Square integrability of a composite fourier transform Let $\mathcal{H} = L^2(\mathbb{R}, d\theta)$. Then for a function $f \in C^{\infty}_0(\mathbb{R}^2)$, define $f^+(\theta):= \tilde{f}(p(\theta))$ (with $\tilde{f}$ the fourier transform of $f$) where $p(\theta) =m\begin{pmatrix} \cosh(\theta) \\ \sinh(\theta)\end{pmatrix}$, $m>0$.
My question:
Is $f^+(\theta) \in \mathcal{H}$?
EDIT:
I should mention the nature of the Fourier transform here:
$$\tilde{f}(p) = \frac{1}{2\pi} \int_{\mathbb{R}} e^{ix\cdot p}f(x) \ dx.$$
There is purposefully a missing negative sign on the exponential because the scalar product $\cdot$ is the Minkowski scalar product given by (in $2$ dimensions) by $x \cdot p = x_0p_0 - x_1p_1.$
 A: First, since 
$$\tilde f(p_0, p_1) = \frac{1}{(2\pi)^2} \iint_{\mathbb R^2} e^{i(x_0 p_0 - x_1 p_1)} \, f(x_0, x_1) \, dx_0 \, dx_1 = \hat f(p_0, -p_1)$$
where (note the change of sign in the exponent!)
$$\hat f(p_0, p_1) = \frac{1}{(2\pi)^2} \iint_{\mathbb R^2} e^{i(x_0 p_0 + x_1 p_1)} \, f(x_0, x_1) \, dx_0 \, dx_1$$
it doesn't matter much that we use the Minkowski scalar product. The result just differs by a reflection.
Then, $f \in C_c^\infty(\mathbb R^2) \subset \mathcal S(\mathbb R^2)$ makes $\tilde f \in \mathcal S(\mathbb R^2) \subset L^2(\mathbb R^2)$ so $\tilde f(p) \to 0$ faster than $|p|^{-N}$ for every $N \in \mathbb N$ as $|p| \to \infty$. Composition with $p(\theta) = (m \cosh\theta, m \sinh\theta) \sim (e^{|\theta|}, e^{|\theta|} \operatorname{sign}(\theta))$ having $|p(\theta)| \sim e^{|\theta|}$ only makes it decay even quicker: $\tilde f(p(\theta)) \to 0$ faster than $e^{-N|\theta|}$ for every $N \in \mathbb N$ as $\theta \to \pm\infty.$
Therefore, $\tilde f \in L^2(\mathbb R).$
