Operator equation has been one of the principal tools in a large area of applied mathematics, and the literature discussing around this topic is vast. In this paper, we will limit our discussion on the compact operator equations of the second kind, which has the form

$ Tx:=(I-K)x=b, \quad x\in X, \label{eqn1.0} $

(1.1)

where $K:X\rightarrow X$ is a compact operator and $b\in X$ is given. Thanks to the First and the Second Riesz Theorem, we know $\dim \mathcal{N}\left( T\right) < \infty $ [6, Theorem 3.1, p.28], and $\mathcal{R}\left( T\right) $ is closed [6, Theorem 3.2, p.29]. We aim to obtain the best-approximate solution of (1.1), which is denoted as $x^{\dagger }:=T^{\dagger }b, $ where $T^\dagger$ is the Moore-Penrose inverse of $T$. Note that as $\mathcal R(T)$ is closed, $T^\dagger$ is naturally guaranteed to be bounded.

Due to the complexity of the specific problems that has form (1.1), it is difficult for us to find a universal solution to all the problems. A more promising strategy is finding the numerical solution, which involve approximating the abstract space and operator with finite freedoms. Let $\left\{ X_{n}\right\} $ be a sequence of finite-dimensional subspaces of $X$ such that

$ X_{0}\subseteq X_{1}\subseteq X_{2}\subseteq\cdots, \quad \overline{\overset{\infty}{\underset{n=0}{\cup}}X_{n}}=X, $

(1.2)

and for each $n\in\mathbb{N}$ set $T_{n}:=TP_{n}\text{, } $ where $P_{n}:=P_{X_{n}}$ is the orthogonal projection from $X$ onto $X_{n}$. Note that (1.2) implies

$ \underset{n\to \infty }{\mathop{s\text{-}\lim }}\,{{P}_{n}}={{I}_{X}},\quad \underset{n\to \infty }{\mathop{s\text{-}\lim }}\,{{P}_{\mathcal{R}\left( {{T}_{n}} \right)}}={{P}_{\mathcal{R}\left( T \right)}}\text{, } $

(1.3)

here, we say $\{(X_{n}, T_{n})\}_{n\in\mathbb{N}}$ is a LSA (least-squares approximation setting) for $T$, and all of our following discussions will be based on this setting. Our target is to find suitable LSA such that

$ \underset{n\to \infty }{\mathop{s-\lim }}\,T_{n}^{\dagger }={{T}^{\dagger }}, $

(1.4)

namely, $x^{\dagger }:=T^{\dagger }b$ can be approximated by $x_{n}^{\dagger }:=T_{n}^{\dagger}b$. There are many works touching upon this problem such as Du [1, 2], Groetsch [4], Groetsch-Neubauer [3], and Seidman [5]. Note that (1.4) does not naturally holds for equation (1.1), as Du's example [2, Example 2.10] shows. To guarantee the convergency of the approximation scheme $\{T_n^\dagger\}$ for $T^\dagger$. Groetsch [4, Proposition 0] and Du [2, Theorem 2.8 (d)] provide the following convergence conditions

$ {\text{(1}}{\text{.4)}} \Leftrightarrow \mathop {\sup }\limits_n \left\| {T_n^\dagger T} \right\| < + \infty \Leftrightarrow {\text{(1}}{\text{.5)}}, $

where (1.5) is the stability condition of LSA $\{(X_{n}, T_{n})\}_{n\in\mathbb{N}}$, that is,

$ \mathop {\sup }\limits_n \left\Vert T_{n}^{\dagger }\right\Vert < +\infty. $

(1.5)

However, as will show in a simple and important example, a direct examination of (1.5) could be difficult, but the examination of some other equivalent condition of (1.5) that we will soon give in our theorem can be very easy.

In this paper, we will give some equivalent characterizations for (1.4) (or (1.5)). These equivalent characterizations can not only increase our understanding on the convergence of this approximation scheme by offering different perspectives, but also provide us with some simple and `easy to check' criteria to examine the convergence. To proceed, we need the following notation

$$ \begin{matrix} \underset{n\to \infty }{\mathop{s-\lim }}\,{{S}_{n}}:=\left\{ \left. x\ \right|\ \text{there is a sequence }\left\{ {{x}_{n}} \right\}\text{ such that }{{S}_{n}}\ni {{x}_{n}}\to x \right\},\text{ } \\ \underset{n\to \infty }{\mathop{\text{w}-\widetilde{\lim }}}\,{{S}_{n}}:=\left\{ \left. x\ \right|\ \text{there is a sequence }\left\{ {{x}_{n}} \right\}\text{ such that }\overset{\infty }{\mathop{\underset{k=n}{\mathop{\cup }}\,}}\,{{S}_{k}}\ni {{x}_{n}}\rightharpoonup x \right\}, \\ \end{matrix} $$

where $\{S_{n}\}$ is a sequence of nonempty subsets of a Banach space. With the above notation the main result obtained in the paper is as below.

Theorem 1.1 For the compact operator equation (1.1) with LSA $\{(X_{n}, T_{n})\}_{n\in\mathbb{N}}$, the following propositions are equivalent:

(a) (1.4) (or (1.5)) holds.

(b) There holds

$$ \underset{n\to \infty }{\mathop{s-\lim }}\,\mathcal{G}\left( T_{n}^{\dagger } \right)=\underset{n\to \infty }{\mathop{\text{w}-\widetilde{\lim }}}\,\mathcal{G}\left( T_{n}^{\dagger } \right)=\mathcal{G}\left( {{T}^{\dagger }} \right). $$

(c) There holds

$$ \mathop {s - \lim }\limits_{n \to \infty } {\mkern 1mu} \mathcal{N}\left( {{T_n}} \right) = \mathop {{\text{w}} - \widetilde {\lim }}\limits_{n \to \infty } {\mkern 1mu} \mathcal{N}\left( {{T_n}} \right) = \mathcal{N}\left( T \right). $$

(d) For any $b, b_{n}\in X$ with $\left\Vert b_{n}-b\right\Vert\rightarrow 0$ ($n\rightarrow \infty $) there holds

$$ \underset{n\to \infty }{\mathop{s\text{-}\lim }}\,T_{n}^{-1}\left( {{P}_{\mathcal{R}\left( {{T}_{n}} \right)}}{{b}_{n}} \right)=\mathop {{\text{w}} - \widetilde {\lim }}\limits_{n \to \infty } {\mkern 1mu}\,T_{n}^{-1}\left( {{P}_{\mathcal{R}\left( {{T}_{n}} \right)}}{{b}_{n}} \right)={{T}^{-1}}\left( Pb \right). $$

(e) There is a $n_{\ast }\in\mathbb{N}$ such that $X_{n_{\ast }}\supseteq \mathcal{N}(T)$.

In Section 2, we will give some lemmas and the proof of Theorem 1.1. In Section 3, we will study some examples to further explain the theorem.

To prove Theorem 1.1 we need to prepare several lemmas.

Lemma 2.1 Let $T\in\mathcal{B}\left(X\right)$ with $\dim\mathcal{N}\left(T\right)< \infty$, and have LSA $\left\{\left( X_{n}, T_{n}\right)\right\}$.

(a) There hold

$ \mathcal{N}\left( T_{n}\right)=\left( \mathcal{N}\left( T\right)\cap X_{n}\right)\overset{\bot}{\oplus}X_{n}^{\, \bot }, \quad P_{\mathcal{N} \left( T_{n}\right)}=P_{\mathcal{N}\left( T\right)\cap X_{n}}+I-P_{n}. $

(b) There is a $n_{\ast }\in\mathbb{N}$ such that

$ \begin{array}{*{20}{c}} {\mathcal{N}(T) \cap {X_n} = \;\overline {\mathop {\mathop \cup \limits_{k = 0} {\mkern 1mu} }\limits^\infty {\mkern 1mu} \left( {\mathcal{N}\left( T \right) \cap {X_k}} \right)} \quad {\text{for}}\;n \geqslant {n_*},{\text{ }}} \\ {\mathop {s - \lim }\limits_{n \to \infty } {\mkern 1mu} {P_{\mathcal{N}\left( {{T_n}} \right)}} = {P_{\overline {\mathop {\mathop \cup \limits_{k = 0} {\kern 1pt} }\limits^\infty {\kern 1pt} \left( {\mathcal{N}\left( T \right) \cap {X_k}} \right)} }} = {P_{\mathcal{N}(T) \cap {X_{{n_*}}}}}.} \end{array} $

(c) If $\mathcal R(T)$ is closed and

$ \overline{\overset{\infty }{\underset{n=0}{\cup }}\left( \mathcal{N}\left( T\right) \cap X_{n}\right) }=\mathcal{N}\left( T\right), $

(2.1)

then (1.5) holds.

Proof (a) It is clear that

$ \mathcal{N}\left( T_{n}\right) =\left\{ \left. x\in X\, \right\vert \, P_{X_{n}}x\in \mathcal{N}\left( T\right) \cap X_{n}\right\} \\ =\left( \mathcal{N}\left( T\right) \cap X_{n}\right) \overset{\bot }{\oplus }X_{n}^{\, \bot }, $

and therefore

$ P_{\mathcal{N}\left( T_{n}\right) }=P_{\mathcal{N}\left( T\right) \cap X_{n}}+I-P_{n}. $

(b) Since (1.2) and $\dim \mathcal{N}\left( T\right) < \infty $ hold, we have

$ {{\mathcal{N}(T)}\cap X_{n}\subseteq {\mathcal{N}(T)}\cap X_{n+1}\ }\left( \forall n\right), \quad \dim \left( {{\mathcal{N}(T)}\cap X_{n}}\right) \leq \dim \mathcal{N}\left( T\right) < \infty, $

and therefore there is a $n_{\ast }\in \mathbb{N}$ such that

$ {{\mathcal{N}(T)}\cap X_{n}=\ }\overline{{\overset{\infty }{\underset{k=0}{\cup }} \left( \mathcal{N}\left( T\right) \cap X_{k}\right) }}\quad \text{for}\ n\geq n_{\ast }. $

This implies that

$ \underset{n\rightarrow \infty }{s\text{-}\lim }P_{\mathcal{N} \left( T_{n}\right) } =\underset{n\rightarrow \infty }{s\text{-} \lim }P_{\mathcal{N}\left( T\right) \cap X_{n}} \\ =P_{\overline{{\overset{\infty }{\underset{k=0}{\cup }}\left( \mathcal{N} \left( T\right) \cap X_{k}\right) }}}=P_{{{\mathcal{N}(T)}\cap X_{n_{\ast }}}}. $

(c) Assume that $\mathop {\sup }\limits_n \left\Vert T_{n}^{\dagger }\right\Vert =+\infty $. Then by the uniform boundedness principle, there is an $u\in X$ such that

$ \mathop {\sup }\limits_n \left\Vert T_{n}^{\dagger }u\right\Vert =+\infty . $

Hence there exists a subsequence $\left\{ T_{n_{k}}^{\dagger }\right\} $ of $\left\{ T_{n}^{\dagger }\right\} $ such that $\mathop {\lim }\limits_{k\rightarrow \infty }\left\Vert T_{n_{k}}^{\dagger }u\right\Vert =+\infty. $ Due to (b), there is a $n_{\ast }\in \mathbb{N}$ such that

$ {{\mathcal{N}(T)}\cap X_{n}=\ }\overline{{\overset{\infty }{\underset{k=0}{\cup }} \left( \mathcal{N}\left( T\right) \cap X_{k}\right) }}\quad \text{for }n\geq n_{\ast }. $

Hence, it follows from (2.1) that $ {{\mathcal{N}(T)}\cap X_{n}={\mathcal{N}(T)\subseteq X}}_{n} \quad \text{for}\ n\geq n_{\ast }, $ and therefore

$ \mathcal{N}\left( T_{n}\right) ^{\bot } =\left( \mathcal{N}\left( T\right) \cap X_{n}\right) ^{\bot }\cap X_{n} \\ ={{\mathcal{N}(T)}}^{\bot }\cap {{X_{n}\subseteq X}}_{n}\quad \text{for } n\geq n_{\ast }. $

Set $v_{k}:=\frac{T_{n_{k}}^{\dagger }u}{\Vert T_{n_{k}}^{\dagger }u\Vert }$. Then $v_{k}\in \mathcal{N}\left( T_{n_{k}}\right) ^{\bot }=X_{n_{k}}\cap {{\mathcal{N}(T)}}^{\perp }$ for $k$ large enough, and therefore

$ \left\{ \begin{array}{l} v_{k}=P_{\mathcal{N}\left( T\right) ^{\bot }}v_{k}=T^{\dagger }Tv_{k}\quad \text{for }k\text{ large enough, } \\ \left\Vert Tv_{k}\right\Vert =\left\Vert T_{n_{k}}v_{k}\right\Vert =\frac{\big\Vert P_{\mathcal{R}\left( T_{n_{k}}\right) }u\big\Vert}{\Vert T_{n_{k}}^{\dagger }u\Vert }\rightarrow 0\ (k\rightarrow \infty ). \end{array} \right. $

This with $T^{\dagger }\in \mathcal{B}\left( X\right) $ (by $\mathcal{R}\left( T\right) $ being closed) implies that

$ \underset{k\rightarrow \infty }{\lim }\left\Vert v_{k}\right\Vert =\mathop {\lim }\limits_{n\rightarrow \infty }\left\Vert T^{\dagger }Tv_{k}\right\Vert =0, $

that contradicts with $\left\Vert v_{k}\right\Vert =1$.

Lemma 2.2 Let $T\in \mathcal{B}\left( X\right) $ have LSA $\left\{ \left( X_{n}, T_{n}\right) \right\} $. Then

$$ \underset{n\rightarrow \infty }{s\text{-}\lim }\mathcal{G}\left( T_{n}\right) =\underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }} \mathcal{G}\left( T_{n}\right) =\mathcal{G}\left( T\right) . $$

Proof It is clear that $$\mathop {s - \lim }\limits_{n \to \infty } \mathcal{G}\left( {{T_n}} \right) \subseteq \mathop {w - \widetilde {\lim }}\limits_{n \to \infty } \mathcal{G}\left( {{T_n}} \right)$$. Hence, we need only to show that

$$ \underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }} \mathcal{G}\left( T_{n}\right) \subseteq \mathcal{G}\left( T\right) \subseteq \underset{n\rightarrow \infty }{{s}\text{-}\lim }\mathcal{G} \left( T_{n}\right) . $$

Let $\left( x, y\right) \in \mathcal{G}\left( T\right) $. Then $\left( x, T_{n}x\right) \in \mathcal{G}\left( T_{n}\right) $, and $\left( x, T_{n}x\right) \rightarrow \left( x, y\right) \ \left( n\rightarrow \infty \right) $. Therefore, $\left( x, y\right) \in {s}$-$\mathop {\lim }\limits_{n \to \infty } \mathcal{G}\left( {{T_n}} \right)$. This gives that $ \mathcal{G}\left( T\right) \subseteq \underset{n\rightarrow \infty } {{s}\text{-}\lim }\mathcal{G}\left( T_{n}\right) .$ Let $\left( x, y\right) \in {w}$-$\mathop {\widetilde {\lim }}\limits_{n \to \infty } \mathcal{G}\left( {{T_n}} \right) $. Then there is a sequence $\left\{ \left( x_{n}, y_{n}\right) \right\} $ such that $\cup _{k=n}^{\infty } \mathcal{G}\left( T_{k}\right) \ni \left( x_{n}, y_{n}\right) \rightharpoonup \left( x, y\right) \ \left( n\rightarrow \infty \right) $. Thus, there is a sequence $\left\{ k_{n}\right\} $ such that

$ k_{n}\geq n, \quad x_{n}\rightharpoonup x\ \left( n\rightarrow \infty \right) , \quad T_{k_{n}}x_{n}=y_{n}\rightharpoonup y\ \left( n\rightarrow \infty \right) . $

Note that for all $v\in X$ there holds

$ \left\vert \left\langle Tx-y, v\right\rangle \right\vert \leq \left\vert \left\langle T\left( x-x_{n}\right), v\right\rangle \right\vert +\left\vert \left\langle \left( T-T_{k_{n}}\right) x_{n}, v\right\rangle \right\vert +\left\vert \left\langle y_{n}-y, v\right\rangle \right\vert \\ =\left\vert \left\langle x-x_{n}, T^{\ast }v\right\rangle \right\vert +\left\vert \left\langle x_{n}, \left( I-P_{k_{n}}\right) T^{\ast }v\right\rangle \right\vert +\left\vert \left\langle y_{n}-y, v\right\rangle \right\vert . $

Thus, we have $\left\langle Tx-y, v\right\rangle =0\ \forall v\in X, \text{ that is, }\left( x, y\right) \in \mathcal{G}\left( T\right) . $$ This gives that $\underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }} \mathcal{G}\left( T_{n}\right) \subseteq \mathcal{G}\left( T\right) . $$

Lemma 2.3 Let $H$ be a Hilbert space and $\left\{ H_{n}\right\} $ a sequence of closed subspaces of $H$. Then

$$ \left\{ P_{H_{n}}\right\} \text{ is strongly convergent } \Longleftrightarrow \, \underset{n\rightarrow \infty }{{s}\text{-} \lim }H_{n}=\, \underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }}H_{n}\, \text{;} $$

in the case that $\left\{ P_{H_{n}}\right\} $ is strongly convergent,

$ \underset{n\rightarrow \infty }{{s}\text{-}\lim }P_{H_{n}}=P_{M}\, , \quad \text{where }M:=\, \underset{n\rightarrow \infty }{{s}\text{-} \lim }H_{n}\, . $

Proof See [1, Lemma 2.13].

Next, we prove Theorem 1.1 as follows.

Proof of Theorem 1.1 Note that, by Lemma 2.1 (a), $P_{\mathcal{N}\left( T_{n}\right) }=P_{\mathcal{N}\left( T\right) \cap X_{n}}+I-P_{n}, $ and there is a $n_{\ast }\in \mathbb{N}$ such that

$ \underset{n\rightarrow \infty }{{s}\text{-}\lim }P_{\mathcal{N} \left( T_{n}\right) }=P_{\mathcal{N}\left( T\right) \cap X_{n_{\ast }}}. $

(2.2)

Therefore it follows that

$ T_{n}^{\dagger }-T^{\dagger } =T_{n}^{\dagger }P_{\mathcal{R}\left( T\right) }-T_{n}^{\dagger }T_{n}T^{\dagger }+T_{n}^{\dagger }T_{n}T^{\dagger }-T^{\dagger } \\ =T_{n}^{\dagger }T\left( I-P_{n}\right) T^{\dagger }+\left( P_{\mathcal{N}\left( T_{n}\right) ^{\bot }}-P_{\mathcal{N}\left( T\right) ^{\bot }}\right) T^{\dagger } \\ =\left( T_{n}^{\dagger }T-I\right) \left( I-P_{n}\right) T^{\dagger }, $

that is,

$ T_{n}^{\dagger }-T^{\dagger }=\left( T_{n}^{\dagger }T-I\right) \left( I-P_{n}\right) T^{\dagger }. $

(2.3)

Due to this with the uniform boundedness principle, it is clear that

$ {\text{(1}}{\text{.4)}} \Leftrightarrow \mathop {\sup }\limits_n \left\| {T_n^\dagger T} \right\| < + \infty \Leftrightarrow {\text{(1}}{\text{.5)}}{\text{.}} $

(2.4)

(a)$\Rightarrow $(b) We need only to show that (a) implies that

$$ \underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }} \mathcal{G}\left( T_{n}^{\, \dagger }\right) \subseteq \mathcal{G}\left( T^{\dagger }\right) \subseteq \underset{n\rightarrow \infty }{{s}\text{-}\lim } \mathcal{G}\left( T_{n}^{\, \dagger }\right). $$

(2.5)

Let (a) be valid, namely, (1.4) and (1.5) hold (by (2.4)). Then for all $x\in X$ there hold

$ \left\Vert T_{n}^{\dagger }T_{n}x-T^{\dagger }Tx\right\Vert \leq \left\Vert T_{n}^{\dagger }\right\Vert \left\Vert \left( T_{n}-T\right) x\right\Vert +\left\Vert \left( T_{n}^{\dagger }-T^{\dagger }\right) Tx\right\Vert \rightarrow 0\ \left( n\rightarrow 0\right), $

$ \left\Vert T_{n}T_{n}^{\dagger }x-TT^{\dagger }x\right\Vert \leq \left\Vert T_{n}\right\Vert \left\Vert \left( T_{n}^{\dagger }-T^{\dagger }\right) x\right\Vert +\left\Vert \left( T_{n}-T\right) T^{\dagger }x\right\Vert \rightarrow 0\ \left( n\rightarrow 0\right) . $

Thus we obtain that

$ \mathop {{\text{s - }}\lim }\limits_{n \to \infty } {P_{\mathcal{N}{{\left( {{T_n}} \right)}^ \bot }}} = {P_{\mathcal{N}{{\left( T \right)}^ \bot }}},\quad \mathop {{\text{s - }}\lim }\limits_{n \to \infty } {P_{\mathcal{R}\left( {{T_n}} \right)}} = {P_{\mathcal{R}\left( T \right)}}. $

(2.6)

Let $\left( y, x\right) \in \mathcal{G}\left( T^{\dagger }\right) $. Then

$ \left( T^{\dagger }y, P_{\mathcal{R}\left( T\right) }y\right) =\left( T^{\dagger }y, TT^{\dagger }y\right) =\left( x, Tx\right) \in \mathcal{G}\left( T\right) . $

By Lemma 2.2, ${s}$-$\mathop {\lim }\limits_{n \to \infty } \mathcal{G}\left( {{T_n}} \right) = \mathcal{G}\left( T \right)$, hence there is a squence $\left\{ x_{n}\right\} $ such that

$ x_{n}\rightarrow x\ \left( n\rightarrow \infty \right), \quad T_{n}x_{n}\rightarrow Tx\ \left( n\rightarrow \infty \right) . $

This with (2.6) implies that

$ \left\{ \begin{array}{l} T_{n}x_{n}\rightarrow Tx\ \left( n\rightarrow \infty \right), \\ P_{\mathcal{N}\left( T_{n}\right) ^{\bot }}x_{n}\rightarrow P_{\mathcal{N} \left( T\right) ^{\bot }}x\ \left( n\rightarrow \infty \right), \\ P_{\mathcal{R}\left( T_{n}\right) ^{\bot }}y\rightarrow P_{\mathcal{R}\left( T\right) ^{\bot }}y\ \left( n\rightarrow \infty \right), \end{array} \right. $

and therefore

$ \left\{ \begin{array}{l} T_{n}x_{n}+P_{\mathcal{R}\left( T_{n}\right) ^{\bot }}y\rightarrow Tx+ P_{\mathcal{R}\left( T\right) ^{\bot }}y=y\ \left( n\rightarrow \infty \right), \\ T_{n}^{\dagger }\left( T_{n}x_{n}+P_{\mathcal{R}\left( T_{n}\right) ^{\bot }}y\right) \rightarrow T^{\dagger }Tx=T^{\dagger }y=x\ \left( n\rightarrow \infty \right). \end{array} \right. $

Thus, $\left( y, x\right) \in {s}$-$\mathop {\lim }\limits_{n \to \infty } \mathcal{G}\left( {T_n^\dagger } \right) $, we get

$ \mathcal{G}\left( T^{\dagger }\right) \subseteq \, \underset{n\rightarrow \infty }{{s}\text{-}\lim }\mathcal{G}\left( T_{n}^{\, \dagger }\right) . $

Let $\left( y, x\right) \in {w}$-$\mathop {\widetilde {\lim }}\limits_{n \to \infty } \mathcal{G}\left( {T_n^{{\kern 1pt} \dagger }} \right)$. Then there is a sequence $\left\{ \left( y_{n}, x_{n}\right) \right\} $ such that

$ \underset{k=n}{\overset{\infty }{\cup }}\mathcal{G}\left( T_{k}^{\, \dagger }\right) \ni \left( y_{n}, x_{n}\right) \rightharpoonup \left( y, x\right) \ \left( n\rightarrow \infty \right) . $

Hence, there is a sequence $\left\{ k_{n}\right\} $ such that

$ k_{n}\geq n, \quad y_{n}\rightharpoonup y\ \left( n\rightarrow \infty \right) , \quad T_{k_{n}}^{\, \dagger }y_{n}=x_{n}\rightharpoonup x\ \left( n\rightarrow \infty \right) . $

This with $T_{k_{n}}^{\, \dagger }y_{n}\in \mathcal{N}\left( T_{k_{n}}\right) ^{\bot }\subseteq X_{k_{n}}$ and (2.6) implies that

$$ x\in \underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }} \mathcal{N}\left( T_{n}\right) ^{\bot }=\mathcal{N}\left( T\right) ^{\bot }\quad \text{(by Lemma 2.3), } $$

$ T_{k_{n}}x_{n}=T_{k_{n}}T_{k_{n}}^{\, \dagger }y_{n}=TT_{k_{n}}^{\, \dagger }y_{n}\rightharpoonup Tx\ \left( n\rightarrow \infty \right), $

and for any $v\in X,$

$ \left\langle T_{k_{n}}x_{n}, v\right\rangle =\left\langle P_{\mathcal{R} \left( T_{k_{n}}\right) }y_{n}, v\right\rangle =\left\langle y_{n}, P_{\mathcal{R}\left( T_{k_{n}}\right) }v\right\rangle \\ =\left\langle y_{n}, \left( P_{\mathcal{R}\left( T_{k_{n}}\right) }- P_{\mathcal{R}\left( T\right) }\right) v\right\rangle +\left\langle y_{n}, P_{\mathcal{R}\left( T\right) }v\right\rangle \\ \rightarrow \left\langle y, P_{\mathcal{R}\left( T\right) }v\right\rangle =\left\langle P_{\mathcal{R}\left( T\right) }y, v\right\rangle \ \left( n\rightarrow \infty \right) . $

Thus we have

$ x\in \mathcal{N}\left( T\right) ^{\bot }, \ Tx=P_{\mathcal{R}\left( T\right) }y, \quad \text{that is, }\left( y, x\right) \in \mathcal{G}\left( T^{\dagger }\right) . $

So, we obtain that $$\underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }} \mathcal{G}\left( T_{n}^{\, \dagger }\right) \subseteq \mathcal{G}\left( T^{\dagger }\right) . $$ Now (2.5) is proved.

(b) $\Rightarrow $(c) Let (b) be valid. By Lemma 2.3 it follows from (2.2) that

$ \underset{n\rightarrow \infty }{{s}\text{-}\lim }\mathcal{N}\left( T_{n}\right) =\underset{n\rightarrow \infty }{{w}\text{-} \widetilde{\lim }}\mathcal{N}\left( T_{n}\right) \subseteq \mathcal{N}\left( T\right) . $

Hence to prove (c) we need only to show that

$ \mathcal{N}\left( T\right) \cap \left( \underset{n\rightarrow \infty } {{s}\text{-}\lim }\mathcal{N}\left( T_{n}\right) \right) ^{\bot }=\left\{ 0\right\}. $

(2.7)

Assume that (2.7) is not valid. Then there is a $x_{0}\in \mathcal{N}\left( T\right) \cap \left( \underset{n\rightarrow \infty }{{s} \text{-}\lim }\mathcal{N}\left( T_{n}\right) \right) ^{\bot }$ with $\left\Vert x_{0}\right\Vert =1$. Note that, by (2.2) with Lemma 2.3,

$ \left. \begin{array}{c} \underset{n\rightarrow \infty }{{s}\text{-}\lim }P_{\mathcal{N} \left( T_{n}\right) }=P_{\mathcal{N}\left( T\right) \cap X_{n_{\ast }}}, \\ \left( \underset{n\rightarrow \infty }{{s}\text{-}\lim }\mathcal{N} \left( T_{n}\right) \right) ^{\bot }=\left( \mathcal{N}\left( T\right) \cap X_{n_{\ast }}\right) ^{\bot } \end{array} \right\} \text{ for some }n_{\ast }\in \mathbb{N}. $

It follows that $x_{0}\in \mathcal{N}\left( T\right) \cap \left( \mathcal{N} \left( T\right) \cap X_{n_{\ast }}\right) ^{\bot }$, and that $x_{0}$ satisfies

$ T_{n}P_{n}x_{0}=TP_{n}x_{0}=T\left( P_{n}-I\right) x_{0}\rightarrow 0\ \left( n\rightarrow \infty \right), $

and (noting $\mathcal{N}\left( T_{n}\right) ^{\bot }=\left( \mathcal{N} \left( T\right) \cap X_{n}\right) ^{\bot }\cap X_{n}^{\, }\subseteq X_{n}$)

$ x_{0}-T_{n}^{\dagger }\left( T_{n}P_{n}x_{0}\right) =x_{0}-P_{\mathcal{N} \left( T_{n}\right) ^{\bot }}x_{0} \\ =P_{\mathcal{N}\left( T_{n}\right) }x_{0}\rightarrow P_{\mathcal{N}\left( T\right) \cap X_{n_{\ast }}}x_{0}=0\ \left( n\rightarrow \infty \right) . $

So we have

$ \left( 0, x_{0}\right) \in \underset{n\rightarrow \infty }{{s}\text{-} \lim }\mathcal{G}\left( T_{n}^{\, \dagger }\right) =\mathcal{G}\left( T^{\dagger }\right) \text{, that is, }x_{0}=T^{\dagger }0=0. $

This contradicts with $\left\Vert x_{0}\right\Vert =1$.

(c)$\Rightarrow $(d) Let (c) be valid. For any $b, b_{n}\in X$ with $\left\Vert b_{n}-b\right\Vert \rightarrow 0$ ($n\rightarrow \infty $) we need only to show that

$$ \underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }} T_{n}^{-1}\left( P_{\mathcal{R}\left( T_{n}\right) }b_{n}\right) \subseteq T^{-1}\left( P_{\mathcal{R}\left( T\right) }b\right) \subseteq \underset{n\rightarrow \infty }{{s}\text{-}\lim }T_{n}^{-1}\left( P_{ \mathcal{R}\left( T_{n}\right) }b_{n}\right). $$

(2.8)

Due to Lemma 2.3, (c) is equivalent to $ \mathop {{\text{s - }}\lim }\limits_{n \to \infty } {P_{\mathcal{N}\left( {{T_n}} \right)}} = {P_{\mathcal{N}\left( T \right)}}.$ By Lemma 2.1 (a) and Lemma 2.3, the above equation is equivalent to (2.1). Then, by Lemma 2.1 (c) we obtain (1.5), that is

$ M: = \mathop {\sup }\limits_n \left\| {T_n^\dagger } \right\| < + \infty . $

(2.9)

Let $x\in {w}$-$\mathop {\widetilde {\lim }}\limits_{n \to \infty } T_n^{ - 1}\left( {{P_{\mathcal{R}\left( {{T_n}} \right)}}{b_n}} \right) $. Then there is a squence $\left\{ x_{n}\right\} $ with an integer sequence $\left\{ k_{n}\right\} $ such that

$ k_{n}\geq n, \quad x_{n}\rightharpoonup x, \quad T_{k_{n}}x_{n}=P_{\mathcal{R} \left( T_{k_{n}}\right) }b_{k_{n}}\rightarrow P_{\mathcal{R}\left( T\right) }b\ \left( n\rightarrow \infty \right) . $

Note that for all $v\in X$ there holds

$ \left\vert \left\langle Tx-P_{\mathcal{R}\left( T\right) }b, v\right\rangle \right\vert \leq \left\vert \left\langle T\left( x-x_{n}\right) , v\right\rangle \right\vert +\left\vert \left\langle \left( T-T_{k_{n}}\right) x_{n}, v\right\rangle \right\vert \\ +\left\vert \left\langle T_{k_{n}}x_{n}-P_{\mathcal{R}\left( T\right) }b, v\right\rangle \right\vert \\ \leq \left\vert \left\langle x-x_{n}, T^{\ast }v\right\rangle \right\vert +\left\Vert x_{n}\right\Vert \left\Vert \left( I-P_{k_{n}}\right) T^{\ast }v\right\Vert \\ +\left\vert \left\langle T_{k_{n}}x_{n}-P_{\mathcal{R}\left( T\right) }b, v\right\rangle \right\vert . $

Hence we have $\left\langle Tx-P_{\mathcal{R}\left( T\right) }b, v\right\rangle =0\text{ } \forall v\in X, \quad \text{that is, }x\in T^{-1}\left( P_{\mathcal{R}\left( T\right) }b\right) . $ This gives that

$$ \underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }} T_{n}^{-1}\left( P_{\mathcal{R}\left( T_{n}\right) }b_{n}\right) \subseteq T^{-1}\left( P_{\mathcal{R}\left( T\right) }b\right) . $$

Let $x\in T^{-1}\left( P_{\mathcal{R}\left( T\right) }b\right) $. Take $x_{n}:=x+T_{n}^{\dagger }\left( P_{\mathcal{R}\left( T_{n}\right) }b_{n}-T_{n}x\right), \quad n\in \mathbb{N}. $ Then

$ T_{n}x_{n}=T_{n}T_{n}^{\dagger }P_{\mathcal{R}\left( T_{n}\right) }b_{n}= P_{\mathcal{R}\left( T_{n}\right) }b_{n}, \text{that is, }x_{n}\in T_{n}^{-1}\left( P_{\mathcal{R}\left( T_{n}\right) }b_{n}\right), $

and, by use of (2.9),

$ \left\Vert x_{n}-x\right\Vert =\left\Vert T_{n}^{\dagger }\left( P_{ \mathcal{R}\left( T_{n}\right) }b_{n}-T_{n}x\right) \right\Vert \\ \leq M\left( \left\Vert b_{n}-b\right\Vert +\left\Vert \left( P_{\mathcal{R}\left( T_{n}\right) }-P_{\mathcal{R}\left( T\right) }\right) b\right\Vert +\left\Vert \left( T-T_{n}\right) x\right\Vert \right) \\ \rightarrow 0\ \left( n\rightarrow \infty \right) . $

That gives $x\in {s}$-$\mathop {\lim }\limits_{n \to \infty } T_n^{ - 1}\left( {{P_{\mathcal{R}\left( {{T_n}} \right)}}{b_n}} \right) $. So we get

$ T^{-1}\left( P_{\mathcal{R}\left( T\right) }b\right) \subseteq \underset{n\rightarrow \infty }{{s}\text{-}\lim }T_{n}^{-1}\left( P_{\mathcal{R}\left( T_{n}\right) }b_{n}\right) . $

Thus we get (d).

(d) $\Rightarrow $ (e) Let (d) be valid. It is clear that

$ \underset{n\rightarrow \infty }{{s}\text{-}\lim }T_{n}^{-1}\left( 0\right) =\underset{n\rightarrow \infty }{{w}\text{-}\widetilde{\lim }}T_{n}^{-1}\left( 0\right) =T^{-1}\left( 0\right) . $

By (2.2) and Lemma 2.3, there is a $n_{\ast }\in \mathbb{N}$ such that

$ P_{\mathcal{N}\left( T\right) \cap X_{n_{\ast }}}=\underset{n\rightarrow \infty }{{s}\text{-}\lim }P_{\mathcal{N}\left( T_{n}\right) }= P_{\mathcal{N}\left( T\right) }, $

That gives (e).

(e) $\Rightarrow $ (a) Let (e) hold. Then (2.1) is valid. Hence we have (1.5) by Lemma 2.1 (c), that is (a) holds.

In Theorem 1.1, one thing worth to notice is the condition (e), which claims that the strong convergence of the LSA $\{(X_{n}, T_{n})\}_{n\in\mathbb{N}}$ is equivalent to

$ \exists\ n_{\ast }\in\mathbb{N}\ {\rm s.t.}\ X_{n_{\ast }}\supseteq \mathcal{N}(T). $

Note that the examination of this condition does not involve any computation of operator norm or generalized-inverse, which are unavoidable in the examination of the stability condition (the condition (a) in Theorem 1.1). Here we will look at a simple example to see how the condition (e) can be used in specific integral equation.

Example 1 Let $X:=L^2[-\pi, \pi]$, and we choose the approximation space as

$ X_{n+1}:=\text{span}\{1, \ \sin x, \ \cos x, \ \sin 2x, \ \cos 2x, ..., \ \sin nx, \ \cos nx\}. $

We consider the below integral equation of the second kind:

$ T\varphi:=\frac{2}{3}\pi^3\varphi-\int xt\varphi(t) dt=f, \quad \varphi\in L^2[-\pi, \pi], $

where $f\in L^2[-\pi, \pi]$ is known, and we want to get $\varphi$. Let $\{X_n, T_n\}$ be the LSA of the equation. It is easy to check that $\mathcal N(T) = \text{span}\{x\}, $ and for $x$, it has the Fourier series

$ x = 2\sin x - 2\times\frac{1}{2}\sin 2x + ... + 2\times\frac{(-1)^{n+1}}{n}{\sin nx}+...\quad {\rm in}\ L^2[-\pi, \pi]. $

As the non-zero coefficients in the series has infinite term, so it is obvious that there is no $n^*\in\mathbf N$ such that $\mathcal N(T)\subseteq X_{n^*}$, namely the convergence condition (e) in Theorem 1.1 does not be satisfied. According to the Theorem 1.1, the stability condition fail in this case, and $\underset{n\rightarrow \infty }{{s}\text{-}\lim }T_{n}^{\, \dagger}\neq T^{\dagger }$.

Here we look again on the stability condition (a), namely

$ \mathop {\sup }\limits_n \left\| {T_n^\dagger } \right\| < + \infty . $

We notice that to examine this condition, we need to compute generalized-inverse and operator norm, the cost of which is almost equal to computing the minimal spectral of $T_n$. Thus, it is hard to find a unified method to achieve this task.

The condition (e) in Theorem 1.1 also give the clue to choose convergent approximation scheme. For Example 1, to guarantee the convergence, we choose the approximation space as

$ X_{n+1} = \text{span}\bigg\{1, \ \frac{x}{\pi}, \ \frac{1}{2}[3 {(\frac{x}{\pi})}^2-1], \ ...\ \frac{\pi^n}{2^n n!}\frac{d^n}{dx^n}\{[{(\frac{x}{\pi})}^2-1]^n\}\bigg\}. $

The above is the subspace spanned by the first $(n+1)$ terms of the sequence of Legendre polynomial on $[-\pi, \pi]$. Now the LSA $\{X_n, T_n\}$ possesses convergency as a result of $X_2\supseteq \mathcal N(T)$.