06.10.E1 · riemann-surfaces / several-variables

Several complex variables exercise pack (Krantz supplement)

shippedIntermediate-onlyLean: nonepending prereqs

Anchor (Master):

Formal definition of the pack Intermediate

This pack supplements Krantz's Function Theory of Several Complex Variables: the rigidity that distinguishes $C^{n}$ ( $n \geq 2$ ) from $C$ , expressed through the Hartogs extension phenomenon, domains of holomorphy and holomorphic convexity, plurisubharmonicity and the Levi form, the $\overset{ˉ}{\partial}$ -equation with Hörmander $L^{2}$ estimates, and the Bergman kernel. Its problems exercise the units on domains of holomorphy and holomorphic convexity 06.10.01, plurisubharmonic functions 06.10.02, pseudoconvexity and the Levi form 06.10.03, the $\overset{ˉ}{\partial}$ -equation and Hörmander $L^{2}$ estimates 06.10.04, the Bergman kernel and metric 06.10.08, and the Hartogs phenomenon 06.07.02.

The pack collects ten problems — two easy, five medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units rather than as a standalone development. The problems are grouped by theme: Hartogs extension and the failure of one-variable intuition (easy/medium), plurisubharmonicity and Levi-form computations on standard domains (medium), the $\overset{ˉ}{\partial}$ - $L^{2}$ existence theorem and its corollaries (hard), and the Bergman kernel's reproducing and transformation properties (medium/hard). Conventions follow Krantz: a domain $Ω \subset C^{n}$ is Levi pseudoconvex at a boundary point if the Levi form $L_{ρ} (p; w) = \sum_{j, k} \frac{\partial ^{2} ρ}{\partial z _{j} \partial z ˉ _{k}} (p) w_{j} \overset{w}{ˉ}_{k}$ of a defining function $ρ$ is $\geq 0$ on the complex tangent space ${w : \sum_{j} \frac{\partial ρ}{\partial z _{j}} (p) w_{j} = 0}$ ; the Bergman space is $A^{2} (Ω) = O (Ω) \cap L^{2} (Ω)$ .

Key theorem with full solution Intermediate

Before the pack proper, we work one problem in full as an exemplar of the format. The remaining nine follow the same structure (problem, hint, full answer in <details> blocks).

Lead problem. Prove the Hartogs extension phenomenon on the Hartogs figure: every $f$ holomorphic on the figure $H = {(z, w) \in Δ^{2} : ∣ z ∣ > 1 - ε} \cup {∣ w ∣ < ε}$ extends holomorphically to the full bidisc $Δ^{2}$ .

Solution. Write $Δ = {∣ z ∣ < 1}$ and fix the Hartogs figure $H \subset Δ^{2}$ as stated: it contains a neighbourhood of the boundary annulus in the first variable together with a thin slab near $w = 0$ . For each fixed $z \in Δ$ , consider the circle $∣ w ∣ = r$ with $1 - ε <$ chosen so that the slice ${z} \times {∣ w ∣ = r}$ lies in $H$ for some admissible $r$ close to $1$ ; concretely, for every $z \in Δ$ the circle ${z} \times {∣ w ∣ = 1 - ε /2}$ lies in $H$ because $∣ z ∣ < 1$ is covered through the $∣ w ∣ < ε$ slab continued by the $∣ z ∣ > 1 - ε$ shell. Define

F (z, w) = \frac{1}{2 π i} \oint_{∣ ζ ∣ = 1 - ε /2} \frac{f ( z , ζ )}{ζ - w} d ζ, ∣ w ∣ < 1 - ε /2.

For each fixed $z$ , $w \mapsto f (z, w)$ is holomorphic on the disc $∣ w ∣ < 1 - ε /2$ where the slice is interior to $H$ , so by the one-variable Cauchy integral formula $F (z, w) = f (z, w)$ there. The integral $F$ is holomorphic in $(z, w)$ jointly: it is holomorphic in $w$ (parameter integral of a Cauchy kernel) and holomorphic in $z$ (differentiate under the integral; $f (z, ζ)$ is holomorphic in $z$ and the contour is fixed). By Osgood's lemma, separate holomorphy plus local boundedness gives joint holomorphy, so $F \in O (Δ^{2})$ .

Finally $F = f$ on the part of $H$ where the Cauchy formula applies, and both are holomorphic, so by the identity theorem $F = f$ on all of $H$ . Thus $F$ is the required extension to $Δ^{2}$ . $□$

This is the engine of the Hartogs phenomenon: holomorphic functions of $n \geq 2$ variables cannot have isolated or compact singularities, and they extend across "holes" that one-variable functions need not. The Cauchy-integral-in-one-slice construction is the cheapest proof; the Bochner-Martinelli kernel gives the coordinate-free version. Krantz Ch. 2.

Exercises Intermediate

Exercise 1 (easy, proof). No isolated zeros, no isolated singularities in $C^{n}$ , $n \geq 2$ .

Show that a holomorphic function on a domain $Ω \subset C^{n}$ with $n \geq 2$ cannot have an isolated zero, and (Hartogs) cannot have an isolated singularity.

Hint

For the zero: if $f (p) = 0$ for isolated $p$ , restrict to a complex line through $p$ to get a one-variable function with an isolated zero — but the zero set of $f$ , where nonempty, has complex codimension at most $1$ , so real dimension $\geq 2 n - 2 \geq 2 > 0$ . For the singularity, use Hartogs extension.

Answer

Isolated zeros. Suppose $f \in O (Ω)$ , $f \neq \equiv 0$ , and $f (p) = 0$ . The zero set $V = f^{- 1} (0)$ is a complex-analytic hypersurface near $p$ (where $f$ is not identically zero), of complex dimension $n - 1$ , hence real dimension $2 (n - 1) = 2 n - 2$ . For $n \geq 2$ this is $\geq 2 > 0$ , so $V$ is positive-dimensional near $p$ and $p$ is not isolated in $V$ . Concretely, by the Weierstrass preparation theorem one may, after a linear change, write $f$ near $p$ as a Weierstrass polynomial in one variable with coefficients holomorphic in the others; its zero set is a branched cover over a polydisc in the remaining $n - 1$ variables — never a single point.

Isolated singularities. If $f$ is holomorphic on $Ω ∖ {p}$ with $p$ isolated, surround $p$ by a Hartogs figure (a small punctured polydisc contains one). By the Hartogs extension theorem (the lead problem), $f$ extends holomorphically across $p$ . So an isolated singularity is removable; equivalently, holomorphic functions of $n \geq 2$ variables have no isolated singularities.

This is the first and most striking rigidity of higher dimensions: the Riemann-sphere picture of isolated poles has no several-variable analogue. Krantz Ch. 2.

Exercise 2 (easy, numeric). The Bergman kernel of the unit disc.

Compute the Bergman kernel $K_{Δ} (z, w)$ of the unit disc $Δ \subset C$ using the orthonormal basis $e_{n} (z) = \frac{n + 1}{π} z^{n}$ of $A^{2} (Δ)$ .

Hint

$K (z, w) = \sum_{n} e_{n} (z) \overline{e_{n} (w)}$ . Sum the resulting geometric-type series $\sum_{n} (n + 1) (z \overset{w}{ˉ})^{n}$ .

Answer

The monomials $z^{n}$ are orthogonal in $A^{2} (Δ)$ , with $∥ z^{n} ∥^{2} = \int_{Δ} ∣ z ∣^{2 n} d A = \int_{0}^{1} \int_{0}^{2 π} r^{2 n} r d θ d r = 2 π \cdot \frac{1}{2 n + 2} = \frac{π}{n + 1}$ . So $e_{n} (z) = \frac{n + 1}{π} z^{n}$ is orthonormal.

The Bergman kernel is

K_{Δ} (z, w) = n = 0 \sum \infty e_{n} (z) \overline{e_{n} (w)} = n = 0 \sum \infty \frac{n + 1}{π} (z \overset{w}{ˉ})^{n} = \frac{1}{π} n = 0 \sum \infty (n + 1) t^{n}, t = z \overset{w}{ˉ} .

Using $\sum_{n = 0}^{\infty} (n + 1) t^{n} = \frac{1}{( 1 - t ) ^{2}}$ for $∣ t ∣ < 1$ ,

K_{Δ} (z, w) = \frac{1}{π} \cdot \frac{1}{( 1 - z w ˉ ) ^{2}} .

So $K_{Δ} (z, w) = \frac{1}{π ( 1 - z w ˉ ) ^{2}}$ . On the diagonal, $K_{Δ} (z, z) = \frac{1}{π ( 1 - ∣ z ∣ ^{2} ) ^{2}} \to \infty$ as $z \to \partial Δ$ — the kernel blows up at the boundary, the source of the Bergman metric's completeness. Krantz, Bergman-kernel chapter.

Exercise 3 (medium, symbolic). Levi form of the ball.

Compute the Levi form of the unit ball $B = {z \in C^{n} : ∣ z ∣^{2} < 1}$ with defining function $ρ (z) = ∣ z ∣^{2} - 1$ , and verify the ball is strongly pseudoconvex.

Hint

$\frac{\partial ^{2} ρ}{\partial z _{j} \partial z ˉ _{k}} = δ_{j k}$ , so the Levi form is $\sum_{j} ∣ w_{j} ∣^{2} = ∣ w ∣^{2}$ on the complex tangent space. It is strictly positive for $w \neq = 0$ .

Answer

With $ρ (z) = ∣ z ∣^{2} - 1 = \sum_{j} z_{j} \overset{z}{ˉ}_{j} - 1$ , the complex Hessian is

\frac{\partial ^{2} ρ}{\partial z _{j} \partial z ˉ _{k}} = \frac{\partial}{\partial z _{j}} \overset{z}{ˉ}_{k} = δ_{j k} .

So the Levi form at a boundary point $p$ (with $∣ p ∣ = 1$ ) is

L_{ρ} (p; w) = j, k \sum δ_{j k} w_{j} \overset{w}{ˉ}_{k} = j \sum ∣ w_{j} ∣^{2} = ∣ w ∣^{2} .

On the complex tangent space $T_{p}^{1, 0} \partial B = {w : \sum_{j} \frac{\partial ρ}{\partial z _{j}} (p) w_{j} = 0} = {w : \sum_{j} \overset{p}{ˉ}_{j} w_{j} = 0}$ , the Levi form is still $∣ w ∣^{2}$ , which is strictly positive for every $w \neq = 0$ in that subspace.

A strictly positive Levi form on the complex tangent space (for all nonzero $w$ ) is exactly the definition of strong (strict) pseudoconvexity. So the ball is strongly pseudoconvex at every boundary point. The ball and the polydisc are the two running examples in Krantz; the ball is strongly pseudoconvex everywhere, while the polydisc is (weakly) pseudoconvex but not strongly so on its distinguished boundary edges. Krantz Ch. 3.

Exercise 4 (medium, proof). $lo g ∣ f ∣$ and $∣ f ∣^{2}$ are plurisubharmonic.

Let $f$ be holomorphic on $Ω \subset C^{n}$ . Show that $lo g ∣ f ∣$ and $∣ f ∣^{2}$ are plurisubharmonic.

Hint

A function is plurisubharmonic if its restriction to every complex line is subharmonic. On a line, $f$ restricts to a one-variable holomorphic function, and $lo g ∣ g ∣$ , $∣ g ∣^{2}$ are subharmonic for one-variable holomorphic $g$ .

Answer

A function $u$ is plurisubharmonic (PSH) iff for every complex line $ℓ : ζ \mapsto a + ζ b$ ( $a \in Ω$ , $b \in C^{n}$ ), the restriction $ζ \mapsto u (a + ζ b)$ is subharmonic on its domain in $C$ .

Restrict $f$ to such a line: $g (ζ) = f (a + ζ b)$ is holomorphic in the single variable $ζ$ .

$lo g ∣ f ∣$ . On the line, $lo g ∣ f ∣ = lo g ∣ g ∣$ . For one-variable holomorphic $g$ , $lo g ∣ g ∣$ is subharmonic: away from zeros of $g$ it is harmonic (real part of the holomorphic $lo g g$ , locally), and at zeros $lo g ∣ g ∣ \to - \infty$ , where subharmonicity is the sub-mean-value inequality $lo g ∣ g (ζ_{0}) ∣ = - \infty \leq$ average — automatic. So $lo g ∣ g ∣$ is subharmonic, hence $lo g ∣ f ∣$ is PSH.

$∣ f ∣^{2}$ . On the line, $∣ f ∣^{2} = ∣ g ∣^{2}$ . Compute the Laplacian: $Δ∣ g ∣^{2} = 4 \partial_{ζ} \partial_{\overset{ˉ}{ζ}} (g \overset{g}{ˉ}) = 4∣ g^{'} ∣^{2} \geq 0$ (since $g$ is holomorphic, $\partial_{\overset{ˉ}{ζ}} g = 0$ and $\partial_{ζ} \overset{g}{ˉ} = 0$ ). A function with $Δ \geq 0$ is subharmonic, so $∣ g ∣^{2}$ is subharmonic, hence $∣ f ∣^{2}$ is PSH.

Plurisubharmonicity of $lo g ∣ f ∣$ is the cornerstone: it makes $- lo g d (\cdot, \partial Ω)$ -type exhaustion functions plurisubharmonic on pseudoconvex domains and drives the Cartan-Thullen and $\overset{ˉ}{\partial}$ - $L^{2}$ machinery. Krantz Ch. 3.

Exercise 5 (medium, proof). Holomorphic convexity of a domain of holomorphy.

Show that if $Ω \subset C^{n}$ is a domain of holomorphy, then it is holomorphically convex: for every compact $K \subset Ω$ , its holomorphic hull $K_{Ω} = {z \in Ω : ∣ f (z) ∣ \leq sup_{K} ∣ f ∣ \forall f \in O (Ω)}$ is compact.

Hint

$K_{Ω}$ is closed in $Ω$ and bounded; the issue is that it might approach $\partial Ω$ . Use that on a domain of holomorphy there is, for each boundary point, a holomorphic function that is unbounded there — so the hull stays away from the boundary.

Answer

$K_{Ω}$ is bounded (it lies in the bounded set ${∣ z_{j} ∣ \leq sup_{K} ∣ z_{j} ∣}$ , taking $f = z_{j}$ ) and relatively closed in $Ω$ (intersection of closed sets ${∣ f ∣ \leq sup_{K} ∣ f ∣}$ ). Compactness in $Ω$ fails only if $K_{Ω}$ accumulates at $\partial Ω$ , i.e. $dist (K_{Ω}, \partial Ω) = 0$ .

Suppose $Ω$ is a domain of holomorphy. The Cartan-Thullen theorem characterises this by: $dist (K_{Ω}, \partial Ω) = dist (K, \partial Ω)$ for every compact $K$ (the hull does not get closer to the boundary than $K$ already is). Granting that characterisation — or arguing directly — fix a sequence $z_{m} \in K_{Ω}$ with $z_{m} \to z_{*} \in \partial Ω$ . Because $Ω$ is a domain of holomorphy, there is $f \in O (Ω)$ that does not extend past $z_{*}$ , and one can arrange $∣ f ∣$ to be unbounded along a sequence approaching $z_{*}$ (functions singular exactly at the boundary exist on a domain of holomorphy). Then $sup_{K} ∣ f ∣ < \infty$ but $∣ f (z_{m}) ∣ \to \infty$ , contradicting $z_{m} \in K_{Ω}$ (which forces $∣ f (z_{m}) ∣ \leq sup_{K} ∣ f ∣$ ).

Hence $K_{Ω}$ stays a positive distance from $\partial Ω$ ; being bounded, closed in $Ω$ , and bounded away from $\partial Ω$ , it is compact. So $Ω$ is holomorphically convex. The converse (holomorphically convex ⟹ domain of holomorphy) together with this is the Cartan-Thullen theorem. Krantz Ch. 2.

Exercise 6 (medium, proof). Bergman kernel transformation under biholomorphism.

Let $Φ : Ω_{1} \to Ω_{2}$ be a biholomorphism with complex Jacobian $J_{Φ} = det Φ^{'}$ . Show the Bergman kernels transform as $K_{Ω_{1}} (z, w) = J_{Φ} (z) K_{Ω_{2}} (Φ z, Φ w) \overline{J_{Φ} (w)}$ .

Hint

The map $f \mapsto (f \circ Φ) \cdot J_{Φ}$ is a unitary isomorphism $A^{2} (Ω_{2}) \to A^{2} (Ω_{1})$ (change of variables: $∣ J_{Φ} ∣^{2}$ is the real Jacobian). A unitary intertwines the reproducing kernels.

Answer

Define $U : A^{2} (Ω_{2}) \to A^{2} (Ω_{1})$ by $(U g) (z) = g (Φ (z)) J_{Φ} (z)$ . This is unitary: for $g \in A^{2} (Ω_{2})$ , by the holomorphic change of variables $w = Φ (z)$ (whose real Jacobian determinant is $∣ J_{Φ} (z) ∣^{2}$ ),

∥ U g ∥_{A^{2} (Ω_{1})}^{2} = \int_{Ω_{1}} ∣ g (Φ z) ∣^{2} ∣ J_{Φ} (z) ∣^{2} d V (z) = \int_{Ω_{2}} ∣ g (w) ∣^{2} d V (w) = ∥ g ∥_{A^{2} (Ω_{2})}^{2},

and $U$ is onto with inverse built from $Φ^{- 1}$ . So $U$ is a unitary isomorphism of Bergman spaces.

Take an orthonormal basis ${e_{k}}$ of $A^{2} (Ω_{2})$ ; then ${U e_{k}}$ is an orthonormal basis of $A^{2} (Ω_{1})$ . The reproducing kernel is the sum $K (z, w) = \sum_{k} e_{k} (z) \overline{e_{k} (w)}$ over any orthonormal basis. Hence

K_{Ω_{1}} (z, w) = k \sum (U e_{k}) (z) \overline{(U e_{k}) (w)} = k \sum e_{k} (Φ z) J_{Φ} (z) \overline{e_{k} (Φ w) J_{Φ} (w)} .

Pulling out the Jacobian factors (independent of $k$ ),

K_{Ω_{1}} (z, w) = J_{Φ} (z) \overline{J_{Φ} (w)} k \sum e_{k} (Φ z) \overline{e_{k} (Φ w)} = J_{Φ} (z) K_{Ω_{2}} (Φ z, Φ w) \overline{J_{Φ} (w)} .

This transformation rule is what makes the Bergman kernel a biholomorphic invariant in the appropriate sense, and it is the reason the Bergman metric (built from $\partial \overset{ˉ}{\partial} lo g K (z, z)$ ) is biholomorphically invariant. Krantz, Bergman-kernel chapter.

Exercise 7 (hard, proof). From $\overset{ˉ}{\partial}$ -solvability to the Cousin I problem.

Assume that on a pseudoconvex $Ω \subset C^{n}$ every $\overset{ˉ}{\partial}$ -closed $(0, 1)$ -form is $\overset{ˉ}{\partial}$ -exact (Hörmander's theorem). Solve the additive Cousin I problem: given an open cover ${U_{i}}$ and holomorphic $f_{ij} \in O (U_{i} \cap U_{j})$ with $f_{ij} + f_{j k} = f_{ik}$ , find holomorphic $g_{i} \in O (U_{i})$ with $g_{i} - g_{j} = f_{ij}$ .

Hint

Use a smooth partition of unity ${φ_{k}}$ to build smooth solutions $h_{i} = \sum_{k} φ_{k} f_{ik}$ with $h_{i} - h_{j} = f_{ij}$ , then correct by solving $\overset{ˉ}{\partial} u = \overset{ˉ}{\partial} h_{i}$ (a globally defined $(0, 1)$ -form) to make the pieces holomorphic.

Answer

Smooth solution. Take a smooth partition of unity ${φ_{k}}$ subordinate to ${U_{k}}$ . Define $h_{i} = \sum_{k} φ_{k} f_{ik} \in C^{\infty} (U_{i})$ (each summand is extended by zero off the support of $φ_{k}$ , legitimate since $f_{ik}$ is defined on $U_{i} \cap U_{k} \supset supp φ_{k} \cap U_{i}$ ). On $U_{i} \cap U_{j}$ , using the cocycle relation $f_{ik} - f_{j k} = f_{ik} + f_{k j} = f_{ij}$ ,

h_{i} - h_{j} = k \sum φ_{k} (f_{ik} - f_{j k}) = k \sum φ_{k} f_{ij} = f_{ij},

since $\sum_{k} φ_{k} = 1$ . So ${h_{i}}$ solves Cousin I smoothly, but the $h_{i}$ are not holomorphic.

Holomorphic correction. Since $f_{ij}$ is holomorphic, $\overset{ˉ}{\partial} h_{i} = \overset{ˉ}{\partial} h_{j}$ on overlaps, so the $(0, 1)$ -forms $\overset{ˉ}{\partial} h_{i}$ glue to a single global form $ω \in Ω^{0, 1} (Ω)$ . It is $\overset{ˉ}{\partial}$ -closed: $\overset{ˉ}{\partial} ω = \overset{ˉ}{\partial} \overset{ˉ}{\partial} h_{i} = 0$ . By the assumed Hörmander solvability on the pseudoconvex $Ω$ , there is $u \in C^{\infty} (Ω)$ with $\overset{ˉ}{\partial} u = ω$ , i.e. $\overset{ˉ}{\partial} u = \overset{ˉ}{\partial} h_{i}$ on each $U_{i}$ .

Set $g_{i} = h_{i} - u$ . Then $\overset{ˉ}{\partial} g_{i} = \overset{ˉ}{\partial} h_{i} - \overset{ˉ}{\partial} u = 0$ , so $g_{i} \in O (U_{i})$ , and on overlaps

g_{i} - g_{j} = (h_{i} - u) - (h_{j} - u) = h_{i} - h_{j} = f_{ij} .

Thus ${g_{i}}$ is a holomorphic Cousin I solution. The same $\overset{ˉ}{\partial}$ -engine, applied to a multiplicative cocycle via $lo g$ , handles Cousin II (subject to a topological obstruction). This is the structural payoff of the $\overset{ˉ}{\partial}$ - $L^{2}$ theory: pseudoconvex ⟹ $\overset{ˉ}{\partial}$ -solvable ⟹ Cousin problems solvable. Krantz Ch. 4.

Exercise 8 (hard, symbolic). Hörmander's weighted estimate forces solvability — the model computation.

On a pseudoconvex domain with strictly plurisubharmonic weight $φ$ , the key inequality is $∥ f ∥_{φ}^{2} \leq C (∥ \overset{ˉ}{\partial}_{φ}^{*} f ∥^{2} + ∥ \overset{ˉ}{\partial} f ∥^{2})$ for $f$ in the domain of the adjoint. Explain, via the basic Hilbert-space duality argument, how this a priori estimate yields a solution $u$ of $\overset{ˉ}{\partial} u = g$ with the $L^{2}$ bound $∥ u ∥_{φ}^{2} \leq C ∥ g ∥_{φ}^{2}$ for $\overset{ˉ}{\partial}$ -closed $g$ .

Hint

Define the linear functional $L (\overset{ˉ}{\partial}_{φ}^{*} f) = ⟨ f, g ⟩_{φ}$ on the range of $\overset{ˉ}{\partial}_{φ}^{*}$ . The estimate bounds it: $∣ L (\overset{ˉ}{\partial}_{φ}^{*} f) ∣ \leq ∥ f ∥_{φ} ∥ g ∥_{φ} \leq C^{1/2} ∥ g ∥_{φ} ∥ \overset{ˉ}{\partial}_{φ}^{*} f ∥$ . Apply Riesz representation / Hahn-Banach.

Answer

Work in the weighted Hilbert spaces $L_{(0, q)}^{2} (Ω, φ)$ . Let $g$ be a $\overset{ˉ}{\partial}$ -closed $(0, 1)$ -form with $∥ g ∥_{φ} < \infty$ . We seek $u \in L^{2} (Ω, φ)$ with $\overset{ˉ}{\partial} u = g$ , characterised weakly by $⟨ u, \overset{ˉ}{\partial}_{φ}^{*} f ⟩_{φ} = ⟨ g, f ⟩_{φ}$ for all $f$ in the domain of $\overset{ˉ}{\partial}_{φ}^{*}$ .

Defining the functional. On $g \in ker \overset{ˉ}{\partial}$ , one splits any $f$ into a part in $ker \overset{ˉ}{\partial}$ and an orthogonal part; the estimate is needed precisely on $ker \overset{ˉ}{\partial}$ . For $f \in Dom (\overset{ˉ}{\partial}_{φ}^{*}) \cap ker \overset{ˉ}{\partial}$ , the a priori inequality reduces (since $\overset{ˉ}{\partial} f = 0$ ) to $∥ f ∥_{φ}^{2} \leq C ∥ \overset{ˉ}{\partial}_{φ}^{*} f ∥^{2}$ . Define

L (\overset{ˉ}{\partial}_{φ}^{*} f) := ⟨ f, g ⟩_{φ} .

This is well-defined on the range ${\overset{ˉ}{\partial}_{φ}^{*} f}$ : if $\overset{ˉ}{\partial}_{φ}^{*} f = 0$ then $∥ f ∥_{φ}^{2} \leq C \cdot 0 = 0$ , so $f = 0$ and $⟨ f, g ⟩ = 0$ .

Boundedness. By Cauchy-Schwarz and the estimate,

∣ L (\overset{ˉ}{\partial}_{φ}^{*} f) ∣ = ∣ ⟨ f, g ⟩_{φ} ∣ \leq ∥ f ∥_{φ} ∥ g ∥_{φ} \leq C^{1/2} ∥ \overset{ˉ}{\partial}_{φ}^{*} f ∥ ∥ g ∥_{φ} .

So $L$ is a bounded linear functional on the subspace ${\overset{ˉ}{\partial}_{φ}^{*} f} \subset L^{2} (Ω, φ)$ with norm $\leq C^{1/2} ∥ g ∥_{φ}$ .

Representation. By Hahn-Banach extend $L$ to all of $L^{2} (Ω, φ)$ preserving the norm, then by Riesz representation there is $u \in L^{2} (Ω, φ)$ with $L (ψ) = ⟨ ψ, u ⟩_{φ}$ and $∥ u ∥_{φ} \leq C^{1/2} ∥ g ∥_{φ}$ . Taking $ψ = \overset{ˉ}{\partial}_{φ}^{*} f$ ,

⟨ \overset{ˉ}{\partial}_{φ}^{*} f, u ⟩_{φ} = L (\overset{ˉ}{\partial}_{φ}^{*} f) = ⟨ f, g ⟩_{φ} \forall f,

which says exactly $\overset{ˉ}{\partial} u = g$ in the weak sense. Squaring the norm bound gives $∥ u ∥_{φ}^{2} \leq C ∥ g ∥_{φ}^{2}$ .

This duality-plus-a-priori-estimate is the entire architecture of Hörmander's $L^{2}$ method: the geometry (pseudoconvexity, strict plurisubharmonicity of $φ$ ) is spent establishing the inequality, and the functional analysis converts it into existence with bounds. Krantz Ch. 4; Hörmander Ch. 4.

Exercise 9 (hard, proof). The Levi problem direction: pseudoconvex ⟹ domain of holomorphy (outline via $\overset{ˉ}{\partial}$ ).

Outline how the $\overset{ˉ}{\partial}$ - $L^{2}$ existence theorem proves that a (smoothly bounded) pseudoconvex domain $Ω$ is a domain of holomorphy, by constructing, for each boundary point, a holomorphic function singular exactly there.

Hint

At a boundary point $p$ use the Levi polynomial to build a local holomorphic function $h$ singular at $p$ and nonvanishing on a neighbourhood intersected with $\overline{Ω} ∖ {p}$ . Globalise $1/ h$ by cutting off and correcting the resulting $\overset{ˉ}{\partial}$ error using Hörmander.

Answer

Fix $p \in \partial Ω$ with defining function $ρ$ . The Levi polynomial of $ρ$ at $p$ is

L_{p} (z) = j \sum \frac{\partial ρ}{\partial z _{j}} (p) (z_{j} - p_{j}) + \frac{1}{2} j, k \sum \frac{\partial ^{2} ρ}{\partial z _{j} \partial z _{k}} (p) (z_{j} - p_{j}) (z_{k} - p_{k}),

a holomorphic polynomial. A Taylor expansion of $ρ$ shows $2 Re L_{p} (z) = - ρ (z) + (Levi form) (z - p) + o (∣ z - p ∣^{2})$ . Where the domain is pseudoconvex (Levi form $\geq 0$ on the complex tangent space), this gives $Re L_{p} (z) \geq 0$ on $\overline{Ω}$ near $p$ with equality only at $p$ — so $L_{p}$ is nonvanishing on $(\overline{Ω} \cap U) ∖ {p}$ for a small neighbourhood $U$ , and $1/ L_{p}$ is holomorphic there and blows up at $p$ .

Globalisation. Let $χ$ be a cutoff supported in $U$ , $\equiv 1$ near $p$ . The function $χ / L_{p}$ is smooth on $Ω$ and holomorphic near $p$ , but $\overset{ˉ}{\partial} (χ / L_{p}) = (\overset{ˉ}{\partial} χ) / L_{p} =: α$ is a smooth $\overset{ˉ}{\partial}$ -closed $(0, 1)$ -form supported away from $p$ (where $\overset{ˉ}{\partial} χ$ lives), so $α$ is bounded — in particular in $L^{2} (Ω, φ)$ for a suitable plurisubharmonic weight. By Hörmander's theorem on the pseudoconvex $Ω$ , solve $\overset{ˉ}{\partial} u = α$ with $u \in L^{2}$ , and the weight can be chosen so that $u$ is forced to vanish to high order at $p$ (or at least to be bounded near $p$ ).

Set $F = χ / L_{p} - u$ . Then $\overset{ˉ}{\partial} F = α - α = 0$ , so $F \in O (Ω)$ , and near $p$ , $F = 1/ L_{p} - u$ is singular (blows up) at $p$ because $1/ L_{p} \to \infty$ while $u$ stays bounded. So $F$ is a holomorphic function on $Ω$ that cannot be continued across $p$ .

Since such an $F$ exists for every boundary point $p$ , no boundary point admits holomorphic extension, i.e. $Ω$ is a domain of holomorphy. This is the solution of the Levi problem via the $\overset{ˉ}{\partial}$ -method (Hörmander). Krantz Ch. 3-4.

Exercise 10 (medium, numeric). Bergman kernel of the polydisc by the product rule.

Compute the Bergman kernel of the bidisc $Δ^{2} = Δ \times Δ \subset C^{2}$ , using the product structure $A^{2} (Ω_{1} \times Ω_{2})$ and the disc kernel from Exercise 2.

Hint

For a product domain, an orthonormal basis is the set of products $e_{m} (z_{1}) e_{n} (z_{2})$ , so $K_{Ω_{1} \times Ω_{2}} ((z_{1}, z_{2}), (w_{1}, w_{2})) = K_{Ω_{1}} (z_{1}, w_{1}) K_{Ω_{2}} (z_{2}, w_{2})$ .

Answer

For a product domain $Ω_{1} \times Ω_{2}$ , if ${e_{m}}$ is an orthonormal basis of $A^{2} (Ω_{1})$ and ${f_{n}}$ of $A^{2} (Ω_{2})$ , then the products ${e_{m} \otimes f_{n} : (z_{1}, z_{2}) \mapsto e_{m} (z_{1}) f_{n} (z_{2})}$ form an orthonormal basis of $A^{2} (Ω_{1} \times Ω_{2})$ (Fubini factorises the inner product, and the products span by holomorphic separation of variables). Therefore the Bergman kernel factorises:

K_{Ω_{1} \times Ω_{2}} ((z_{1}, z_{2}), (w_{1}, w_{2})) = m, n \sum e_{m} (z_{1}) f_{n} (z_{2}) \overline{e_{m} (w_{1}) f_{n} (w_{2})} = K_{Ω_{1}} (z_{1}, w_{1}) K_{Ω_{2}} (z_{2}, w_{2}) .

Using the disc kernel $K_{Δ} (z, w) = \frac{1}{π ( 1 - z w ˉ ) ^{2}}$ from Exercise 2,

K_{Δ^{2}} ((z_{1}, z_{2}), (w_{1}, w_{2})) = \frac{1}{π ( 1 - z _{1} w ˉ _{1} ) ^{2}} \cdot \frac{1}{π ( 1 - z _{2} w ˉ _{2} ) ^{2}} = \frac{1}{π ^{2} ( 1 - z _{1} w ˉ _{1} ) ^{2} ( 1 - z _{2} w ˉ _{2} ) ^{2}} .

The polydisc kernel is the product of disc kernels. Contrast this with the ball $B^{2}$ , whose Bergman kernel $K_{B^{2}} (z, w) = \frac{2}{π ^{2} ( 1 - z \cdot w ˉ ) ^{3}}$ does not factorise — reflecting that the ball is not biholomorphic to the bidisc (Poincaré's theorem). Krantz, Bergman-kernel chapter.

Exercise pack supplementing Krantz's Function Theory of Several Complex Variables: the Hartogs extension phenomenon, domains of holomorphy and holomorphic convexity, plurisubharmonicity and the Levi form, the $\overset{ˉ}{\partial}$ -equation with Hörmander $L^{2}$ estimates, and the Bergman kernel and its transformation rule.

Prerequisites

06.10.01 pending
06.10.02 pending
06.10.03 pending
06.10.04 pending
06.10.08 pending
06.07.02 pending

Tier anchors

beginner
intermediate: Krantz *Function Theory of Several Complex Variables* (2nd ed.) exercises across Ch. 2 (Hartogs phenomenon, domains of holomorphy), Ch. 3 (plurisubharmonic functions, the Levi form, pseudoconvexity), Ch. 4 (the $\bar\partial$-equation and Hörmander $L^2$ estimates on pseudoconvex domains), Ch. 1 §1.4 and the Bergman-kernel chapter (reproducing kernel of $A^2(\Omega)$, transformation rule)
master

References

Krantz — Function Theory of Several Complex Variables (2nd ed., AMS Chelsea 2001) · Ch. 2 (Hartogs extension, domains of holomorphy, holomorphic convexity), Ch. 3 (plurisubharmonic functions, Levi form, pseudoconvexity, Cartan-Thullen), Ch. 4 (Hörmander $L^2$ solution of $\bar\partial$), and the Bergman-kernel material (reproducing property, transformation rule under biholomorphism)
Hörmander — An Introduction to Complex Analysis in Several Variables (3rd ed., North-Holland 1990) · Ch. 2 (domains of holomorphy, pseudoconvexity), Ch. 4 ($L^2$ estimates for $\bar\partial$) — companion problems
Range — Holomorphic Functions and Integral Representations in Several Complex Variables (GTM 108, Springer 1986) · Ch. 2-4 (pseudoconvexity, integral kernels, Bergman theory)

Estimated time

intermediate: 6h