Extending linear continuous operator on subspace of $L^\infty$?

Question

Consider $S$ to be the subspace of $L^\infty(\mathbb R)$ consisting of all functions supported on finite measure sets and $T:S\to B$ is a linear continuous operator between $S$ and some Banach space $B$, is there always a linear continuos mapping $F:L^\infty(\mathbb R)\to B$ extending $T$?

Note that $S$ is not dense in $L^\infty(\mathbb R)$ and if $B=\mathbb R$, this is just Hahn-Banach. But what if $B$ is a general Banach space? What if $B=L^p(\mathbb R)$ for any $p\in [1,\infty]$? I believe we should use Hahn-Banach, but I don't know how, could any one hint?

Answer 1

I am not sure about the general case, but suspect it impossible.

If $B$ admits a Markushevich basis $\{(f_\alpha,x_\alpha)\}_\alpha$, then, for each $\alpha$, the functional $f_\alpha T$ admits a norm-preserving extension $g_{\alpha}\in L^{\infty}(\mathbb{R})\to B$. Moreover, since $\sum_\alpha{f_\alpha T}=T$ converges in the strong operator topology, by the principle of uniform boundedness, $\sum_\alpha{g_\alpha}$ converges in that same topology. It is easy to see that said sum is the desired extension of $T$.

The above almost entirely handle the particular cases of $B=L^p(\mathbb{R})$. For $p<\infty$, the Haar wavelets on $[n,n+1]$ (for all $n$ and all frequencies) form a Schauder (and thus Markushevich) basis for $L^p$.