On the structure of invertible elements of the convolution algebra \({\mathcal S}^ \prime_ k\)
The setting is the Schwartz space \({\mathcal S}_k\) on \(\mathbb{R}^n\) for \(k\) a negative integer; that is, the set of \(\varphi\in C^{( \infty)} (\mathbb{R}^n)\) such that \((1+|x|^2 )^k \varphi^{( \alpha)} (x)\) vanishes at infinity for each \(\alpha\in (\mathbb{Z}_+ )^n\). Since \(e^{ix \cdot y}\in {\mathcal S}_k\) as a function of \(x\in \mathbb{R}^n\) for each fixed \(y\in \mathbb{R}^n\), the Fourier transform of \(f\) in the dual space \({\mathcal S}'_k\) can be defined by the pairing \(F(y):= \langle f(x), e^{ix \cdot y}\rangle\). The authors show that there exists a polynomial \(P\) such that \(|\partial^m F(y) |\leq P(|y|)\) for all \(y\in \mathbb{R}^n\) and all \(m\in (\mathbb{Z}_+ )^n\) with \(|m|\leq |2k|-1\). As a partial converse they prove that if \(F\in C^{(|2k|+ 2n)} (\mathbb{R}^n)\) satisfies \(|\partial^m F(y) |\leq P(|y|)\) for some polynomial \(P\), all \(y\in \mathbb{R}^n\) and all \(m\in (\mathbb{Z}_+)^n\) with \(|m|\leq |2k|+ 2n\), then \(F\) is the Fourier transform of some \(f\in {\mathcal S}'_k\). They use this to produce (non-zero) divisors of zero in the convolution multiplication of \({\mathcal S}'_k\) which they had previously investigated [same Journal 190, No. 3, 829-843 (1995; Zbl 0827.46035)], and to find sufficient conditions on \(g,h\in {\mathcal S}'_k\) which insure that the convolution equation \(h*f =g\) has a unique solution \(f\in {\mathcal S}'_k\). Because \(P(\partial) u=(P (\partial) \delta) *u\) for any polynomial \(P\), this result supplies a sufficient condition for the solvability of partial differential equations \(P(\partial) u=v\), namely, that none of the zeros of \(P\) in \(\mathbb{C}^n\) lie in \(i\mathbb{R}^n\). [The proof of Lemma 2.1 begins with the sentence: ``Since \(g(t)\) is continuous on \(\mathbb{R}^n\), the function \(G(y)\) is integrable.'' But the hypotheses and conclusion of the lemma make clear that the authors intend this sentence to read: ``The function \(g(t)\) is continuous on \(\mathbb{R}^n\) because \(G(y)\) is integrable.''].
scientific article; zbMATH DE number 861778