Villaretq0

2022-06-25

I would like to know, why:
If $X$ is a subspace of ${L}^{p}\left(G\right)$ such that $\overline{X}\ne {L}^{p}\left(G\right)$, then there exists $g\in {L}^{q}\left(G\right)$, $\frac{1}{p}+\frac{1}{q}=1$ such that
${\int }_{G}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}g\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx=0;\phantom{\rule{1em}{0ex}}\mathrm{\forall }f\in \overline{X}.$
Where $G$ is a locally compact group with Haar measure $dx$.

svirajueh

Since $\overline{X}$ is a proper closed subspace of the Banach space ${L}^{p}\left(G\right)$, the quotient space ${L}^{p}\left(G\right)/\overline{X}$ is a nonzero Banach space. So there is some nonzero linear functional $\varphi$ on ${L}^{p}\left(G\right)/\overline{X}$, whose composition with the quotient map is a nonzero linear functional on ${L}^{p}\left(G\right)$ that vanishes on $\overline{X}$. Finally that functional corresponds to some $g\in {L}^{q}\left(G\right)$ since ${L}^{q}$ is the dual of ${L}^{p}$.