Dual space of weighted $L^p(\omega)$ Let $\omega \in A_p$, where $A_p$ is the family of Muckenhoupt weights. I'm wondering what is the topological dual space of $L^p(\omega)$. Is it isometrically isomorphic to $L^q(\omega)$? (1/p + 1/q = 1) I have the feeling that the answer would be something like $L^q(\omega ^{1/q})$ but I don't know how to check this.
My attemp: If $f \in L^p(\omega)$ and $g \in L^q(\omega)$ then $f \omega^{1/p} \in L^p$ and $g \omega^{1/q} \in L^q$ and hence, by Hölder inequality, $fg \in L^1(\omega)$. I'm not sure what this means. Should $fg$ belong to unweighted $L^1$ of is this right? Does this mean $<f,g>$ can be interpreted as a weighted pair of duality in some sense? I think I have some confussion with the concept "pair of duality".
Any clarification or help with all this confusion would be really really appreciated. Thanks.
 A: The duality between $L^p$ and $L^q$ is true for any measure space . In particular, it holds for $L^p\bigl(\mathbb{R}^n,w\,dx)$. The fact that $w\in A_p$ is unnecessary. The pairing between $L^p(w)$ and $L^q(w)$ is given by
$$
\langle f,g\rangle=\int_{\mathbb{R}^n}f\,g\,w,\quad f\in L^p(w),g\in L^q(w).
$$
A: This is a natural question. It depends on how people define and use the inner product.
Given a Borel measure $w$, if we look at the inner product
$$<f,g>_w  = \int f(x)g(x)w(x)dx$$
Then the duality of $L^p(w)$ is $L^q(w)$, when $1/p + 1/q =1$.
But if we look at $Ap$ weight directly, then the $L^p(w)$ is closely related to the index of $w$, i.e., when w is in $Ap$, then we will only consider $L^p(w)$, but not $L^q(w)$ for any other $q$. For example, to show the Hilbert transform or Riesz transform to be bounded on $L^p(w)$, $w\in Ap$, et al.
In this case, we need to consider the usual inner product:
$$<f,g> = \int f(x)g(x)dx$$
So, we will obtain that for $w\in Ap$,  $( L^p(w) )'    = L^q (w')$, where $w'$ is the conjugate of $w$,  and $1/p +1/q =1$.
