H¨ ormander’s ¯ ∂-estimate,
Some Generalizations, and New Applications
Zbigniew Błocki
(Uniwersytet Jagielloński, Kraków, Poland) http://gamma.im.uj.edu.pl/eblocki
Analysis, Complex Geometry, and Mathematical Physics:
A Conference in Honor of Duong H. Phong Columbia University, May 7-11, 2013
We will discuss applications of H¨ormander’s L2-estimate for ¯∂ in the following problems:
1. Suita Conjecture (1972) from potential theory
2. Optimal constant in the Ohsawa-Takegoshi extension theorem (1987) 3. Mahler Conjecture (1938) from convex analysis
Suita Conjecture
Green function for bounded domain D in C:
(∆GD(·, z) = 2πδz
GD(·, z) = 0 on ∂D (if D is regular) cD(z) := exp lim
ζ→z(GD(ζ, z) − log |ζ − z|)
(logarithmic capacity of C \ D w.r.t. z) cD|dz| is an invariant metric (Suita metric)
CurvcD|dz|= −(log cD)z ¯z
c2D
Suita Conjecture (1972): CurvcD|dz|≤ −1
• “=” if D is simply connected
• “<” if D is an annulus (Suita)
• Enough to prove for D with smooth boundary
• “=” on ∂D if D has smooth boundary
2 4 6 8 10
-7 -6 -5 -4 -3 -2 -1
CurvcD|dz|for D = {e−5< |z| < 1} as a function of t = −2 log |z|
5 10 15 20
-3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5
CurvKD|dz|2 for D = {e−10< |z| < 1} as a function of t = −2 log |z|
1 2 3 4 5
-6 -5 -4 -3 -2 -1
Curv(log KD)z ¯z|dz|2 for D = {e−5< |z| < 1} as a function of t = −2 log |z|
∂2
∂z∂ ¯z(log cD) = πKD (Suita) where KDis the Bergman kernel on the diagonal:
KD(z) := sup{|f (z)|2: f ∈ O(D), Z
D
|f |2dλ ≤ 1}.
Therefore the Suita conjecture is equivalent to c2D≤ πKD.
It is thus an extension problem: for z ∈ D find holomorphic f in D such that f (z) = 1 and
Z
D
|f |2dλ ≤ π (cD(z))2.
Ohsawa (1995), using the methods of the Ohsawa-Takegoshi extension theorem, showed the estimate
c2D≤ CπKD with C = 750.
C = 2 (B., 2007)
C = 1.95388 . . . (Guan-Zhou-Zhu, 2011)
Ohsawa-Takegoshi Extension Theorem (1987)
Ω - bounded pseudoconvex domain in Cn, ϕ - psh in Ω H - complex affine subspace of Cn
f - holomorphic in Ω0:= Ω ∩ H
Then there exists a holomorphic extension F of f to Ω such that Z
Ω
|F |2e−ϕdλ ≤ C Z
Ω0
|f |2e−ϕdλ0, where C depends only on n and the diameter of Ω.
Siu / Berndtsson (1996): If Ω ⊂ Cn−1× {|zn< 1} and H = {zn= 0}
then C = 4π.
Problem.Can we improve to C = π?
B.-Y. Chen (2011): Ohsawa-Takegoshi extension theorem can be proved using directly H¨ormander’s estimate for ¯∂-equation!
Mahler Conjecture
K - convex symmetric body in Rn
K0:= {y ∈ Rn: x · y ≤ 1 for every x ∈ K}
Mahler volume := λ(K)λ(K0)
Santaló Inequality (1949): Mahler volume ismaximizedby balls.
Mahler Conjecture (1938):Mahler volume isminimizedby cubes.
True for n = 2:
@@
Bourgain-Milman (1987): There exists c > 0 such that λ(K)λ(K0) ≥ cn4n
n!. Mahler Conjecture: c = 1
G. Kuperberg (2006):c = π/4
Equivalent SCV formulation (Nazarov, 2012) For u ∈ L2(K0) we have
|u(0)|b 2= Z
K0
u dλ
2
≤ λ(K0)||u||2L2(K0)= (2π)−nλ(K0)||bu||2L2(Rn)
with equality for u = χK0. Therefore λ(K0) = (2π)nsup
f ∈P
|f (0)|2
||f ||2
L2(Rn)
,
where P = {u : u ∈ Lb 2(K0)} ⊂ O(Cn). By Paley-Wiener thm the Mahler Conjecture is equivalent to the following SCV problem: find f ∈ O(Cn) with exponential growth (|f (z)| ≤ CeC|z|) s.th. f (0) = 1,
|f (iy)| ≤ CeqK(y), (qK is Minkowski function for K),
and Z
Rn
|f (x)|2dλ(x) ≤ n!
π 2
n
λ(K).
Nazarov: One can show the Bourgain-Milman inequality with c = (π/4)3 using H¨ormander’s estimate.
H¨ormander’s Estimate (1965)
Ω - pseudoconvex in Cn, ϕ - smooth, strongly psh in Ω α =P
jαjd¯zj∈ L2loc,(0,1)(Ω), ¯∂α = 0
Then one can find u ∈ L2loc(Ω) with ¯∂u = α and Z
Ω
|u|2e−ϕdλ ≤ Z
Ω
|α|2
i∂ ¯∂ϕe−ϕdλ.
Here |α|2
i∂ ¯∂ϕ=P
j,kϕj ¯kα¯jαk, where (ϕj ¯k) = (∂2ϕ/∂zj∂ ¯zk)−1is the length of α w.r.t. the K¨ahler metric i∂ ¯∂ϕ.
The estimate also makes sense for non-smooth ϕ: instead of |α|2
i∂ ¯∂ϕone has to take any nonnegative H ∈ L∞loc(Ω) with
i ¯α ∧ α ≤ H i∂ ¯∂ϕ (B., 2005).
Donnelly-Fefferman (1982) Ω, α, ϕ as before ψ psh in Ω s.th. | ¯∂ψ|2
i∂ ¯∂ψ≤ 1 (that is i∂ψ ∧ ¯∂ψ ≤ i∂ ¯∂ψ) Then one can find u ∈ L2loc(Ω) with ¯∂u = α and
Z
Ω
|u|2e−ϕdλ ≤ C Z
Ω
|α|2i∂ ¯∂ψe−ϕdλ, where C is an absolute constant.
Berndtsson (1996) Ω, α, ϕ, ψ as before
Then, if 0 ≤ δ < 1, one can find u ∈ L2loc(Ω) with ¯∂u = α and Z
Ω
|u|2eδψ−ϕdλ ≤ 4 (1 − δ)2
Z
Ω
|α|2
i∂ ¯∂ψeδψ−ϕdλ.
The above constant was obtained in B. 2004 and is optimal (B. 2012).
Therefore C = 4 is optimal in Donnelly-Fefferman.
Berndtsson’s estimate is not enough to obtain Ohsawa-Takegoshi (it would be if it were true for δ = 1).
Berndtsson’s Estimate Ω - pseudoconvex α ∈ L2loc,(0,1)(Ω), ¯∂α = 0 ϕ, ψ - psh, | ¯∂ψ|2i∂ ¯∂ψ≤ 1
Then, if 0 ≤ δ < 1, one can find u ∈ L2loc(Ω) with ¯∂u = α and Z
Ω
|u|2eδψ−ϕdλ ≤ 4 (1 − δ)2
Z
Ω
|α|2
i∂ ¯∂ψeδψ−ϕdλ.
Theorem. Ω, α, ϕ, ψ as above
Assume in addition that | ¯∂ψ|2i∂ ¯∂ψ≤ δ < 1 on supp α.
Then there exists u ∈ L2loc(Ω) solving ¯∂u = α with Z
Ω
|u|2(1 − | ¯∂ψ|2i∂ ¯∂ψ)eψ−ϕdλ ≤ 1 (1 −√
δ)2 Z
Ω
|α|2i∂ ¯∂ψeψ−ϕdλ.
From this estimate one can obtain Ohsawa-Takegoshi and Suita with C = 1.95388 . . . (obtained earlier by Guan-Zhou-Zhu).
Theorem. Ω - pseudoconvex in Cn, ϕ - psh in Ω α ∈ L2loc,(0,1)(Ω), ¯∂α = 0
ψ ∈ Wloc1,2(Ω) locally bounded from above, s.th.
| ¯∂ψ|2i∂ ¯∂ϕ
(≤ 1 in Ω
≤ δ < 1 on supp α.
Then there exists u ∈ L2loc(Ω) with ¯∂u = α and Z
Ω
|u|2(1 − | ¯∂ψ|2i∂ ¯∂ϕ)e2ψ−ϕdλ ≤1 +√ δ 1 −√
δ Z
Ω
|α|2
i∂ ¯∂ϕe2ψ−ϕdλ.
Proof. (Some ideas going back to Berndtsson and B.-Y. Chen.) By approximation we may assume that ϕ is smooth up to the boundary and strongly psh, and ψ is bounded.
u - minimal solution to ¯∂u = α in L2(Ω, eψ−ϕ)
⇒ u ⊥ ker ¯∂ in L2(Ω, eψ−ϕ)
⇒ v := ueψ⊥ ker ¯∂ in L2(Ω, e−ϕ)
⇒ v - minimal solution to ¯∂v = β := eψ(α + u ¯∂ψ) in L2(Ω, e−ϕ) By H¨ormander’s estimate
Z
Ω
|v|2e−ϕdλ ≤ Z
Ω
|β|2
i∂ ¯∂ϕe−ϕdλ.
Therefore Z
Ω
|u|2e2ψ−ϕdλ ≤ Z
Ω
|α + u ¯∂ψ|2i∂ ¯∂ϕe2ψ−ϕdλ
≤ Z
Ω
|α|2i∂ ¯∂ϕ+ 2|u|
√
H|α|i∂ ¯∂ϕ+ |u|2H
e2ψ−ϕdλ, where H = | ¯∂ψ|2
i∂ ¯∂ϕ. For t > 0 we will get Z
Ω
|u|2(1 − H)e2ψ−ϕdλ
≤ Z
Ω
|α|2
i∂ ¯∂ϕ
1 + t−1 H 1 − H
+ t|u|2(1 − H)
e2ψ−ϕdλ
≤
1 + t−1 δ 1 − δ
Z
Ω
|α|2i∂ ¯∂ϕe2ψ−ϕdλ
+ t Z
Ω
|u|2(1 − H)e2ψ−ϕdλ.
We will obtain the required estimate if we take t := 1/(δ−1/2+ 1).
Theorem. Ω - pseudoconvex in Cn, ϕ - psh in Ω α ∈ L2loc,(0,1)(Ω), ¯∂α = 0
ψ ∈ Wloc1,2(Ω) locally bounded from above, s.th.
| ¯∂ψ|2i∂ ¯∂ϕ
(≤ 1 in Ω
≤ δ < 1 on supp α.
Then there exists u ∈ L2loc(Ω) with ¯∂u = α and Z
Ω
|u|2(1 − | ¯∂ψ|2i∂ ¯∂ϕ)e2ψ−ϕdλ ≤1 +√ δ 1 −√
δ Z
Ω
|α|2
i∂ ¯∂ϕe2ψ−ϕdλ.
Remarks.1. Setting ψ ≡ 0 we recover the H¨ormander estimate.
2. This theorem implies Donnelly-Fefferman and Berndtsson’s estimates with optimal constants: for psh ϕ, ψ with | ¯∂ψ|2
i∂ ¯∂ψ≤ 1 and δ < 1 set ϕ := ϕ + ψ and ee ψ =1+δ2 ψ.
Then 2 eψ −ϕ = δψ − ϕ and | ¯e ∂ eψ|2
i∂ ¯∂ϕe≤ (1+δ)4 2 =: eδ.
We will get Berndtsson’s estimate with the constant 1 +
p eδ (1 −p
eδ)(1 − eδ)
= 4
(1 − δ)2.
Theorem (Ohsawa-Takegoshi with optimal constant) Ω - pseudoconvex in Cn−1× D, where 0 ∈ D ⊂ C, ϕ - psh in Ω, f - holomorphic in Ω0:= Ω ∩ {zn= 0}
Then there exists a holomorphic extension F of f to Ω such that Z
Ω
|F |2e−ϕdλ ≤ π (cD(0))2
Z
Ω0
|f |2e−ϕdλ0.
(For n = 1 and ϕ ≡ 0 we obtain the Suita Conjecture.)
Sketch of proof. By approximation may assume that Ω is bounded, smooth, strongly pseudoconvex, ϕ is smooth up to the boundary, and f is holomorphic in a neighborhood of Ω0.
ε > 0
α := ¯∂ f (z0)χ(−2 log |zn|), where χ(t) = 0 for t ≤ −2 log ε and χ(∞) = 1.
G := GD(·, 0)
ϕ := ϕ + 2G + η(−2G)e ψ := γ(−2G)
F := f (z0)χ(−2 log |zn|) − u, where u is a solution of ¯∂u = α given by the previous thm.
Crucial ODE Problem
Find g ∈ C0,1(R+), h ∈ C1,1(R+) such that h0< 0, h00> 0,
t→∞lim(g(t) + log t) = lim
t→∞(h(t) + log t) = 0 and
1 −(g0)2
h00
e2g−h+t≥ 1.
Crucial ODE Problem
Find g ∈ C0,1(R+), h ∈ C1,1(R+) such that h0< 0, h00> 0,
t→∞lim(g(t) + log t) = lim
t→∞(h(t) + log t) = 0 and
1 −(g0)2
h00
e2g−h+t≥ 1.
Solution:
h(t) := − log(t + e−t− 1)
g(t) := − log(t + e−t− 1) + log(1 − e−t).
Another approach: general lower bound for the Bergman kernel
KΩ(w) = sup{|f (w)|2: f ∈ O(Ω), R
Ω|f |2dλ ≤ 1} (Bergman kernel) GΩ(·, w) = sup{v ∈ P SH−(Ω), lim
z→w(v(z) − log |z − w|) < ∞}
(pluricomplex Green function)
Theorem. Assume Ω is pseudoconvex in Cn. Then for a ≥ 0 and w ∈ Ω
KΩ(w) ≥ 1
e2naλ({GΩ(·, w) < −a}).
For n = 1 letting a → ∞ this gives the Suita Conjecture:
KΩ(w) ≥cΩ(w)2
π .
Theorem. Assume Ω is pseudoconvex in Cn. Then for a ≥ 0 and w ∈ Ω
KΩ(w) ≥ 1
e2naλ({GΩ(·, w) < −a}).
Proof.May assume that Ω is bounded, smooth and strongly pseudoconvex.
G := GΩ,w. Will use Donnelly-Fefferman with ϕ := 2nG, ψ := − log(−G),
α := ¯∂(χ ◦ G) = χ0◦ G ¯∂G, (χ will be determined later).
i ¯α ∧ α ≤ (χ0◦ G)2i∂G ◦ ¯∂G ≤ G2(χ0◦ G)2i∂ ¯∂ψ We will find u ∈ L2loc(Ω) with ¯∂u = α and
Z
Ω
|u|2dλ ≤ Z
Ω
|u|2e−ϕdλ ≤ C Z
Ω
G2(χ0◦ G)2e−2nGdλ.
With χ(t) :=
0 t ≥ −a,
Z −t a
e−ns
s ds t < −a, we thus get Z
Ω
|u|2dλ ≤ C λ({G < −a}).
f := χ ◦ G − u ∈ O(Ω) satisfies f (w) = χ(−∞) =
Z ∞ na
e−s
s ds = Ei(na) (because e−ϕis not integrable near w). Also
||f || ≤ ||χ ◦ G|| + ||u|| ≤ (χ(−∞) +√ C)p
λ({G < −a}).
Therefore
KΩ(w) ≥|f (w)|2
||f ||2 ≥ cn,a
λ({G < −a}), where
cn,a= Ei(na)2 (Ei(na) +√
C)2.
Tensor power trick. Ω := Ωe m⊂ Cnm,w := (w, . . . , w), m 0e KΩe(w) = (Ke Ω(w))m, λ2nm({G
Ω,ewe< −a}) = (λ2n({G < −a})m. (KΩ(w))m≥ cnm,a
(λ2n({G < −a}))m but
m→∞lim c1/mnm,a= e−2na.
Application to the Bourgain-Milman Inequality K - convex symmetric body in Rn
Nazarov: consider the tube domain TK:= intK + iRn⊂ Cn. Then
(1) π
4
2n 1
(λn(K))2 ≤ KTK(0) ≤ n!
πn λn(K0)
λn(K). Therefore
λn(K)λn(K0) ≥π 4
3n4n n!.
To show the lower bound in (1) we can use the previous estimate:
KΩ(w) ≥ 1
e2naλ2n({GΩ(·, w) < −a}), w ∈ Ω, a ≥ 0.
By Lempert’s theorem we will get as a → ∞
Theorem. If Ω is a convex domain in Cnthen for w ∈ Ω KΩ(w) ≥ 1
λ2n(IΩ(w)),
where IΩ(w) = {ϕ0(0) : ϕ ∈ O(∆, Ω), ϕ(0) = w} (Kobayashi indicatrix).
Proposition (Nazarov). ITK(0) ⊂ 4
π(K + iK) Sketch of proof. For y ∈ K0consider
F (z) = Φ(z · t) ∈ O(Ω, ∆),
where Φ : {|Re ζ| < 1} → ∆ is conformal with Φ(0) = 0. By the Schwarz lemma we will get
ITK(0) ⊂4
π{z ∈ Cn: |z · y| ≤ 1 for every y ∈ K0}.
Corollary. λ2n(ITK(0)) ≤ 4 π
2n
(λn(K))2
Conjecture. λ2n(ITK(0)) ≤ 4 π
n
(λn(K))2 KTK(0) ≥π
4
n 1
(λn(K))2. (equality for cubes)
Lempert (1981)
Ω - bounded strongly convex domain in Cnwith smooth boundary ϕ ∈ O(∆, Ω) ∩ C( ¯∆, ¯Ω) is a geodesic if and only if ϕ(∂∆) ⊂ ∂Ω and there exists h ∈ O(∆, Cn) ∩ C( ¯∆, Cn) s.th. the vector eith(eit) is outer normal to ∂Ω at ϕ(eit) for every t ∈ R.
There exists F ∈ O(Ω, ∆), a left-inverse to ϕ (i.e. F ◦ ϕ = id∆) s.th.
(z − ϕ(F (z))) · h(F (z)) = 0, z ∈ Ω.
Lempert’s Theory for Tube Domains (S. Zaja¸c)
Ω = TK = intK + iRn, where K is smooth and strongly convex in Rn Since Im (eith(eit)) = 0, h must be of the form
h(ζ) = ¯w + ζb + ζ2w for some w ∈ Cnand b ∈ Rn. Therefore
Re ϕ(eit) = ν−1
b + 2Re (eitw)
|b + 2Re (eitw)|
, where ν : ∂K → Sn−1is the Gauss map.
By the Schwarz formula ϕ(ζ) = 1
2π Z 2π
0
eit+ ζ eit− ζν−1
b + 2Re (eitw)
|b + 2Re (eitw)|
dt + iIm ϕ(0).
If K is in addition symmetric then all geodesics in TK with ϕ(0) = 0 are of the form
ϕ(ζ) = 1 2π
Z 2π 0
eit+ ζ eit− ζν−1
Re (eitw)
|Re (eitw)|
dt for some w ∈ (Cn)∗. Then
ϕ0(0) = 1 π
Z 2π 0
eitν−1
Re (eitw)¯
|Re (eitw)|¯
dt parametrizes ∂ITK(0) for w ∈ S2n−1.
Conjectureλ2n(ITK(0)) ≤ 4 π
n
(λn(K))2
