I'm currently studying unbounded operators on hilbert spaces and wanted to show that the multiplication operator $$ T: D(T) \subset L^2(\mathbb{R}) \to L^2(\mathbb{R}) $$$$ f(t) \mapsto tf(t) $$ is self-adjoint. There i was faced with the following problem: If $$ \int (fg)(t) dt = 0 \quad \forall f\in D(T), $$ $D(T)$ being dense in $L^2(\mathbb{R})$, can i say that $g=0$ almost everywhere, even though $g$ is not necesseraliy in $L^2(\mathbb{R})$? Sure, if $g \in L^2(\mathbb{R})$ would hold i could just argue with the scalar product but that's basically what i want to prove. Am i just missing something?
EDIT: Here the situation:
$$ \int f(t)(tg(t)-h(t)) = 0 \quad \forall f\in D(T), $$ Here $g,h \in L^2(\mathbb{R})$ and i want to show that $tg(t) = h(t)$ so that $tg(t) \in L^2(\mathbb{R})$. $D(T)$ is defined as the space of all $f \in L^2(\mathbb{R})$ that satisfy: $tf(t) \in L^2(\mathbb{R})$.