measures on R n µ = e − V dx , V is convex . Ω convex is achieved in the limit: V ( x ) = ∞ , x ∈ R n \ Ω . µ = 1 Ω dx ; V ( x ) = 0 , x ∈ Ω , KLS Hyperplane Conjecture. There is an absolute constant c ∗ > 0 such that if µ is log-concave on R n and E is isoperimetric with µ ( E ) = µ ( R n ) / 2, then there is a half space H for which P µ ( E ) ≥ c ∗ P µ ( H ) , µ ( H ) = µ ( E ) . David Jerison The Two Hyperplane Conjecture
Proposition (E. Milman) “Two implies One”. Suppose that µ is log-concave, E is an isoperimetric set with µ ( E ) = µ ( R n ) / 2, and there are half spaces H i such that µ ( H i ) ≥ b ∗ > 0 , H 2 ⊂ R n \ E . H 1 ⊂ E , Let H 0 be the translate of H 1 such that µ ( H 0 ) = µ ( E ). Then 1 c ∗ = P µ ( E ) ≥ c ∗ P µ ( H 0 ) , 4log(1 / b ∗ ) . David Jerison The Two Hyperplane Conjecture
Conjecture 3. (Half space conjecture) If ∇ 2 V >> 0 and E ⊂ R n is isoperimetric for µ = e − V dx , with µ ( E ) = µ ( R n ) / 2, then there are convex sets K 1 and K 2 such that K 2 ⊂ R n \ E , K 1 ⊂ E , µ ( K i ) ≥ c > 0 for an absolute constant c . Moreover, at least one of the two sets K i can be taken to be a half space. David Jerison The Two Hyperplane Conjecture
First Variation for µ = w dx w = e − V . H µ = ( n − 1) H − n · ∇ V , Second Variation (stability) for S = ∂ E � � S ( | A | 2 + ∇ 2 V ( n , n )) f 2 w d σ ≤ S | ∇ S f | 2 w d σ � provided S f w d σ = 0 . David Jerison The Two Hyperplane Conjecture
Symmetry Breaking Proposition (variant of Sternberg-Zumbrun) If E is isoperimetric for µ = e − V dx , ∇ 2 V >> 0, and V ( − x ) = V ( x ) , − E = E , then E is not stable. Proof: Take f j = e j · n (orthonormal basis e j ). ∑ | ∇ f j | 2 = | A | 2 , ∑ f 2 j = 1 Rediscovered by Rosales, Ca˜ nete, Bayle and Morgan in radial case. David Jerison The Two Hyperplane Conjecture
Let C ( µ ) be the best constant in Poincar´ e’s inequality � � � | f | 2 d µ ≤ C ( µ ) | ∇ f | 2 d µ f d µ = 0 . (*) (KLS ⇐ ⇒ linear test functions suffice.) When µ = 1 Ω dx , extremals u are Neumann eigenfunctions for λ = 1 / C ( µ ): ∆ u = − λ u in Ω , ν · ∇ u = 0 on ∂ Ω , David Jerison The Two Hyperplane Conjecture
Level sets of the first nonconstant Neumann eigenfunction are analogous to isoperimetric sets. “Hot Spots” Conjecture (J. Rauch) First Neumann eigenfunctions usually achieve their maximum on the boundary. David Jerison The Two Hyperplane Conjecture
Level sets of the first nonconstant Neumann eigenfunction are analogous to isoperimetric sets. “Hot Spots” Conjecture (J. Rauch) First Neumann eigenfunctions usually achieve their maximum on the boundary. Our version for today: If Ω ⊂ R n is convex, open, bounded, and − Ω = Ω, then each first Neumann eigenfunction is monotone: e · ∇ u > 0 in Ω (for some direction e ). David Jerison The Two Hyperplane Conjecture
Two axes of symmetry: J-, Nadirashvili 2000. “lip domains”, obtuse triangles: Atar-Burdzy 2004, acute triangles: Judge-Mondal (preprint). N. B. With monotonicity (and strict convexity) the level sets are topologically trivial, smooth graphs. Some extra hypothesis like − Ω = Ω is needed to get monotonicity. Already for many acute triangles some level sets are disconnected. David Jerison The Two Hyperplane Conjecture
Deformation approach to hot spots Consider u t , Ω t , 0 ≤ t ≤ 1, and G = { t ∈ [0 , 1] : e · ∇ u t ( x ) > 0 in Ω t } 0 ∈ G , G is open and closed = ⇒ G = [0 , 1]. Show G is closed by showing that the level sets { x ∈ Ω t : u t ( x ) = c } are Lipschitz graphs with vertical direction e . David Jerison The Two Hyperplane Conjecture
A priori Lipschitz bounds Theorem (Bombieri, De Giorgi, Miranda) If ϕ ∈ C ∞ ( B 1 ) satisfies � � ∇ϕ ∇ · = 0 � 1+ | ∇ϕ | 2 with | ϕ | ≤ M , then there is C = C ( n , M ) s. t. | ∇ϕ | ≤ C in B 1 / 2 . David Jerison The Two Hyperplane Conjecture
A priori Lipschitz bounds Theorem (Bombieri, De Giorgi, Miranda) If ϕ ∈ C ∞ ( B 2 ) satisfies � � ∇ϕ ∇ · = 0 � 1+ | ∇ϕ | 2 with | ϕ | ≤ M , then there is C = C ( n , M ) s. t. | ∇ϕ | ≤ C in B 1 . David Jerison The Two Hyperplane Conjecture
Proof (De Silva, J-) S r := { ( x , ϕ ( x )) : x ∈ B r } We want to prove dist( S 1 , S 1 +(0 , ε )) ≥ c ε Step 1. The normal distance is ≥ c 1 ε on a “good” set G of large measure because � | ∇ϕ | d σ ≤ CM 2 . B 3 / 2 David Jerison The Two Hyperplane Conjecture
Step 2. Define the normal variation ψ ( X ) by (∆ S + | A | 2 ) ψ = 0 on S 2 \ G . ψ = 1 on G , ψ = 0 on ∂ S 2 . Our goal S ( t ) = { X + t ψ ( X ) ν : X ∈ S 1 } cannnot touch S 1 +(0 , ε ) for 0 ≤ t ≤ c 1 ε for X ∈ G . David Jerison The Two Hyperplane Conjecture
Step 3. ∆ S w ≤ 0 (supersolution) Bombieri-Giusti Harnack inequality: � inf w ≥ c 2 w d σ ≥ c 2 σ ( G ) S 1 S 1 Hence, ψ ≥ c > 0 on S 1 . Hence we have normal separation by c ε on all of S 1 . David Jerison The Two Hyperplane Conjecture
Free boundary gradient bound Thm (De Silva, J-) If u is positive, harmonic in { ( x , y ) : y > ϕ ( x ) } , | ∇ u | = 0 on { y = ϕ ( x ) } , and | ϕ | ≤ M on | x | < 2, then | ∇ϕ | ≤ C M on | x | < 1 . Proof: y > ϕ ( x ) is an NTA domain and the boundary Harnack inequality plays the role analogous to the Bombieri-Giusti intrinsic Harnack inequality. David Jerison The Two Hyperplane Conjecture
Bombieri-Giusti Harnack is deduced via Moser type argument from De Georgi local isoperimetric inequality: � � | f − k | n / ( n − 1) d σ ≤ C min | ∇ S f | d σ . k B β R B R Proof is via blow up and compactness from qualitative (measure-theoretic) connectivity of area minimizing cones due to Almgren and De Giorgi. David Jerison The Two Hyperplane Conjecture
Proposition Let S ⊂ R n be an area minimizer, then S divides R n into two NTA domains. Lemma If E = A 1 ∪ A 2 , | A 1 ∩ A 2 | = 0, then for some β > 0, i ( | A i ∩ B β r | ) ( n − 1) / n P ( A i , E ∩ B r ) ≥ c min The lemma is proved by the method of Almgren-De Giorgi. Then methods of G. David and S. Semmes yield the proposition. David Jerison The Two Hyperplane Conjecture
Recommend
More recommend
