C\|f\|_{X} , \quad f\in X.\end{aligned}$$ In an important special case when $T$ is the identity operator, we say that $X$ is *embedded* into $Y$ and write $X\hookrightarrow Y$. If $T'$ is another operator defined at least on $Y'$ with values in $\mathcal{M}(R,\mu )$ and such that $$\begin{aligned}
\label{E:duality-of-operators}
\int _{R}(Tf) g\,{\mathrm{d}}\mu =\int _{S}f(T'g)\,\mathrm{d}\nu\end{aligned}$$ for every $f\in X$ and $g\in Y'$, then $T\colon X\to Y$ is equivalent to $T'\colon Y'\to X'$.

Let $P$ and $Q$ be the integral operators defined by $$\begin{aligned}
(Pf)(t)=\frac{1}{t}\int _{0}^{t} f(s)\,{\mathrm{d}}s,\quad t\in (0,\infty ),\end{aligned}$$ and $$\begin{aligned}
(Qf)(t)=\int _{t}^{\infty }f(s)\frac{{\mathrm{d}}s}{s}, \quad t\in (0,\infty ),\end{aligned}$$ for those functions on $f\in \mathcal{M}_{0}(0,\infty )$ for which the respective integrals have sense. As an interchange of integration shows, $$\begin{aligned}
\int _{0}^{\infty }(Pf)(t)g(t)\,{\mathrm{d}}t=\int _{0}^{\infty }f(t)(Qg)(t)\,{\mathrm{d}}t,\end{aligned}$$ for all $f$ and $g$ for which the integrals make sense. Hence, the operators $P$ and $Q$ are formally adjoint with respect to the $L^{1}$-pairing and therefore satisfy a relation in the spirit of . As a consequence, one has the equivalence $$\begin{aligned}
\label{E:equivalence-of-P-Q}
P\colon X\to Y\quad \Leftrightarrow \quad Q\colon Y'\to X'\end{aligned}$$ for every pair of r.i. spaces $X,Y$ over $(0,\infty )$ (with the same operator norm). Another important example is that when $(R,\mu )$ is arbitrary and 