Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the Bergman kernel $K_\Omega$ associated to a domain $\Omega \subseteq \mathbb C^n$. By the reproducing property, it is easy to show that $$K_\Omega(z,\zeta) = \sum_{n=1}^\infty \varphi_k(z) \overline{\varphi_k(\zeta)},\qquad(z,\zeta\in\Omega)$$ where $\{\varphi_k\}_{k=1}^\infty$ is any orthonormal basis of the Bergman space $A^2(\Omega)$ of Lebesgue square-integrable holomorphic functions on $\Omega$.

This series representation converges at least pointwise, since the Bergman kernel's Fourier series, $K_\Omega(\cdot,\zeta) = \sum_{k=1}^\infty \langle K_\Omega(\cdot,\zeta), \varphi_k \rangle \varphi_k$ with $\langle K_\Omega(\cdot,\zeta), \varphi_k \rangle = \overline{\varphi_k(\zeta)}$ converges in norm which implies uniform convergence in the first argument for fixed $\zeta \in \Omega$.

Now in Books such as Function Theory of Several Complex Variables by S. Krantz, it is shown that the series is uniformly bounded on compact sets, namely $$ \sum_{k=1}^\infty \big| \varphi_k(z) \overline{\varphi_k(\zeta)} \big| \leq \bigg(\sum_{k=1}^\infty |\varphi_k(z)|^2 \bigg)^{1/2} \bigg(\sum_{k=1}^\infty |\varphi_k(\zeta)|^2 \bigg)^{1/2} \leq C(K)^2,\qquad(z,\zeta \in K)$$ where $C(K)$ is a constant depending only on the compact set $K\subseteq \Omega$.

My question is this: Why does this imply uniform convergence on compact sets in $\Omega \times \Omega$? This is claimed in several sources, but just stated and not proven. Am I missing something obvious here?

One book which is a bit more specific is Holomorphic Functions and Integral Representations in Several Complex Variables by M. Range. There it is written that uniform convergence on compact subsets of $\Omega \times \Omega$ follows from the uniform bound and a "normality argument", which I take as referring to Montel's theorem. Does anyone know the details on how this argument works?

Any help is appreciated.

share|cite|improve this question
up vote 2 down vote accepted

Unless I misunderstand, I think the following statement may answer your question.

Let $U\subseteq\mathbb{C}^n$ be a domain, and let $f_n\colon U\to \mathbb{C}$ be a sequence of holomorphic functions converging pointwise to a function $f\colon U\to \mathbb{C}$. Suppose that for each compact set $K\subset U$ the family $\{f_n|_K\}$ is uniformly bounded. Then $f_n$ converges uniformly on compact sets to $f$.

Proof: By looking locally, we can assume without loss of generality that the $f_n$ are uniformly bounded on $U$. By Montel's theorem, every subsequence of $f_n$ has a subsequence converging uniformly on compact sets.

Let $\mathcal{L}$ denote the set of limits of subsequences of $f_n$ in the topology of uniform convergence on compact sets. On the one hand, $\mathcal{L}$ is nonempty, as we have just noted by Montel's theorem. Suppose that $g\in \mathcal{L}$. Then there is a subsequence $f_{n_k}$ converging uniformly on compact sets to $g$. However, since $f_n$ was assumed to have converged pointwise to $f$, it must be that $g = f$. Thus $\mathcal{L} = \{f\}$. This proves that the sequence $f_n$ has exactly one limit point in the topology of uniformly convergence on compact sets, which exactly means that $f_n\to f$ in this topology. This proves the statement.

To finish out the answer to your question, you would like to just take $f_n$ to be the sequence on $U = \Omega\times\Omega$ given by $f_n(z,\zeta) = \sum_{k=1}^n \varphi_k(z)\overline{\varphi_k(\zeta)}$. This technically doesn't fit into the hypotheses of the statement, since these $f_n$ are not holomorphic. However, if $\overline{\Omega}$ denotes $\Omega$ with the conjugate complex structure, then $f_n\colon \Omega\times\overline\Omega\to \mathbb{C}$ given by $f_n(z,\zeta) = \sum_{k=1}^n \varphi_k(z)\overline{\varphi_k(\zeta)}$ is holomorphic, so you can apply the above here.

I hope that works!

share|cite|improve this answer
Thank you for your answer, froggie! I've almost forgotten about this question, and came up with something similar in the meantime; I'll post the details in an extra answer. – fbg Oct 17 '12 at 15:05

The following is a slight variation of a problem posed in the book on functional analysis by Reed & Simon (page 35, problem 33(b)).

Consider the same statement as in froggie's answer, with the same assumptions:

Theorem: Let $U\subseteq\mathbb{C}^n$ be a domain, and let $f_n\colon U\to \mathbb{C}$ be a sequence of holomorphic functions converging pointwise to a function $f\colon U\to \mathbb{C}$. Suppose that for each compact set $K\subset U$ the family $\{f_n|_K\}$ is uniformly bounded. Then $f_n$ converges uniformly on compact sets to $f$.

Proof: Apply Montel's theorem to every sub-sequence of $(f_n)_n$ and get that every subsequence now has a uniformly convergent (on compact subsets) sub-sub-sequence. Since $(f_n)_n$ converges pointwise, these limits must coincide with $f$.

But then, the orignal sequence must converge to $f$ too (in the topology of uniform convergence on compact subsets), for assume otherwise, then there exists $\varepsilon > 0$ such that for all $n$ there exists $k(n)\geq n$ such that $d(f_{k(n)},f) > \varepsilon$ (the metric is the one from $H(U)$). This contradicts the fact that $(f_{k(n)})_n$ should have a subsequence that converges to $f$. $\square$

I also seem to have come up with another way to show this, without using Montel's theorem:

Proof: Let $K \subseteq U$ be compact and choose $V\subseteq U$ open such that $\overline V$ is compact and $K\subseteq V \subseteq \overline V \subseteq U$.

Since $\overline V$ is compact, we have $$ |f_n(z)| \leq c(\overline{V}), \qquad(z \in V) $$ where $c(\overline V)$ is the uniform bound of the $f_n$ on $\overline V$. The constant function $z \mapsto c(\overline V)$ is Lebesgue-integrable on $V$ and dominates $(f_n)_n$ on $V$. Since $f_n \to f$ pointwise, we can apply the dominated convergence theorem to obtain $f_n \to f$ in $L^1(V)$. In particular, $(f_n)_n$ is Cauchy in $L^1(V)$.

Now consider the Bergman space $A^1(V)$ of absolutely integrable holomorphic functions on $V$. Since $A^1(V)$ has the same norm as $L^1(V)$ and all $f_n$ are holomorphic, we get that $(f_n)_n$ is a Cauchy sequence in $A^1(V)$. But then, by the fundamental estimate for Bergman spaces (see e.g. the book of Krantz linked in the question; the proof there works exactly the same way for $A^1$ instead of $A^2$ and doesn't need the assumption of connectedness), there is a constant $\tilde C(K)$ such that $$ \sup_{z \in K} |f_n(z) - f_m(z)| \leq \tilde{C}(K) \|f_n - f_m\|_{A^1(V)}.$$ (Such an estimate holds for all compact sets in $V$.) Thus, $(f_n)_n$ is also Cauchy wrt. uniform convergence on $K$, hence convergent by completeness. $\square$

So basically, this replaces Montel's theorem with the dominated convergence theorem and some facts about Bergman spaces as the key ingredient.

To apply this to my original problem, one would pick $\tilde K \subseteq \Omega \times \Omega$ compact and choose $V$ as above such that $(\mathrm{pr}_1(\tilde K)\cup \mathrm{pr}_2(\tilde K)) \subseteq V \subseteq \overline V \subseteq \Omega$, where $\mathrm{pr}_i : \mathbb C^{2n} \to \mathbb{C}^n$, $i=1,2$ are the projections of the first, respectively second $n$ components to $\mathbb C^n$. Then take $\tilde V := V \times V$. This is necessary since the uniform bound of the Bergman kernel's series representation works only on sets of the form $K \times K$ for $K$ compact in $\Omega$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.