Background: I'm working on a project where the beloved Hilbert space $L^2$ is replaced by $L^1$ and I'm searching for a suitable generalisation of the inner product using duality.
The space of Lebesgue-integrable functions $L^1$ is separable and thus I can choose a (Schauder) basis $(u_\alpha)_{\alpha\in\mathbb{N}}$. I know that the dual space of $L^1$ is isometrically isomorphic to $L^\infty$, which is not separable. In particular, the dual set $(u_\alpha^*)_{\alpha\in\mathbb{N}}$ is too small to span $L^\infty$. (I do deliberately identify functionals in $(L^1)^*$ with functions in $L^\infty$ by Riesz theorem.)
My question is: what is $\overline{\mathrm{span}} (u_\alpha^*)_{\alpha\in\mathbb{N}}$? How do I need to restrict $L^\infty$ so that the elements of this subspace can be expanded in $(u_\alpha^*)$?
After long research in text books and on the web, some thoughts are:
- one has to restrict to the continuous dual space, i.e., bounded continuous functions $C_b \subset L^\infty$. But this space is still not separable, so too large
- one may need to remove null functionals from $(L^1)^*$, i.e., divide $L^\infty$ by the annihilator of the set $(u_\alpha) \in L^1$. But I'm not so familiar with factor spaces and have no idea how to proceed ...
- ... ?
After all, would such a separable subspace of $L^\infty$ depend on the choice of $(u_\alpha)$? Even if this is not the case, there may be a "sensible" subspace that is relevant for applications.
If needed, the domain $X$ of the functions in $L^1$ can be assumed to be bounded. However, I believe this is irrelevant as the Lebesgue measure is $\sigma$-finite.