Introduction

Abstract

Let g ∈ L²([0,1]) and fix 2 < r < ∞. The corresponding multiplication operator T : L^r([0,1]) → L¹([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

$$ \begin{gathered} \left\| {Tf} \right\|_{L^1 ([0,1])} = \int_0^1 {\left| {f(t)} \right| \cdot \left| {g(t)} \right|dt} \hfill \\ \leqslant \left\| g \right\|_{L^{r'} ([0,1])} \left\| f \right\|_{L^r ([0,1])} , \hfill \\ \end{gathered} $$

for f ∈ L^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 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 L^p([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 L^p([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 ∈ L^p([0, 1]), in which case T : f ↦ gf is continuous by the Closed Graph Theorem. Since ξ:h↦∫ ¹₀ h(t) is a continuous linear functional on L¹([0, 1]) it follows that the composition ξ∘T:f↦∫ ¹₀ 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 L²([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 L¹([0, 1])-valued extension to Z? If so, then we have just seen that Z cannot be an L^p([0, 1])-space.