Uniform convergence of the Bergman kernel's orthonormal basis representation on compact subsets

Question

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.

Answer 1

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!

Answer 2

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 $f$. $\square$

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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 $\Omega$.