Notes for "Deep null space learn

作者: jjx323 | 来源:发表于2020-02-05 17:22 被阅读0次

J. Schwab, S. Antholzer, M. Haltmeier, Deep null space learning for inverse problems: convergence analysis and rates, Inverse Problems, 35, 2019, 025008.

Page 8, proof of Theorem 2.8

We have $P_{ran(A^{\dagger})}\mathcal{M}_{\mu,\rho,\Phi} = (A^{*}A)^{\mu}\left( \overline{B_{\rho}(0)} \right)$ and $P_{ran(A^{\dagger})}R_{\alpha} = g_{\alpha}(A^{*}A)A^{*}$

Notes:
In the following, we assume $A$ is a compact, bounded linear operator. Hence, we immediately have
$ran(A^{\dagger}) = ker(A)^{\perp} = \overline{ran(A^{*})} = \overline{ran((A^{*}A)^{1/2})}.$
Since $g_{\alpha}$ is a piecewise continuous function, according to formula (2.43) in

H. W. Engl, M. Hanke, A. Neubauer, Regularization of Inverse Problems, vol 375, 1996

we obtain
$ran(g_{\alpha}(A^{*}A)A^{*}) = ran(A^{*}g_{\alpha}(AA^{*}))\subset ran(A^{*}) \subset ran(A^{\dagger}).$
Hence, we find that $P_{ran(A^{\dagger})}g_{\alpha}(A^{*}A)A^{*} = g_{\alpha}(A^{*}A)A^{*}$ which implies $P_{ran(A^{\dagger})}R_{\alpha} = g_{\alpha}(A^{*}A)A^{*}$ . Similarly, we need to show $ran((A^* A)^{1/2})\subset ran(A^{\dagger}) = ker(A)^{\perp}$ . Assume $A$ has an eigen-system $\{ (\sigma_n, v_n, u_n)\}_{n=1}^{\infty}$ . For each $y\in ran((A^* A)^{\mu})$ , there exists a $x$ such that $y = (A^* A)^{\mu}x = \sum_{n=1}^{\infty}\sigma_{n}^{2\mu}\langle x, v_n \rangle v_n$ . Let $z\in ker(A)$ , we obtain
$\|Az\|^2 = \sum_{n=1}^{\infty}\sigma_n^2\langle z, v_n\rangle^2 = 0,$
which implies $\langle z, v_n \rangle = 0$ for every $n=1,2,\cdots$ . Now, we can give the following calculations
$\langle z, y \rangle = \langle z, \sum_{n=1}^{\infty}\sigma_{n}^{2\mu}\langle x, v_n \rangle v_n \rangle = \sum_{n=1}^{\infty}\sigma_n^{2\mu}\langle x, v_n \rangle \langle z, v_n \rangle = 0.$
The above equality implies $y\in ker(A)^{\perp}$ . Obviously, we already proved $ran((A^* A)^{\mu})\subset ker(A)^{\perp} = ran(A^{\dagger})$ .

Page 7, proof of Proposition 2.7

We have $R_{\alpha}=(Id_X + \Phi) \circ P_{ran(A^{\dagger})} \circ R_{\alpha}$ and, $\cdots$

Notes:
In general, I can not understand that why this equality holds true. If $R_{\alpha} := (Id_X + \Phi)\circ B_{\alpha}$ with $B_{\alpha}:=g_{\alpha}(A^{*}A)A^{*}$ , I can deduce this equality by using similar deductions in note "Page 8, proof of Theorem 2.8 ".

Notes for "Deep null space learn

Page 8, proof of Theorem 2.8

Page 7, proof of Proposition 2.7

相关文章

网友评论

延伸阅读

深度阅读

栏目导航

热点阅读