Villaretq0

2022-06-25

I would like to know, why:

If $X$ is a subspace of ${L}^{p}(G)$ such that $\overline{X}\ne {L}^{p}(G)$, then there exists $g\in {L}^{q}(G)$, $\frac{1}{p}+\frac{1}{q}=1$ such that

${\int}_{G}f(x)\phantom{\rule{thinmathspace}{0ex}}g(x)\phantom{\rule{thinmathspace}{0ex}}dx=0;\phantom{\rule{1em}{0ex}}\mathrm{\forall}f\in \overline{X}.$

Where $G$ is a locally compact group with Haar measure $dx$.

If $X$ is a subspace of ${L}^{p}(G)$ such that $\overline{X}\ne {L}^{p}(G)$, then there exists $g\in {L}^{q}(G)$, $\frac{1}{p}+\frac{1}{q}=1$ such that

${\int}_{G}f(x)\phantom{\rule{thinmathspace}{0ex}}g(x)\phantom{\rule{thinmathspace}{0ex}}dx=0;\phantom{\rule{1em}{0ex}}\mathrm{\forall}f\in \overline{X}.$

Where $G$ is a locally compact group with Haar measure $dx$.

svirajueh

Beginner2022-06-26Added 29 answers

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

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