I am currently working on the Exercise 4.10 in Stein's Complex Analysis:
Suppose $f(z)$ is an entire function that satisfies $$|f(x+iy)|\leq ce^{-ax^2+by^2}$$ for some $a,b,c>0$. Let $$\hat{f}(\zeta)=\int_{-\infty}^\infty f(x)e^{-2\pi i x\zeta}dx$$ Prove that $\hat{f}$ is an entire function of $\zeta$ that satisfies $$|\hat{f}(\xi+i\eta)| \leq c'e^{-a'\zeta^2+b'\eta^2}$$ for some $a',b',c'>0$.
I was able to prove that $|\hat{f}(\zeta)|\leq c'e^{-a'\zeta+b'\eta^2}$. However, I'm having a hard time showing that $\hat{f}$ is entire. Any hints or suggestions would be greatly appreciated! Having thought about it some more, how about the following by the Liebniz integral rule:
$$\hat{f}(\zeta) = \int_{-\infty}^\infty f(x)e^{-2\pi i x\zeta} dx= \int_{-\infty}^\infty f(x)e^{2\pi x\eta}\cos(2\pi z \xi)dx - i\int_{-\infty}^\infty f(x)e^{2\pi x\eta}\sin(2\pi x\xi)dx$$ Then, $$\frac{\partial}{\partial \xi}\hat{f}(\zeta) = -\int_{-\infty}^\infty f(x)e^{2\pi x \eta}\cdot 2\pi x\sin(2\pi x\xi)dx-i\int_{-\infty}^\infty f(x)e^{2\pi x\eta}\cdot 2\pi x\cos(2\pi x \xi)dx$$ and $$i\frac{\partial}{\partial \eta}\hat{f}(\zeta) = i\int_{-\infty}^\infty 2\pi xf(x)e^{2\pi x \eta}\cdot \cos(2\pi x\xi)dx+\int_{-\infty}^\infty 2\pi xf(x)e^{2\pi x\eta}\sin(2\pi x \xi)dx$$ It follows that $\frac{\partial}{\partial \overline{z}}\hat{f}(\zeta) = 0$. Is this sufficient to show that $\hat{f}$ is entire?