Math 650-600: Several Complex Variables
Harold P. Boas
boas@tamu.edu
Recap from last time
We reduced the solution of the ∂-equation on (0, 1)-forms to the following.
Key estimate. If Ω is a pseudoconvex domain in C n , then there exist smooth weight functions
∗
ϕ1 , ϕ2 , and ϕ3 such that when f = ∑nj=1 f j dz j is in the intersection of the domains of ∂ and ∂ ,
then
∗
k f k ϕ2 ≤ k∂ f k ϕ1 + k∂ f k ϕ3 .
To prove the key estimate, we need (a) to compute the terms on the right-hand side and (b) to
choose suitable weight functions.
Math 650-600
April 26, 2005 — slide #2
Computation of the norm of ∂ f
A clever technical point: if the weight functions ϕ 1 , ϕ2 , and ϕ3 are chosen to grow suitably
∗
fast at bΩ, then the C ∞ functions with compact support in Ω will be dense in Dom ∂ ∩ Dom ∂ .
Assuming this density for now, we may compute in C0∞ (Ω).
¯
¯
´
³ ∂f
¯ ∂ f j ∂ f k ¯2
∂ fk
j
n
2
If f = ∑ j=1 f j dz j , then ∂ f = ∑ j<k ∂z − ∂z dzk ∧ dz j , so |∂ f | = ∑ j<k ¯ ∂z − ∂z ¯ =
j
j
k
¶ k
µ¯ ¯
¯2
¯
2
∂ f j ∂ fk
¯ ∂ fj ¯
¯ ∂ f j ∂ fk ¯
1
2 ∑ j,k ¯ ∂zk − ∂z j ¯ = ∑ j,k ¯ ∂zk ¯ − ∂zk ∂z j . Therefore
k∂ f k2ϕ3
¯
¶
Z µ¯
¯ ∂ f j ¯2 ∂ f j ∂ f k
−ϕ3
¯
¯
=∑
¯ ∂z ¯ − ∂z ∂z e
j
k
k
Ω
j,k
and we need to handle the term with the minus sign.
Math 650-600
April 26, 2005 — slide #3
Computation of ∂
∗
The choice of weight functions (to be made later) guarantees that C0∞ (Ω) is dense in Dom ∂,
∗
so to compute ∂ we may pair with a smooth, compactly supported function and integrate by
parts without boundary terms.
n
∗
n
∂g
h∂ f , gi ϕ1 = h f , ∂gi ϕ2 = ∑ h f j , ∂z i ϕ2 = − ∑ he ϕ1 ∂z∂ (e−ϕ2 f j ), gi ϕ1 .
j
j
n
j=1
j=1
n ¡
∗
So ∂ f = − ∑ e ϕ1 ∂z∂ (e−ϕ2 f j ). Set δj f j := e ϕ3 ∂z∂ (e−ϕ3 f j ) and rewrite ∂ f = −e ϕ1 −ϕ2 ∑ δj f j −
j
j
j=1
j=1
¢
∂
f j ∂z (ϕ2 − ϕ3 ) .
∗
j
We do not need three independent weight functions. It is convenient to use two
independent
¶
n µ
∂ψ
∗
−ψ
functions ϕ := ϕ3 and ψ satisfying ϕ3 − ϕ2 = ψ = ϕ2 − ϕ1 . Then ∂ f = −e ∑ δj f j + f j
.
∂z
j
j=1
Math 650-600
April 26, 2005 — slide #4
Approaching the key estimate
∗
By the triangle inequality, ke −ψ ∑nj=1 δj f j k ϕ1 ≤ k∂ f k ϕ1 + ke−ψ ∑nj=1 f j ∂z k ϕ1 .
j
Since 2ψ + ϕ1 = ϕ3 = ϕ, squaring both sides gives
Z
Z n
¯
¯
¯ ∑ δj f j ¯2 e−ϕ ≤ 2k∂∗ f k2ϕ + 2
| f |2 |∂ψ|2 e−ϕ .
1
∂ψ
Ω
Ω j=1
Adding the previously computed term k∂ f k 2ϕ3 yields
∑
j,k
Z Ã
Ω
∂ f j ∂ fk
(δj f j )(δk f k ) −
∂zk ∂z j
!
e
−ϕ
≤
k∂ f k2ϕ3
+ 2k∂
∗
f k2ϕ1
+2
Z
| f |2 |∂ψ|2 e−ϕ .
Ω
Because δj is adjoint to −∂/∂z j on C0∞ (Ω), the left-hand side simplifies after two integrations by
parts.
Math 650-600
April 26, 2005 — slide #5
The punch line
The commutator δj ∂z∂ −
k
Z
∂
∂zk δj
=
∂2 ϕ
∂z j ∂zk .
Therefore
∂2 ϕ
∗
f j f k e−ϕ ≤ k∂ f k2ϕ3 + 2k∂ f k2ϕ1 + 2
∑
Ω j,k ∂z j ∂zk
Z
| f |2 |∂ψ|2 e−ϕ .
Ω
If we can choose ϕ sufficiently strongly plurisubharmonic that for every vector w in C n we have
∂2 ϕ
∑ ∂z j ∂zk w j wk ≥ 2|w|2 |∂ψ|2 + 2|w|2 eψ
j,k
then we will have the key estimate, and hence the solution of the ∂-equation on pseudoconvex
domains.
Math 650-600
April 26, 2005 — slide #6