\documentclass[12pt, a4paper,twoside,openright]{article} \usepackage{latexsym} \usepackage{indentfirst} \usepackage{graphicx} \usepackage{amsmath,amssymb,amsfonts,amsthm,amstext,amscd} \usepackage{mathrsfs} \usepackage{fancyhdr} \usepackage{times} \usepackage{enumerate} \usepackage{natbib} \usepackage[hang,flushmargin]{footmisc} \newtheorem{defi}{Definition}[section] \newtheorem{exam}{Example}[section] \newtheorem{lem}{Lemma}[section] \newtheorem{rem}{Remark}[section] \newtheorem{thm}{Theorem}[section] \newtheorem{prop}{Proposition}[section] \newtheorem{coro}{Corolary}[section] \newtheorem{nota}{Notation}[section] \setcounter{page}{7} %Must start on an odd page! \fancyhf{} \fancyfoot[C]{\thepage } \fancyhead[CE]{\emph{E.SHINE LAL, P.RAMYA}} %Fill in author's name \fancyhead[RE]{} %Fill in abbreviated title \fancyhead[CO]{\emph{Tikhonov Type Regularization for Unbounded operators}} %Fill in abbreviated title \fancyhead[RO]{} \renewcommand{\headrulewidth}{0pt} \fancypagestyle{plain}{% \fancyhf{}% \fancyhead[LO]{\textbf{Ratio Mathematica} } \fancyhead[RO]{Volume 40, 2021, pp. 7-xx } \fancyfoot[C]{\thepage } %\fancyfoot[LE,RO]{ \vspace{1cm} doi:xxxx-xxxxx-xxx-x } %\fancyfoot[LO,CE]{\vspace{1cm} received on: 15-12-2016\\ %accepted on: 15-01-2017\\ %published on: 22-02-2017} %\fancyfoot[CO,RE]{} \renewcommand{\headrulewidth}{0pt}} \pagestyle{fancy} \addtolength{\oddsidemargin}{+1,3cm} %inversione destra/sinistra tradizionale \addtolength{\evensidemargin}{-1,3cm} %inversione destra/sinistra tradizionale \begin{document} \title{\textbf{Tikhonov Type Regularization for Unbounded operators}} \author{E.SHINE LAL\footnote{First Author's Affiliation (Dept. of Mathematics, University College, Thiruvananthapuram, Kerala, India -695034); shinelal.e@gmail.com}\\ P.RAMYA\footnote{Second Author's Affiliation ( Dept. of Mathematics, N,S.S College, Nemmara, Kerala, India-678508); ramyagcc@gmail.com} } \date{} \maketitle \begin{abstract} \begin{normalsize} \noindent In this paper, we introduce a Tikhonov type regularization method for an ill-posed operator equation $ Tx=y, $ where $ T $ is a closed densely defined unbounded operator on a Hilbert space $ H $.\\ \textbf{Keywords}: densely defined operator, closed operator, Tikhonov type regularization.\\ \textbf{2020 AMS subject classifications}: 47A10, 47A52 \footnote{\noindent Received on January 12th, 2021. Accepted on May 12th, 2021. Published on June 30th, 2021. doi: 10.23755/rm.v39i0.5xx. ISSN: 1592-7415. eISSN: 2282-8214. \copyright The Authors. This paper is published under the CC-BY licence agreement.} \end{normalsize} \end{abstract} % % %\begin{figure}[h] %\centering %\includegraphics[height=0.7cm, width=1.9cm]{copy.jpg} %\end{figure} \newpage \section{Introduction} Most of the problems arise in the field of science and engineering can be modelled as an operator equation \begin{equation} Tx=y \end{equation} where $ T:X\rightarrow Y $ is a bounded linear map from a normed linear space $ X $ to a normed linear space $ Y $. In most of the cases (1) is Ill-posed. Certain regularization procedures are known for solving ill-posed operator equation (1). For example Tikhonov regularization, Mollifier method, Ritz method \cite{louis,MTN}. In this paper we introduce a Tikhonov type regularization method for solving an ill-posed operator equation (1), where T is a closed densely defined operator on a Hilbert space $ H $ and we study the order of convergence. \section{Preliminaries} Let $ L(H), C(H) $ and $ B(H) $ denote the space of all linear, closed linear and bounded linear operators on a Hilbert space $H$ respectively. For $ T \in L(H), $ the domain, range of $ T $ are denoted by $ D(T), N(T) $ respectively. An operator $ T \in L(H) $ is said to be densely defined if $ \overline{D(T)} = H. $ \\ For example let $ T : l^{2}(\mathbb{N}) \rightarrow l^{2}(\mathbb{N}) $ defined by $$ T(x_{1},x_{2},x_{3},....,x_{n},....) =(x_{1},2x_{2},3x_{3},.....,nx_{n},....) $$ with domain $$D(T) =\lbrace (x_{1},x_{2},x_{3},....,x_{n},....)\in H : \Sigma_{j=1}^{\infty}\vert j x_{j}\vert^{2} < \infty\rbrace.$$ Then $T $ is closed and unbounded. Since $ c_{00} \subseteq D(T) $ and $ c_{00} $ is dense in $ l^{2}(\mathbb{N}) $, $ D(T) $ is dense in $ l^{2}(\mathbb{N}). $ \begin{prop} Let $ T \in C(H) $ be a densely defined operator. Then there exist a unique operator $T ^{*}\in C(H) $ such that $$\langle Tx,y \rangle = \langle x,T^{*}y \rangle~~ \forall x \in D(T), ~ \forall y \in D(T^{*}).$$ \end{prop} \begin{proof} Let $ D(T^{*}) = \{ y \in H : \langle Tx,y \rangle $ is continuous for every $ x \in D(T) $ \}. For $ y \in D(T) $, define $f : D(T)\rightarrow \mathbb{C}$ by $ f(x) = \langle Tx,y \rangle ~\forall x \in D(T) $. Extend $ f $ to $ f_{0}: H \rightarrow \mathbb{C} $ by $ f_{0}(x) = \lim\limits_{ n \to \infty} \langle Tx_{n},y \rangle $ where $ (x_{n}) $ is a sequence in $ D(T) $ such that $ x_{n}\rightarrow x$. Next we prove that $ f_{0} $ is well defined. For, let $ (x_{n}) $ and $ (y_{n}) $ be two sequences in $ D(T) $ converges to $ x $. Since $ T $ is closed, $ T(x_{n}-y_{n})\rightarrow 0 $. If $ \langle Tx_{n},y \rangle \rightarrow \langle x,y \rangle $, then \begin{align*} \vert \langle Ty_{n}, y \rangle - \langle x,y \rangle \vert&=\vert \langle Ty_{n}-Tx_{n}+Tx_{n}-x,y \rangle \vert \\ &\leq \Vert T(y_{n}-x_{n})\Vert \Vert y \Vert + \vert \langle Tx_{n}-x,y \rangle \vert \\ &\rightarrow 0 ~\mbox{ as} ~ n \rightarrow \infty \end{align*} Hence $ f_{0} $ is well defined. Since $ f_{0} $ is a bounded linear functional on the Hilbert space $ H, $ by Riesz representation theorem there exist a unique $ y^{*}\in H $ such that $ f_{0}(x) = \langle x,y^{*} \rangle$. Thus $ \langle Tx,y \rangle = \langle x,y^{*}\rangle ~\forall x \in D(T) $. Define $ T^{*}:D(T^{*})\rightarrow H $ by $ T^{*}y = y^{*} $. Then $ T^{*} $ is well-defined. Also $ \langle Tx,y \rangle = \langle x,T^{*} y\rangle ~ \forall x \in D(T), ~\forall y\in D(T^{*})$. \end{proof} Consider an ill-posed operator equation \begin{equation} Tx=y \end{equation} where $ T $ is a closed densely defined operator on $ H. $ \begin{defi}\cite{rame} Let $ T \in C(H) $ be densely defined. Then there exist a unique densely defined operator $ T^{\dagger} \in C(H) $ with domain $ D(T^{\dagger}) = R(T) \oplus R(T)^{\perp} $ satisfies the following properties (i) $ TT^{\dagger}y = P_{\overline {R(T)}} ~y $ for all $ y \in D(T^{\dagger}),$ (ii) $ T^{\dagger}Tx = Q_{N(T)^{\perp}}x $ for all $ x \in D(T). $ (iii) $ N(T^{\dagger}) =R(T)^{\perp}. $ where $ P $ and $ Q $ are the orthogonal projection on to $ \overline {R(T)} $ and $ N(T^{\perp}) $ respectively. The operator $ T^{\dagger} $ is called the Moore-Penrose inverse of $ T $. \end{defi} For $ y \in D(T^{\dagger}), $ let $ S_{y} =\lbrace x\in D(T): \Vert Tx-y \Vert \leq \Vert Tu-y \Vert ~\forall u \in D(T) \rbrace. $ Then $ u \in S_{y} $ is called least square solution of the operator equation (2). Note that $ \Vert T^{\dagger}y \Vert \leq \Vert x \Vert ~\forall x \in S_{y},$ is called least square solution of minimal norm and is denoted by $ \widehat{x}$ \cite{rame}. If $ R(T) $ is not closed, then $ T^{\dagger} $ is not continuous. Now we introduce a Tikhonov type regularization procedure for finding an approximate solution for $ T^{\dagger}y. $ \section{Tikhonov type regularization} In this section we introduce a Tikhonov type regularization procedure for solving (2). \begin{lem} Let $ T \in C(H) $ be densely defined and $ \alpha >0 $. Then $ T^{*}T+\alpha I $ and $ TT^{*}+\alpha I $ are bijective closed densely defined operators on $ H $. Also $ (TT^{*}+\alpha I)^{-1} $ and $ (T^{*}T+\alpha I)^{-1} $ are bounded, self adjoint operators on $ H $. \end{lem} \begin{proof} Let $ T \in C(H)$ and $ \alpha >0 $. By proposition 2.1, we have $ T^{*} \in C(H). $ Hence, $ (TT^{*}+ \alpha I) $ and $ (T^{*}T+\alpha I ) $ are closed densely defined operators on $ H $. Since $ \langle(TT^{*}+\alpha I)x,x \rangle = \langle T^{*}x,T^{*}x \rangle + \alpha \langle x,x \rangle \geq 0, \forall x \in D(T^{*}), $ we have $ (TT^{*}+\alpha I) $ is a positive operator. Similarly $ (T^{*}T +\alpha I) $ is also a positive operator. Since $ T^{*}T +\alpha I $ is positive, \begin{align*} \Vert ( T^{*}T+ \alpha I)x \Vert \Vert x \Vert & \geq \langle(T^{*}T +\alpha I)x,x \rangle \\ & = \langle T^{*}Tx,x \rangle+\alpha \Vert x \Vert^{2}\\ &\geq \alpha \Vert x \Vert^{2} ~\forall x\in H \end{align*} Thus \begin{equation} \Vert (T^{*}T+\alpha I)x\Vert \geq\alpha \Vert x \Vert ~\forall x \in H \end{equation} Since $ T^{*}T+\alpha I $ is bounded below, it is one-one and its inverse from the range is continuous. Also $ R(T^{*}T+\alpha I) $ is closed. Since $ T^{*}T+\alpha I $ is also self adjoint, $ R(T^{*}T+\alpha I)= N(T^{*}T+\alpha I)^{\perp} = H $. Hence $ T^{*}T+\alpha I $ is onto. Therefore $ (T^{*}T+\alpha I)^{-1}\in B(H) $. Similary $(TT^{*}+\alpha I)^{-1}\in B(H) $. From (3), \newline $ \Vert (T^{*}T+\alpha I)^{-1}\Vert\leq \dfrac{1}{\alpha}. $ \end{proof} \begin{thm} \label{bdd} Let $ T \in C(H) $ be densely defined. Then $ T^{*}( TT^{*}+ \alpha I)^{-1}$ and $ T(T^{*}T+\alpha I)^{-1} $ are bounded operators on $ H $. Also $ \parallel T^{*}(TT^{*} + \alpha I)^{-1} \parallel \leq \frac{1}{\displaystyle\sqrt{\alpha}} $ and $ \parallel T (T^{*}T+\alpha I)^{-1}\parallel \leq \frac{1}{\displaystyle\sqrt{\alpha}}. $ \end{thm} \begin{proof} We have $ (T^{*}T+\alpha I)^{-1}T^{*}T = I-\alpha (T^{*}T+\alpha I)^{-1}$ Since $ \langle(T^{*}T+\alpha I)^{-1}x,x \rangle \geq 0 $ $ \forall x\in H, $ \begin{align*} \langle(T^{*}T+\alpha I)^{-1} T^{*}T x,x \rangle &=\langle I-\alpha (T^{*}T+\alpha I)^{-1} x,x \rangle\\ &= \langle x,x \rangle-\alpha \langle (T^{*}T+\alpha I)^{-1} x,x \rangle \leq \langle x,x \rangle. \end{align*} Since $ (T^{*}T+\alpha I)^{-1}T^{*}T $ self adjoint, $ \Vert (T^{*}T+\alpha I)^{-1}T^{*}T \Vert \leq 1 $. Let $ x\in H $.\begin{align*} \Vert T^{*}(TT^{*}+\alpha I)^{-1}x \Vert^{2} &= \langle T^{*}( TT^{*}+\alpha I)^{-1} x,T^{*} (TT^{*}+\alpha I)^{-1}x \rangle \\ &= \langle TT^{*}(TT^{*}+\alpha I)^{-1}x, (TT^{*}+\alpha I)^{-1}x \rangle\\ &=\langle (TT^{*}+\alpha I)^{-1}TT^{*}x, (TT^{*}+\alpha I)^{-1}x \rangle\\ &\leq \Vert(TT^{*}+\alpha I)^{-1}TT^{*}x\Vert \Vert(TT^{*}+\alpha I)^{-1}x \Vert \\ &\leq \dfrac{1}{\alpha}\Vert x \Vert^{2} \end{align*} we have $ \Vert T^{*}( TT^{*}+\alpha I)^{-1} x \Vert ^{2} \leq \dfrac{1}{\alpha} \Vert x \Vert ^{2} ~\forall x \in H $.\\Thus $ \Vert T^{*}(TT^{*}+\alpha I)^{-1}\Vert \leq \dfrac{1}{\sqrt{\alpha}} $. Hence $ T^{*}(TT^{*}+\alpha I)^{-1} $ is bounded. Similarly $ T(T^{*}T+\alpha I)^{-1} $ is bounded. \end{proof} \begin{lem}\cite{rame} \label{rame} Let $ T \in C(H) $ be densely defined. Then \\(i) $( TT^{*}+ I)^{-1}T \subseteq T (T^{*}T+ I)^{-1} $\\ (ii) $ (T^{*}T +I)^{-1}T^{*} \subseteq T^{*}(TT^{*}+ I)^{-1} $ \end{lem} \begin{rem} From Theorem \ref{bdd} , we have $ T^{*}(TT^{*}+\alpha I)^{-1} $ and $ T(T^{*}T+\alpha I)^{-1} $ are bounded. Therefore from Lemma \ref{rame}, we have $( TT^{*}+ \alpha I)^{-1}T $ and \\$ (T^{*}T + \alpha I)^{-1}T^{*} $ are bounded. \end{rem} \begin{lem} \label{conver} Let $ T \in C(H) $ be densely defined. For every $ x \in D(T)\cap N(T)^{\perp} $ $ \Vert \alpha (T^{*}T + \alpha I)^{-1} x \Vert \longrightarrow 0, as~~ \alpha\rightarrow 0 $. \end{lem} \begin{proof} Let $ T_{\alpha} = \alpha (T^{*}T + \alpha I)^{-1}, $ $ \alpha > 0.$\\ From (3.1) we have $ \Vert (T^{*}T + \alpha I)^{-1} \Vert \leq \dfrac{1}{\alpha}. $ Hence $\Vert T_{\alpha} \Vert \leq 1 $ for every $ \alpha > 0. $ Let $ u \in R(T^{*}T) $ then there exist $ v \in D(T^{*}T) $ such that $ T^{*}Tv = u. $ \begin{align*} \Vert T_{\alpha} u \Vert &= \Vert T_{\alpha} T^{*}Tv \Vert \\ &= \alpha \Vert (T^{*}T + \alpha I)^{-1} T^{*}Tv \Vert \\ &\leq \alpha \Vert (T^{*}T + \alpha I)^{-1} T^{*}T \Vert \Vert v \Vert \\ &\leq \alpha \Vert v \Vert \end{align*} Hence $ \Vert T_{\alpha} u \Vert \leq \alpha \Vert v \Vert $ $ \forall u \in R(T^{*}T). $\newline Thus for every $ u \in R(T^{*}T), $ $ \Vert \alpha (T^{*}T + \alpha I)^{-1}u \Vert \longrightarrow 0 $ as $ \alpha \longrightarrow 0 $. Since $ \overline{R(T^{*}T)} = N(T)^{\perp} $, $ \Vert \alpha (T^{*}T + \alpha I)^{-1}x \Vert \longrightarrow 0 $, $ \forall x \in D(T)\cap N(T)^{\perp}. $ \end{proof} \begin{thm} Let $ T \in C(H) $ be densely defined and $ R_{\alpha} = (T^{*}T + \alpha I)^{-1} T^{*} $. Then $ \lbrace R_{\alpha} \rbrace_{\alpha > 0} $ is a regularization family for (2). \end{thm} \begin{proof} Let $ y \in D(T^{*}). $ Then $ (T^{*}T+\alpha I)\widehat{x} = T^{*}y+\alpha \widehat{x}.$ \newline Hence $ \widehat{x}=(T^{*}T + \alpha I)^{-1}(T^{*}y+\alpha \widehat{x}). $ Thus \begin{align*} T^{\dagger}y-R_{\alpha}y &= \widehat{x} - (T^{*}T + \alpha I)^{-1}T^{*}y \\ & = (T^{*}T + \alpha I)^{-1}(T^{*}y+\alpha \widehat{x})-(T^{*}T + \alpha I)^{-1}T^{*}y \\ & =(T^{*}T + \alpha I)^{-1}\alpha \widehat{x} \end{align*} Hence $ \Vert T^{\dagger}y-R_{\alpha}y \Vert = \alpha \Vert (T^{*}T + \alpha I)^{-1} \widehat{x} \Vert. $ Since $ \widehat{x} \in D(T) \cap N(T)^{\perp} $, by Lemma \ref{conver}, $ \Vert T^{\dagger}y - R_{\alpha}y \Vert \longrightarrow 0 $ as $ \alpha \longrightarrow 0. $ Thus $ \lbrace R_{\alpha}\rbrace_{\alpha > 0} $ is a regularization family for (2). \end{proof} \section{Order estimate} In this section we find an error estimate for the regularization family \\$ R_{\alpha} = (T^{*}T + \alpha I)^{-1} T^{*} ,$ where $ T $ is a closed densely defined operator. We use the following lemmas. \begin{lem}\cite{rame} For $ T \in C(H) $ we have the following (i) If $ \mu \in \mathbb{C} $ and $ \lambda \in \sigma(T)$ then $ \lambda + \mu \in \sigma (T+\mu I) $ (ii) If $ \alpha \in \mathbb{C} $ and $ \lambda \in \sigma (T) $ then $\alpha \lambda \in \sigma (\alpha T) $ (iii) $ \sigma (T^{2}) = \lbrace \lambda^{2}: \lambda \in \sigma (T)\rbrace $ \end{lem} \begin{lem}\cite{rame} Let $ T \in L(H) $ be a positive operator. Then the following results bold. (i) $ T^{\dagger} $ is positive. (ii) $ \sigma (T) = \sigma_{a}(T) $ (iii) $ 0 \notin \sigma (I+T) $ that is $ (I+T)^{-1} \in B(H) $ (iv) If $ 0 \notin \sigma (T) $ then $ 0 \neq \lambda \in \sigma (T) $ if and only if $ \dfrac{1}{\lambda} \in \sigma(T^{-1}) $ \end{lem} \begin{thm} Suppose $ T \in C(H) $ is densely defined positive operator. Then for every $ \alpha > 0 $ $ \sigma \Big((T+\alpha I)^{-2}T \Big) = \Big\lbrace \dfrac{\lambda}{(\lambda+\alpha)^{2}} : \lambda \in \sigma (T) \Big\rbrace $ \end{thm} \begin{proof} Since $ T $ is positive, $ T+\alpha I $ is bijective.\\ Also $ (T+\alpha I)^{-2}T = (T+\alpha I)^{-1}-\alpha (T+\alpha I)^{-2}. $\newline From Lemmas 4.1, 4.2 for $ \alpha, \lambda > 0, $ we have \\$ \lambda \in \sigma (T) $ if and only if $ (\lambda+\alpha)^{-1} \in \sigma \Big((T+\alpha I)^{-1}\Big) $.\newline Hence \begin{align*} \sigma \Big((T+\alpha I)^{-2}T \Big) &= \Big\lbrace \mu - \alpha \mu^{2} : \mu \in \sigma \Big((T+\alpha I)^{-1}\Big)\Big\rbrace \\ & = \Big\lbrace \dfrac{1}{\lambda + \alpha}- \dfrac{\alpha}{(\lambda+\alpha)^{2}}:\lambda \in \sigma (T) \Big\rbrace \\ & = \Big\lbrace \dfrac{\lambda}{(\lambda+\alpha)^{2}}: \lambda \in \sigma (T)\Big\rbrace . \end{align*} \end{proof} \begin{coro} Let $ T \in C(H) $ be densely defined and $ \alpha > 0 .$ Then $ \Vert (T^{*}T+\alpha I)^{-1}T^{*}\Vert = $ sup $ \Big\lbrace \dfrac{\sqrt{\lambda}}{\lambda+\alpha} : \lambda \in \sigma (T^{*}T) \Big\rbrace \leqslant \dfrac{1}{2 \sqrt{\alpha}} $ \end{coro} \begin{proof} We have $ R_{\alpha} = (T^{*}T+\alpha I)^{-1}T^{*}.$ Hence $ R_{\alpha}^{*}R_{\alpha} = T (T^{*}T+\alpha I)^{-2}T^{*}. $ From Lemma 2.2 in \cite{groet}, we have $ R_{\alpha}^{*}R_{\alpha} = (TT^{*}+\alpha I)^{-2}TT^{*}.$ Since $ R_{\alpha}^{*}R_{\alpha} $ is self adjoint and bounded, $ \Vert R_{\alpha} \Vert^{2} = \Vert R_{\alpha}^{*}R_{\alpha} \Vert $ \hspace{1.25cm}= Sup $ \Big\lbrace \vert k \vert : k \in \sigma (R_{\alpha}^{*}R_{\alpha}) \Big\rbrace $ \hspace{1.25cm}= Sup $ \Big\lbrace \dfrac{\lambda}{(\lambda+\alpha)^{2}} : \lambda \in \sigma(TT^{*})\Big\rbrace $ \hspace{0.2cm}$ \Vert R_{\alpha} \Vert $ = Sup $ \Big\lbrace \dfrac{\sqrt{\lambda}}{\lambda+\alpha} : \lambda \in \sigma (TT^{*})\Big\rbrace. $ Since $ 2 \sqrt{\alpha \lambda} (\lambda + \alpha)^{-1} \leqslant 1 $ for $ \lambda , \alpha > 0,$ we have $ \Vert R_{\alpha} \Vert \leqslant \dfrac{1}{2 \sqrt{\alpha}}. $ \end{proof} Now we find an order estimate for $ R_{\alpha}. $ \begin{coro} Let $ T \in C(H) $ is densely defined and $ R_{\alpha} = (T^{*}T+\alpha I)^{-1}T^{*}. $ For every $ \alpha > 0 $ and $ \delta > 0, $ let $ y^{\delta}\in H $ be such that $ \Vert y-y^{\delta} \Vert \leqslant \delta $. Then $ \Vert R_{\alpha}y - R_{\alpha}y^{\delta}\Vert \leqslant \dfrac{\delta}{2\sqrt{\alpha}}. $ \end{coro} \begin{proof} For $ \Vert y-y^{\delta} \Vert \leqslant \delta $,\begin{align*} \Vert R_{\alpha} y - R_{\alpha}y^{\delta} \Vert &\leqslant \Vert R_{\alpha}\Vert \Vert y-y^{\delta} \Vert\\ & \leqslant \dfrac{1}{2 \sqrt{\alpha}}\Vert y-y^{\delta}\Vert \\ & \leqslant\dfrac{\delta}{2\sqrt{\alpha}} \end{align*} \end{proof} \begin{thm} Let $ T \in C(H) $ is densely defined and $ R_{\alpha} = (T^{*}T+\alpha I)^{-1}T^{*} $. Then $ \Vert \widehat{x} - R_{\alpha}y^{\delta}\Vert \leqslant\Vert \widehat{x} - R_{\alpha}y \Vert + \dfrac{\delta}{2\sqrt{\alpha}} $ . If $ \alpha = \alpha(\delta) $ is chosen such that $ \alpha(\delta)\longrightarrow 0 $ and $ \dfrac{\delta}{\sqrt{\alpha}(\delta)}\longrightarrow 0 $ as $ \delta \longrightarrow 0 $, then $ \Vert \widehat{x} - R^{\delta}_{\alpha (\delta)} \Vert \longrightarrow 0 $ as $ \delta \longrightarrow 0 .$ \end{thm} \begin{proof} $ \Vert \widehat{x} - R_{\alpha} y^{\delta} \Vert \leq \Vert \widehat{x} - R_{\alpha} y \Vert + \Vert R_{\alpha} y - R_{\alpha} y^{\delta} \Vert $ \hspace{3cm}$ \leq \Vert \widehat{x} - R_{\alpha}y \Vert + \dfrac{\delta}{2\sqrt{\alpha}} $ by Theorem 3.6, $ \Vert \widehat{x} - R_{\alpha}y \Vert \longrightarrow 0 $ as $ \alpha \longrightarrow 0. $ \end{proof} \bibliographystyle{plain} \bibliography{s} \nocite{*} \end{document}