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.


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.


Unable to display preview. Download preview PDF.

Unable to display preview. Download preview PDF.

Copyright information

© Birkhäuser Verlag AG 2008

Personalised recommendations