The integral operator of the form $$\begin{aligned} \bigl (Nu\bigr )(x)=\sum _{k=1}^\infty e^{i\langle \omega _k,x\rangle } \int _{\mathbb {R}^c}n_k(x-y)\,u(y)\,dy \end{aligned}$$ ( N u ) ( x ) = ∑ k =… Click to show full abstract
The integral operator of the form $$\begin{aligned} \bigl (Nu\bigr )(x)=\sum _{k=1}^\infty e^{i\langle \omega _k,x\rangle } \int _{\mathbb {R}^c}n_k(x-y)\,u(y)\,dy \end{aligned}$$ ( N u ) ( x ) = ∑ k = 1 ∞ e i ⟨ ω k , x ⟩ ∫ R c n k ( x - y ) u ( y ) d y acting in $$L_p(\mathbb {R}^c)$$ L p ( R c ) , $$1\le p\le \infty $$ 1 ≤ p ≤ ∞ , is considered. It is assumed that $$\omega _k\in \mathbb {R}^c$$ ω k ∈ R c , $$n_k\in L_1(\mathbb {R}^c)$$ n k ∈ L 1 ( R c ) , and $$\begin{aligned} \sum _{k=1}^\infty \Vert n_k\Vert _{L_1}<\infty . \end{aligned}$$ ∑ k = 1 ∞ ‖ n k ‖ L 1 < ∞ . We prove that if the operator $$\mathbf {1}+N$$ 1 + N is invertible, then $$(\mathbf {1}+N)^{-1}=\mathbf {1}+M$$ ( 1 + N ) - 1 = 1 + M , where M is an integral operator possessing the analogous representation.
               
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