and inverse Radon transform on our data, asymptotically. Let us define the estimator: $$\label{estimator}
    \widehat{W}^\eta_{h}(q,p) = \frac{1}{\pi n} \sum_{\ell=1}^n
    K_h^\eta \left(q \cos \Phi_\ell + p \sin \Phi_\ell
        -\frac{Y_\ell}{\sqrt{\eta}} \right),$$ where $0<\eta<1$ is a fixed parameter, and the kernel is defined by $$\label{noisy.reg.operator}
    K_h^\eta (u) = \frac{1}{4 \pi}\int_{-1/h}^{1/h}
    \frac{\exp(-iut)|t|}{\widetilde N^\eta (t/\sqrt{\eta})} dt, \quad
    \widetilde K_h^\eta\left(t \right) = \frac{1}{2} \frac{|t|}{\widetilde
      N^\eta (t/\sqrt{\eta})}I(|t|\leq 1/h),$$ and $h>0$ tends to $0$ when $n \to \infty$ in a proper way to be chosen later. For simplicity, let us denote $z=(q,p)$ and $ [z,\phi]
=q \cos \phi + p \sin \phi$, then the estimator can be written: $$\widehat{W}^\eta_{h}(z) = \frac{1}{\pi n} \sum_{\ell=1}^n
    K_h^\eta \left([ z,\Phi_\ell ] -\frac{Y_\ell}{\sqrt{\eta}}\right).$$

This is a one-step procedure for treating two successive inverse problems. The main difference with the noiseless problem treated by  is that the deconvolution is more ‘difficult’ than the inverse Radon transform. In the literature on inverse problems, this problem would be qualified as severely ill-posed, meaning that the noise is dramatically (exponentially) smooth and makes the estimation problem much harder.

$\mathbb{L}_2$ risk estimation {#sec.l2.est}
------------------------------

We establish next the rates of estimation of $W_\rho$ from i.i.d. observations $(Y_\ell,\Phi_\ell), \, \ell=1,\ldots,n$ when the quality of estimation is measured in $\mathbb{L}_2$ 