Unbounded Versions of Two Old Summability Theorems

Abstract

In this note, we obtain extensions of a theorem of Meyer-König and Zeller and a theorem of Wilansky in that the given results do not require a summability matrix to be a bounded operator from the convergent sequences into themselves. The culmination of the results in this note is that a triangle matrix method T with null columns maps a bounded divergent sequence to a null sequence if and only if the range of T is not closed in the null sequences.

1. Introduction

This paper has roots in two, somewhat distinct, areas of summability theory: generalizations of statistical convergence and the applications of functional analysis to summability theory. For the most part, the first applications of functional analysis to summability theory pertained to the action of regular (or conservative) summability methods which can be studied as bounded linear operators from the convergent sequences, a Banach space when given the supremum norm, into themselves. For instance, Banach’s seminal monograph Theory of Linear Operators [1] includes a proof of the Silverman–Toeplitz Theorem using the Principle of Uniform Boundedness, the Bounded Consistency Theorem for regular summability methods, and, via the Hahn–Banach Theorem, the existence of Banach limits. Two texts on the interplay between functional analysis and summability theory are Wilansky’s book Summability through Functional Analysis [2] and Boos’ text Classical and Modern Methods in Summability [3], which provides a comprehensive review of the application of functional analytic techniques to summability theory.

This note partially extends two older results from the realm of bounded matrix summability methods to unbounded matrix summability methods (terms will be defined below). In particular:

Theorem 1

(Meyer-König and Zeller; Wilansky, 6.1.1 [2]). Let T be a conservative summability matrix (and hence a bounded linear operator from the convergent sequences into themselves). If the range of T is not closed in c, then there is a bounded divergent sequence x such that $Tx$ is convergent.

Theorem 2

(Wilansky, 17.6.11 [2]). Suppose that $T:c\to c$ is a bounded triangle and that the range of T is closed. If x is a bounded sequence such that $Tx$ is convergent, then x is convergent.

In essence, the closure of the range of T determines whether or not T is stronger than ordinary convergence on the bounded sequences. The Meyer-König and Zeller result guarantees T is stronger and Wilansky’s result says nearly the opposite for bounded sequences. Both theorems also assume that the matrix maps can be viewed as bounded operators.

The motivation for considering the behavior of unbounded matrix maps comes from some recent developments in the theory of statistical convergence (or convergence with respect to an ideal). Statistical convergence has been an active research area since the 1980s, when Šalát [4] and Fridy [5] published their often cited papers and has only made little use of functional analysis.

The original definition of a statisically convergent sequence had an intimate connection to convergence in arithmetic mean or, alternatively described, the Cesàro summability transformation. A real-valued sequence $x=\left(x_k\right)$ is said to be statistically convergent to L provided

$$\begin{array}{c}\hfill \underset{n\to +\infty }{lim}\frac{1}{n}\left|\right\{k\le n:|x_k-L|\ge \epsilon |=0\end{array}$$

(1)

for all $\epsilon >0$, where $\left|A\right|$ denotes the cardinality of the set A. This concept has been extended several times. Recently, Çolak [6] introduced the notion of statistical convergence of order $\alpha$: a real-valued sequence is convergent of order $\alpha$ to L, $0<\alpha \le 1$, provided

$$\begin{array}{c}\hfill \underset{n\to +\infty }{lim}\frac{1}{n^{\alpha }}\left|\right\{k\le n:|x_k-L|\ge \epsilon |=0\end{array}$$

(2)

for all $\epsilon >0$. Again, this can be associated with a summability matrix, but when $0<\alpha <1$, the method is unbounded.

2. Notation, Definitions and Examples

2.1. Notation and Definitions

Throughout this note, we will be working with real-valued sequences, and we will let $c_0,c$, and ${\mathsf{\ell }}_{\infty }$ denote the sequences convergent to 0, the convergent sequences, and the bounded sequences; we will also let $\parallel x\parallel ={sup}_n\left|x_n\right|$ denote the supremum norm and recall that $c_0,c$, and ${\mathsf{\ell }}_{\infty }$ are all Banach spaces with respect to the supremum norm.

A summability matrix T is an infinite array of real numbers $T={\left[t_{\begin{array}{c}\hfill n,k\end{array}}\right]}_{n,k=1}^{\begin{array}{c}\hfill +\infty \end{array}}$ and, for a sequence $x=\left(x_k\right)$, we let $Tx$ be the sequence $(T{x}_n$) where $T{x}_n={\sum }_{k=1}^{\begin{array}{c}\hfill +\infty \end{array}}{t}_{n,k}{x}_{\begin{array}{c}\hfill k\end{array}}$. The domain of T is the set of all sequences for which $T{x}_n$ is defined for all $n\in \mathbb{N}$. A summability matrix is conservative if convergent sequences are mapped to convergent sequences, and it is called regular if, for any convergent real-valued sequence x, one has that ${lim}_n{x}_n={lim}_n{\sum }_{k=1}^{\begin{array}{c}\hfill +\infty \end{array}}{t}_{\begin{array}{c}\hfill n,k\end{array}}{x}_k$.

If T is a conservative matrix, the Principle of Uniform Boundedness or a classical “sliding hump” argument shows that ${sup}_n{\sum }_{k=1}^{\begin{array}{c}\hfill +\infty \end{array}}\left|t_{\begin{array}{c}\hfill n,k\end{array}}\right|<\begin{array}{c}\hfill +\infty \end{array}$. It is worth appreciating that if a conservative matrix T is viewed as a linear operator from c to c, it is a bounded linear operator and $\parallel T\parallel ={sup}_n{\sum }_{k=1}^{\begin{array}{c}\hfill +\infty \end{array}}\left|t_{\begin{array}{c}\hfill n,k\end{array}}\right|$ is the norm of T (1.3.6, [2]).

A summability matrix $T=\left[t_{n,k}\right]$ is said to be triangular if $k>n$ implies $t_{n,k}=0$ and triangle if it is triangular and $t_{n,n}\ne 0$ for all $n\in \mathbb{N}$. Perhaps the best-known example of a matrix summability method is the Cesàro matrix $M_1=\left[m_{n,k}\right]$ where $m_{n,k}=1/n$ for $k\le n$ and $m_{n,k}=0$ otherwise. A sequence x is statistically convergent to L when for any $\epsilon >0$, the characteristic function of the set $V\left(\epsilon \right)=\{k:|x_k-L|\ge \epsilon \}$ is mapped to a null sequence by the Cesàro matrix; observe that $M_1$ is a triangle and that $\parallel M_1\parallel =1$. Çolak’s statistical convergence of order $\alpha$ can be described by a triangle matrix summability matrix $M_{\alpha }$ in a similar manner. In this case, the matrix is given by $M_{\alpha }=m_{n,k}$ where $m_{n,k}=1/n^{\alpha }$ for $k\le n$ and $m_{n,k}=0$ otherwise. Observe that $0<\alpha <1$ implies $\parallel M_{\alpha }\parallel =\begin{array}{c}\hfill +\infty \end{array}$.

We now define some notation needed for our discussion of unbounded summabililty matrices and introduce some examples. The notation follows Goldberg’s text [7], which includes many examples of unbounded operators, e.g., differentiation as an operator from the continous functions on $[0,1]$ into themselves. The interested reader may enjoy following up with the work of R.W. Cross ([8,9]), who has conducted an extensive study of the second adjoints of unbounded closed operators, especially related to Tauberian operators as defined, for bounded operators, by Kalton and Wilansky in [10].

Definition 1.

Let X and Y be normed linear spaces and let $T:X\to Y$ be linear mapping, not necessarily defined on all of X.

• The domain of T is $D\left(T\right)=\left\{x\in X:Tx\in Y\right\}\begin{array}{c}\hfill .\end{array}$
• The range of T is $R\left(T\right)=\left\{Tx:x\in D\left(T\right)\right\}\begin{array}{c}\hfill .\end{array}$
• The mapping T is a closed linear operator if for $\left(x^l\right)\subset D\left(T\right)$, $x^l\to x$ and $T{x}^l\to y$ implies that $Tx=y$ and $x\in D\left(T\right),$ i.e., the graph of T is closed in $X\times Y$ with respect to the norm $\parallel (x,y)\parallel =max(\parallel x\parallel ,\parallel y\parallel )$.
• An operator T is said to have a dense domain if the $D\left(T\right)$ is dense in X.

In the special case that X and Y are complete normed linear spaces, i.e., Banach spaces, T is a closed linear operator, and $D\left(T\right)=X$, the Closed Graph Theorem yields the that T is bounded, which is one of the key properties of conservative matrices. In this note, we will remove the constraint that $D\left(T\right)=X$ and only assume that T is a closed operator and, hence, possibly, an unbounded summability matrix.

2.2. Some Elementary Examples

The matrices we will be considering will be closed operators with a dense domain in $c_0$. Let us consider a couple of examples. This first example is a very simple unbounded matrix. Define $L:c\to c$ by ${\left(Lx\right)}_n=n{x}_n$ for all $n\in \mathbb{N}$.

$$Lx=\left[\begin{array}{ccccc}1& 0& 0& 0& \dots \\ 0& 2& 0& 0& \dots \\ 0& 0& 3& \dots & \dots \\ ⋮& ⋮& ⋮& ⋮\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ ⋮\end{array}\right]=\left[\begin{array}{c}{x}_1\\ 2x_2\\ 3x_3\\ ⋮\end{array}\right]$$

The domain of L is $D\left(L\right)=\left\{x\in c:{lim}_{n}n{x}_n\mathrm{exists}\right\}\subset c_0$, $∥L∥={sup}_n\left|n\right|=\begin{array}{c}\hfill +\infty \end{array}$, and $R\left(L\right)=c$. Observe that $R\left(L\right)$ is closed and the inverse of L is given by ${\left(L^{-1}x\right)}_n=x_n/n,$ which is a bounded operator of L and that for any bounded divergent sequence z, the sequence $Lz$ is divergent. This is analogous to Wilansky’s Theorem in that if $z\in {\mathsf{\ell }}_{\infty }$ and $Lz\in c$, then $z\in c$. It will be shown, by a general argument below, that $D\left(L\right)$ is $\parallel \parallel$-dense in $c_0$.

The next example, $M=M_{1/2}$ as defined above, suggests there may be an analog to the Meyer–König, Zeller Theorem. Define $M:c\to c$ by ${\left(Mx\right)}_n=\frac{1}{\sqrt{n}}{\sum }_{k=1}^n{x}_k$

$$Mx=\left[\begin{array}{ccccc}1& 0& 0& 0& \dots \\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0& 0& \dots \\ \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}& 0& \dots \\ ⋮& ⋮& ⋮& ⋮\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ ⋮\end{array}\right]=\left[\begin{array}{c}{x}_1\\ \frac{1}{\sqrt{2}}\sum _{k=1}^2{x}_k\\ \frac{1}{\sqrt{3}}\sum _{k=1}^3{x}_k\\ ⋮\end{array}\right]$$

In this example, $∥M∥={sup}_n\left|\sqrt{n}\right|=+\infty$ and $D\left(M\right)\subset c_0$. In order to see that the domain is contained in $c_0$, suppose that the sequence x converges to L and, without loss of generality, that $L>0$. Let $x_k=L+{\xi }_k$ for a null sequence $\left({\xi }_k\right)$. Pick $K\in \mathbb{N}$ such that $k>K$ implies $x_k>\frac{3}{4}L$. Now, for $n>K$,

$$\begin{array}{c}\hfill {\left(Tx\right)}_n=\frac{1}{\sqrt{n}}\sum _{k=1}^{n-K}(L-{\xi }_k)+\frac{1}{\sqrt{n}}\sum _{k=K+1}^n(L-{\xi }_k).\end{array}$$

(3)

Now, note that for sufficiently large values of n, we have that $T{x}_n>\frac{3(n-K)L}{4\sqrt{n}}$ and hence $Tx$ is unbounded.

The inverse M is given by ${\left(M^{-1}x\right)}_n=\sqrt{n}{x}_n-\sqrt{n-1}{x}_{n-1},$ which is an unbounded operator and hence cannot have a closed range (IV.1.1, [7]). Also note that if z has bounded partial sums, $Mz$ is a null sequence; hence, it maps a divergent sequence to a null sequence.

The next lemma ensures that the results of the next two sections can be applied to a variety of summability methods.

Lemma 1.

Let T be a triangle, which may or may not be bounded, and regard T as a mapping from $c_0$ into c. Then, T is a closed linear operator with a dense domain in $c_0$.

Proof. 

Let $D\left(T\right)=\{x\in c_0:Tx\in c\}$ and let $R\left(T\right)=\{Tx:x\in D(T\left)\right\}$. We need to show that if $x^n\to x\mathrm{in}{c}_0$ and $T{x}^n\to y\mathrm{in}c$, then $Tx=y$.

Suppose $x^n\to x$ in $c_0$ and $T{x}^n\to y$ in c. Now, for every $k\in \mathbb{N}$, we have that $x^l\to x\mathrm{in}{c}_0\mathrm{implies}\mathrm{that}{x}_k^l\to x_k$ and hence, for each $n\in \mathbb{N}$,

$$\begin{array}{c}\hfill {\left(T{x}^l\right)}_n=\sum _{k=1}^n{t}_{n,k}{x}_k^l\to \sum _{k=1}^n{t}_{n,k}{x}_k={\left(Tx\right)}_n.\end{array}$$

(4)

But ${\left(T{x}^l\right)}_n\to y_n$ for all n, hence ${\left(Tx\right)}_n=y_n$ for all n, thus $Tx=y$. Hence, T is a closed operator.

Now, since T has null columns, it follows that $D\left(T\right)$ contains the finitely nonzero sequences, which are a norm-dense subset of $c_0$, and hence, T has a dense domain.  □

3. A “Bigness” Theorem for Unbounded Matrices

This section extends the Meyer–König and Zeller result to include unbounded triangle matrix summability methods. The proofs of the following two results are fairly standard arguments. For instance, one may wish to compare the proof of Theorem 3 with the proof of the earlier quoted Meyer–König, Zeller Theorem (e.g., see the proof of 6.1.1, [2]). It will be convenient to introduce some notation for the next lemma. If $l\in \mathbb{N}$ and $x=\left(x_k\right)$ is a sequence, we let

$$\begin{array}{c}\hfill P(x_1,x_2,\dots ,x_{l-1},x_l,x_{l+1},\dots )=(x_1,x_2,\dots ,x_{l-1},x_l,0,0,\dots )\end{array}$$

(5)

and $Q_l\left(x\right)=\left(I-P_l\right)\left(x\right);$ we also let $E_l=P_l\left(c_0\right)$ and $F_l=Q_l\left(c_0\right)$.

Lemma 2.

Suppose that T is a closed linear operator from $c_0$ into c with a dense domain, and $l\in \mathbb{N}$. If $R\left(T\right)$ is not closed, then T does not have a bounded inverse on $T{F}_l$ for all l. Hence, for all $\epsilon >0$, there is an $x^l\in F_l$, $∥x^l∥=1$ and $∥T{x}^l∥<\epsilon$.

Proof. 

Suppose there exists an l such that T has a bounded inverse on $F=F_l$. First, we establish that $T\left(F\right)$ is closed. Suppose that $y\in \overline{T\left(F\right)}$. Then, there exists $x^p\in F$ (as F is a closed subspace of $c_0$) such that $T{x}^p\to y$ and hence $\left(T{x}^p\right)$ is Cauchy. Now

$$∥x^p-x^q∥\le ∥T^{-1}∥∥T{x}^p-T{x}^q∥$$

and hence, there is an $x\in F$ such that $x^p\to x.$ Now, T closed implies that $x\in D\left(T\right)$ and $Tx=y.$

Now, $c_0=E_l\oplus F_l$ and $dim\left(E_l\right)<\begin{array}{c}\hfill +\infty \end{array},$ hence $R\left(T\right)=T\left(E_l\right)\oplus T\left(F_l\right).$ As $T\left(E_{\begin{array}{c}\hfill l\end{array}}\right)$ and $T\left(F_{\begin{array}{c}\hfill l\end{array}}\right)$ are both closed, one has that $R\left(T\right)$ is closed. □

Theorem 3.

Let $T:c_0\to c$ be a triangle with null columns. If $R\left(T\right)$ is not closed, then there is a bounded divergent sequence $\begin{array}{c}\hfill x\end{array}$ such that $Tx$ is a convergent sequence.

Proof. 

First, we construct a candidate for $x.$ Let $\delta =\left({\delta }_j\right)$ be an absolutely convergent sequence of positive numbers. Select $x^1\in c_0$ such that $∥x^1∥=1$ and $∥T{x}^1∥<{\delta }_1$. Now, as $x^1\in c_0$, there is a natural number $K\left(1\right)$ such that $k>K\left(1\right)$ implies that $\left|x_k^1\right|<1/8.$ Note that there is an integer $p_1\le K\left(1\right)$ such that $\left|x_{p_1}^1\right|=1$. Now, select $K\left(2\right)>K\left(1\right)$ and observe that $|x_k^1|<1/8$ for $K\left(1\right)<k\le K\left(2\right)$.

Now, consider T restricted to $F_{K\left(2\right)}$. By the lemma, there is an $x^2\in F_{K\left(2\right)}$ such that $∥x^2∥=1$ and $∥T{x}^2∥<{\delta }_2$. As before, as $x^2\in c_0$, there is a natural number $K\left(3\right)$ such that $k>K\left(3\right)$ implies that $\left|x_k^2\right|<1/16$. Note that there is an integer $K\left(2\right)<$$p_2\le K\left(3\right)$ such that $\left|x_{p_2}^2\right|=1$. Select $K\left(4\right)>K\left(3\right)$.

Proceed inductively: Suppose that we have selected $x^{n-1}\in F_{K\left(2(n-1)\right)}$ such that $∥x^{n-1}∥=1$ and $∥T{x}^{n-1}∥<{\delta }_{n-1}$; the natural number $K\left(2n-1\right)>K\left(2n-2\right)$ such that $k>K\left(2n-1\right)$ implies that $\left|x_k^{n-1}\right|<\begin{array}{c}\hfill 1/2^{n+1}\end{array}$ and $\begin{array}{c}{p}_{n-1}\end{array}$, $K\left(2n-2\right)<\begin{array}{c}\hfill p_{n-1}\end{array}\le K\left(2n-1\right)$, such that $\left|x_{p_{n-1}}^{n-1}\right|=1$. Now, by the lemma, there exists $x^n\in F_{K\left(2n\right)}$ such that $∥x^n∥=1$ and $∥T{x}^n∥<{\delta }_n$. As $x^n\in c_0$, there is a natural number $K\left(2n+1\right)>K\left(2n\right)$ such that $k>K\left(2n+1\right)$ implies that $\left|x_k^n\right|<1/2^{n+2}$. As $x^n\in F_{K\left(2n\right)}$ and $∥x^n∥=1$, there is an integer $p_n$, $K\left(2n\right)<p_n\le K\left(2n+1\right)$ such that $\left|x_{p_n}^n\right|=1.$

Define the sequence x by $x_k={\sum }_{n=1}^{\begin{array}{c}\hfill +\infty \end{array}}{x}_k^n$ for each $k\in \mathbb{N}.$ To see that x is defined, let $l\in \mathbb{N}$ and suppose $K\left(2r-2\right)\le l<K\left(2r\right).$ As $x_l^n=0$ for all $n\ge 2\begin{array}{c}\hfill r\end{array},$

$$x_l=\sum _{n=1}^r{x}_l^n=x_l^{r-1}+\sum _{n=1}^{r-2}{x}_l^n$$

which is a finite sum, hence defined, and

$$\left|x_l\right|=\left|x_l^{r-1}\right|+\sum _{n=1}^{r-2}\left|x_l^n\right|<1+\sum _{n=1}^{r-2}\frac{1}{2^{n+2}}<\frac{5}{4}$$

which yields that x is bounded. Also, as

$$\left|x_{p_n}\right|=\left|x_{p_n}^r\right|+\sum _{n=1}^{r-1}\left|x_{p_n}^n\right|>1-\sum _{n=1}^{r-1}\frac{1}{2^{n+2}}>\frac{3}{4}$$

one has that x is not a null sequence. Morever, since $K(2r-1)<l<K\left(2r\right)$ implies $|x_l|<1/4$ for all $r\in \mathbb{N}$, we have that x is a bounded divergent sequence.

Next, we show that $Tx$ is convergent. Let $y^l={\sum }_{n=1}^{l}T{x}^n$. Observe that as $x^l\in D\left(T\right),$ one has that $y^l\in c$. Also

$$\begin{array}{ccc}\hfill ∥y^l-y^m∥& =& ∥\sum _{n=1}^{l}T{x}^n-\sum _{n=1}^{m}T{x}^n∥=∥\sum _{n=m}^{l}T{x}^n∥\hfill \\ & \le & \sum _{n=m}^l∥T{x}^n∥<\sum _{n=m}^l{\delta }_n\hfill \end{array}$$

and hence $\left(y^l\right)$ is Cauchy in c. Hence, there is a $y\in c$ such that $y^l\to y$ in the supremum norm.

Now, since T is a triangle and

$$\begin{array}{c}\hfill x_k=\sum _{n=1}^{\begin{array}{c}\hfill +\infty \end{array}}{x}_k^n,{\left(Tx\right)}_k=T\left(\underset{l}{lim}\sum _{n=1}^l{x}_k^n\right),\mathrm{and}\underset{l}{lim}{\left(\sum _{n=1}^{l}T{x}^n\right)}_k=y_k.\end{array}$$

(6)

Since T is a closed operator, we have that $Tx=y$. □

4. A Tauberian Theorem

The next result is of the form of a Tauberian theorem. Briefly, a Tauberian theorem is one for which $Tx\in c$ and x satisfies some additional condition G is a sufficient condition to guarantee $x\in c$. The condition G is called a Tauberian condition for T, after Tauber, who proved that if a sufficiently slowly oscillating sequence is convergent in arithmetic mean, then it must be convergent [11]. Wilansky’s Theorem follows the same pattern, the Tauberian condition being that T has a closed range. As before, the main result of this section seeks to replace the condition that T is bounded with the condition T as a closed range.

As the proof of the next theorem will take us out of the realm of sequence spaces, its proof will be given in the next section. The proof of Theorem 5 is just a matter of applying Theorem 4 to some familiar sequence spaces.

Theorem 4.

Let $T:X\to Y$ be an injective closed operator with a dense domain and a closed range. If $T^{**}{x}^{**}=j_y\in j\left(Y\right)$, then there is a $z\in D\left(T\right)$ such that $Tz=y$.

We can now remove the boundedness restriction from Wilansky’s Theorem. In the following result, T is injective since it is a triangle, and since T has null columns, the domain of T contains the finitely nonzero sequences and hence the domain of T is dense in $c_0$. The continuous bidual of c is the set of bounded sequences and so $T^{**}$ maps ${\mathsf{\ell }}_{\infty }$ to ${\mathsf{\ell }}_{\infty }$. Abusing the notation in the usual way (substituting y for $j_y$), it follows that if T maps a bounded sequence x to a convergent sequence y, then $Tz=y$ for a convergent sequence z. Hence:

Theorem 5.

Suppose T is a triangle with null columns and that the range of T is closed. If x is a bounded sequence and $Tx\in c_0$, then $x\in c_0.$

The Proof of Theorem 4

First, we establish some further notation.

Definition 2.

Let X and Y be a normed linear space and let T be a linear operator from X into Y. We assume the domain of T is dense in X.

• We let $X^{*}$ denote the continuous dual of X and $X^{**}$ denote the continuous bidual of X. The canonical mapping $j:X\to X^{**}$ is defined by $x\to j_x$ where $j_x\left(x^{*}\right)=x^{*}\left(x\right)$.
• We let $\sigma (E,F)$ denote the weak topology induced on E by F.
• As before, $D\left(T\right)=\{x\in X:Tx\in Y\}$ and $R\left(T\right)=\{Tx:x\in D(T\left)\right\}$.
• The adjoint of T is a mapping $T^{*}$ from $Y^{*}$ to $X^{*}$ that has the domain

  $$\begin{array}{c}\hfill D\left(T^{*}\right)=\{\begin{array}{c}\hfill y^{*}\end{array}\in Y^{*}:T^{*}{y}^{*}\mathit{is}\mathit{continuous}\mathit{on}D\left(T\right)\}\end{array}$$

  (7)

  and range $R\left(T^{*}\right)=\{T^{*}{y}^{*}:y^{*}\in D\left(T^{*}\right)\}$ where $T^{*}{y}^{*}\left(x\right)$ is defined to be the continuous extension of the map defined by $T^{*}{y}^{*}\left(x\right)=y^{*}\left(Tx\right)$ for $x\in D\left(T\right)$ and $y^{*}\in Y^{*}$ to the closure of $R\left(T\right)$.
• The adjoint of $T^{*}$ is denoted $T^{**}$, where

  $$\begin{array}{c}\hfill D\left(T^{**}\right)=\{x^{**}\in X^{**}:T^{**}{x}^{**}\mathit{is}\mathit{continuous}\mathrm{on}D\left(T^{*}\right)\}\end{array}$$

  (8)

  where $T^{**}{x}^{**}$ is the continuous extension of the mapping defined by $T^{**}{x}^{**}\left(y^{*}\right)=x^{**}\left(T^{*}{y}^{*}\right)$ to the closure of $R\left(T^{*}\right)$.

Observe that $T^{**}{j}_x\left(y^{*}\right)=j_x\left(T^{*}{y}^{*}\right)=T^{*}{y}^{*}\left(x\right)=y^{*}\left(Tx\right)$. These concepts are developed more completely in [7,8,9].

Lemma 3.

Let T be a linear operator from X into Y. Then, y is in the norm closure of $R\left(T\right)$ if and only if $j_y$ is in the $\sigma (\overline{jR\left(T\right)},D\left(T^{*}\right))$ closure of $\{j_{Tx}:x\in D\left(T\right)\}.$

Proof. 

First, we establish the necessity. Let $y\in \overline{R\left(T\right)}$ and let $\left(x_n\right)$ be a sequence in $D\left(T\right)$ such that $T{x}_n\to y$. Now, if $y^{*}\in D\left(T^{*}\right)$, we have that $y^{*}\left(T{x}_n\right)\to y^{*}\left(y\right)$ and hence $\left(T^{**}{j}_{x_n}\right)\left(y^{*}\right)\to j_y\left(y^{*}\right)$ for all $y^{*}\in D\left(T^{*}\right)$. Thus, $j_y$ is in the $\sigma (j\overline{R\left(T\right)},D\left(T^{*}\right))$ closure of $\{j_{Tx}:x\in D\left(T\right)\}.$

Next, we establish the converse. Suppose that $j_y\in$${\overline{\{j_{Tx}:x\in D\left(T\right)\}}}^{\sigma (j\overline{R\left(T\right)},D\left(T^{*}\right))}$ and hence, there is a net $\left(x_{\alpha }\right)\in D\left(T\right)$ such that $j_{T_{x_{\alpha }}}\left(y^{*}\right)\to j_y\left(y^{*}\right)$ for all $y^{*}\in D\left(T^{*}\right)$ and thus, y is contained in the $\sigma (R\left(T\right),D\left(T^{*}\right))$ closure of $R\left(T\right)$.

Now, for the sake of a contradiction, suppose that $y\notin \overline{R\left(T\right)}$. Then, there is a $z^{*}\in Y^{*}$ such that $z^{*}{|}_{R\left(T\right)}=0$ and $z^{*}\left(y\right)\ne 0$. However, if $T{x}_{\alpha }\to y$ in $R\left(T\right)$, then $z^{*}\left(T{x}_{\alpha }\right)\to z^{*}\left(y\right)=0$ as $T{x}_{\alpha }=0$ for all $\alpha$, which contradicts $z^{*}\left(y\right)\ne 0$.

Hence, $y\in \overline{R\left(T\right)}$. □

Theorem 6.

Suppose that $T:X\to Y$ is range closed and that $T^{**}{x}^{**}\in j\left(Y\right)$. Then, there is a $z\in X$ such that $T^{**}{x}^{**}=T^{**}{j}_z$.

Proof. 

Suppose that $T^{**}{x}^{**}=j_y$ for some $y\in Y$. Then, $y\in R\left(T\right)=\overline{R\left(T\right)}$ and hence, there is a $z\in D\left(T\right)$ such $Tz=y.$ Now, since, for all $y^{*}\in D\left(T^{*}\right)$,

$$T^{**}{j}_z\left(y^{*}\right)=j_z\left(T^{*}{y}^{*}\right)=T^{*}{y}^{*}\left(z\right)=y^{*}\left(Tz\right)=y^{*}\left(y\right)=j_y\left(y^{*}\right)=T^{**}{x}^{**}\left(y^{*}\right),$$

we have that $T^{**}{j}_z=T^{**}{x}^{**}$. □

Proof of Theorem 4

Since T is injective, we have that the kernel of T, and hence the kernel of $T^{**}$, is the null space. Now, by the preceding lemma, there is a $z\in D\left(T\right)$ such that $T^{**}(j_z-x^{**})=0$, and hence $j_z=x^{**}$. Thus, $T^{**}{j}_z=j_y$. It follows that if $y^{*}\in D\left(T^{*}\right)$ and by previous calculations, we have that $y^{*}\left(Tz\right)=y^{*}\left(y\right)$, hence, as $D\left(T^{*}\right)$ separates the points of $D\left(T\right)$, $Tz=y$. □

5. Conclusions

The above work suggests that some of the functional analytic results related to summability theory can be extended to unbounded matrices. For instance, the results of Theorems 3 and 5 yield the following corollary:

Corollary 1.

Let T be an unbounded triangle with null columns. Then, T maps a bounded divergent sequence to a null sequence if and only $R\left(T\right)$ is not closed.

This closely parallels a 1963 result of Wilanksy and Zeller (Theorem 7, [12]), which states that a conservative (hence bounded) matrix that is one-to-one on c sums no bounded divergent sequence if and only if $A\left[c\right]$ is closed in c.

More work could be conducted on this topic; this article is only a short exploration into unbounded summability methods, and there has been no attempt to be either comprehensive or to obtain the results in their full generality. The author took some delight, and hopes others take some pleasure, in seeing two old theorems being extended to a broader class of summability methods.