Bergman Kernel in Convex Domains

Bergman Kernel in Convex Domains

Zbigniew B locki

Uniwersytet Jagiello´ nski, Krak´ ow, Poland http://gamma.im.uj.edu.pl/ eblocki

Several Complex Variables Symposium Tsinghua Sanya International Mathematics Forum

May 12–16, 2014

Ω ⊂ C

, w ∈ Ω

K

(w ) = sup{|f (w )|

: f ∈ O(Ω), Z

Ω

|f |

d λ ≤ 1}

(Bergman kernel on the diagonal)

G

(z) = G

(z, w )

= sup{u(z) : u ∈ PSH

(Ω) : lim

z→w

u(z) − log |z − w |
< ∞}

(pluricomplex Green function)

Theorem 0 Assume Ω is pseudoconvex in C

. Then for w ∈ Ω and t ≤ 0

K

(w ) ≥ 1

e

λ({G

< t}) .

Optimal constant: “=” if Ω = B(w , r ).

Proof 1 Using Donnelly-Fefferman’s estimate for ¯ ∂ one can prove

K

(w ) ≥ 1

c(n, t)λ({G

< t}) , (1) where

c(n, t) =



1 + C

Ei (−nt)



, Ei (a) = Z

a

ds se

(B. 2005). Now use the tensor power trick: e Ω = Ω × · · · × Ω ⊂ C

, w = (w , . . . , w ) for m  0. Then e

K

( w ) = (K e

(w ))

, λ({G

we

< t}) = (λ({G

< t}))

, and by (1) for e Ω

K

(w ) ≥ 1

c(nm, t)

λ({G

< t}) . But lim

m→∞

c(nm, t)

= e

.

Proof 2 (Lempert) By Berndtsson’s result on log-(pluri)subharmonicity of the Bergman kernel for sections of a pseudoconvex domain it follows that log K

(w ) is convex for t ∈ (−∞, 0]. Therefore

t 7−→ 2nt + log K

(w )

is convex and bounded, hence non-decreasing. It follows that K

(w ) ≥ e

K

(w ) ≥ e

λ({G

< t}) .

Berndtsson: This method can be improved to show the

Ohsawa-Takegoshi extension theorem with optimal constant.

Theorem 0 Assume Ω is pseudoconvex in C

. Then for w ∈ Ω and t ≤ 0

K

(w ) ≥ 1

e

λ({G

< t}) .

What happens when t → −∞? For n = 1 Theorem 0 immediately gives:

Theorem (Suita conjecture) For a domain Ω ⊂ C one has

K

(w ) ≥ c

(w )

/π, w ∈ Ω, (2) where c

(w ) = exp lim

(G

(z, w ) − log |z − w |)

(logarithmic capacity of C \ Ω w.r.t. w ).

Theorem (Guan-Zhou) Equality holds in (2) iff Ω ' ∆ \ F , where ∆ is

the unit disk and F a closed polar subset.

-10 -8 -6 -4 -2

1 2 3 4 5 6 7

πK

c

for Ω = {e

< |z| < 1} as a function of 2 log |w |

What happens with e

λ({G

< t}) as t → −∞ for arbitrary n? For convex Ω using Lempert’s theory one can get

Proposition If Ω is bounded, smooth and strongly convex in C

then for w ∈ Ω

t→−∞

lim e

λ({G

< t}) = λ(I

(w )),

where I

(w ) = {ϕ

(0) : ϕ ∈ O(∆, Ω), ϕ(0) = w } (Kobayashi indicatrix).

Corollary If Ω ⊂ C

is convex then K

(w ) ≥ 1

λ(I

(w )) , w ∈ Ω.

For general Ω one can prove

Theorem (B.-Zwonek) If Ω is bounded and hyperconvex in C

and w ∈ Ω then

t→−∞

lim e

λ({G

< t}) = λ(I

(w )), where I

(w ) = {X ∈ C

: lim

G

(w + ζX ) − log |ζ|
≤ 0}

(Azukawa indicatrix)

Corollary (SCV version of the Suita conjecture) If Ω ⊂ C

is pseudoconvex and w ∈ Ω then

K

(w ) ≥ 1 λ(I

(w )) .

Conjecture 1 For Ω pseudoconvex and w ∈ Ω the function t 7−→ e

λ({G

< t})

is non-decreasing in t.

It would easily follow from the following:

Conjecture 2 For Ω pseudoconvex and w ∈ Ω the function t 7−→ log λ({G

< t})

is convex on (−∞, 0].

Theorem (B.-Zwonek) Conjecture 1 is true for n = 1.

Proof It is be enough to prove that f

(t) ≥ 0 where f (t) := log λ({G

< t}) − 2t and t is a regular value of G

. By the co-area formula

λ({G

< t}) = Z

−∞

Z

{Gw=s}

d σ

|∇G

| ds and therefore

f

(t) = Z

{Gw=t}

d σ

|∇G

| λ({G

< t}) − 2.

By the Schwarz inequality Z

{Gw=t}

d σ

|∇G

| ≥ (σ({G

= t}))

Z

{Gw=t}

|∇G

|d σ

= (σ({G

= t}))

2π .

The isoperimetric inequality gives

(σ({G

= t}))

≥ 4πλ({G

< t}) and we obtain f

(t) ≥ 0.

Conjecture 1 for arbitrary n is equivalent to the following pluricomplex isoperimetric inequality for smooth strongly pseudoconvex Ω

Z

∂Ω

d σ

|∇G

| ≥ 4nπλ(Ω).

Conjecture 1 also turns out to be closely related to the problem of

symmetrization of the complex Monge-Amp` ere equation.

What about corresponding upper bound in the Suita conjecture?

Not true in general:

Proposition (B.-Zwonek) Let Ω = {r < |z| < 1}. Then K

( √

r ) (c

( √

r ))

≥ −2 log r π

.

It would be interesting to find un upper bound of the Bergman kernel for domains in C in terms of logarithmic capacity which would in particular imply the ⇒ part in the well known equivalence

K

> 0 ⇔ c

> 0

(c

≤ πK

being a quantitative version of ⇐).

The upper bound for the Bergman kernel holds for convex domains:

Theorem (B.-Zwonek) For a convex Ω and w ∈ Ω set F

(w ) := K

(w )λ(I

(w ))

.

Then F

(w ) ≤ 4. If Ω is in addition symmetric w.r.t. w then F

(w ) ≤ 16/π

= 1.621 . . . .

Sketch of proof Denote I := int I

(w ) and assume that w = 0. One can show that I ⊂ 2 Ω (I ⊂ 4/π Ω if Ω is symmetric). Then

K

(0)λ(I ) ≤ K

(0)λ(I ) = λ(I )

λ(I /2) = 4

.

For convex domains F

is a biholomorphically invariant function satisfying 1 ≤ F

≤ 4. Can we find an example with F

(w ) > 1? Using Jarnicki-Pflug-Zeinstra’s formula for geodesics in convex complex ellipsoids (which is based on Lempert’s theory) one can show the following

Theorem (B.-Zwonek) Define

Ω = {z ∈ C

: |z

| + · · · + |z

| < 1}.

Then for w = (b, 0, . . . , 0), where 0 < b < 1, one has

K

(w )λ(I

(w )) = 1 + (1 − b)

(1 + b)

− (1 − b)

− 4nb 4nb(1 + b)

= 1 + (1 − b)

(1 + b)

n−1

X

j =1

1 2j + 1

2n − 1 2j



b

.

0.2 0.4 0.6 0.8 1.0 1.001

1.002 1.003 1.004

F

(b, 0, . . . , 0) in Ω = {|z

| + · · · + |z

| < 1} for n = 2, 3, . . . , 6.

Theorem (B.-Zwonek) For m ≥ 1/2 set Ω = {|z

|

+ |z

|

< 1} and w = (b, 0), 0 < b < 1. Then

K

(w )λ(I

(w )) = P m(1 − b

) + 1 + b

2(1 − b

)

(m − 2)m

(m + 1)(3m − 2)(3m − 1) , where

P =b

−m

+ 2m

+ m − 2
+ b

−27m

+ 54m

− 33m + 6
+ b

m

3m

+ 2m − 1
+ 6b

m

3m

− 5m

− 4m + 4

+ b

−36m

+ 81m

+ 10m

− 71m

+ 32m − 4
+ 2m

9m

− 27m

+ 20m − 4
.

In this domain all values of F

are attained for (b, 0), 0 < b < 1.

0.2 0.4 0.6 0.8 1.0 1.002

1.004 1.006 1.008 1.010

F

(b, 0) in Ω = {|z

|

+ |z

|

< 1} for m = 4, 8, 16, 32, 64, 128.

sup

0<b<1

F

(b, 0) → 1.010182 . . . as m → ∞

What is the highest value of F

for convex Ω?

What can be said the function w 7−→ − log λ(I

(w ))?

Is it plurisubharmonic?

It does not have to be C

:

Theorem (B.-Zwonek) If Ω = {|z

| + |z

| < 1} and 0 < b ≤ 1/4, λ(I

((b, b)))

= π

6 30b

− 64b

+ 80b

− 80b

+ 76b

− 16b

− 8b

+ 1
. λ(I

((b, b))) is not C

at b = 1/4.

It is known (Hahn-Pflug) that for 0 < b < 1/2:

K

((b, b)) = 2 8b

− 6b

+ 3

π

(1 − 4b

)

.

0.05 0.10 0.15 0.20 0.25 1.002

1.004 1.006 1.008

F

(b, b) in Ω = {|z

| + |z

| < 1} for 0 < b ≤ 1/4.

Thank you!