2.5: Hartogs Pseudoconvexity

It is worth mentioning explicitly that by the exercises of the previous section, plurisubharmonicity is preserved under holomorphic mappings. That is, if \(g\) is holomorphic and \(f\) is plurisubharmonic, then \(f \circ g\) is plurisubharmonic. In particular, if \(\varphi \colon \mathbb{D} \to \mathbb{C}^n\) is an analytic disc and \(f\) is plurisubharmonic in a neighborhood of \(\varphi(\mathbb{D})\), then \(f \circ \varphi\) is subharmonic.

Definition: Word

Let \(\mathcal{F}\) be a class of (extended\(^{1}\))-real-valued functions defined on an open \(U \subset \mathbb{R}^n\). If \(K \subset U\), define \(\widehat{K}\), the hull of \(K\) with respect to \(\mathcal{F}\), as the set \[\widehat{K} \overset{\text{def}}{=} \Bigl\{ x \in U : f(x) \leq \sup_{y\in K} f(y) \text{ for all $f \in \mathcal{F}$ } \Bigr\} .\]

An open set \(U\) is said to be convex with respect to \(\mathcal{F}\) if for every \(K \subset \subset U\), the hull \(\widehat{K} \subset \subset U\).\(^{2}\)

Clearly \(K \subset \widehat{K}\). The key is to show that \(\widehat{K}\) is not “too large” for \(U\). Keep in mind that the functions in \(\mathcal{F}\) are defined on \(U\), so \(\widehat{K}\) depends on \(U\) not just on \(K\). An easy mistake is to consider functions defined on a larger set, obtaining a smaller \(\mathcal{F}\) and hence a larger \(\widehat{K}\). Sometimes it is useful to write \(\widehat{K}_{\mathcal{F}}\) to denote the dependence on \(\mathcal{F}\), especially when talking about several different hulls.

For example, if \(U=\mathbb{R}\) and \(\mathcal{F}\) is the set of real-valued smooth \(f \colon \mathbb{R} \to \mathbb{R}\) with \(f''(x) \geq 0\), then for any \(a,b \in \mathbb{R}\) we have \(\widehat{\{ a, b \}} = [a,b]\). In general, if \(\mathcal{F}\) was the set of convex functions, then a domain \(U \subset \mathbb{R}^n\) is geometrically convex if and only if it is convex with respect to convex functions, although let us not define what that means except for smooth functions in exercises below.

Exercise \(\PageIndex{1}\)

Suppose \(U \subset \mathbb{R}^n\) is a domain.

1. Show that \(U\) is geometrically convex if and only if it is convex with respect to the affine linear functions.
2. Suppose \(U\) has smooth boundary. Show that \(U\) is convex if and only if it is convex with respect to the smooth convex functions on \(U\), that is, with respect to smooth functions with positive semidefinite Hessian.

Exercise \(\PageIndex{2}\)

Show that every open set \(U \subset \mathbb{R}^n\) is convex with respect to real polynomials.

Theorem \(\PageIndex{1}\): Kontinuitätssatz—Continuity Principle

Suppose an open set \(U \subset \mathbb{C}^n\) is convex with respect to plurisubharmonic functions, then given any collection of closed analytic discs \(\Delta_\alpha \subset U\) such that \(\bigcup_\alpha \partial \Delta_\alpha \subset \subset U\), we have \(\bigcup_\alpha \Delta_\alpha \subset \subset U\).

Various similar theorems are named the continuity principle. Generally what they have in common is the family of analytic discs whose boundaries stay inside a domain, and whose conclusion has to do with extension of holomorphic functions, or with domains of holomorphy.

Proof

Let \(f\) be a plurisubharmonic function on \(U\). If \(\varphi_\alpha \colon \overline{\mathbb{D}} \to U\) is the holomorphic (in \(\mathbb{D}\)) mapping giving the closed analytic disc, then \(f \circ \varphi_\alpha\) is subharmonic. By the maximum principle, \(f\) on \(\Delta_\alpha\) must be less than or equal to the supremum of \(f\) on \(\partial \Delta_\alpha\), so \(\overline{\Delta_\alpha}\) is in the hull of \(\partial \Delta_\alpha\). In other words, \(\bigcup_\alpha \Delta_\alpha\) is in the hull of \(\bigcup_\alpha \partial \Delta_\alpha\) and therefore \(\bigcup_\alpha \Delta_\alpha \subset \subset U\) by convexity.

Let us illustrate the failure of the continuity principle. If you have discs (denoted by straight line segments) that approach the boundary as in Figure \(\PageIndex{1}\), then the domain is not not convex with respect to plurisubharmonic functions. In the diagram, the boundaries of the discs are denoted by the dark dots at the end of the segments. In fact, if we replace discs with line segments, this is the standard convexity, see the exercises below.

Figure \(\PageIndex{1}\)

Exercise \(\PageIndex{3}\)

Suppose \(U \subset \mathbb{C}^n\) is a domain and \(K \subset \subset U\) is a nonempty compact subset. Prove that \(U \setminus K\) is not convex with respect to plurisubharmonic functions.

Exercise \(\PageIndex{4}\)

Suppose \(U \subset \mathbb{C}^n\) is a domain with smooth boundary, \(p \in \partial U\), and \(\Delta\) is an affine linear analytic disc with \(p \in \Delta\), but \(\Delta \setminus \{ p \} \subset U\). Prove that \(U\) is not convex with respect to the plurisubharmonic functions.

Exercise \(\PageIndex{5}\)

Prove the corresponding Kontinuitätssatz, and its converse, for geometric convexity: Prove that a domain \(U \subset \mathbb{R}^n\) is geometrically convex if and only if whenever \([x_\alpha,y_\alpha] \subset U\) is a collection of straight line segments such that \(\bigcup_{\alpha} \{ x_\alpha,y_\alpha \} \subset \subset U\) implies \(\bigcup_{\alpha} [ x_\alpha,y_\alpha ] \subset \subset U\).

Definition: Exhaustion Function

Let \(U \subset \mathbb{C}^n\) be open. An \(f \colon U \to \mathbb{R}\) is an exhaustion function for \(U\) if \[\bigl\{ z \in U : f(z) < r \bigr\} \subset \subset U \qquad \text{for every $r \in \mathbb{R}$.}\]

A domain \(U \subset \mathbb{C}^n\) is Hartogs pseudoconvex if there exists a continuous plurisubharmonic exhaustion function. The set \(\{ z \in U : f(z) < r \}\) is called the sublevel set of \(f\), or the \(r\)-sublevel set.

Example \(\PageIndex{1}\)

The unit ball \(\mathbb{B}_{n}\) is Hartogs pseudoconvex. The continuous function \[z \mapsto - \log \bigl( 1-||z|| \bigr)\] is an exhaustion function, and it is easy to check directly that it is plurisubharmonic.

Example \(\PageIndex{2}\)

The entire \(\mathbb{C}^n\) is Hartogs pseudoconvex as \(||z||^2\) is a continuous plurisubharmonic exhaustion function. Also, because \(||z||^2\) is plurisubharmonic, then given any \(K \subset \subset \mathbb{C}^n\), the hull \(\widehat{K}\) with respect to plurisubharmonic functions must be bounded. In other words, \(\mathbb{C}^n\) is convex with respect to plurisubharmonic functions.

Theorem \(\PageIndex{2}\)

Suppose \(U \subsetneq \mathbb{C}^n\) is a domain. The following are equivalent:

1. \(-\log \rho(z)\) is plurisubharmonic, where \(\rho(z)\) is the distance from \(z\) to \(\partial U\).
2. \(U\) has a continuous plurisubharmonic exhaustion function, that is, \(U\) is Hartogs pseudoconvex.
3. \(U\) is convex with respect to plurisubharmonic functions defined on \(U\).

Proof

(i) \(\Rightarrow\) (ii): If \(U\) is bounded, the function \(-\log \rho(z)\) is clearly a continuous exhaustion function. If \(U\) is unbounded, take \(z \mapsto \max \{ -\log \rho(z) , ||z||^2 \}\).

(ii) \(\Rightarrow\) (iii): Suppose \(f\) is a continuous plurisubharmonic exhaustion function. If \(K \subset \subset U\), then for some \(r\) we have \(K \subset \{ z \in U : f(z) < r \} \subset \subset U\). But then by definition of the hull \(\widehat{K}\) we have \(\widehat{K} \subset \{ z \in U : f(z) < r \} \subset \subset U\).

(iii) \(\Rightarrow\) (i): For \(c \in \mathbb{C}^n\) with \(||c||=1\), let \(\rho_c(z)\) be the radius of the largest affine disc centered at \(z\) in the direction \(c\) that still lies in \(U\). That is, \[\rho_c(z) = \sup \bigl\{ \lambda > 0 : z+ \zeta c \in U \text{ for all $\zeta \in \lambda\mathbb{D}$} \bigr\} .\] As \(\rho(z) = \inf_c \rho_c(z)\), \[- \log \rho(z) = \sup_{||c||=1} \bigl(-\log \rho_c(z)\bigr) .\] If we prove that for any \(a, b \in \mathbb{C}^n\) the function \(\xi \mapsto -\log \rho_c(a+b\xi)\) is subharmonic, then \(\xi \mapsto - \log \rho(a+b\xi)\) is subharmonic, and we are done. See for the setup:

Figure \(\PageIndex{2}\)

Suppose \(\Delta \subset \mathbb{C}\) is a disc such that \(a+b\xi \in U\) for all \(\xi \in \overline{\Delta}\). If \(u\) is a harmonic function on \(\Delta\) continuous on \(\overline{\Delta}\) such that \(- \log \rho_c(a+b\xi) \leq u(\xi)\) on \(\partial \Delta\), we must show that the inequality holds on \(\Delta\). By Exercise 2.4.13 we may assume \(u\) is harmonic on a neighborhood of \(\overline{\Delta}\) and so let \(u = \Re f\) for a holomorphic function \(f\). Fix a \(\xi \in \partial \Delta\). We have \(- \log \rho_c(a+b\xi) \leq \Re f(\xi)\), or in other words \[\rho_c(a+b\xi) \geq e^{-\Re f(\xi)} = \left|e^{-f(\xi)}\right|.\] Using \(\zeta = t e^{-f(\xi)}\) in the definition of \(\rho_c(a+b\xi)\), the statement above is equivalent to saying that \[(a+b\xi)+te^{-f(\xi)}c \in U \quad \text{for all $t \in \mathbb{D}$}.\] This statement holds whenever \(\xi \in \partial \Delta\). We must prove that it also holds for all \(\xi \in \Delta\).

The function \(\varphi_t(\xi) = (a+b\xi)+te^{-f(\xi)}c\) gives a closed analytic disc with boundary inside \(U\). We have a family of analytic discs, parametrized by \(t\), whose boundaries are in \(U\) for all \(t\) with \(|t| < 1\). For \(t=0\) the entire disc is inside \(U\). As \(\varphi_t(\xi)\) is continuous in both \(t\) and \(\xi\) and \(\overline{\Delta}\) is compact, \(\varphi_t(\Delta) \subset U\) for \(t\) in some neighborhood of \(0\). Take \(0 < t_0 < 1\) such that \(\varphi_t(\Delta) \subset U\) for all \(t\) with \(|t| < t_0\). Then \[\bigcup_{|t| < t_0} \varphi_t(\partial \Delta) \subset \bigcup_{|t| \leq t_0} \varphi_t(\partial \Delta) \subset \subset U ,\] because continuous functions take compact sets to compact sets. Kontinuitätssatz implies \[\bigcup_{|t| < t_0} \varphi_t(\Delta) \subset \subset U .\] By continuity again \(\bigcup_{|t| \leq t_0} \varphi_t(\Delta) \subset \subset U\), and so \(\bigcup_{|t| < t_0+\epsilon} \varphi_t(\Delta) \subset \subset U\) for some \(\epsilon > 0\). Consequently \(\varphi_t(\Delta) \subset U\) for all \(t\) with \(|t| < 1\). Thus \((a+b\xi)+te^{-f(\xi)}c \in U\) for all \(\xi \in \Delta\) and all \(|t| < 1\). This implies \(\rho_c(a+b\xi) \geq e^{-\Re f(\xi)}\) for all \(\xi \in \Delta\), which in turn implies \(-\log \rho_c(a+b\xi) \leq \Re f(\xi) = u(\xi)\) for all \(\xi \in \Delta\). Therefore, \(-\log \rho_c(a+b\xi)\) is subharmonic.

Exercise \(\PageIndex{6}\)

Show that if \(U_1 \subset \mathbb{C}^n\) and \(U_2 \subset \mathbb{C}^n\) are Hartogs pseudoconvex domains, then so are all the topological components of \(U_1 \cap U_2\).

Exercise \(\PageIndex{7}\)

Show that if \(U \subset \mathbb{C}^n\) and \(V \subset \mathbb{C}^m\) are Hartogs pseudoconvex domains, then so is \(U \times V\).

Exercise \(\PageIndex{8}\)

Show that every domain \(U \subset \mathbb{C}\) is Hartogs pseudoconvex.

Exercise \(\PageIndex{9}\)

Consider the union \(U = \bigcup_k U_k\) of a nested sequence of Hartogs pseudoconvex domains, \(U_{k-1} \subset U_k \subset \mathbb{C}^n\). Show that \(U\) is Hartogs pseudoconvex.

Exercise \(\PageIndex{10}\)

Let \(\mathbb{R}^2 \subset \mathbb{C}^2\) be naturally embedded (that is, it is the set where \(z_1\) and \(z_2\) are real). Show that the set \(\mathbb{C}^2 \setminus \mathbb{R}^2\) is not Hartogs pseudoconvex.

Exercise \(\PageIndex{11}\)

Let \(U \subset \mathbb{C}^n\) be a domain and \(f \in \mathcal{O}(U)\). Prove that \(U' = \bigl\{ z \in U : f(z) \not= 0 \bigr\}\) is a Hartogs pseudoconvex domain. Hint: See also Exercise 1.6.2.

Exercise \(\PageIndex{12}\)

Suppose \(U,V \subset \mathbb{C}^n\) are biholomorphic domains. Prove that \(U\) is Hartogs pseudoconvex if and only if \(V\) is Hartogs pseudoconvex.

Exercise \(\PageIndex{13}\)

Let \(U = \bigl\{ z \in \mathbb{C}^2 : |z_1| > |z_2| \bigr\}\).

1. Prove that \(U\) is a Hartogs pseudoconvex domain.
2. Find a closed analytic disc \(\Delta\) in \(\mathbb{C}^2\) such that \(0 \in \Delta\) (\(0 \notin U\)) and \(\partial \Delta \subset U\).
3. What do you think would happen if you tried to move \(\Delta\) a little bit to avoid the intersection with the complement? Think about the continuity principle (Theorem \(\PageIndex{1}\)). Compare with Exercise \(\PageIndex{4}\).

Exercise \(\PageIndex{14}\)

Let \(U \subset \mathbb{C}^n\) be a domain, and \(f \colon \overline{U} \to \mathbb{R}\) be a continuous function, plurisubharmonic and negative on \(U\), and \(f=0\) on \(\partial U\). Prove that \(U\) is Hartogs pseudoconvex.

The statement corresponding to Exercise \(\PageIndex{9}\) on nested unions for domains of holomorphy is the Behnke-Stein theorem, which follows using this exercise and the solution of the Levi problem. Although historically Behnke–Stein was proved independently and used to solve the Levi problem.

Exercise \(\PageIndex{12}\) says that (Hartogs) pseudoconvexity is a biholomorphic invariant. That is a good indication that we are looking at a correct notion. It also allows us to change variables to more convenient ones when proving a specific domain is (Hartogs) pseudoconvex.

It is not immediately clear from the definition, but Hartogs pseudoconvexity is a local property.

Lemma \(\PageIndex{1}\)

A domain \(U \subset \mathbb{C}^n\) is Hartogs pseudoconvex if and only if for every point \(p \in \partial U\) there exists a neighborhood \(W\) of \(p\) such that \(W \cap U\) is Hartogs pseudoconvex.

Proof

One direction is trivial, so consider the other. Suppose \(p \in \partial U\), and let \(W\) be such that \(U \cap W\) is Hartogs pseudoconvex. Intersection of Hartogs pseudoconvex domains is Hartogs pseudoconvex, so assume \(W = B_r(p)\). Let \(B = B_{r/2}(p)\). If \(q \in B \cap U\), the distance from \(q\) to the boundary of \(W \cap U\) is the same as the distance to \(\partial U\). The setup is illustrated in Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\)

The part of the boundary \(\partial U\) in \(W\) is marked by a thick black line, the part of the boundary of \(\partial (W \cap U)\) that arises as the boundary of \(W\) is marked by a thick gray line. A point \(q \in B\) is marked and a ball of radius \(\frac{r}{2}\) around \(q\) is dotted. No point of distance \(\frac{r}{2}\) from \(q\) is in \(\partial W\), and the distance of \(q\) to \(\partial U\) is at most \(\frac{r}{2}\) as \(p \in \partial U\) and \(p\) is the center of \(B\). Let \(\operatorname{dist}(x,y)\) denote the euclidean distance function\(^{3}\). Then for \(z \in B \cap U\) \[- \log \, \operatorname{dist}(z, \partial U) = - \log \, \operatorname{dist}\bigl(z, \partial (U \cap W)\bigr).\] The right-hand side is plurisubharmonic as \(U \cap W\) is Hartogs pseudoconvex. Such a ball \(B\) exists around every \(p \in \partial U\), so near the boundary, \(- \log \, \operatorname{dist}(z, \partial U)\) is plurisubharmonic.

If \(U\) is bounded, then \(\partial U\) is compact. So there is some \(\epsilon > 0\) such that \(- \log \, \operatorname{dist}(z, \partial U)\) is plurisubharmonic if \(\operatorname{dist}(z, \partial U) < 2\epsilon\). The function \[\varphi(z) = \max \bigl\{ - \log \, \operatorname{dist}(z, \partial U) , - \log \epsilon \bigr\}\] is a continuous plurisubharmonic exhaustion function. The proof for unbounded \(U\) is left as an exercise.

Exercise \(\PageIndex{15}\)

Finish the proof of the lemma for unbounded domains. See Exercise \(\PageIndex{9}\).

It may seem that we defined a totally different concept, but it turns out that Levi and Hartogs pseudoconvexity are one and the same on domains where both concepts make sense. As a consequence of the following theorem we say simply “pseudoconvex” and there is no ambiguity.

Theorem \(\PageIndex{3}\)

Let \(U \subset \mathbb{C}^n\) be a domain with smooth boundary. Then \(U\) is Hartogs pseudoconvex if and only if \(U\) is Levi pseudoconvex.

Proof

Suppose \(U \subset \mathbb{C}^n\) is a domain with smooth boundary that is not Levi pseudoconvex at \(p \in \partial U\). As in Theorem 2.3.2, change coordinates so that \(p=0\) and \(U\) is defined by \[\Im z_n > - |z_1|^2 + \sum_{j=2}^{n-1} \epsilon_j |z_j|^2 + O(3) .\] For a small fixed \(\lambda > 0\), the closed analytic discs defined by \(\xi \in \overline{\mathbb{D}} \mapsto (\lambda \xi, 0, \cdots, 0, is)\) are in \(U\) for all small enough \(s > 0\). The origin is a limit point of the insides, but not a limit point of their boundaries. Kontinuitätssatz (Theorem \(\PageIndex{1}\)) is not satisfied, and \(U\) is not convex with respect to the plurisubharmonic functions. Therefore, \(U\) is not Hartogs pseudoconvex.

Next suppose \(U\) is Levi pseudoconvex. Take any \(p \in \partial U\). After translation and rotation by a unitary, assume \(p=0\) and write a defining function \(r\) as \[r(z,\bar{z}) = \varphi(z',\bar{z}',\Re z_n) - \Im z_n ,\] where \(z' = (z_1,\ldots,z_{n-1})\). Levi pseudoconvexity says \[\label{eq:4} \sum_{j=1,\ell=1}^n \bar{a}_j a_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q \geq 0 \quad \text{whenever} \quad \sum_{j=1}^n a_j \frac{\partial r}{\partial z_j} \Big|_q = 0 ,\] for all \(q \in \partial U\) near \(0\). Let \(s\) be a small real constant, and let \(\widetilde{q} = (q_1,\ldots,q_{n-1},q_n + is)\). None of the derivatives of \(r\) depends on \(\Im z_n\), and therefore \(\frac{\partial r}{\partial z_\ell} \big|_{\widetilde{q}} = \frac{\partial r}{\partial z_\ell} \big|_{q}\) and \(\frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \big|_{\widetilde{q}} = \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \big|_{q}\) for all \(j\) and \(\ell\). So condition \(\eqref{eq:4}\) holds for all \(q \in U\) near \(0\). We will use \(r\) to manufacture a plurisubharmonic exhaustion function, that is one with a semidefinite Hessian. Starting with \(r\), we already have what we need in all but one direction.

Let \(\nabla_z r|_q = \bigl( \frac{\partial r}{\partial z_1}\big|_q,\ldots, \frac{\partial r}{\partial z_n}\big|_q \bigr)\) denote the gradient of \(r\) in the holomorphic directions only. Given \(q \in U\) near \(0\), decompose an arbitrary \(c \in \mathbb{C}^n\) as \(c = a+b\), where \(a = (a_1,\ldots,a_n)\) satisfies \[\sum_{j=1}^n a_j \frac{\partial r}{\partial z_j} \Big|_q = \langle a,\overline{\nabla_z r|_q}\rangle = 0 .\] Taking the orthogonal decomposition, \(b\) is a scalar multiple of \(\overline{\nabla_z r|_q}\). By Cauchy–Schwarz, \[\left| \sum_{j=1}^n c_j \frac{\partial r}{\partial z_j} \Big|_q\right| = \left| \sum_{j=1}^n b_j \frac{\partial r}{\partial z_j} \Big|_q\right| = \left|\langle b, \overline{\nabla_z r|_q}\rangle \right| = ||b||\: ||\nabla_z r|_q|| .\] As \(\nabla_z r|_0 = (0,\ldots,0,-\frac{1}{2i})\), then for \(q\) sufficiently near \(0\) we have that \(||\nabla_z r|_q|| \geq \frac{1}{3}\), and \[||b|| = \frac{1}{||\nabla_z r|_q||} \left|\sum_{j=1}^n c_j \frac{\partial r}{\partial z_j} \Big|_q\right| \leq 3 \left|\sum_{j=1}^n c_j \frac{\partial r}{\partial z_j} \Big|_q\right| .\] As \(c = a+b\) is the orthogonal decomposition, \(||c|| \geq ||b||\).

The complex Hessian matrix of \(r\) is continuous, and so let \(M \geq 0\) be an upper bound on its operator norm for \(q\) near the origin. Again using Cauchy–Schwarz \[\begin{align}\begin{aligned} \sum_{j=1,\ell=1}^n \bar{c}_j c_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q & = \sum_{j=1,\ell=1}^n ( \bar{a}_j + \bar{b}_j ) (a_\ell + b_\ell) \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q \\ & = \sum_{j=1,\ell=1}^n \bar{a}_j a_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q \\ & \phantom{=} \quad + \sum_{j=1,\ell=1}^n \bar{b}_j c_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q + \sum_{j=1,\ell=1}^n \bar{c}_j b_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q + \sum_{j=1,\ell=1}^n \bar{b}_j b_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q \\ & \geq \sum_{j=1,\ell=1}^n \bar{a}_j a_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q - M||b||\: ||c|| - M||c||\: ||b|| - M||b||^2 \\ & \geq - 3 M ||c||\: ||b|| . \end{aligned}\end{align}\] Together with what we know about \(||b||\), for \(q \in U\) near the origin \[\sum_{j=1,\ell=1}^n \bar{c}_j c_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q \geq -3M ||c||\: ||b|| \geq -3^2 M ||c|| \left| \sum_{j=1}^n c_j \frac{\partial r}{\partial z_j} \Big|_q\right| .\] For \(z \in U\) sufficiently close to \(0\) define \[f(z) = -\log \bigl(-r(z)\bigr) + A ||z||^2 ,\] where \(A > 0\) is some constant we will choose later. The log is there to make \(f\) blow up as we approach the boundary. The \(A ||z||^2\) is there to add a constant diagonal matrix to the complex Hessian of \(f\), which we hope is enough to make it positive semidefinite at all \(z\) near \(0\). Compute: \[\frac{\partial^2 f}{\partial \bar{z}_j \partial z_\ell} = \frac{1}{r^2} \frac{\partial r}{\partial \bar{z}_j} \frac{\partial r}{\partial z_\ell} - \frac{1}{r} \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} + A\delta_{j}^{\ell} ,\] where \(\delta_j^\ell\) is the Kronecker delta\(^{4}\). Apply the complex Hessian of \(f\) to \(c\) at \(q \in U\) near the origin (recall that \(r\) is negative on \(U\) and so for \(q \in U\), \(-r = |r|\)): \[\begin{align}\begin{aligned} \sum_{j=1,\ell=1}^n \bar{c}_j c_\ell \frac{\partial^2 f}{\partial \bar{z}_j \partial z_\ell} \Big|_q & = \frac{1}{r^2} \left| \sum_{\ell=1}^n c_\ell \frac{\partial r}{\partial z_\ell} \Big|_q \right|^2 + \frac{1}{|r|} \sum_{j=1,\ell=1}^n \bar{c}_j c_\ell \frac{\partial^2 r}{\partial \bar{z}_j \partial z_\ell} \Big|_q + A ||c||^2 \\ & \geq \frac{1}{r^2} \left| \sum_{\ell=1}^n c_\ell \frac{\partial r}{\partial z_\ell} \Big|_q \right|^2 - \frac{3^2 M}{|r|} ||c|| \left|\sum_{j=1}^n c_j \frac{\partial r}{\partial z_j} \Big|_q\right| + A ||c||^2 . \end{aligned}\end{align}\] Now comes a somewhat funky trick. As a quadratic polynomial in \(||c||\), the right-hand side of the inequality is always nonnegative if \(A > 0\) and if the discriminant is negative or zero. Let us set the discriminant to zero: \[0 = {\Biggl( \frac{3^2 M}{|r|} \left|\sum_{j=1}^n c_j \frac{\partial r}{\partial z_j} \Big|_q\right| \Biggr)}^2 - 4A \frac{1}{r^2} \left| \sum_{\ell=1}^n c_\ell \frac{\partial r}{\partial z_\ell} \Big|_q \right|^2 .\] All the nonconstant terms go away and \(A=\frac{3^4 M^2}{4}\) makes the discriminant zero. Thus for that \(A\), \[\sum_{j=1,\ell=1}^n \bar{c}_j c_\ell \frac{\partial^2 f}{\partial \bar{z}_j \partial z_\ell} \Big|_q \geq 0.\] In other words, the complex Hessian of \(f\) is positive semidefinite at all points \(q \in U\) near \(0\). The function \(f(z)\) goes to infinity as \(z\) approaches \(\partial U\). So for every \(t \in \mathbb{R}\), the \(t\)-sublevel set (the set where \(f(z) < t\)) is a positive distance away from \(\partial U\) near \(0\).

We have constructed a local continuous plurisubharmonic exhaustion function for \(U\) near \(p\). If we intersect with a small ball \(B\) centered at \(p\), then we get that \(U \cap B\) is Hartogs pseudoconvex. This is true at all \(p \in \partial U\), so \(U\) is Hartogs pseudoconvex.

Footnotes

[1] By extended reals we mean \(\mathbb{R} \cup \{ -\infty,\infty\}\).

[2] Recall that \(\subset \subset\) means relatively compact, that is, the closure in the relative (subspace) topology is compact.

[3] If either \(x\) and/or \(y\) are sets of points, we take the infimum of the euclidean distance over all the points.

[4] Recall \(\delta_j^\ell = 0\) if \(j\not= \ell\) and \(\delta_j^\ell = 1\) if \(j = \ell\).