Exponential inequalities for a class of operators (Q1607945)

The paper is a very interesting contribution to the study of exponential-logarithmic type integral operators \[ E_kf(x):=\exp(K\log f)(x), \] where \(K\) is the positive integral operator, given by \[ (Kf)(x)=\int_0^\infty k(x,y)f(y) dy, \] where \(f\) is a non-negative function and the kernel \(k\) is assumed to meet a few quite natural restrictions. Specific examples of such operators have been studied in various situations by many authors, in particular, for example, when \(K\) is the Hardy averaging operator or some of its modifications. It is known (cf., for instance, [\textit{H. P. Heinig, R. Kerman} and \textit{M. Krbec} [Georgian Math. J. 8, 69-86 (2001; Zbl 1009.26014)] and the references therein) that the condition \[ \sup_{y>0}\int_{0}^{\infty}\left[\frac{y}{x}k(x,y)\right]w(x) dx<\infty \leqno {(1)} \] is sufficient for the weighted Knopp-type inequality \[ \int_{0}^{\infty}u(x)(E_kf)(x) dx\leq C \int_{0}^{\infty}v(x)f(x) dx,\leqno {(2)} \] here \(u,v\) are weights and \(w(x)=u(x)\exp (K\log(1/v))(x)\). In specific situations (usually involving operators of Hardy-type), it is also known that the converse implication holds, that is, (1) is also necessary for (2). In the relevant papers, however, it was assumed, sometimes somewhat implicitly, that \(E_k f\) and/or \(E_u(1/v)\) exist almost everywhere and are Lebesgue measurable. The author demonstrates with examples that this is not necessarily the case and gives necessary and sufficient conditions of \(f\) (and \(v\)) for this to be satisfied. He also shows that (1) is equivalent to (2) in fairly general context (the main innovation being, of course, the `only if' part, i.e., the necessity of (2) for (1)). These two main results are then illustrated with non-trivial examples involving the power-weight Hardy averaging operator \[ (P_\beta f)(x):=\beta x^{-\beta}\int_{0}^{x}t^{\beta-1}f(t) dt,\qquad \beta>0, \] the Riemann-Liouville operator \[ (I_\alpha f)(x):=\frac{\alpha}{x^\alpha}\int_{0}^{x}(x-y)^{\alpha-1}f(y) dy,\qquad \alpha\geq 1,\;x>0, \] and the generalized Laplace transform \[ (L_a f)(x):=\frac{1}{x\Gamma(1+1/a)} \int_{0}^{\infty}e^{-(y/x)^a}f(y) dy, \qquad a\geq 1, x\geq 0. . \]