inverse Fourier transform of compactly supported function is in $L^1$ Let $N > d/2$, and $N$ is chosen such that it is an integer. Let $f\in C^N(\mathbb{R^d})$ and $f$ has compact support. I want to show that $f$ is the Fourier transform of a function $g\in L^1(\mathbb{R^d})$.
I really don't know where to start with this, other than just writing down the definition of $\check{f}$, but that doesn't get me very far. I don't know why we're given the information that $f\in C^N$.
 A: Claim: if $f \in C^N(\mathbb R^d)$, then $\int_{\mathbb R^d} |\check f (\xi)|^2 (1+\xi^2)^N d \xi < \infty$. 
Suppose for a moment that the claim is true. Then, by Holder's inequality, $$\|\check f \|_{L^1} \leq \|\check f(\xi)(1+|\xi|^2)^{N/2}\|_{L^2} \|(1+|\xi|^2)^{-N/2}\|_{L^2}.$$
Since $N>d/2$, it is easy to verify that $(1+|\xi|^2)^{-N/2} \in L^2$, and so $\check f \in L^1$.
To see the claim, note that since $f$ and all of its partial derivatives of order up to $N$ are continuous with compact support, they are in $L^2$, and therefore $$(\sum\limits_{|\alpha| \leq N} \|\partial^\alpha f\|^2_{L^2})^{1/2} < \infty.$$ On the other hand, by Plancharel's theorem and the fact that (up to some $i$'s and some $2 \pi$'s that don't matter for the argument and depend on how you define the Fourier transform) $\mathcal F^{-1} (\partial^\alpha f) (\xi) = \xi^\alpha \check f (\xi)$, we have
$$\sum\limits_{|\alpha| \leq N} \| \partial^\alpha f\|_{L^2}^2 = \sum\limits_{|\alpha| \leq N} \int |\check f(\xi)|^2 |\xi^\alpha|^2 d\xi,$$ so proving the claim amounts to showing that $\sum\limits_{|\alpha| \leq N} |\xi^\alpha|^2 \geq C (1 + |\xi|^2)^{N}$ for some constant C.
To see this, note that by binomial expansion, 
$$(1 + |\xi|^2)^N \leq 2^N \max (1,|\xi|^{2N}) \leq 2^N(1 + |\xi|^{2N}).$$ Now, $|\xi|^{2N}$ and $\sum_{i=1}^d |\xi_j^N|^2$ are both homogeneous of degree $2N$, and are therefore comparable by comparing their max and min values on the unit sphere. Thus $(1 + |\xi|^2)^N \leq 2^N C (1 + \sum_{i=1}^d |\xi_i^N|^2)$. For $|\xi| \geq 1$, this last quantity is controlled by $\sum_{|\alpha|=N} |\xi^N|^2$. On the other hand, for $|\xi| < 1$, $(1 + |\xi|^2)^N < 2^N$ and $|\xi^0|^2 = 1$.
The main part of this argument was cribbed from Folland's pde book.
