Aronszajn's Theorem and the Bergman Kernel

Theorem

Aronszajn's Theorem [^1]

A Hilbert space $X$ has a reproducing kernel $K$ iff there exists, for any $y_0 \in A$, a positive constant $C_{y_0}$, depending upon $y_0$, such that

$$ |f (y_0) | \leq C_{y_0} |f| \quad \forall f \in X $$

Proof

The "only if" part is proved by applying Schwarz' inequality to $f (y_0) = (f, K (\cdot, y_0))$:

$$ |f (y_0) | \leq |f| \cdot |K (\cdot, y_0)|, $$

thus

$$ |f (y_0) |^2 \leq |f|^2 \cdot |K (\cdot, y_0)|^2 = |f|^2 \cdot (K (y_0, y_0)), $$

so

$$ |f (y_0) | \leq |f| \sqrt{K (y_0, y_0)} $$

The "if" part is proved by applying F. Riesz' representation theorem to the linear functional $F_{y_0} (f) = f (y_0)$ of $f \in X$. Thus, there exists a uniquely determined vector $g_{y_0} (x)$ of $X$ such that

$$ f (y_0) = F_{y_0} (f) = (f, g_{y_0} (x)) \forall f \in X $$

and so $g_{y_0} (x) = K (x, y_0)$ is a reproducing kernel of $X$. The proof shows that the reproducing kernel is uniquely determined.

Corollary

We have

$$ \sup |f (y_0) | = \sqrt{K (y_0, y_0)}, $$

the supremum being attained by

$$ f_0 (x) = \frac{K (x, y_0)}{|K (\cdot, y_0)|} \quad \text{with } |f_0 | = 1. $$

Example

Consider the Hilbert space $A^2 (G)$. For any $f \in A^2 (G)$ and $z \in G$, we have (see (4) in Chapter I, 9)

$$ |f (z_0) |^2 \leq \frac{\int_{|z - z_0 | \leq r} |f (z) |^2 \hspace{0.17em} dx dy}{\pi r^2} \quad (z = x + iy) $$

Thus $A^2 (G)$ has the reproducing kernel which will be denoted by $K_G (z, z')$. This $K_G (z, z')$ is called Bergman's kernel of the domain $G$ of the complex plane.

Bibliography

[^1]: Kōsaku Yosida. Functional Analysis. Classics in Mathematics. Springer Berlin Heidelberg, Reprint of the 1980 Edition edition, 1995.