# Introduction

Part of the Operator Theory: Advances and Applications book series (OT, volume 180)

## Abstract

Let g ∈ L2([0,1]) and fix 2 < r < ∞. The corresponding multiplication operator T : Lr([0,1]) → L1([0,1]) defined by T : f ↦ gf, for f ∈ Lr([0,1]), is surely continuous. Indeed, if $$\frac{1} {r} + \frac{1} {{r'}} = 1$$, then g ∈ Lŕ ([0,1]) and so, by Hölder’s inequality
$$\begin{gathered} \left\| {Tf} \right\|_{L^1 ([0,1])} = \int_0^1 {\left| {f(t)} \right| \cdot \left| {g(t)} \right|dt} \hfill \\ \leqslant \left\| g \right\|_{L^{r'} ([0,1])} \left\| f \right\|_{L^r ([0,1])} , \hfill \\ \end{gathered}$$
for f ∈ Lr([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 L2([0, 1]), that is, still as f ↦ gf (with values in L1([0, 1])) but now for all f ∈ L2([0, 1]). Again Hölder’s inequality ensures that the extended operator T : L2([0, 1]) → L1([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 L1([0, 1])-valued operator T as being defined on Lp([0, 1]). In the event that g∉∪2<q<∞L q ([0,1]), the space L2([0, 1]) is the “best choice” of domain space for T amongst all Lp([0, 1])-spaces in the sense that it is the largest one that T maps into L1([0, 1]). For, suppose that there exists p ∈ [1, 2] such that gf ∈ L1([0, 1]) for all f ∈ Lp([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 L1([0, 1]) it follows that the composition ξ∘T:f 0 1 f(t)g(t) is a continuous linear functional on Lp([0, 1]) and so gLp([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 L2([0, 1]), which contains Lr([0, 1]) continuously and such that the initial operator T:L t ([0,1])→1([0,1]) has a continuous L1([0, 1])-valued extension to Z? If so, then we have just seen that Z cannot be an Lp([0, 1])-space.

## Keywords

Banach Space Function Space Integration Operator Banach Lattice Vector Measure
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