# Entire extension of the Fouier transform of a smooth compactly supported function

The statement which is not very clear is this: if $\phi \in C^{\infty}_0$ then its Fourier transform extends to a complex-analytic function.

Of course the candidate extension is $\hat{\phi}(z)=\int e^{-i<x,\xi+i\eta>}\phi(x) dx$ where $x,\xi,\eta$ are in $\mathbb{R}^n$ and $z=\xi+i\eta$. First of all this integral is absolutely convergent since $\phi$ has compact support, and this is ok. Then we prove estimates on the derivatives: $$\partial_{z_j}\hat{\phi}(z)= \int e^{-i<x,z>}(-i)x_j\phi(x)dx$$ and supposing the support of $\phi$ included in the ball centered at the origin with radius $A$ we can prove the estimate $$|\partial_{z_j}\hat{\phi}(z)|\leq c_je^{A |Im(z)|}.$$ And analogously

$$|\partial^{\alpha}_{z}\hat{\phi}(z)|\leq c_{\alpha}e^{A |Im(z)|}$$ where $\alpha$ is a multindex. I don't understand how these estimates prove the analyticity of the function $\hat{\phi}(z)$. Shouldn't we prove estimates of the type $|\partial^{\alpha}f| \leq |\alpha !|C^{|\alpha|}$?

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It's a general fact that if $f(z,t)$ is analytic in $z$ for every $t$ and the integral $F(z)=\int_X f(z,t)\,d\mu(t)$ converges locally uniformly wrt $z$, then $F$ is analytic. Probably the easiest way to show this is to use Morera's theorem. It is generally more pleasant to integrate an integral than to differentiate it.
Thank you. What is the variabile $t$ in this example? Isn't there any way to compute the classical estimates (if they are correct also in several complex variable)? – balestrav Jun 27 '12 at 20:03
@balestrav $t$ can be any variable of integration for an arbitrary measure space. Higher dimensions should not be a problem: as far as I remember, a separately analytic function is analytic. – user31373 Jun 27 '12 at 20:16