# Bergman norm on a bigger domain

Let $D$ be a unit disc (I am actually interested in a much more general setting, but let's start with explicit examples). Let $E$ be an open subset of $D$. Consider the functional on the space of all holomorphic functions on $D$ defined by $\|f\|^2_E=\int_{E}|f(z)|^{2}dA(z)$. I need to find out for which $E$ this norm defines a Banach space of holomorphic functions. I've reduced this problem to the following: does there exist a sequence $f_1,f_2,...$ of holomorphic functions on $D$ and a sequence of points $z_1,z_2,...$, converging in $D\backslash E$, such that $\|f\|_E\le 1$, but $|f_n(z_n)|\to\infty$?

Off-course, if $D=E$, then the space is the classical Bergman Space, which is a Banach Space. The situation is similar in the case when $D\backslash E$ is a compact, i.e. $E$ "covers" the boundary.

I have very limited knowledge of these kind of matters, but I believe, that the answer depends on how $D\backslash E$ "intersects" with the unit circle. For example, it would be interesting to consider $D\backslash E$ to be a horde in $D$, solid triangle with a vertex on $\partial D$ or a disc of a smaller radius touching $\partial D$. My guess is that the first two give rise to a Banach space, while the last one doesn't.

Thank you.

Consider the sequence of functions $$f_n(z):=e^{n((1-z)^{-3}-10)},$$ and take $E:=\{z:\Re\mathfrak{e}((1-z)^{-3}-10)<0\}$, $z_n\equiv 0.9$. This shows that the assertion might fail even if $D\setminus E$ is contained in a solid (euclidean) triangle with one vertex on the boundary.

On the other hand, it is obvious that if the intersection of $D\setminus E$ with some neighborhood of the boundary has zero area, then $$|f(z)|\leq\frac{1}{2\pi \epsilon}\int_{1-\epsilon}^1dr\int_{|\zeta|=r}\left|\frac{f(\zeta)}{z-\zeta}\right|\leq\frac{1}{2\sqrt{\pi \epsilon}(\epsilon-|z|)}||f||_{L_2(E)},$$ and the assertion follows.

Update: You can take the vertices of the triangle to be $1,\frac{i}{2},-\frac{i}{2}$. If $z$ satisfies $\frac{\pi}{2}<\mathrm{arg}(z-1)<\frac{5\pi}{6}$ or $\frac{7\pi}{6}<\mathrm{arg}(z-1)<\frac{3\pi}{2}$, then $z\in E$, because $\Re\mathfrak{e}(1-z)^{-3}<0$. If $\Re\mathfrak{e} z<0$, then $\Re\mathfrak{e}(1-z)^{-3}-10<|1-z|^{-3}-10<-9$.

• I'm sorry, could you please elaborate, why in your example $D\backslash E$ is contained in a triangle with one vertex on the boundary? I can't imagine this figure at all.
– erz
Feb 25 '16 at 3:31
• @erz: see update Feb 25 '16 at 7:54
• thank you very much; apparently $D\backslash E$ cannot reach the boundary at all; I'm going to think a bit, and then edit the question accordingly.
– erz
Feb 25 '16 at 18:49

If I understand your question properly, then your space is always a Banach space. This follows from the abstract result that if $B$ is a closed, bounded, absolutely convex subset of a complete locally convex space, then the linear space spanned by it is a Banach space with $B$ as unit ball. In your case, the locally space is that of the holomorphic functions on $E$ and $B$ is the set of such functions for which your expression is at most $1$. We note that this set is bounded in $H(E)$. This follows from the well-known and easily proved fact that a subset of the latter space is bounded if and only it is locally bounded, i.e., every point in $E$ is the centre of a disc on which the family is uniformly bounded.

• Well, this is exactly how I thought; what I need to understand if $B$ is bounded.
– erz
Feb 24 '16 at 17:25