# Introduction

Chapter

## Abstract

Let g ∈ L for f ∈ L

^{2}([0,1]) and fix 2 < r < ∞. The corresponding*multiplication operator*T : L^{r}([0,1]) → L^{1}([0,1]) defined by T : f ↦ gf, for f ∈ L^{r}([0,1]), is surely continuous. Indeed, if \( \frac{1} {r} + \frac{1} {{r'}} = 1 \), then g ∈ L^{ŕ}([0,1]) and so, by Hölder’s inequality$$
\begin{gathered}
\left\| {Tf} \right\|_{L^1 ([0,1])} = \int_0^1 {\left| {f(t)} \right| \cdot \left| {g(t)} \right|dt} \hfill \\
\leqslant \left\| g \right\|_{L^{r'} ([0,1])} \left\| f \right\|_{L^r ([0,1])} , \hfill \\
\end{gathered}
$$

^{r}([0, 1]). Accordingly, \( \left\| T \right\| \leqslant \left\| g \right\|_{L^{r'} ([0,1])} \). Somehow, this estimate awakens a “feeling of discontent” concerning*T*. It seems more natural to interpret*T*as being defined on the larger domain space L^{2}([0, 1]), that is, still as f ↦ gf (with values in L^{1}([0, 1])) but now for all f ∈ L^{2}([0, 1]). Again Hölder’s inequality ensures that the extended operator T : L^{2}([0, 1]) → L^{1}([0, 1]) is continuous with \( \left\| T \right\| = \left\| g \right\|_{L^2 ([0,1])} \geqslant \left\| g \right\|_{L^{r'} ([0,1])} \). Of course, for each 2 ≤ p ≤ r, we can also consider the L^{1}([0, 1])-valued operator*T*as being defined on L^{p}([0, 1]). In the event that*g*∉∪_{2<q<∞}*L*^{ q }([0,1]), the space L^{2}([0, 1]) is the “best choice” of domain space for*T*amongst all L^{p}([0, 1])-spaces in the sense that it is the*largest*one that*T*maps into L^{1}([0, 1]). For, suppose that there exists p ∈ [1, 2] such that gf ∈ L^{1}([0, 1]) for all f ∈ L^{p}([0, 1]), in which case T : f ↦ gf is continuous by the Closed Graph Theorem. Since ξ:*h*↦*∫*_{0}^{1}*h*(*t*) is a continuous linear functional on L^{1}([0, 1]) it follows that the composition ξ∘*T*:*f*↦*∫*_{0}^{1}*f*(*t*)*g*(*t*) is a continuous linear functional on L^{p}([0, 1]) and so*g*∈*L*^{p′}([0,1]) with \( \frac{1} {p} + \frac{1} {{p'}} = 1 \). Since 2≤*p*′≤∞, this is only possible if p = 2 (by the assumption on*g*). Can there exist another function space*Z*(over [0, 1]), even larger than L^{2}([0, 1]), which contains L^{r}([0, 1]) continuously and such that the initial operator*T*:*L*^{ t }([0,1])→^{1}([0,1]) has a continuous L^{1}([0, 1])-valued extension to*Z?*If so, then we have just seen that*Z*cannot be an L^{p}([0, 1])-space.## Keywords

Banach Space Function Space Integration Operator Banach Lattice Vector Measure
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

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© Birkhäuser Verlag AG 2008