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I know that the fourier transform $\hat f$ of a $L^1$ function $f$ is (even uniformly) continuous by estimating $\lvert \hat f(x+ h) - \hat f(x)|$ and using DCT.

Now given a function $f\in C_c^\infty(\mathbb R^n)$ (smooth and compactly supported in the ball $\{\lvert x \rvert \leq R\}$) and $\zeta \in \mathbb C^n$, the expression $$\hat f(\zeta) := (2\pi)^{-n/2} \int_{\mathbb R^n} e^{-i\zeta\cdot x} f(x) \, dx$$ is well-defined. Is there a similar way to show that this function is (uniformly?) continuous? I ultimately want to prove analyticity by Osgood's lemma but I need continuity first. So far I couldn't apply the technique used for the real case since I end up with something like$$\lvert \hat f(\zeta) - \hat f(\zeta + h)| \leq (2\pi)^{-n/2}\int\lvert f(x) \rvert e^{R \cdot\text{ Im } \zeta}\lvert1-e^{-ih\cdot x} \rvert dx.$$

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For continuity, let $\zeta_k \to \zeta_0.$ Then $\{\zeta_k: k=0,1,\dots\}$ is a bounded set, say bounded by $M$ in absolute value. So

$$|e^{-i\zeta \cdot x}| \le e^{|\zeta| |x|} \le e^{MR},$$

for all $\zeta \in \{\zeta_k: k=0,1,\dots\}$ and $|x|\le R.$ The DCT argument you already know will now work.

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  • $\begingroup$ Nice method! Thank you :) $\endgroup$ – Staki42 Sep 1 '18 at 16:23
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Hint

You can use the Continuity under Integral Sign theorem or mimic its demonstration using DCT theorem.

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  • $\begingroup$ I will try proving it using this theorem, thanks! $\endgroup$ – Staki42 Sep 1 '18 at 16:22

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