$\begingroup$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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$\begingroup$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.
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