The uniform boundedness theorem, the closed graph theorem and the open mapping theorem are usually referred to as fundamental principles in functional analysis (see e.g. ^[5]). In this paper, based upon the Baire-Hausdorff theorem, we prove a theorem which contains the above principal theorems as its simple corollaries (see Theorem 1.1). As a more profound application of the theorem, a useful result in numerical analysis is established, which may be viewed as an abstract generalization of the well-known Lax-Richtmyer equivalene theorem ^[3] (see Theorem 1.2).

We need to recall the definition of $F$-spaces and to illustrate the notation used in the paper. A linear space $X$ is called a quasi-normed linear space, if for every $x\in X$ there is associated a real number $\left\Vert x\right\Vert $, the quasi-norm of the vector $x$, which satisfies

$\begin{equation*} \begin{array}{l} \left\Vert x\right\Vert \geq 0\ \text{and }\left\Vert x\right\Vert =0\ \text{if and only if}\ \ x=0,\\ \left\Vert x+y\right\Vert \leq \left\Vert x\right\Vert +\left\Vert y\right\Vert \quad \text{(triangle inequality),} \\ \left\Vert -x\right\Vert =\left\Vert x\right\Vert ,\quad \lim\limits_{\alpha _{n}\rightarrow 0}\left\Vert \alpha _{n}x\right\Vert =0\quad \text{and } \lim\limits_{\left\Vert x_{n}\right\Vert \rightarrow 0}\left\Vert \alpha x_{n}\right\Vert =0. \end{array} \end{equation*}$

A quasi-normed linear space $X$ is called an $F$-space if it is complete. Next, let $X$ be an $F$-space. For a point $x\in X$ and a real number $r>0$, by $B_{X}\left( x,r\right) $ we denote the open ball in $X$ with the center at $x$ and the radius $r$, namely,

$\begin{equation*} B_{X}\left( x,r\right) :=\left\{ y\in X\mid \left\Vert y-x\right\Vert <r\right\} . \end{equation*}$

For a subset $A$ of $X$, let the symbols $A^{\circ }$ and $\overline{A}$ denote the interior and the closure of $A$ respectively. The set $A$ is said to be bounded if it is absorbed by any open ball $B_{X}\left( 0,\varepsilon \right) $ with center at $0$, i.e., if there exists a positive constant $ \alpha $ such that $\alpha ^{-1}A\subseteq B_{X}\left( 0,\varepsilon \right) $, where

$\begin{equation*} \alpha ^{-1}A:=\left\{ x\in X\mid x=\alpha ^{-1}a,a\in A\right\} . \end{equation*}$

For a sequence of nonempty subsets $\left\{ A_{n}\right\} $ of $X$, we set

$\begin{equation*} \underset{n\rightarrow \infty }{ {s}\text{-}\lim A_{n}}:=\left\{ x\in X\mid \lim\limits_{n\rightarrow \infty }{\rm dist}\left( x,A_{n}\right) =0\right\} , \end{equation*}$

where

$\begin{equation*} {\rm dist}\left( x,A_{n}\right) :=\inf \left\{ \left\Vert y-x\right\Vert \mid y\in A_{n}\right\} . \end{equation*}$

Now, let $Y$ be an $F$-space on the same scalar field as the $F$-space $X$, $ T:D\subset X\rightarrow Y$ a linear mapping from the subspace $D$ of $X$ into $Y$. By $\mathcal{D}\left( T\right) $, $\mathcal{R}\left( T\right) $, $ \mathcal{N}\left( T\right) $ and $\mathcal{G}\left( T\right) $, we denote the domain, the range, the null space, and the graph of $T$, respectively, i.e.,

$\begin{equation*} \begin{array}{l} \mathcal{D}\left( T\right) :=D,\quad \mathcal{R}\left( T\right) :=\left\{ Tx\mid x\in D\right\} ,\\ \mathcal{N}\left( T\right) :=\left\{ x\in D\mid Tx=0\right\} ,\quad \mathcal{ G}\left( T\right) :=\left\{ \left( x,Tx\right) \mid x\in D\right\} . \end{array} \end{equation*}$

For an $y\in Y,$ the preimage of the point $y$ is denoted by $T^{-1}(y)$, namely,

$\begin{equation*} T^{-1}(y):=\left\{ x\in \mathcal{D}\left( T\right) \mid Tx=y\right\} . \end{equation*}$

In addition, $X\times Y$ is also an $F$-space by the algebraic operations

$\begin{equation*} (x_{{\small 1}},y_{{\small 1}})+(x_{{\small 2}},y_{{\small 2}}):=(x_{{\small 1}},y_{{\small 1}})+(x_{{\small 2}},y_{{\small 2}}),\quad \alpha (x,y):=(\alpha x,\alpha y), \end{equation*}$

and the quasi-norm

$\begin{equation*} \left\Vert (x,y)\right\Vert :=\left( \left\Vert x\right\Vert ^{{\small 2} }+\left\Vert y\right\Vert ^{{\small 2}}\right) ^{{\small 1/2}}. \end{equation*}$

The main results in this paper are the two following theorems.

Theorem 1.1  Let $X$ be an $F$-space. Let $p:X\rightarrow \mathbb{R}$ be a real-valued function on $X$ with the following properties:

a) $p\left( x\right) \geq 0$ for all $x\in X$ (nonnegativity);

b) $p\left( -x\right) =p\left( x\right) $ for every $x\in X$ (symmetry);

c) $\underset{n\in \mathbb{N}\rightarrow \infty }{\underline{\lim }}p\left( n^{-1}x\right) =0$ for each $x\in X$ (absorbability);

d) $p\left( \sum\limits_{1}^{\infty }x_{n}\right) \leq \sum\limits_{1}^{\infty }p\left( x_{n}\right) $ if $\sum\limits_{1}^{\infty }\left\Vert x_{n}\right\Vert <\infty$ (countable subadditivity).

Then $p$ is continuous on $X$.

Theorem 1.2  Let $X$ and $Y$ be $F$-spaces. Let $\left\{ T_{n}:\mathcal{D} \left( T_{n}\right) \subset X\rightarrow Y\right\} $ be a sequence of closed operators with closed ranges. Then the following three properties of $ \left\{ T_{n}\right\} $ are equivalent:

A) if $x_{n}\in \mathcal{D}\left( T_{n}\right) $ ($n\in \mathbb{N}$) with $\lim\limits_{n\rightarrow \infty }\left\Vert T_{n}x_{n}\right\Vert =0,$

$\begin{equation*} \underset{k\in \mathbb{N}\rightarrow \infty }{\underline{\lim }}\,\underset{ n\in \mathbb{N}}{\sup }\,{\rm dist}\ \left( k^{-1}x_{n},\mathcal{N}\left( T_{n}\right) \right) =0\text{;} \end{equation*}$

B) for every $\varepsilon >0$ there exists a $\delta >0$ such that

$\begin{equation*} B_{Y}\left( 0,\delta \right) \cap \mathcal{R}\left( T_{n}\right) \subseteq T_{n}\left( B_{X}\left( 0,\varepsilon \right) \cap \mathcal{D}\left( T_{n}\right) \right) \quad \text{for all }n\in \mathbb{N}\mathbf{;} \end{equation*}$

C) if $y_{n}\in \mathcal{R}\left( T_{n}\right) $ ($n\in \mathbb{N}$) with $\lim\limits_{n\rightarrow \infty }y_{n}=y,$

$\begin{equation*} \underset{n\rightarrow \infty }{{s}\text{-}\lim }T_{n}^{-1}(y_{n})= \left\{ x\in X\mid (x,y)\in \,\underset{n\rightarrow \infty }{{s} \text{-}\lim }\mathcal{G}\left( T_{n}\right) \right\} . \end{equation*}$

Theorem 1.1 may be regarded as the basic tool in the theory of $F$-spaces, for it implies some fundamental principles as simple corollaries, such as the uniform boundedness theorem, the open mapping theorem, and the closed graph theorem. It must be noted that Theorem 1.1 contains Theorem 1.2 as a more profound application.

Theorem 1.2 can be interpreted as a generalization of the well-known Lax-Richtmyer equivalence theorem ^[3] (which states that, for linear well-posed initial value problems, a consistent difference scheme is convergent if and only if it is stable): for an oprater equation of the first kind $Tx=y,$ where $T:\mathcal{D}\left( T\right) \subset X\rightarrow Y$ is a closed linear operator, we assume that the equation is with a consistent approximation scheme $\left\{ T_{n}\right\} $ as given in Theorem 1.2, here "consistent" means

$\begin{equation*} \mathcal{G}\left( T\right) =\,\underset{n\rightarrow \infty }{{s} \text{-}\lim }\mathcal{G}\left( T_{n}\right) . \end{equation*}$

Then condition A or B stands for stability of the scheme $\left\{ T_{n}\right\} $, condition C stands for convergence of $\left\{ T_{n}\right\} $, and the conclusion is

$\begin{equation*} \mbox{convergence}\text{ C (}\underset{n\rightarrow \infty }{{s} \text{-}\lim }T_{n}^{-1}(y_{n})=T^{-1}(y)\text{)}\Longleftrightarrow \mbox{ stability}\text{ A}\Longleftrightarrow \mbox{stability}\text{ B.} \end{equation*}$

This is also a generalization of [1, Theorem 2.1] and [2, Theorem 1].

The paper is organized as follows: in Section 2, we present the proof of Theorem 1.1, and in Section 3, we show that some fundamental principles as simple corollaries of the theorem. In Section 4, we present the proof of Theorem 1.2 with remarks and examples of application.

We recall the Baire-Hausdorff theorem before proving Theorem 1.1 : A non-void complete metric space is of the second category (see, e.g., ^[5]). As is well known, the completeness of an $F$-space enables us to apply the Baire-Hausdorff theorem and to obtain such fundamental principles in functional analysis as the uniform boundedness theorem, the closed graph theorem and open mapping theorem. Here, we are to apply the Baire-Hausdorff theorem to establish a more general principle, Theorem 1.1.

Proof of Theorem 1.1 The proof is to be carried out in three steps.

Step 1 Prove that for any $\varepsilon >0$ there exists a sequence $\left\{ \delta _{n}\right\}$ of positive real numbers such that

$\begin{equation} B_{X}\left( 0,\delta _{n}\right) \subset \overline{p^{{\small -1}}\left( \left[0,\varepsilon /2^{n}\right] \right) }\quad \text{ for each } n\in \mathbb{N} \end{equation}$

(2.1)

and

$\begin{equation} \sum\limits_{n=1}^{\infty }\delta _{n}<+\infty . \end{equation}$

(2.2)

It is obvious by properties a) and c) of $p$ that

$\begin{equation*} X=\overset{\infty }{\underset{n=1}{\cup }}np^{-1}\left( \left[ 0,\varepsilon /2\right] \right) \quad \text{for every }\varepsilon >0\text{.} \end{equation*}$

Since $X$ is a non-void complete metric space, $X$ is of the second category by the Baire-Hausdorff theorem, so that there must be a natural number $k$ such that $\overline{kp^{-1}\left( \left[ 0,\varepsilon /2\right] \right) }^{\circ }\neq \varnothing.$ Hence, there exist $x_{0}\in X$ and $r>0$ such that $B_{X}\left( x_{{\small 0}},r\right) \subset \overline{kp^{-1}\left( \left[0,\varepsilon /2\right] \right) },$ that is,

$\begin{equation} k^{-1}B_{X}\left( x_{{\small 0}},r\right) \subset \overline{p^{-1}\left( \left[0,\varepsilon /2\right] \right) }. \end{equation}$

(2.3)

Note that, by properties b) and d) of $p$, the set $\overline{ p^{-1}\left( \left[0,\varepsilon /2\right] \right) }$ is with the following two properties

$\begin{equation} \left. \begin{array}{l} x\in \overline{p^{-1}\left( \left[0,\varepsilon /2\right] \right) } \Longrightarrow \quad -x\in \overline{p^{-1}\left( \left[0,\varepsilon /2 \right] \right) }; \\ \left\{ x,y\right\} \subset \overline{p^{-1}\left( \left[0,\varepsilon /2 \right] \right) }\quad \Longrightarrow \quad x+y\in \overline{p^{-1}\left( \left[0,\varepsilon \right] \right) }. \end{array} \right\} \end{equation}$

(2.4)

Therefore from (2.3) and (2.4), we obtain $B_{X}\left( 0,r/k\right) \subset \overline{p^{-1}\left( \left[0,\varepsilon \right] \right) }.$ Thus for every $\varepsilon >0$, there exists $\delta >0$ such that

$\begin{equation} B_{X}\left( 0,\delta \right) \subset \overline{p^{-1}\left( \left[ 0,\varepsilon \right] \right) }. \end{equation}$

(2.5)

From (2.5), we obtain that for any $\varepsilon >0$ there exists a sequence $\left\{ \delta _{n}\right\} $ of positive real numbers such that (2.1) and (2.2) hold.

Step 2 Prove that for any $\varepsilon >0$ there exists $\delta >0$ such that

$\begin{equation} B_{X}\left( 0,\delta \right) \subset p^{-1}\left( \left[0,\varepsilon \right] \right). \end{equation}$

(2.6)

By the conclusion of Step 1, for any $\varepsilon >0$ there exists a sequence $\left\{ \delta _{n}\right\} $ of positive real numbers such that (2.1) and (2.2) hold. Let $\delta =\delta _{1}$ and $ x\in B_{X}\left( 0,\delta \right) $. Then we have that $x\in \overline{ p^{-1}\left( \left[0,\varepsilon /2\right] \right) }$ from (2.1), and hence there exists $x_{1}\in p^{-1}\left( \left[ 0,\varepsilon /2\right] \right) $ such that $\left\Vert x-x_{{\small 1}}\right\Vert <\delta _{{\small 2}},$ i.e., $x-x_{{\small 1}}\in B\left( 0,\delta _{{\small 2}}\right).$ Assume $n\geq 1$ and $x_{k}\in p^{-1}\left( \left[0,\varepsilon /2^{k} \right] \right) $ $\left( k=1,\cdots ,n\right) $ are chosen to satisfy

$\begin{equation*} \left\Vert x-\sum\limits_{j=1}^{k}x_{j}\right\Vert <\delta _{k+1},\quad \text{i.e.,} \quad x-\overset{k}{\underset{j=1}{\sum }}x_{j}\in B_{X}\left( 0,\delta _{k+1}\right) \text{.} \end{equation*}$

Then by (2.1), there exists $x_{n+1}\in p^{-1}\left( \left[ 0,\varepsilon /2^{n+1}\right] \right) $ such that

$\begin{equation*} \left\Vert x-\sum\limits_{j=1}^{n+1}x_{j}\right\Vert <\delta _{n+{\small 2}},\quad \text{i.e.,}x-\sum\limits_{j=1}^{n+1}x_{j}\in B\left( 0,\delta _{n+{\small 2} }\right) . \end{equation*}$

Thus we obtain a sequence $\left\{ x_{n}\right\} $ of $X$ such that

$\begin{equation} x_{n}\in p^{-1}\left( \left[0,\varepsilon /2^{n}\right] \right) \end{equation}$

(2.7)

and

$\begin{equation} \left\Vert x-\sum\limits_{k=1}^{n}x_{k}\right\Vert <\delta _{n+{\small 1}}\quad \text{and }\left\Vert x\right\Vert <\delta _{{\small 1}}. \end{equation}$

(2.8)

Note that (2.8) with (2.2) implies that

$\begin{equation} \sum\limits_{n=1}^{\infty }\left\Vert x_{n}\right\Vert \leq 2\sum\limits_{n=1}^{\infty }\delta _{n}<+\infty \end{equation}$

(2.9)

and

$\begin{equation} x=\sum\limits_{n=1}^{\infty }x_{n}. \end{equation}$

(2.10)

Therefore, from (2.7), (2.9), (2.10), and the property d) of $p$, we conclude that

$\begin{equation*} p\left( x\right) \leq \sum\limits_{n=1}^{\infty }p\left( x_{n}\right) \leq \sum\limits_{n=1}^{\infty }\frac{\varepsilon }{2^{n}}=\varepsilon . \end{equation*}$

Thus (2.6) follows.

Step 3 Prove that $p$ is continuous on $X$, we need only to show that $p$ is continuous at $0$. This is obviously true by the conclusion of Step 2.

In this section, we will see that the uniform boundedness theorem, the closed graph theorem, and the open mapping theorem can be proved as simple corollaries of Theorem 1.1. In order to emphasize the role of Theorem 1.1, the above three theorems are stated in a little more generality than is usually needed. By doing that, we get more interesting versions.

Corollary 3.1 (The Uniform Boundedness Theorem) Let $X$ be an $F$-space, $Y$ a quasi-normed linear space and $\left\{ T_{\lambda }\right\} _{\lambda \in \Lambda }$ a family of continuous mappings defined on $X$ into $Y$. Assume that

1) $\left\Vert T_{\lambda }\left( x+y\right) \right\Vert \leq \left\Vert T_{\lambda }\left( x\right) \right\Vert +\left\Vert T_{\lambda }\left( y\right) \right\Vert $ for any $\lambda \in \Lambda $ and $ x,y\in X$, and

2) $\lim\limits_{n\in\mathbb{N}\rightarrow \infty }\sup\limits_{\lambda \in \Lambda }\,\left\Vert T_{\lambda }\left( n^{-1}x\right) \right\Vert =0 $for any $x\in X.$

Then $\lim\limits_{x\rightarrow 0}\sup\limits_{\lambda \in \Lambda }\,\left\Vert T_{\lambda }\left( x\right) \right\Vert =0.$

Proof  By Assumption 1), we have

$\begin{equation*} \frac{1}{n}\left\Vert T_{{\small \lambda }}\left( x\right) \right\Vert \leq \left\Vert T_{{\small \lambda }}\left( \frac{1}{n}x\right) \right\Vert \quad \text{for each }n\in \mathbb{N}\mathbf{.} \end{equation*}$

Therefore, by Assumption 2), we obtain

$\begin{equation} \underset{n\rightarrow \infty }{\lim }\left\{ \frac{1}{n}\underset{{\small \lambda }\in \Lambda }{\sup }\left\Vert T_{{\small \lambda }}\left( x\right) \right\Vert \right\} =0\quad \text{for any }x\in X. \end{equation}$

(3.1)

It follows from (3.1) that

$\begin{equation} \underset{{\small \lambda }\in \Lambda }{\sup }\left\Vert T_{{\small \lambda }}\left( x\right) \right\Vert <\infty \quad \text{for any }x\in X. \end{equation}$

(3.2)

Now, define $p:X\rightarrow \mathbb{R}$ as follows

$\begin{equation*} p\left( x\right) :=\max \left\{ \underset{{\small \lambda }\in \Lambda }{ \sup }\left\Vert T_{{\small \lambda }}\left( x\right) \right\Vert,\underset{{\small \lambda }\in \Lambda }{\sup }\left\Vert T_{{\small \lambda }}\left( -x\right) \right\Vert \right\} . \end{equation*}$

It is clear that $p$ is well-defined (by (3.2)) and is with nonnegativity, symmetry, absorbability and countable subadditivity. Hence, by Theorem 1.1, $p$ is continuous on $X,$ which implies

$\begin{equation*} \underset{x\rightarrow 0}{\lim }\,\underset{{\small \lambda }\in \Lambda }{ \sup }\,\left\Vert T_{{\small \lambda }}\left( x\right) \right\Vert =0. \end{equation*}$

Corollary 3.2 (The Generalized Closed Graph Theorem) Let $X$ and $Y$ be $F$-spaces. Let $T:\mathcal{D}\left( T\right) \subset X\rightarrow Y$ be a mapping which satisfies the following conditions

1) $\mathcal{D}\left( T\right) $ is a closed subspace of $X;$

2) $\left\Vert T\left( -x\right) \right\Vert =\left\Vert T\left( x\right) \right\Vert \quad $for every $x\in \mathcal{D}\left( T\right) ;$

3) $\underset{n\rightarrow \infty }{\underline{\lim }}\left\Vert T\left( n^{-1}x\right) \right\Vert =0\quad $for every $x\in \mathcal{D}\left( T\right) ;$ and

4) $\left\Vert T\left( \sum\limits_{1}^{\infty }x_{n}\right) \right\Vert \leq \sum\limits_{1}^{\infty }\left\Vert T\left( x_{n}\right) \right\Vert \quad $for any sequence $\left\{ x_{n}\right\} $ of $\mathcal{D}\left( T\right) $ with

$\begin{equation*} \sum\limits_{1}^{\infty }\left\Vert x_{n}\right\Vert <\infty \text{ and } \sum\limits_{1}^{\infty }\left\Vert T\left( x_{n}\right) \right\Vert <\infty . \end{equation*}$

Then

(a) $\lim\limits_{x\rightarrow 0}T\left( x\right) =0;$

(b) $\left\Vert T\left( x\right) \right\Vert $ is continuous on $ \mathcal{D}\left( T\right).$

Proof  Define $p:\mathcal{D}\left( T\right) \rightarrow \mathbb{R}$ as follows $p\left( x\right) :=\left\Vert T\left( x\right) \right\Vert.$ It is not difficult to see that $p$ satisfies all the conditions in Theorem 1.1. Hence, by Theorem 1.1, $p$ is continuous on $\mathcal{D}\left( T\right) ,$ which implies (a) and (b).

Corollary 3.3 (The Open Mapping Theorem) Let $X$ and $Y$ be $F$-spaces. Let $T:\mathcal{D}\left( T\right) \subset X\rightarrow Y$ be a closed linear operator with $\mathcal{R}\left( T\right) =Y.$ Then $T$ is an open mapping, i.e., $T\left( U\right) $ is open in $Y$ whenever $U$ is open in $\mathcal{D}\left( T\right) $.

Proof  Define $p:Y\rightarrow \mathbb{R}$ as follows

$\begin{equation} p\left( y\right) :=\underset{x\in T^{-1}\left( y\right) }{\inf }\left\Vert x\right\Vert . \end{equation}$

(3.3)

Then $p$ is obviously with nonnegativity, symmetry, and absorbability on $Y$. We now verify that $p$ is with countable subadditivity. Let a sequence $\left\{ y_{n}\right\}$ of $Y$ satisfy $\sum\limits_{1}^{\infty }\left\Vert y_{n}\right\Vert <+\infty $.

Then, by the completeness of $Y$, there exists a vector $y_{\infty }\in Y$ such that $y_{\infty }=\sum\limits_{1}^{\infty }y_{n}$ in $Y$. We need to show that

$\begin{equation} p\left( y_{\infty }\right) \leq \sum\limits_{1}^{\infty }p\left( y_{n}\right) . \end{equation}$

(3.4)

If $\sum\limits_{1}^{\infty }p\left( y_{n}\right) =+\infty $, (3.4) is automatically satisfied. Assume $\sum\limits_{1}^{\infty }p\left( y_{n}\right) <+\infty $. For every $\varepsilon >0$ and every $n\in \mathbb{N}$, by (3.3), there exists a vector $x_{n}\in T^{-1}\left( y_{n}\right) $ such that $\left\Vert x_{n}\right\Vert <p\left( y_{n}\right) +\frac{\varepsilon }{2^{n}},$ and therefore

$\begin{equation} \sum\limits_{1}^{\infty }\left\Vert x_{n}\right\Vert <\sum\limits_{1}^{\infty }p\left( y_{n}\right) +\varepsilon . \end{equation}$

(3.5)

Since $X$ is an $F$-space, $\sum\limits_{1}^{\infty }x_{n}$ is convergent to a vector $x_{\infty }$ of $X$. By the closedness of $T$, we have that $x_{\infty }\in \mathcal{D}\left( T\right) $ and $Tx_{\infty }=y_{\infty }$. Hence, by (3.5),

$\begin{equation} p\left( y_{\infty }\right) \leq \left\Vert x_{\infty }\right\Vert \leq \sum\limits_{1}^{\infty }\left\Vert x_{n}\right\Vert <\sum\limits_{1}^{\infty }p\left( y_{n}\right) +\varepsilon . \end{equation}$

(3.6)

Note that the $\varepsilon $ is arbitrary, so (3.6) implies (3.4). Now, by Theorem 1.1, $p$ is continuous on $Y$.

To prove that $T$ is an open mapping, we have only to show that for every $\varepsilon >0$, there exists a $\delta >0$ such that

$\begin{equation} B_{Y}\left( 0,\delta \right) \subset T\left( \mathcal{D}\left( T\right) \cap B_{X}\left( 0,\varepsilon \right) \right) . \end{equation}$

(3.7)

By the continuity of $p$, for any $\varepsilon >0$, there is a $\delta >0$ such that

$\begin{equation*} p\left( y\right) <\varepsilon \quad \text{for all}\ y\in B_{Y}\left( 0,\delta \right) . \end{equation*}$

Hence, for every $y\in B_{Y}\left( 0,\delta \right),$ there exists a vector $ x\in T^{-1}\left( y\right) \cap B_{X}\left( 0,\varepsilon \right) $ by the definition of $p$, and therefore (3.5) is true.

To prove Theorem 1.2, we prepare the following lemma.

Lemma 4.1  Let $Y$ be an $F$-space, $\left\{ Y_{n}\right\} $ a sequence of closed subspaces of $Y$. Let $\mathfrak{Y}$ be the set of all vectors $ \mathbf{ \pmb{\mathsf{ ξ}} =}\left\{ y_{n}\right\} $ with $y_{n}\in Y_{n}$ (for every $n\in \mathbb{N}$) and $\lim\limits_{n\rightarrow \infty }y_{n}=0$, i.e.,

$\begin{equation*} \mathfrak{Y}=\left\{ \mathbf{ \pmb{\mathsf{ ξ}} }\mid \mathbf{ \pmb{\mathsf{ ξ}} =}\left\{ y_{n}\right\} ,\ y_{n}\in Y_{n}\ \left( \forall n\in \mathbb{N}\right) ,\ \underset{ n\rightarrow \infty }{\lim }y_{n}=0\right\}. \end{equation*}$

Then $\mathfrak{Y}$ is an $F$-space by the algebrac operations $\left\{ y_{n}^{\prime }\right\} +\left\{ y_{n}^{\prime \prime }\right\} :=\left\{ y_{n}^{\prime }+y_{n}^{\prime \prime }\right\},\alpha \left\{ y_{n}\right\} :=\left\{ \alpha y_{n}\right\}$ and the quasi-norm $\left\Vert \left\{ y_{n}\right\} \right\Vert :=\ \underset{n\in \mathbb{N}}{ \sup }\left\Vert y_{n}\right\Vert.$

Proof  It is easy to show that $\mathfrak{Y}$ is a linear space and

$\begin{equation*} \begin{array}{c} \left\Vert \mathbf{ \pmb{\mathsf{ ξ}} }\right\Vert \geq 0\quad \text{and }\left\Vert \mathbf{ \pmb{\mathsf{ ξ}} }\right\Vert =0\ \Longleftrightarrow \ \mathbf{ \pmb{\mathsf{ ξ}} =0,} \quad \left\Vert \mathbf{ \pmb{\mathsf{ ξ}} + \pmb{\mathsf{ η}} }\right\Vert \leq \left\Vert \mathbf{ \pmb{\mathsf{ ξ}} } \right\Vert +\left\Vert \mathbf{ \pmb{\mathsf{ η}} }\right\Vert ,\quad \left\Vert -\mathbf{ \pmb{\mathsf{ ξ}} }\right\Vert =\left\Vert \mathbf{ \pmb{\mathsf{ ξ}} }\right\Vert . \end{array} \end{equation*}$

To prove that $\left\Vert \mathbf{\cdot }\right\Vert $ is a quasi-norm on $ \mathfrak{Y}$, we need only to show that

$\begin{equation} \lim\limits_{\alpha _{k}\rightarrow 0}\left\Vert \alpha _{k}\mathbf{ \pmb{\mathsf{ ξ}} } \right\Vert =0 \end{equation}$

(4.1)

and

$\begin{equation} \lim\limits_{\left\Vert \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right\Vert \rightarrow 0}\left\Vert \alpha \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right\Vert =0. \end{equation}$

(4.2)

Noting that $\mathbf{ \pmb{\mathsf{ ξ}} =}\left\{ y_{n}\right\} ,$ as a subset of $Y,$ is bounded and that $\alpha _{k}\rightarrow 0$, we obtain (4.1). Put $ \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }:=\left\{ y_{n}^{\left( k\right) }\right\} _{n=1}^{\infty }$ ($k\in \mathbb{N}$). Then $\left\Vert \mathbf{ \pmb{\mathsf{ ξ}} } ^{\left( k\right) }\right\Vert \rightarrow 0$ implies $\lim\limits_{k\rightarrow \infty }\sup\limits_{n\in \mathbb{N}}\left\Vert y_{n}^{\left( k\right) }\right\Vert =0$ and therefore

$\begin{equation*} \lim\limits_{k\rightarrow \infty }\sup\limits_{n\in \mathbb{N}}\left\Vert \alpha y_{n}^{\left( k\right) }\right\Vert =0, \end{equation*}$

since $\lim\limits_{\left\Vert y\right\Vert \rightarrow 0}\left\Vert \alpha y\right\Vert =0$. Thus, (4.2) holds.

Next, we show that $\mathfrak{Y}$ is complete. Let $ \lim\limits_{k,l\rightarrow \infty }\left\Vert \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }- \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( l\right) }\right\Vert =0$ in $\mathfrak{Y}$. Put $ \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }:=\left\{ y_{n}^{\left( k\right) }\right\} $ ($k\in \mathbb{N}$). Then $\lim\limits_{k,l\rightarrow \infty }\sup\limits_{n\in \mathbb{N}}\left\Vert y_{n}^{\left( k\right) }-y_{n}^{\left( l\right) }\right\Vert =0.$ Hence there exists a sequence $\left\{ y_{n}^{\left( \infty \right) }\right\} $ of $Y$ such that

$\begin{equation} \lim\limits_{k\rightarrow \infty }\sup\limits_{n\in \mathbb{N}}\left\Vert y_{n}^{\left( k\right) }-y_{n}^{\left( \infty \right) }\right\Vert =0 \end{equation}$

(4.3)

by the completeness of $Y$. Note that (4.3) implies that $\lim\limits_{n\rightarrow \infty }\left\Vert y_{n}^{\left( \infty \right) }\right\Vert =0.$ Put $\mathbf{ \pmb{\mathsf{ ξ}} }^{\left( \infty \right) }:=\left\{ y_{n}^{\left( \infty \right) }\right\} $, then $\mathbf{ \pmb{\mathsf{ ξ}} }^{\left( \infty \right) }\in \mathfrak{Y}$ and $\mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\rightarrow \mathbf{ \pmb{\mathsf{ ξ}} } ^{\left( \infty \right) }$ as $k\rightarrow \infty $ in $\mathfrak{Y.}$

Proof of Theorem 1.2 The proof will be carried out in three steps.

Step 1 A) $\Rightarrow $ B): Assume A) and put

$\begin{equation*} \mathfrak{Y}=\left\{ \mathbf{ \pmb{\mathsf{ ξ}} }\mid \mathbf{ \pmb{\mathsf{ ξ}} =}\left\{ y_{n}\right\} ,\ y_{n}\in \mathcal{R}\left( T_{n}\right) \!\ \left( \forall n\in \mathbb{N} \right) ,\ \underset{n\rightarrow \infty }{\lim }y_{n}=0\right\} . \end{equation*}$

Then

$\begin{equation*} \underset{k\rightarrow \infty }{\underline{\lim }}\!\,\underset{n\in \mathbb{ N}}{\sup }\!\,{\rm dist}\left( k^{-1}x_{n},\mathcal{N}\left( T_{n}\right) \right) =0 \end{equation*}$

for any $\mathbf{ \pmb{\mathsf{ ξ}} =}\left\{ y_{n}\right\} \in \mathfrak{Y}$ and $\left\{ x_{n}\right\} $ with $x_{n}\in T_{n}^{-1}\left( y_{n}\right) $ $\left( n\in \mathbb{N}\right) $. Since

$\begin{equation*} {\rm dist}\left( k^{-1}x_{n},\mathcal{N}\left( T_{n}\right) \right) = \underset{x\in T_{n}^{-1}\left( y_{n}\right) }{\inf }\left\Vert k^{-1}x\right\Vert \text{,}\quad \text{where }y_{n}=T_{n}\left( x_{n}\right) \text{,} \end{equation*}$

and since

$\begin{equation*} \underset{x\in T_{n}^{-1}\left( y_{n}\right) }{\inf }\left\Vert x\right\Vert \leq k\underset{x\in T_{n}^{-1}\left( y_{n}\right) }{\inf }\left\Vert k^{-1}x\right\Vert \quad \left( n,k\in \mathbb{N}\right) , \end{equation*}$

we have that

$\begin{equation} \underset{n\in \mathbb{N}}{\sup }{\tiny \!\,}\underset{x\in T_{n}^{-1}\left( y_{n}\right) }{\inf }\left\Vert x\right\Vert <+\infty \quad \text{for any } \mathbf{ \pmb{\mathsf{ ξ}} =}\left\{ y_{n}\right\} \in \mathfrak{Y} \end{equation}$

(4.4)

and that

$\begin{equation} \underset{k\rightarrow \infty }{\underline{\lim }}{\tiny \!\,}\underset{n\in \mathbb{N}}{\sup }{\tiny \!\,}\underset{x\in T_{n}^{-1}\left( y_{n}\right) }{ \inf }\left\Vert k^{-1}x\right\Vert =0\text{ for any }\mathbf{ \pmb{\mathsf{ ξ}} =} \left\{ y_{n}\right\} \in \mathfrak{Y}\text{.} \end{equation}$

(4.5)

Noting that $\mathcal{R}\left( T_{n}\right) \!$ $\left( n\in \mathbb{N} \right) $ are all closed, by Lemma 4.1, $\mathfrak{Y}$ is an $F$-space by the algebrac operations $\left\{ y_{n}^{\prime }\right\} +\left\{ y_{n}^{\prime \prime }\right\} :=\left\{ y_{n}^{\prime }+y_{n}^{\prime \prime }\right\} ,\alpha \left\{ y_{n}\right\} :=\left\{ \alpha y_{n}\right\}$ and the quasi-norm $\left\Vert \left\{ y_{n}\right\} \right\Vert :=\ \underset{n\in \mathbb{N}}{ \sup }\left\Vert y_{n}\right\Vert.$ Define $p:\mathfrak{Y}\rightarrow \mathbb{R}$ as follows

$\begin{equation} p\left( \mathbf{ \pmb{\mathsf{ ξ}} }\right) :=\ \underset{n\in \mathbb{N}}{\sup }{\tiny \!\, }\underset{x\in T_{n}^{-1}\left( y_{n}\right) }{\inf }\left\Vert x\right\Vert \quad \text{for any }\mathbf{ \pmb{\mathsf{ ξ}} =}\left\{ y_{n}\right\} \in \mathfrak{Y}\text{.} \end{equation}$

(4.6)

By (4.4), (4.6) and (4.5), $p$ is well-defined on $ \mathfrak{Y}$, and is with nonnegativity, symmetry and absorbability. Next we verify that $p$ is with countable subadditivity.

Let $\left\{ \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right\} _{k=1}^{\infty }$ be a sequence of $\mathfrak{Y}$ with $\sum\limits_{k=1}^{\infty }\left\Vert \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right\Vert <+\infty $. Then, by the completeness of $\mathfrak{Y}$, there exists a vector $\mathbf{ \pmb{\mathsf{ ξ}} }^{\left( \infty \right) }\in \mathfrak{Y}$ such that

$\begin{equation} \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( \infty \right) }=\sum\limits_{k=1}^{\infty }\mathbf{ \pmb{\mathsf{ ξ}} } ^{\left( k\right) }\quad \text{in }\mathfrak{Y}\text{.} \end{equation}$

(4.7)

We have to show that

$\begin{equation} p\left( \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( \infty \right) }\right) \leq \sum\limits_{k=1}^{\infty }p\left( \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right) . \end{equation}$

(4.8)

If $\sum\limits_{k=1}^{\infty }p\left( \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right) =+\infty $, (4.8) holds automatically. Now, assume $ \sum\limits_{k=1}^{\infty }p\left( \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right) <+\infty $ and put

$\begin{equation} \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }=\left\{ y_{n}^{\left( k\right) }\right\} ,\quad \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( \infty \right) }=\left\{ y_{n}^{\left( \infty \right) }\right\} . \end{equation}$

(4.9)

By the definition of $p$ (see (4.6)),

$\begin{equation*} \underset{x\in T_{n}^{-1}\left( y_{n}^{\left( k\right) }\right) }{\inf } \left\Vert x\right\Vert \leq p\left( \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right) \quad \text{ for any }k\in \mathbb{N}\text{ and }n\in \mathbb{N}\text{.} \end{equation*}$

Therefore, for every $\varepsilon >0$, every $k\in \mathbb{N}$ and every $ n\in \mathbb{N}$, there exists a point $x_{n}^{\left( k\right) }\left( \varepsilon \right) \in T_{n}^{-1}\left( y_{n}^{\left( k\right) }\right) $ such that

$\begin{equation*} \left\Vert x_{n}^{\left( k\right) }\left( \varepsilon \right) \right\Vert <p\left( \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right) +\frac{\varepsilon }{2^{k}}. \end{equation*}$

Hence, for every $\varepsilon >0$ and every $n\in \mathbb{N}$, there is a sequence $\left\{ x_{n}^{\left( k\right) }\left( \varepsilon \right) \right\} _{k=1}^{\infty }$ of $\mathcal{D}\left( T_{n}\right) $ such that

$\begin{equation} T_{n}\left( x_{n}^{\left( k\right) }\left( \varepsilon \right) \right) =y_{n}^{\left( k\right) }\quad \text{for every }k\in \mathbb{N} \end{equation}$

(4.10)

and

$\begin{equation} \sum\limits_{k=1}^{\infty }\left\Vert x_{n}^{\left( k\right) }\left( \varepsilon \right) \right\Vert <\sum\limits_{k=1}^{\infty }p\left( \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right) +\varepsilon . \end{equation}$

(4.11)

By (4.11) and the completeness of $X$, there exists a point $ x_{n}^{\left( \infty \right) }\left( \varepsilon \right) \in X$ such that

$\begin{equation} x_{n}^{\left( \infty \right) }\left( \varepsilon \right) =\sum\limits_{k=1}^{\infty }x_{n}^{\left( k\right) }\left( \varepsilon \right) \quad \text{in }X. \end{equation}$

(4.12)

On the other hand, (4.10) with (4.7) and (4.9) implies that for every $n\in \mathbb{N}$, there holds

$\begin{equation} y_{n}^{\left( \infty \right) }=\sum\limits_{k=1}^{\infty }y_{n}^{\left( k\right) }=\sum\limits_{k=1}^{\infty }T_{n}\left( x_{n}^{\left( k\right) }\left( \varepsilon \right) \right) \quad \text{in }Y. \end{equation}$

(4.13)

Since, for every $n\in \mathbb{N}$, $T_{n}$ is closed, it follows from (4.12) and (4.13) that $x_{n}^{\left( \infty \right) }\left( \varepsilon \right) \in \mathcal{D}\left( T_{n}\right) $ and $x_{n}^{\left( \infty \right) }\left( \varepsilon \right) \in T_{n}^{-1}\left( y_{n}^{\left( \infty \right) }\right) $. Hence, by (4.12) and (4.11), for every $\varepsilon >0$ and every $n\in \mathbb{N}$, we have that

$\begin{equation*} \begin{array}{l} \underset{x\in T_{n}^{-1}\left( y_{n}^{\left( \infty \right) }\right) }{\inf }\left\Vert x\right\Vert \leq \left\Vert x_{n}^{\left( \infty \right) }\left( \varepsilon \right) \right\Vert \leq \sum\limits_{k=1}^{\infty }\left\Vert x_{n}^{\left( k\right) }\left( \varepsilon \right) \right\Vert <\sum\limits_{k=1}^{\infty }p\left( \mathbf{ \pmb{\mathsf{ ξ}} } ^{\left( k\right) }\right) +\varepsilon. \end{array} \end{equation*}$

So for every $\varepsilon >0$, we have

$\begin{equation*} p\left( \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( \infty \right) }\right) \leq \sum\limits_{k=1}^{\infty }p\left( \mathbf{ \pmb{\mathsf{ ξ}} }^{\left( k\right) }\right) +\varepsilon . \end{equation*}$

Thus (4.8) is true.

We have proved that $p$ satisfies all the conditions of Theorem 1.1. Hence, $p$ is continuous on $\mathfrak{Y}$, and is continuous at $0$. This implies that for every $\varepsilon >0$, there exists a $\delta >0$ such that

$\begin{equation*} \underset{x\in T_{n}^{-1}\left( y\right) }{\inf }\left\Vert x\right\Vert <\varepsilon \quad \text{for every }n\in \mathbb{N}\text{ and every }y\in \mathcal{R}\left( T_{n}\right) \text{ with }\left\Vert y\right\Vert <\delta . \end{equation*}$

Therefore we obtain that for every $\varepsilon >0$ there exists a $\delta >0$ such that

$\begin{equation*} B_{Y}\left( 0,\delta \right) \cap \mathcal{R}\left( T_{n}\right) \subseteq T_{n}\left( B_{X}\left( 0,\varepsilon \right) \cap \mathcal{D}\left( T_{n}\right) \right) \quad \text{for all }n\in \mathbb{N}\mathbf{.} \end{equation*}$

That shows that A) implies B).

Step 2 B) $\Rightarrow $ C) Assume that $ \left\{ T_{n}\right\} $ satisfies B) and that $y_{n}\in \mathcal{R}\left( T_{n}\right) $ ($n\in \mathbb{N}$) and $\lim\limits_{n\rightarrow \infty }y_{n}=y$. We have to show that

$\begin{equation} \underset{n\rightarrow \infty }{{s}\text{-}\lim }T_{n}^{-1}(y_{n})= \left\{ x\in X\mid (x,y)\in \,\underset{n\rightarrow \infty }{{s} \text{-}\lim }\mathcal{G}\left( T_{n}\right) \right\}. \end{equation}$

(4.14)

If $x\in {s}$-$\lim\limits_{n\rightarrow \infty }T_{n}^{-1}(y_{n})$, there exists a sequence $\left\{ x_{n}\right\} $ with $x_{n}\in T_{n}^{-1}(y_{n})$ ($n\in \mathbb{N}$) such that $x_{n}\rightarrow x,\ T_{n}x_{n}\rightarrow y,\text{ as }n\rightarrow \infty.$ This implies that $\left( x,y\right) \in {s}$-$\lim\limits_{n\rightarrow \infty }\mathcal{G}\left( T_{n}\right) $. Hence we obtain

$\begin{equation} \underset{n\rightarrow \infty }{{s}\text{-}\lim }T_{n}^{-1}(y_{n}) \subseteq \left\{ x\in X\mid (x,y)\in \,\underset{n\rightarrow \infty }{ {s}\text{-}\lim }\mathcal{G}\left( T_{n}\right) \right\} . \end{equation}$

(4.15)

On the other hand, if $x\in X$ such that $\left( x,y\right) \in {s}$-$\lim\limits_{n\rightarrow \infty }\mathcal{G}\left( T_{n}\right)$, then there exists a sequence $\left\{ x_{n}\right\} $ of $X$ with $x_{n}\in \mathcal{D} \left( T_{n}\right) $ ($n\in \mathbb{N}$) such that $\underset{n\rightarrow \infty }{\lim }x_{n}=x, \underset{n\rightarrow \infty }{\lim }T_{n}x_{n}=y,$ and therefore

$\begin{equation} \underset{n\rightarrow \infty }{\lim }\left( y_{n}-T_{n}x_{n}\right) =0. \end{equation}$

(4.16)

Since $\left\{ T_{n}\right\} $ satisfies B), i.e., for every $\varepsilon >0$, there exists a $\delta >0$ such that

$\begin{equation*} B_{Y}\left( 0,\delta \right) \cap \mathcal{R}\left( T_{n}\right) \subseteq T_{n}\left( B_{X}\left( 0,\varepsilon \right) \cap \mathcal{D}\left( T_{n}\right) \right) \quad \text{for all }n\in \mathbb{N}\mathbf{,} \end{equation*}$

it follows from (4.16) that for every $\varepsilon >0$, there exists a natural number $N$ such that

$\begin{equation*} y_{n}-T_{n}x_{n}\in T_{n}\left( B_{X}\left( 0,\varepsilon \right) \cap \mathcal{D}\left( T_{n}\right) \right) \quad \text{for all }n>N. \end{equation*}$

This implies that $\underset{n\rightarrow \infty }{\lim }\underset{v\in T_{n}^{-1}\left( y_{n}-T_{n}x_{n}\right) }{\inf }\left\Vert v\right\Vert =0.$ So we obtain that

$\begin{eqnarray*} &{\rm dist}\left( x,T_{n}^{-1}\left( y_{n}\right) \right) =\underset{v\in T_{n}^{-1}\left( y_{n}-T_{n}x_{n}\right) }{\inf }\left\Vert v+x_{n}-x\right\Vert \\ \leq& \left\Vert x_{n}-x\right\Vert +\underset{v\in T_{n}^{-1}\left( y_{n}-T_{n}x_{n}\right) }{\inf }\left\Vert v\right\Vert \rightarrow 0,\quad \text{as }n\rightarrow \infty. \end{eqnarray*}$

Thus $x\in {s}$-$\lim\limits_{n\rightarrow \infty }T_{n}^{-1}(y_{n})$. Hence we have that

$\begin{equation} \left\{ x\in X\mid (x,y)\in \,\underset{n\rightarrow \infty }{{s} \text{-}\lim }\mathcal{G}\left( T_{n}\right) \right\} \subseteq \,\underset{ n\rightarrow \infty }{{s}\text{-}\lim }T_{n}^{-1}(y_{n}). \end{equation}$

(4.17)

(4.14) follows from (4.15) and (4.17).

Step 3 C) $\Rightarrow $ A) Assume C) and let $\left\{ x_{n}\right\} $ be a sequence of $X$ with $x_{n}\in \mathcal{D} \left( T_{n}\right) $ ($n\in \mathbb{N}$) and $\lim\limits_{n\rightarrow \infty }T_{n}x_{n}=0$. Then

$\begin{equation*} \underset{n\rightarrow \infty }{{s}\text{-}\lim } T_{n}^{-1}(T_{n}x_{n})=\left\{ x\in X\mid (x,0_{Y})\in \,\underset{ n\rightarrow \infty }{{s}\text{-}\lim }\mathcal{G}\left( T_{n}\right) \right\} \ni 0_{X}\,. \end{equation*}$

That implies that$ \underset{n\rightarrow \infty }{\lim }\underset{x\in T_{n}^{-1}\left( T_{n}x_{n}\right) }{\inf }\left\Vert x\right\Vert =0.$ Hence, there exists a sequence $\left\{ x_{n}^{\prime }\right\} $ of $X$ with $x_{n}^{\prime }\in T_{n}^{-1}\left( T_{n}x_{n}\right) $ ($n\in \mathbb{ N}$) such that $\lim\limits_{n\rightarrow \infty }x_{n}^{\prime }=0\,;$ therefore, $ \left\{ x_{n}^{\prime }\right\} $ is a bounded subset of $X$ and this impies that $\underset{k\rightarrow \infty }{\underline{\lim }}\,\underset{n\in \mathbb{N} }{\sup }\left\Vert k^{-1}x_{n}^{\prime }\right\Vert =0.$ Noting that

$\begin{equation*} \begin{array}{c} {\rm dist}\left( k^{-1}x_{n},\mathcal{N}\left( T_{n}\right) \right) = {\rm dist}\left( k^{-1}x_{n}^{\prime },\mathcal{N}\left( T_{n}\right) \right) \leq \left\Vert k^{-1}x_{n}^{\prime }\right\Vert \quad \text{for all }n\in \mathbb{N}\text{,} \end{array} \end{equation*}$

we obtain that $\underset{k\rightarrow \infty }{\underline{\lim }}\,\underset{n\in \mathbb{N} }{\sup } {\rm dist}\left( k^{-1}x_{n},\mathcal{N}\left( T_{n}\right) \right) =0,$ that is, A) holds.

Remark 4.1 In Theorem 1.2, if $X$ and $Y$ are Banach spaces, then the following three properties of $\left\{ T_{n}\right\} $ are equivalent:

A) For any $x_{n}\in \mathcal{D}\left( T_{n}\right) $ ($n\in \mathbb{N}$) with $\lim\limits_{n\rightarrow \infty }\left\Vert T_{n}x_{n}\right\Vert =0$, there holds

$\begin{equation*} \underset{n\in \mathbb{N}}{\sup }\,{\rm dist}\ \left( x_{n},\mathcal{N}\left( T_{n}\right) \right) <+\infty \text{.} \end{equation*}$

B) There exists a $\delta >0$ such that

$\begin{equation*} B_{Y}\left( 0,\delta \right) \cap \mathcal{R}\left( T_{n}\right) \subseteq T_{n}\left( B_{X}\left( 0,1\right) \cap \mathcal{D}\left( T_{n}\right) \right) \quad \text{for all }n\in \mathbb{N}\mathbf{.} \end{equation*}$

C) If $y_{n}\in \mathcal{R}\left( T_{n}\right) $ ($n\in \mathbb{N}$) with $\lim\limits_{n\rightarrow \infty }y_{n}=y,$

$\begin{equation*} \underset{n\rightarrow \infty }{{s}\text{-}\lim }T_{n}^{-1}(y_{n})= \left\{ x\in X\mid (x,y)\in \,\underset{n\rightarrow \infty }{{s} \text{-}\lim }\mathcal{G}\left( T_{n}\right) \right\} . \end{equation*}$

Remark 4.2  The following case is of practical interest for the applications of Theorem 1.2: There exists a closed linear operator $T:\mathcal{D}\left( T\right) \subset X\rightarrow Y$ such that $\left\{ T_{n}\right\} $ is its consistent approximation scheme, that is, $\mathcal{G}\left( T\right) =\,\underset{n\rightarrow \infty }{{s} \text{-}\lim }\mathcal{G}\left( T_{n}\right).$ Here we provide two examples with the above case.

Example 4.1  Let $T$, $T_{n}$ be all self-adjoint operators in a Hilbert space $H$, and let

$\begin{equation*} \underset{n\rightarrow \infty }{{s}\text{-}\lim }R_{{\small \lambda } }\left( T_{n}\right) =R_{{\small \lambda }}\left( T\right) \ \left( \forall {\small \lambda }\in \mathbb{C}\mathbf{\setminus }\mathbb{R}\right) , \end{equation*}$

where $R_{\lambda }\left( T\right) $ and $R_{\lambda }\left( T_{n}\right) $ denote the resolvent operator of $T$ and $T_{n}$, respectively. Then $ \left\{ T_{n}\right\} $ is a consistent approximation scheme of $T$. See [6, pp.152{153] or [4, pp. 148{149] for the proof of the statement. Now, by Theorem 1.2 we conclude that: if $\mathcal{R} \left( T_{n}\right) $ ($n\in \mathbb{N}$) are closed, and if

$\begin{equation*} \underset{n\in \mathbb{N}}{\sup }\left\Vert T_{n}^{\dagger }\right\Vert <\infty \text{ (where }T_{n}^{\dagger }\text{ is the Moore-Penrose inverse of }T_{n}\text{),} \end{equation*}$

then for any $y_{n}\in \mathcal{R}\left( T_{n}\right) $ ($n\in \mathbb{N}$) with $\lim\limits_{n\rightarrow \infty }y_{n}=y$, there holds $\underset{n\rightarrow \infty }{{s}\text{-}\lim } T_{n}^{-1}(y_{n})=T^{-1}\left( y\right);$ especially, there holds $\underset{n\rightarrow \infty }{{s}\text{-}\lim }\mathcal{N}\left( T_{n}\right) =\mathcal{N}\left( T\right).$

Example 4.2  Let $A,A_{n}\in \mathcal{B}\left( X,Y\right) $, where $X$ and $Y$ be Banach spaces, and let $\underset{n\rightarrow \infty }{{s}\text{-}\lim }A_{n}=A.$ Then $\left\{ A_{n}\right\} $ is a consistent approximation scheme of $A$. By Theorem 1.2, we conclude that: If $\mathcal{R}\left( A_{n}\right) $ ($n\in \mathbb{N}$) are closed, and if for any $x_{n}\in \mathcal{D}\left( A_{n}\right) $ ($n\in \mathbb{N}$) with $\lim\limits_{n\rightarrow \infty }\left\Vert A_{n}x_{n}\right\Vert =0$, there holds $\underset{n\in \mathbb{N}}{\sup }\,{\rm dist}\ \left( x_{n},\mathcal{N}\left( A_{n}\right) \right) <+\infty,$ then the conclusion of Example 4.1 holds.