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I am currently working on a project relating to the topic of Wavelets and Shearlets. I am wanting to estimate a Shearlet transform of the indicator function of the ball of radius $r$ in $\mathbb{R}^2$.

This boils down to computing the following integral:

$$ \int_{B_{a,s,t}} \psi_{a,s,t}(x)dx. $$ Where we have the following notation:

  • $a,s\in \mathbb{R}$ and $t \in \mathbb{R}^2$.
  • $U_{\psi} = B_\rho$ is the support of the shearlet $\psi$.
  • $B_{a,s,t} := (S_sA_a(B_\rho)+t)\cap B_r$, where $$ S_s = \begin{bmatrix} 1& s\\ 0 & 1\end{bmatrix} \ \ \ \text{and} \ \ \ A_a = \begin{bmatrix} a& 0\\ 0 & a^{1/2}\end{bmatrix}. $$
  • $\psi_{a,s,t}(x) = a^{-3/4}\psi(A_a^{-1}S_s^{-1}(x-t))$

Somethings to know are:

  • $\int_{U_\psi}\psi(x)dx=0$.
  • We can parametrize $\partial (S_sA_a(B_\rho)+t)$ as the curve $$\gamma(\theta)= \begin{bmatrix} r(a\cos\theta + a^{1/2}s\sin\theta) +t_1\\ ra^{1/2}\sin\theta +t_2 \end{bmatrix} \ \ \ \theta\in[0,2\pi). $$ The approach I have tried is the following: Cauchy-Schwartz gives us $$ \left|\int_{B_{a,s,t}} \psi_{a,s,t}(x)dx\right| \leq \lambda(B_{a,s,t})^{1/2}||\psi||_{L^2}. $$ Where $\lambda$ is the 2-dimensional Lebesgue Measure. Then since $\partial B_{a,s,t}$ can be seen as a simple closed piecewise smooth plane curve the Isoperimetric inequality tells us that $$ \lambda(B_{a,s,t}) \leq \frac{1}{4\pi}l(\partial B_{a,s,t})^2 $$ where $l(\partial B_{a,s,t})$ is the arc-length of the curve $\partial B_{a,s,t}$.

Intuitively and pictorially this approach is very simple, however it leads to having to solve a 2-dimensional quadratic equation with no coefficients zero. One could do this numerically but then one could also just compute the original integral numerically which would be easier. However I am not so much interested in exact numeric values as I am in analytic estimates.

Can you think of any other approach? Thanks in advance for any help.

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