S is a dense subset of $L^{p'}$, $\int_{E}fg = 0$ for all $f \in S$, then $g= 0$

Question

Problem:

$E$ is a measurable set and $1 \leq p < \infty$. Let $p′$ be the conjugate of $p$, and $S$ is a dense subset of $L^{p′}(E)$. Show that if $g \in L^p(E)$ and $\int_{E}fg = 0$ for all $f \in S$, then $g= 0$.

Definition of Density: $S$ is dense in $L^{p'}(E)$ if $\forall h \in L^{p'}(E), \forall \epsilon > 0, \exists f \in S$ s.t. $\left| \left| f-h \right| \right|_{p'} < \epsilon$ or equivalently $\exists (f_n)$ in $S$ s.t. $\lim_{{n}\to{\infty}}f_n=h$ a.e. on $E$.

Idea?: As p and p' are conjugates, I was thinking to use Holder's Inequality.

$\int_{E}\left| fg \right| \leq \left| \left| f \right| \right|_{p'} \left| \left| g \right| \right|_p$

Answer 1

This problem was probably designed for you to use duality, namely the fact that if $1 \leq p < \infty$ and $g \in L^{p}(E)$, then $$ \| g \|_{L^p(E)} = \sup_{\| h\|_{L^{p’}(E)}=1} \left| \int_E gh \right|. \tag{1} $$ (See, e.g. Grafakos Volume I, Chapter 1.) We will henceforth use $\| \cdot \|_p$ and $\| \cdot \|_{p’}$ to denote the norms $\| \cdot \|_{L^p(E)}$ and $\| \cdot \|_{L^{p’}(E)}$, respectively.

To prove that $\|g \|_{p} = 0$, let $\varepsilon > 0$ be fixed. For any $h \in L^{p’}(E)$, we can find $f \in S$ with $$ \| h - f \|_{p'} \leq \frac{\varepsilon}{\| g\|_{p} + 1}. $$ (We have put $\| g\|_p + 1$ in the denominator to cover the case that $\| g \|_p = 0$. We could have assumed that $\| g \|_p \neq 0$, but that would have turned the proof into a proof by contradiction, which I find significantly less appealing.) For this $h$ and $f$, it follows from Hölder that $$ \begin{split} \left| \int_E gh \right| &\leq \left| \int_E gf \right| + \left| \int_E g(h-f) \right| = \left| \int_E g(h-f) \right| \\ &\leq \| g \|_{p} \, \| h-f \|_{p'} \leq \varepsilon. \end{split} $$ Since $\varepsilon$ was arbitrary, we conclude that $\|g \|_{p} = 0$.

Answer 2

Notice that $$ X=\{f \in L^{p'}\ |\ \int_E fg=0\} $$ is a closed subspace as a kernel of the continuous linear functional $L:f\mapsto \int_E fg$. By the assumption that $S\subset X$ and $\overline{S}=L^{p'}$, we have that$$ L^{p'} =\overline{S} \subset X, $$ that is, $X=L^{p'}$. This gives $L\equiv 0$ and by Riesz representation theorem, we have $$ \|g\|_{L^{p}}=\|L\|=0, $$ i.e. $g =0$ in $L^{p}$.