Pseudoconvexity

In mathematics, more precisely in the theory of functions of several complex variables, a pseudoconvex set is a special type of open set in the n-dimensional complex space Cn. Pseudoconvex sets are important, as they allow for classification of domains of holomorphy.

Let

G\subset {\mathbb{C}}^n

be a domain, that is, an open connected subset. One says that G is pseudoconvex (or Hartogs pseudoconvex) if there exists a continuous plurisubharmonic function \varphi on G such that the set

\{ z \in G \mid \varphi(z) < x \}

is a relatively compact subset of G for all real numbers x. In other words, a domain is pseudoconvex if G has a continuous plurisubharmonic exhaustion function. Every (geometrically) convex set is pseudoconvex.

When G has a C^2 (twice continuously differentiable) boundary, this notion is the same as Levi pseudoconvexity, which is easier to work with. More specifically, with a C^2 boundary, it can be shown that G has a defining function; ie., that there exists \rho: \mathbb{C}^n \to  \mathbb{R} which is C^2 so that G=\{\rho <0 \}, and \partial G =\{\rho =0\}. Now, G is pseudoconvex iff for every p \in \partial G and w in the complex tangent space at p that is,

 \nabla \rho(p) w = \sum_{i=1}^n \frac{\partial \rho (p)}{ \partial z_j }w_j =0 we have
\sum_{i,j=1}^n \frac{\partial^2 \rho(p)}{\partial z_i \partial \bar{z_j} } w_i \bar{w_j} \geq 0.

If G does not have a C^2 boundary, the following approximation result can come in useful.

Proposition 1 If G is pseudoconvex, then there exist bounded, strongly Levi pseudoconvex domains G_k \subset G with C^\infty (smooth) boundary which are relatively compact in G, such that

G = \bigcup_{k=1}^\infty G_k.

This is because once we have a \varphi as in the definition we can actually find a C exhaustion function.

The case n = 1

In one complex dimension, every open domain is pseudoconvex. The concept of pseudoconvexity is thus more useful in dimensions higher than 1.

See also

References


This article incorporates material from Pseudoconvex on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.