Understanding the dual of a weighted Fourier transform space I'm studying Tao's Dispersive PDE book. In section 2.6, he discusses $X^{s,b}_{\tau=h(\xi)}$ spaces, which are defined by the norm
$$ \| u\|_{X^{s,b}_{\tau=h(\xi)}} = \| \langle \xi \rangle^s \langle \tau -
 h(\xi) \rangle^b \hat{u}(\xi,\tau) \|_{L^2_\xi L^2_\tau}.$$
Here $u: \mathbb{R}^d \times \mathbb{R} \rightarrow \mathbb{C}$, $\xi \in \mathbb{R}^d$, $\tau \in \mathbb{R}$, and $\langle y \rangle = \sqrt{1+y^2}$. 
The book says it's an application of Parseval and Cauchy-Schwarz to show that the dual of $X^{s,b}_{\tau = h(\xi)}$ is 
$$ X^{-s,-b}_{\tau = -h(-\xi)}.$$
I can see where the negative $s$ and $b$ come from, but I don't see where the change $h \rightarrow -h(- \cdot)$ comes from. For simplicity, I assume $s=0$ and $b=1$. Then Parseval gives
$$ \iint u \bar{v} \; dx dt = \iint \hat{u} \bar{\hat{v}} \; d\xi d \tau = \iint \Bigl(\langle \tau - h(\xi)\rangle\hat{u} \Bigr)  \Bigl(\langle \tau - h(\xi) \rangle^{-1} \bar{\hat{v}} \Bigl) d \xi d \tau.$$
Cauchy-Schwarz bounds this by 
$$ \| u\|_{X^{0,1}_{\tau=h(\xi)}} \| \langle \tau - h(\xi) \rangle^{-1} \bar{\hat{v}}\|_{L^2_\xi L^2_\tau} = \| u\|_{X^{0,1}_{\tau=h(\xi)}} \| \langle \tau + h(-\xi) \rangle^{-1} \bar{\hat{v}}(-\xi, -\tau)\|_{L^2_\xi L^2_\tau}.$$
But I can't see where to go from here. Any advice would be appreciated.
 A: You introduced $\int u \overline{v}$ which is not a bilinear form since Bourgain's spaces $X_{\tau = h(\xi)}^{s,b}$ are complex vector spaces. Thus, you should rather consider
$$\begin{cases}\mathcal{B}:  &S(\mathbb{R}^{d+1})\times S(\mathbb{R}^{d+1}) &\to &\mathbb{C} \\
&(u,v) &\mapsto &\int u v .\end{cases}$$
which defines a bilinear form. By Plancherel's formula we have 
\begin{aligned}
\left\vert  \int u(t,x) v(t,x)dtdx  \right\vert &= \left\vert \int \mathcal{F}u(\tau,\xi)\overline{\mathcal{F}\overline{v}}(\tau,\xi) d\tau d\xi\right\vert \\
&= \left\vert \int \mathcal{F}u(\tau,\xi)\mathcal{F}v(-\tau,-\xi)d\tau d\xi \right\vert \\
&\leq \left\Vert\left<\tau - h(\xi)\right> \mathcal{F}u(\tau,\xi) \right\Vert_{L_\tau^2L_\xi^2} \left\Vert\left<\tau - h(\xi)\right>^{-1} \mathcal{F}v(-\tau,-\xi) \right\Vert_{L_\tau^2L_\xi^2} \\
&= \left\Vert\left<\tau - h(\xi)\right> \mathcal{F}u(\tau,\xi) \right\Vert_{L_\tau^2L_\xi^2} \left\Vert\left<\tau + h(-\xi)\right>^{-1} \mathcal{F}v(\tau,\xi) \right\Vert_{L_\tau^2L_\xi^2}
\end{aligned}
which is the desired estimate. It allows us to extend the bilinear form $\mathcal{B}$ to $X_{\tau = h(\xi)}^{s,b}\times X_{\tau = -h(-\xi)}^{-s,-b} $. Thanks to Riesz representation theorem, we conclude by proving that for every linear functional $L$ on $X_{\tau = h(\xi)}^{s,b}$, there is a unique $v$ in $X_{\tau = -h(-\xi)}^{-s,-b}$ such that for all $u$ in $X_{\tau = h(\xi)}^{s,b}$,
$$\left< L, u\right> = \mathcal{B(u,v)}.$$
Moreover, $$\left\Vert L\right\Vert = \left\Vert v \right\Vert_{X_{\tau = -h(-\xi)}^{-s,-b}}.$$
