Compact Fourier multiplier? Let us take some continuous $m\in L^\infty(\mathbb R^d)$ and suppose $m(\xi)\to 0$ as $|\xi|\to \infty$. Let $T_m:L^2(\mathbb R^d)\to L^2(\mathbb R^d)$ be the associated Fourier multiplier operator, i.e. $\widehat{T_m f} = m \widehat{f}$. As we know in general these Fourier multipliers are NOT compact operators. The reason for this is that $\mathsf R(m)$ might contain some open set, and for compact operators we require the spectrum to have at most one accumulation point, namely zero.
Set $\mathcal X := [-\frac{1}{2},\frac{1}{2}]^d$. Now I am trying to find a way to make compact the restriction of $T_m$ to the subspace $L^2(\mathcal X) := \{ f \in L^2(\mathbb R^d) : \mathrm{supp}(f) \subseteq \mathcal X\}$. I have a few reasons to believe it might be compact on this subspace, but I feel like I am missing some crucial background knowledge. My reasons are the following:

*

*We can represent any $f\in L^2(\mathcal X)$, for $x\in\mathcal X$ by

$$ f(x) = \sum_{k\in\mathbb Z^d} \hat{f}(k) e^{2\pi i x \cdot k}, $$
with $\hat{f}(k) = \int_{\mathcal X} f(x) e^{-2\pi i x\cdot k} \ dx$. Now for Fourier series (on the torus), compactness of Fourier multipliers is much easier: $T_m$ is compact whenever $m\to 0$.  This looks a lot like our assumption above. However, it is not immediately clear to me how we can use the assumption above.


*If we had the additional assumption that $m \in L^2(\mathbb R^d)$, then we could take the IFT of $m$ to find that in fact $\mathcal F^{-1} m \in L^2(\mathbb R^d)$, so the operator $T_m$ would act on $L^2(\mathcal X)$ like a convolution with an $L^2(\mathcal X\times\mathcal X)$ function. This operator is then a Hilbert-Schmidt operator, since we are only convolving with functions with support on $\mathcal X$. Note however that this assumption is not necessary, as this approach only would require $\mathcal F^{-1} m \in L^2_{\mathrm{loc}}(\mathbb R^d)$, not $L^2(\mathbb R^d)$.


*If instead we had the additional assumption that $m \in L^1(\mathbb R^d)$, we could say $\mathcal F^{-1} m\in L^\infty(\mathbb R^d)$, and we could again view $T_m$ on $L^2(\mathcal X)$ as a convolution with an $L^2(\mathcal X \times \mathcal X)$ kernel. Again this assumption is too strong, as we would in this case only need $\mathcal F^{-1}m\in L^\infty_{\mathrm{loc}}(\mathbb R^d)$.
Now, looking at $[\mathcal F^{-1} m] 1_{\mathcal X}$ also does not give anything useful (as far as I can tell). If we take the FT of $[\mathcal F^{-1} m] 1_{\mathcal X}$, we end up with
$$ (m \star \widehat{1_{\mathcal X}})(\xi) = \int_{\mathbb R^d} m(x)\prod_{j=1}^d \mathrm{sinc}(\xi_j - x_j) \ dx. $$
And these sinc integrals are terrible to deal with, because $\mathrm{sinc}$ is not $L^1$, there is almost no way to estimate these integrals without losing important information. I have tried searching on the internet and in the literature for ways to solve these integrals and the approach people use is... a Fourier transform to get the indicator function...
I really am at a loss here and feel like I am missing something. Surely someone must have tried something similar before?
Any help is appreciated, thanks.
 A: The operator is indeed compact on $\mathcal X$. To prove this, we need to use the $L^2$ version of the theorem of Ascoli and Arzelà, and while answering this question I found a nice little paper of Robert Pego with it;
https://www.ams.org/journals/proc/1985-095-02/S0002-9939-1985-0801333-9/S0002-9939-1985-0801333-9.pdf

Let us see the details. As in the question, let $m\colon \mathbb R^d \to \mathbb C$ be such that $m(\xi)\to 0$ as $\lvert \xi \rvert \to \infty$, and define
$$
T_m f = \mathcal F^{-1}( m \hat f),$$
where $\mathcal F$ and $\widehat{}$ denote the Fourier transform. We claim that this operator of $L^2(\mathbb R^d)$ into itself is compact on
$$\mathcal X=\{f\in L^2\, :\, \text{supp}(f)\subset B_1\}.$$
(Here $B_1$ denotes the unit ball of $\mathbb R^d$, but any subspace of functions supported in some common bounded set will yield the same result).
To prove this, we consider a sequence $f_n\in \mathcal X$ such that $\lVert f_n\rVert_2\le 1$. As stated in the linked paper, to prove that $\{T_m f_n\}$ is a relatively compact subset of $L^2$ we need to check two conditions;

*

*Equiboundedness. the norm $\lVert T_m f_n \rVert_2$ must be uniformly bounded;

*Equicontinuity. $$\lim_{R\to \infty}\sup_{n\in \mathbb N}\int_{\lvert \xi\vert \ge R} \lvert \widehat{T_m f_n}(\xi)\rvert^2\,d\xi=0.$$
(There is also a third condition - equivanishing -, but it is automatically satisfied since all $f_n$ are supported in the same bounded set.)
Now, the first condition is obviously satisfied, since $T_m$ is bounded. The second is satisfied because of the assumption on $m$, since
$$\int_{\lvert \xi\vert \ge R} \lvert \widehat{T_m f_n}(\xi)\rvert^2\,d\xi = \int_{\lvert \xi\vert \ge R} \lvert m(\xi)\rvert^2 \lvert \hat f_n(\xi)\rvert^2\,d\xi\le \sup_{\lvert \xi \rvert \ge R} \lvert m(\xi)\rvert^2, $$
and the right-hand side tends to $0$. $\Box$
