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Let $\Omega$ be a bounded domain of $\mathbb{C}^n$ and $f$ be a holomorphic function defined on $\Omega$.

Is it possible that $L^2$-norm of $f$ is bounded but $f$ itself is unbounded?

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My stomach feeling is no! In the sense that I don´t see how an holomorphic function on a bounded domain can be unbounded, a priori of any examination of the $L^2$-norm. EDIT: Now I see it... – Giovanni De Gaetano Feb 9 '12 at 14:43
@Student73 Yes it can, you may very well have singularities at the boundary of the domain. See theory of univalent functions, $H^p$-spaces, Bergman spaces, Dirichlet spaces etc... and my answer below. – AD. Feb 10 '12 at 8:54

Yes there it is often the case. The space you look at is called the Bergman space often denoted by $A^2(\Omega)$ or $L^2_a(\Omega)$.

To give a counter-example let us consider a simple case say $\Omega=\mathbb{D}=\{z\in\mathbb{C}: \,|z|<1\}$ - the unit disc.

  1. Note that the Hardy space $H^2=\{f\in L^2(\mathbb{T}):\, f$ is analytic in $\mathbb{D}\}$ (here $\mathbb{T}=\partial\mathbb{D}$ is the unit circle), normed by the $L^2$-norm (see remark below) is included in the Bergman space because $$\|f\|_{H^2}^2=\sup_{0 \lt r \lt 1}\int_0^{2\pi}|f(re^{it})|^2\frac{dt}{2\pi}\geq \int_0^1\int_0^{2\pi}|f(re^{it})|^2\frac{dt}{2\pi} dr \qquad\qquad\qquad\qquad\qquad$$ $$\qquad\qquad\qquad\geq \int_0^1\int_0^{2\pi}|f(re^{it})|^2\frac{dt}{2\pi} rdr =\|f\|_{A^2}^2$$

  2. By Parseval's formula an analytic function $f(z)=\sum a_nz^n$ is in $H^2$ precisely when the sequence $(a_n)$ is in $\ell^2$ and then $$\|f\|_{H^2}^2=\sum |a_n|^2$$

  3. Combining 1. and 2. shows that $z\mapsto\log(1-z)=\sum_{n\geq1} \frac{z^n}{n}$ belongs to $A^2$, but is unbounded at $z=1$.

Remark: For each $f \in H^2$ there is a boundary function $\tilde{f}\in L^2(\mathbb{T})$ such that $\lim_{r\to1}f(re^{it})=\tilde{f}(e^{it})$, and the $H^2$-norm of $f$ coinside with the $L^2$-norm of $\tilde{f}$.

Remark 2: The observation 1. above can be improved to show $H^p\subset A^{2p}$, which is one of the few Hardy-Littlewood theorems (see 5.11 in P. Duren: Theory of $H^p$-spaces, 2nd ed 2000).

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A simple example would be the following: Let $\Omega$ be the right half of the unit disk and consider the function $$f(z):={\rm Log}(z) =\log r + i\phi\ ,\qquad z=r e^{i\phi}, \ 0<r<1, \ |\phi|<{\pi\over2}\ .$$

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Yes, there are such functions (even if $n = 1$).

The space of square integrable holomorphic functions is usually called the Bergman space, often denoted $A^2$. For a concrete example of an unbounded function in $A^2$, let $\mathbb{D}$ be the unit disc, and choose $$ f(z) = \log(1-z) $$ (where $\log$ denotes the principal branch of the complex logarithm). Then, clearly $f$ is unbounded near $z = 1$, and holomorphic on $\mathbb{D}$. It is not too hard to show that $f \in A^2$. (The only problem is close to $z = 1$, and there $|f| \approx \ln|1-z|$. I'll leave the details as an exercise.)

In the argument given by user15464 (local $L^2$ estimates), the bound on $|f(z)|$ depends on $b$ -- which is the distance from $z$ to $\partial\Omega$. This estimate blows up as $z \to \partial\Omega$. However, these estimates show that point evaluations, i.e. the functionals $$f \mapsto f(z) $$ are continuous for each $z \in \Omega$. This observation is usually the starting point in the study of Bergman spaces. For example, the Bergman space is a Hilbert space (under the natural inner product), and by Riesz' representation formula, for each $z \in \Omega$, there is a $B_z \in A^2$ such that $$f(z) = \int_\Omega f(w)\overline{B_z(w)}\,dw$$ The function $$B(z,w) = \overline{B_z(w)}$$ is called the Bergman kernel of $\Omega$. It is a nice exercise to compute this for the unit disc. The result turns out to be $$B(z,w) = \frac{1}{(1-z\bar w)^2}$$

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I believe the answer to this is no. Let $p\in \Omega$ and let $D(p,b)$ be a disk contained in $\Omega$ centered at $p$. Switching to polar coordinates, we have

$\|f\|_2 ^2 \geq \int_{D(p,b)} |f(x+iy)|^2 dxdy = \int_0 ^b \int_0 ^{2\pi} |f(re^{i\theta}+p)|^2 r d\theta dr $

$.\geq \int_0 ^b |\int_0 ^{2\pi} f(re^{i\theta}+p)^2 d\theta|r dr$

Notice that

$\int_0 ^{2\pi} f(re^{i\theta}+p)^2 d\theta = \frac{1}{i}\int_0 ^{2\pi} \frac{f(re^{i\theta}+p)^2}{re^{i\theta} + p - p} ire^{i\theta} d\theta = \frac{1}{i} \int_{D(p,r)} \frac{f(z)^2}{z-p} dz = 2\pi f(p)^2$

by the Cauchy integral theorem. Therefore

$\int_{D(p,b)} |f(x+iy)|^2 dxdy \geq \pi b^2 |f(p)^2|.$

So, $|f(p)|$ is bounded by $\frac{1}{\sqrt\pi b}\|f\|_2$.

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If I understood correctly you are bounding the function with the "local" $L^2$-norm (i.e. the integral of the square in the disc) and not with the (larger) "global" $L^2$-norm. Since any holomorphic function has locally finite $L^2$-norm that would imply that you are avoiding the last hypothesis of the question. Far from being critic I´m just curious, is this true? – Giovanni De Gaetano Feb 9 '12 at 15:27
The local $L^2$ norms are each bounded by the global $L^2$ norm since the integral of the non-negative function $|f|^2$ over a subset is at most the integral over the entire domain. – user15464 Feb 9 '12 at 15:47
Yes! I gave you a +1 because I think your proof works! I was just pointing out that it seems it uses weaker hypothesis that the ones proposed by the OP! – Giovanni De Gaetano Feb 9 '12 at 16:43
Is it true that for bounded domain $\Omega$, we can take $b(p)$ for each $p\in M$ such that $\operatorname{inf}_{p\in \Omega} b(p)>0$? I guess for an unbounded domain, there exists a domain such that choosing such $b(p)$ is impossible. – holomorphicfunction22 Feb 10 '12 at 2:35
What happens with the radius $b$ when $p$ is close to the boundary? – AD. Feb 10 '12 at 8:55

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