The main result of the paper mentioned in the title says that a reflexive subspace contained in the Fourier image of \(L^1({\mathbb {R}}^n)\) and closed in the sup-norm is of finite dimension. If \(n=1\), the result was proved earlier by K. Karlander (see [10] in the reference list to the above paper).

Let me note that the phenomenon is in fact of a more general nature: the group structure and, consequently, the Fourier transformation are not quite relevant.

Specifically, let \(X\) and \(Y\) be two metrizable locally compact spaces first countable at infinity. We denote by \(C_b(Y)\) the space of all bounded continuous functions on \(Y\) with the sup-norm, and by \(C_0(Y)\) its subspace that consists of functions vanishing at infinity. Next, let \(\mu \) be a nonnegative (not necessarily finite) Radon measure on \(X\), and let \(A:X\times Y\rightarrow {\mathbb {C}}\) be a bounded continuous function. Define an operator \(T:L^1(\mu )\rightarrow C_b(Y)\) by

$$\begin{aligned} Tg(y)=\int _X g(x)A(x,y)d\mu (x). \end{aligned}$$

(This operator takes values in \(C_b(Y)\) indeed by the Lebesgue dominated convercence theorem, see III.6.16 in [1]). I shall give a short proof of the following statement, which includes the result mentioned above as a partial case.

FormalPara Proposition

Let \(E\) be a reflexive subspace of \(L^1(\mu )\) such that the space \(F=T(E)\cap C_0(Y)\) is closed in the sup-norm. Then \(F\) is of finite dimension.

For the proof, we need a lemma. For \(K\subset Y\), we denote by \(R_K\) the operator taking each function on \(Y\) to its restriction to \(K\).

FormalPara Lemma

If \(F\) is a reflexive subspace of \(C_0(Y)\), then for some compact subset \(K\) of \(Y\) the operator \(R_K:F\rightarrow C(K)\) is an isomorphic embedding.

FormalPara Proof

If not, it is easy to construct a sequence of “almost disjoint” functions in \(F\). They span a subspace of \(F\) isomorphic to \(c_0(=C_0({\mathbb {N}}))\), which is not reflexive, see Table IV A in [1]. Since \(F\) is reflexive, this is impossible (see II.3.23 in [1]), and the lemma follows. \(\square \)

Now, we prove the proposition. We claim that the composition \(R_KT\) restricted to \(E\) is a compact operator for every compact subset \(K\) of \(Y\).

Suppose the claim is proved. We observe that \(F\) in the proposition is reflexive. Indeed, the closed space \(E_1=E\cap T^{-1}(F)\) is reflexive as a subspace of the reflexive space \(E\); see II.3.23 in [1]. Next, \(T\) maps \(E_1\) onto \(F\), and by the open mapping principle (see II.2 in [1]) the image of the closed unit ball of \(E_1\) contains a closed ball centered at zero in \(F\). It remains to use the fact that a Banach space is reflexive if and only if some closed ball of it centered at zero is weakly compact, see V.5.7 in [1].

Since \(F\) is reflexive, the lemma applies to it. Combined with the above claim, it implies that the closed unit ball of \(F\) is compact. Hence, \(F\) is finite-dimensional, see IV.3.5 in [1].

It remains to prove the claim. Let \(B\) denote the closed unit ball of \(E\). By the Arzela–Ascoli theorem (see IV.6.7 in [1]), it suffices to show that the functions in \(TB\) are equicontinuous on an arbitrary compact subset \(K\) of \(Y\). Let \(\varepsilon >0\). Since \(B\) is weakly compact in \(E\) and hence in \(L^1\), there is a compact set \(S\subset X\) such that \(\int _{X\setminus S}|f|d\mu \le \varepsilon \) for all \(f\in B\); see IV.8.9 in [1]. Next, the function \(A\) is uniformly continuous on \(S\times K\) (see I.6.18 in [1]), hence there exists \(\delta >0\) such that \(|A(x,y_1)-A(x,y_2)|\le \varepsilon \) whenever \(x\in S\), \(y_1,y_2\in K\), and \({\mathrm {dist}}(y_1,y_2)<\delta \). Now, let \(C\) be an upper bound for the modulus of the function \(A\). If \({\mathrm {dist}}(y_1,y_2)<\delta \) and \(y_1,y_2\in K\), then for every \(f\in B\) we have

$$\begin{aligned} |(Tf)(y_1)-Tf(y_2)|\le \int _S|f(x)||A(x,y_1)-A(x,y_2)|d\mu (x)+2C\varepsilon \le (1+2C)\varepsilon , \end{aligned}$$

and the claim follows.