Abstract:
Let $(\Omega, \mu)$ be a measurable space with $\sigma$-finite continuous measure, $\mu(\Omega) = \infty.$ A linear operator $T: L_1(\Omega) + L_\infty(\Omega)\to L_1(\Omega) + L_\infty(\Omega)$ is called the Dunford–Schwartz operator if $\|T(f)\|_1 \leqslant \|f\|_1$ (respectively, $\|T(f)\|_{\infty} \leqslant \|f\|_{\infty}$) for all $f\in L_1(\Omega)$ (respectively, $f\in L_\infty(\Omega)$). If $\{T_t\}_{t\geqslant 0} $ is a strongly continuous in $L_1(\Omega)$ semigroup of Dunford–Schwartz operators, then each operator $A_t(f) = \dfrac1t \int\limits_0^tT_s(f)ds \in L_1(\Omega),$ $f\in L_1(\Omega)$ has a unique extension to the Dunford–Schwartz operator, which is also denoted by $A_t,$$t>0.$ It is proved that in the completely symmetric space $E(\Omega) \nsubseteq L_1$ of measurable functions on $(\Omega, \mu)$ the means $A_t$ converge strongly as $t\to +\infty$ for each strongly continuous in $L_1(\Omega)$ semigroup $\{T_t\}_{t\geqslant 0}$ of Dunford–Schwartz operators if and only if the norm $\|\cdot\|_{E(\Omega)} $ is order continuous.