Bounded composite integral operators (Q2906715)

Let \((X,{\mathcal X},\mu)\) be a \(\sigma\)-finite measure space. The set of real numbers is denoted by \(\mathbb R\), and an increasing convex function \(m:\mathbb R\to(0,\infty)\) is said to be a Young function if \(m(-x)= m(x)\), \(m(0)= 0\), \(m(x)\to \infty\) (as \(x\to\infty\)). The Orlicz space \((L^m(X),\|\cdot\|_m)\) with norm \(\|\cdot\|_m\) is defined by NEWLINE\[NEWLINEL^m(X)= \{f: X\to\mathbb R\text{ measurable:} \int_Xm(\alpha f)\,d\mu<\infty\text{ for some }\alpha> 0\},NEWLINE\]NEWLINE NEWLINE\[NEWLINE\| f\|_m= \text{inf}\Biggl\{a> 0: \int_Xm(|f|/a)\,d\mu\leq 1\Biggr\}.NEWLINE\]NEWLINE The Young function \(m\) is said to satisfy the \(\Delta_2\)-condition if the numbers \(x_0> 0\), \(k> 0\), may be determined such that \(m(2x)\leq km(x)\) for \(x> x_0\).NEWLINENEWLINE If \(1\leq p<\infty\) and \(m_p(x)= x^p\), then \(L^{m_p}(X)\) is denoted by \(L^p(X)\). If \(K: X\times X\times\mathbb R\to\mathbb R\) is such that \(K(\cdot ,\cdot ,u)\) is measurable on \(X\times X\) for \(u\in\mathbb R\), if the integral operator \(T_K\) is defined by NEWLINE\[NEWLINET_K(f)(x)= \int_X K(x,y,f(y))\,d\mu(y),NEWLINE\]NEWLINE and \(T_K: L^p(X)\to L^p(X)\) is bounded, then the composite integral operator \(T_{K,\phi}\) is defined in terms of transformation \(\phi: X\to X\) by NEWLINE\[NEWLINET_{K,\phi}(f)(x)= \int_X K(x,\phi(y), f(\phi(y))\,d\mu(y).NEWLINE\]NEWLINE The main theorems of this paper indicate conditions on the measure \(m\) or kernel \(K\) to ascertain that \(T_{K,\phi}: L^m(X)\to L^m(X)\) is bounded. Continuity results are derived when \(K\) satisfies Lipschitz conditions of the form NEWLINE\[NEWLINE|K(s,t,u)-K(s,t,v)|\leq {\mathcal L}\in L^2(X\times X).NEWLINE\]