Show that f is a polynomial

Suppose $f$ is an entire function on $\mathbb{C}^n$ that satisfies for every $\epsilon>0$ a growth-condition $$|f(z)|\leq C_{\epsilon}(1+|z|)^{N_{\epsilon}}e^{\epsilon | \text{Im}\,z|}$$

Show that $f$ is a polynomial. (Hint: study $\hat{f} = \mathcal{F}(f)$ the Fourier-transform).

I know I'm supposed to apply the Paley-Wiener-Schwartz Theorem, but not sure how.;.

Any suggestions and/or tips are greatly appreciated. Thnx.

• The theorem implies that the distribution $\hat{f}$ is supported in EVERY open ball containing $0$. Therefore evaluating $\hat{f}$ on a function $g$ must depend locally on $g$ at $0$. Can you show that the only such distributions are linear combinations of $D^n\vert_0$? – Aaron Apr 25 '13 at 21:46
• Hm. I dont quite know were the proof should go. But I do know that $F(z^n) = (-2\pi)^n D^{n}\delta$. So we might want to prove that we are dealing with linear combinations of dirac-delta's...but how is this Theorem's gonna help us with that. I think Im missing some basic understanding of what this theorem implies. – DinkyDoe Apr 29 '13 at 11:14
• So if we prove that $F(f)$ satisfies another grow-condition that somehow shows that $F(f)$ is a distribution being supported in $\{0\}$ then we're done I guess. – DinkyDoe Apr 29 '13 at 11:17

Well, I think it should be clear as suggested by Aaron that the Schwartz's Theorem in your link implies that for every $\epsilon >0$ you have a distribution $v$ supported in the closed ball $B(0,\epsilon)$, and $f = \hat{v}$. So we have the distribution $v$ with $f = \hat{v}$ and $supp(v) = \{0\}$. Hence it is a well known result that for some $k$ $$v = \sum_{|\alpha | = k}C_\alpha D^\alpha \delta$$ Hence $f(\xi) = \hat{v}(\xi) = \sum_{|\alpha | = k}C_\alpha \hat{(D^\alpha \delta)}(\xi)= \sum_{|\alpha | = k}C_\alpha (2\pi i \xi)^\alpha$ is indeed a polynomial.