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Consider a function $f\in L^2(\mathbb{R}^d)$ with $\|f\|_{2}=1$ and such that $\hat{f}$ is supported on the ball $B(0,1)$. I am wondering which is the best decay that $f$ can have.

I read on "G. Björck: Linear partial differential operators and generalized distributions" that you can construct an $f$ for which $$|f(x)|\leqslant Ce^{-|x|^\alpha}, x\in\mathbb{R}^d$$ where $0<\alpha<1$. Are there better estimates? And in particular for this estimate, which is the minimal value we can take of $C$ (provided that $\|f\|_{2}=1$)?

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Regarding the first question, the answer is basically no. You cannot have exponential decay, because that would imply $\hat f$ is analytic, which is inconsistent with compact support.

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Fourier transform of a function with bounded support is an entire function of exponential type. Even in dimension 1, such function cannot have decay as $e^{-\epsilon x}$. In fact, $$\int_{-\infty}^\infty\frac{\log|f(x)|}{1+|x|^2}dx<\infty$$ for every such $f\in L^2$. This condition is almost optimal, at least in dimension $1$. How close it is to being optimal, is the subject of famous (and difficult) theorems of Beurling-Mailliavin. The simplest of these theorems says that if $\omega>0$ is uniformly continulus and $$\int_{-\infty}^\infty\frac{\omega(x)}{1+|x|^2}dx,$$ then for every $\delta>0$ there exists an function $f\in L^2$ whose Fourier transform is supported on $[-\delta,\delta]$ and $\log|f|\leq\omega$.

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