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One of my homework problems this week is to "characterize all holomorphic functions in $L^2(\Bbb C^n)$". I'm sorry for not being able to provide much work on my progress, but that is because I really don't know where to begin. Any help would be greatly appreciated!

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Write down some holomorphic functions (say when $n = 1$ for simplicity) and check whether they lie in $L^2$. Repeat. Make a conjecture. See if you can prove it. –  Qiaochu Yuan Apr 10 '12 at 4:30
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Well if a function is in $L^2(\mathbb{C}^n)$, what can you say about it as $|z| \rightarrow \infty$? Then ask yourself what holomorphic functions can satisfy such a condition. –  Michael Joyce Apr 10 '12 at 4:33
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@MichaelJoyce: I'm not sure what you're getting at but I don't think there is a simple condition satisfied by $L^2$ functions at infinity. –  Eric O. Korman Apr 10 '12 at 4:50
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The mean value problem for holomorphic functions will probably be useful. –  Eric O. Korman Apr 10 '12 at 4:51
    
@Eric: Oops, I initially thought $L^2$ + continuous would imply bounded, but that's not the case. –  Michael Joyce Apr 10 '12 at 5:01
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2 Answers 2

This question is a little old, but for some reason it was bumped to the front page and I thought I'd provide another answer, demonstrating the applicability of some basic techniques from the theory of distributions. The conclusion obtained by Davide is a corollary to the slightly stronger statement that the polynomials are the only holomorphic functions on $\mathbb{C}^n$ which are also tempered distributions. The argument is based on the fact that a holomorphic function $f$ is necessarily a solution to an elliptic PDE, namely $\Delta f = 0$, and the benefit of this approach lies in the fact that it generalizes easily to any solution $f$ of an equation $Lf = 0$, where $L$ is an elliptic differential operator. Denote by $\mathscr{S}$ the Schwartz space in $\mathbb{C}^n \cong \mathbb{R}^{2n}$, and by $\mathscr{S}'$ the space of tempered distributions.

For the proof, let $f$ be a holomorphic function on $\mathbb C^n$ which also lies in $\mathscr{S}'$. One can express the Laplacian $\Delta$ as $$ \Delta = 4\sum_{j = 1}^n\partial_{z_j}\partial_{\bar{z}_j}, \tag{1} $$ where the differential operators $\partial_{z_j}$ and $\partial_{\bar{z}_j}$ are defined by $$ \partial_{z_j} = \frac{1}{2}\left(\frac{\partial}{\partial x_j} + i\frac{\partial}{\partial y_j}\right) \quad \text{and} \quad \partial_{\bar{z}_j} = \frac{1}{2}\left(\frac{\partial}{\partial x_j} - i\frac{\partial}{\partial y_j}\right). $$ By assumption $\partial_{\bar{z}_j}f = 0$ for $1 \leq j \leq n$, so we see by $(1)$ that $\Delta f = 0$ as well.

Because $f\in \mathscr{S}'$, we can apply the Fourier transform to the equation $\Delta f = 0$, finding that $$ c |z|^2\hat{f} = \widehat{\Delta f} = 0 \tag{2} $$ as tempered distributions; here $c$ is a nonzero constant which depends on the particular normalization of the Fourier transform used, and $|z|^2\hat{f}$ is the distribution defined for $\varphi \in \mathscr{S}$ by $|z|^2\hat{f}(\varphi) = \hat{f}(|z|^2\varphi)$. It follows at once from $(2)$ that $\hat{f}$ is supported at the origin: If $\varphi \in \mathscr{S}$ is supported on the open set $A_r = \{z:|z| > r\}$ for some $r > 0$, then $|z|^{-2}\varphi(z) \in \mathscr{S}$, so $$ \hat{f}(\varphi) = |z|^2\hat{f}(|z|^{-2}\varphi) = 0. $$ Taking the union of the sets $A_r$, we see that $\hat{f}$ is supported at the origin as claimed. Any distribution supported at the origin can be represented as a linear combination of the $\delta$-function and its derivatives (there is a bit of work hiding in this statement), so in particular we can find a positive integer $N$ and constants $c_\alpha$ such that $$ \hat{f} = \sum_{|\alpha| \leq N} c_\alpha \partial_x^\alpha \delta. $$ Now by applying the inverse Fourier transform we see that $f$ is a polynomial. If $p < \infty$, the only polynomial in $L^p(\mathbb{C}^n)$ is the zero polynomial, and the only polynomials in $L^\infty(\mathbb{C}^n)$ are constants, so we get the same result as Davide.

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It's a part of the exercise 1.4 in R. Michael Range's book, Holomorphic functions and integral representation in several complex variables, Springer-Verlag.

We will see that holomorphic is too restrictive for integrable functions.

We use Cauchy's integral formula. Let $f\in\mathcal O(\mathbb C^n)\cap L^p(\mathbb C^n)$, with $1\leq p<\infty$. Fix $a=(a_1,\ldots,a_n)\in\mathbb C^n$. We define $\gamma^r(\theta):=(\gamma_j^r(\theta_j))_{j=1}^n$, where $\gamma_j^r(\theta)=a_j+r_je^{i\theta}$ and denote $\partial_j$ the partial derivative with respect to $z_j$. We get after passing to a volume integral
$$|\partial_jf(a)|\rho^{e_j+1}\leq \frac 1{(2\pi)^n}\int_{[0,2\pi]^n}|f(\gamma^{r\rho}(\theta))|\rho_1\ldots\rho_nd\theta_1d\theta_2$$ and integrating over $\rho_1,\ldots,\rho_n$ from $0$ to $r$, we get $$|\partial_jf(a)|\leq \frac 3{(2\pi)^nr^3}\lVert f\rVert_{L^p(\mathbb C^n)}\operatorname{vol}([0,2\pi]^n)^{\frac p{p-1}},$$ so letting $r\to +\infty$ it gives $\partial_jf(a)=0$ so $f$ is constant. Since it's $p$-integrable, $f$ needs to be $0$.

For the case $p=+\infty$, we have $|\partial_jf(a)|\leq \frac 3{(2\pi)^nr^3}\lVert f\rVert_{L^{\infty}(\mathbb C^n)}\operatorname{vol}([0,2\pi]^n)$ so $f$ is constant.

Conclusion: $\mathcal O(\mathbb C^n)\cap L^p(\mathbb C^n)=\begin{cases}\{0\}&\mbox{ if }1\leq p<\infty\\\ \{z\mapsto c, c\in\mathbb C^n\}&\mbox{ if }p=+\infty. \end{cases}$

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