We define in this section a multiplication operator on \(C(p)\) with a pre-quasi-norm and give the necessity and sufficient conditions on the multiplication operator to be bounded, approximable, invertible, Fredholm and closed range operator.
Definition 4.1
Let \(\beta \in \mathbb{C}^{\mathbb{N}}\) be a bounded sequence and \(E_{\varrho }\) be a pre-quasi-normed (sss), the multiplication operator is defined as \(T_{\beta }:E\rightarrow E\), where \(T_{\beta }x=\beta x= (\beta _{k}x_{k} )_{k=0}^{\infty }\), for all \(x\in E\). If \(T_{\beta }\) is continuous, we call it a multiplication operator induced by β.
Theorem 4.2
If \(\beta \in \mathbb{C}^{\mathbb{N}}\)and \((p_{n})\)is a bounded sequence, then \(\beta \in \ell _{\infty }\)if and only if \(T_{\beta }\in L((C(p))_{\varrho })\), where \(\varrho (x)=\sum^{\infty }_{i=0} ( \frac{\sum_{j=2^{i}-1}^{2^{i+1}-2}|x_{j}|}{2^{i}} )^{p_{i}}\)for all \(x\in (C(p) )_{\varrho }\).
Proof
Let \(\beta \in \ell _{\infty }\). Then there exists \(C > 0\) such that \(|\beta _{n}|\leq C\), for all \(n\in \mathbb{N}\). For \(x\in (C(p))_{\varrho }\), since \((p_{n})\) is a bounded sequence, we have
$$ \begin{aligned} \varrho (T_{\beta }x)&=\varrho (\beta x)= \varrho \bigl( (\beta _{k}x_{k} )_{k=0}^{\infty } \bigr)=\varrho \bigl( \bigl( \vert \beta _{k} \vert \vert x_{k} \vert \bigr)_{k=0}^{\infty }\bigr) \\ &= \sum^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j} \vert \vert x_{j} \vert }{2^{i}} \biggr)^{p_{i}} \leq \sum ^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2}C \vert x_{j} \vert }{2^{i}} \biggr)^{p_{i}} \leq A\varrho (x), \end{aligned} $$
where \(A=\sup_{n}C^{p_{n}}\), this implies that \(T_{\beta }\in L((C(p))_{\varrho })\).
Conversely, let \(T_{\beta }\in L((C(p))_{\varrho })\). To show that \(\beta \in \ell _{\infty }\). For, if \(\beta \notin \ell _{\infty }\), then, for every \(n\in \mathbb{N}\), there exists some \(i_{n}\in \mathbb{N}\) such that \(\beta _{i_{n}}> n\). Now
$$ \varrho (T_{\beta }e_{i_{n}}) =\varrho (\beta e_{i_{n}})= \sum^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j} \vert (e_{i_{n}})_{j}}{2^{i}} \biggr)^{p_{i}} = \biggl( \frac{ \vert \beta _{i_{n}} \vert }{2^{i_{0}}} \biggr)^{p_{i_{0}}} > \biggl(\frac{n}{2^{i_{0}}} \biggr)^{p_{i_{0}}}=n^{p_{i_{0}}}\varrho (e_{i_{n}}), $$
where \(i_{0}\in \mathbb{N}\) be such that \(2^{i_{0}}-1\leq i_{n}\leq 2^{i_{0}+1}-2\). This shows that \(T_{\beta }\) is not a bounded operator. So, β must be a bounded sequence. □
Theorem 4.3
Let \(\beta \in \mathbb{C}^{\mathbb{N}}\)and \((C(p))_{\varrho }\)be a pre-quasi-normed (sss), where \(\varrho (x)= \sum^{\infty }_{i=0} ( \frac{\sum_{j=2^{i}-1}^{2^{i+1}-2}|x_{j}|}{2^{i}} )^{p_{i}}\), for all \(x\in (C(p))_{\varrho }\). \(|\beta _{n}|=1\), for all \(n\in \mathbb{N}\)if and only if \(T_{\beta }\)is an isometry.
Proof
Let \(|\beta _{n}|=1\), for all \(n\in \mathbb{N}\). Then
$$ \begin{aligned} \varrho (T_{\beta }x)&=\varrho (\beta x)= \varrho \bigl( ( \beta _{k} x_{k} )_{k=0}^{\infty } \bigr)= \sum^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j} \vert \vert x_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ &= \sum ^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert x_{j} \vert }{2^{i}} \biggr)^{p_{i}} =\varrho (x), \end{aligned} $$
for all \(x\in (C(p))_{\varrho }\). Hence \(T_{\beta }\) is an isometry.
Conversely, let \(T_{\beta }\) be an isometry and \(|\beta _{n}|<1\) for all \(n\in \mathbb{N}\). Then
$$ \begin{aligned} \varrho (T_{\beta }x)&=\varrho (\beta x)= \varrho \bigl( ( \beta _{k}x_{k} )_{k=0}^{\infty } \bigr)= \sum^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j} \vert \vert x_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ & < \sum ^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert x_{j} \vert }{2^{i}} \biggr)^{p_{i}} =\varrho (x). \end{aligned} $$
Also if \(|\beta _{n}|>1\), then we get \(\varrho (\beta x)>\varrho (x)\). In both cases, we get a contradiction. Hence, \(|\beta _{n}|=1\), for all \(n\in \mathbb{N}\). □
Theorem 4.4
Let \(\beta \in \mathbb{C}^{\mathbb{N}}\), \((p_{n})\)be a bounded sequence and \(T_{\beta }\in L((C(p))_{\varrho })\), where \(\varrho (x)=\sum^{\infty }_{i=0} ( \frac{\sum_{j=2^{i}-1}^{2^{i+1}-2}|x_{j}|}{2^{i}} )^{p_{i}}\)for all \(x\in (C(p))_{\varrho }\). Then \(T_{\beta }\in \Psi ((C(p))_{\varrho })\)if and only if \((\beta _{n})_{n=0}^{\infty }\in C_{0}\).
Proof
Let \(T_{\beta }\) be an approximable operator. Hence \(T_{\beta }\) is compact operator. We have to show that \((\beta _{n})_{n=0}^{\infty }\in C_{0}\). For if this is not true, then there exists \(\delta > 0\) such that \(B_{\delta }=\{r\in \mathbb{N}: |\beta _{r}|\geq \delta \}\) is an infinite set. Let \(d_{1}, d_{2}, \ldots , d_{n}, \ldots \) be in \(B_{\delta }\). Then \(\{e_{d_{n}}:d_{n}\in B_{\delta }\}\) is an infinite bounded set in \((C(p) )_{\varrho }\). We have
$$ \begin{aligned} \varrho (T_{\beta }e_{d_{n}}-T_{\beta }e_{d_{m}})&= \varrho (\beta e_{d_{n}}-\beta e_{d_{m}})=\varrho \bigl( \bigl( \beta _{k}\bigl((e_{d_{n}})_{k}-(e_{p_{m}})_{k} \bigr) \bigr)_{k=0}^{\infty } \bigr) \\ &= \sum ^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j}((e_{d_{n}})_{j}-(e_{p_{m}})_{j}) \vert }{2^{i}} \biggr)^{p_{i}} \\ & \geq \sum^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \delta ((e_{d_{n}})_{j}-(e_{p_{m}})_{j}) \vert }{2^{i}} \biggr)^{p_{i}} \\ &\geq \Bigl(\inf _{n}\delta ^{p_{n}} \Bigr) \sum ^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert ((e_{d_{n}})_{j}-(e_{p_{m}})_{j}) \vert }{2^{i}} \biggr)^{p_{i}} \\ & = \Bigl(\inf_{n}\delta ^{p_{n}} \Bigr)\varrho (e_{d_{n}}-e_{d_{m}}), \end{aligned} $$
for all \(d_{n}, d_{m}\in B_{\delta }\). This proves \(\{e_{d_{n}}: d_{n}\in B_{\delta }\}\) is a bounded sequence, which cannot have a convergent subsequence under \(T_{\beta }\). This shows that \(T_{\beta }\) cannot be a compact, hence is not approximable operator, which is a contradiction. Hence, \(\lim_{n\rightarrow \infty }\beta _{n}=0\).
Conversely, suppose \(\lim_{n\rightarrow \infty }\beta _{n}=0\). Then, for every \(\delta > 0\), the set \(B_{\delta }=\{n\in \mathbb{N}:|\beta _{n}|\geq \delta \}\) is finite. Then
$$ \bigl( \bigl(C(p) \bigr)_{\varrho } \bigr)_{B_{\delta }}= \bigl\{ x=(x_{n})\in \bigl(C(p) \bigr)_{\varrho } :n\in B_{\delta } \bigr\} $$
is a finite dimensional space for each \(\delta > 0\). Therefore, \(T_{\beta }| ( (C(p) )_{\varrho } )_{B_{\delta }}\) is a finite rank operator. For each \(n\in \mathbb{N}\), define \(\beta _{n}:\mathbb{N}\rightarrow \mathbb{C}\) by
$$ (\beta _{n})_{m} = \textstyle\begin{cases} \beta _{m}, & m\in B_{\frac{1}{n}}, \\ 0, & \text{otherwise}. \end{cases} $$
Clearly, \(T_{\beta _{n}}\) is a finite rank operator as the space \(( (C(p) )_{\varrho } )_{B_{\frac{1}{n}}}\) is finite dimensional for each \(n\in \mathbb{N}\). Now, by using \((p_{n})\) is a bounded sequence, we have
$$\begin{aligned} \varrho \bigl((T_{\beta }-T_{\beta _{n}})x \bigr)&=\varrho \bigl( \bigl(\bigl(\beta _{m}-(\beta _{n})_{m}\bigr)x_{m} \bigr)_{m=0}^{\infty } \bigr) \\ & = \sum^{\infty }_{i=0, i\in B_{\frac{1}{n}}} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert (\beta _{j}-(\beta _{n})_{j})x_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ &\quad {}+ \sum ^{\infty }_{i=0, i\notin B_{ \frac{1}{n}}} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert (\beta _{j}-(\beta _{n})_{j})x_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ & = \sum^{\infty }_{i=0, i\in B_{\frac{1}{n}}} \biggl( \frac{ \sum_{j=2^{i}-1, j\notin B_{\frac{1}{n}}}^{2^{i+1}-2} \vert \beta _{j}x_{j} \vert }{2^{i}} \biggr)^{p_{i}} + \sum ^{\infty }_{i=0, i\notin B_{ \frac{1}{n}}} \biggl( \frac{ \sum_{j=2^{i}-1, j\notin B_{\frac{1}{n}}}^{2^{i+1}-2} \vert \beta _{j}x_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ & < \biggl(\frac{1}{n}\biggr)^{\inf _{n} p_{n}} \sum ^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert x_{j} \vert }{2^{i}} \biggr)^{p_{i}}=\biggl(\frac{1}{n}\biggr)^{\inf _{n} p_{n}} \varrho (x). \end{aligned}$$
This proves that \(\|T_{\beta }-T_{\beta _{n}}\|\leq (\frac{1}{n})^{\inf _{n} p_{n}}\) and that \(T_{\beta }\) is a limit of finite rank operators and hence, \(T_{\beta }\) is an approximable operator. □
Theorem 4.5
Let \(\beta \in \mathbb{C}^{\mathbb{N}}\), \((p_{n})\)be a bounded sequence and \(T_{\beta }\in L((C(p))_{\varrho })\), where \(\varrho (x)=\sum^{\infty }_{i=0} ( \frac{\sum_{j=2^{i}-1}^{2^{i+1}-2}|x_{j}|}{2^{i}} )^{p_{i}}\)for all \(x\in (C(p))_{\varrho }\). Then \(T_{\beta }\in L_{c}((C(p))_{\varrho })\)if and only if \((\beta _{n})_{n=0}^{\infty }\in C_{0}\).
Proof
It is easy so omitted. □
Corollary 4.6
Let \((p_{n})\)be a bounded sequence, we have
$$ L_{c} \bigl( \bigl(C(p) \bigr)_{\varrho } \bigr) \varsubsetneqq L \bigl( \bigl(C(p) \bigr)_{\varrho } \bigr), $$
where \(\varrho (x)=\sum^{\infty }_{i=0} ( \frac{\sum_{j=2^{i}-1}^{2^{i+1}-2}|x_{j}|}{2^{i}} )^{p_{i}}\)for all \(x\in C(p)\).
Proof
Since the identity operator I on \((C(p) )_{\varrho }\) is a multiplication operator induced by the sequence \(\beta =(1, 1,\ldots )\), we have \(I\notin L_{c} ( (C(p) )_{\varrho } )\) and \(I\in L ( (C(p) )_{\varrho } )\). □
Theorem 4.7
Let \((C(p) )_{\varrho }\)be a pre-quasi-Banach (sss), where \(\varrho (x)=\sum^{\infty }_{i=0} ( \frac{\sum_{j=2^{i}-1}^{2^{i+1}-2}|x_{j}|}{2^{i}} )^{p_{i}}\)for all \(x\in C(p)\)and \(T_{\beta }\in L ( (C(p))_{\varrho } )\). Then β is bounded away from zero on \(\mathbb{N}\setminus \ker (\beta ):= \ker (\beta )^{c}\)if and only if \(T_{\beta }\)has closed range.
Proof
Let β be bounded away from zero on \(\ker (\beta )^{c}\). Then there exists \(\epsilon >0\) such that \(|\beta _{n}|\geq \epsilon \), for all \(n\in \ker (\beta )^{c}\). We have to prove that range \((T_{\beta })\) is closed. Let f be a limit point of range \((T_{\beta })\). Then there exists a sequence \(T_{\beta }x_{n}\) in \((C(p) )_{\varrho }\), for all \(n\in \mathbb{N}\) such that \(\lim_{n\rightarrow \infty }T_{\beta }x_{n}=f\). Clearly, the sequence \(T_{\beta }x_{n}\) is a Cauchy sequence. Now, since ϱ is non-decreasing, we have
$$\begin{aligned} &\varrho (T_{\beta }x_{n}-T_{\beta }x_{m}) \\ &\quad = \varrho \bigl( \bigl(\beta _{j}(x_{n})_{j}- \beta _{j}(x_{m})_{j} \bigr)_{j=0}^{\infty } \bigr) = \sum^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j}(x_{n})_{j}-\beta _{j}(x_{m})_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ &\quad = \sum^{\infty }_{i=0, i\in \ker (\beta )^{c}} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j}(x_{n})_{j} -\beta _{j}(x_{m})_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ &\qquad {}+ \sum ^{\infty }_{i=0, i\notin \ker ( \beta )^{c}} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j}(x_{n})_{j}-\beta _{j}(x_{m})_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ &\quad \geq \sum^{\infty }_{i=0, i\in \ker (\beta )^{c}} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j}(x_{n})_{j}-\beta _{j}(x_{m})_{j} \vert }{2^{i}} \biggr)^{p_{i}} = \sum ^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j}(y_{n})_{j}-\beta _{j}(y_{m})_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ &\quad > \sum^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \epsilon (y_{n})_{j}-\epsilon (y_{m})_{j} \vert }{2^{i}} \biggr)^{p_{i}} =\varrho \bigl(\epsilon (y_{n}-y_{m}) \bigr), \end{aligned}$$
where
$$ (y_{n})_{k}= \textstyle\begin{cases} (x_{n})_{k}, & k\in \ker (\beta )^{c}, \\ 0, & k\notin \ker (\beta )^{c}. \end{cases} $$
This proves that \((y_{n})\) is a Cauchy sequence in \((C(p) )_{\varrho }\). But \((C(p) )_{\varrho }\) is complete. Therefore, there exists \(x\in (C(p) )_{\varrho }\) such that \(\lim_{n\rightarrow \infty }y_{n}=x\). In view of the continuity of \(T_{\beta }\), hence \(\lim_{n\rightarrow \infty }T_{\beta }y_{n}=T_{\beta }x\). But \(\lim_{n\rightarrow \infty }T_{\beta }x_{n}=\lim_{n\rightarrow \infty }T_{ \beta }y_{n}=f\). Therefore, \(T_{\beta }x=f\). Hence \(f\in \operatorname{range}(T_{ \beta })\). This proves that \(T_{\beta }\) has closed range.
Conversely, suppose that \(T_{\beta }\) has closed range. Then \(T_{\beta }\) is bounded away from zero on \(( (C(p) )_{\varrho } )_{\ker (\beta )^{c}}\). That is, there exists \(\epsilon >0\) such that \(\varrho (T_{\beta }x)\geq \varrho (\epsilon x)\), for all \(x\in ( (C(p) )_{\varrho } )_{\ker (\beta )^{c}}\). Let \(D= \{k\in \ker (\beta )^{c}:|\beta _{k}|<\epsilon \}\). If \(D\neq \phi \), then, for \(n_{0}\in D\), since ϱ is non-decreasing, we have
$$ \begin{aligned} \varrho (T_{\beta }e_{n_{0}})&=\varrho \bigl( \bigl(\beta _{k}(e_{n_{0}})_{k}\bigr) \bigr)_{k=0}^{\infty } )=\varrho \bigl( \bigl( \vert \beta _{k} \vert \bigl\vert (e_{n_{0}})_{k} \bigr\vert \bigr)_{k=0}^{\infty } \bigr) = \sum ^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2} \vert \beta _{j} \vert \vert (e_{n_{0}})_{j} \vert }{2^{i}} \biggr)^{p_{i}} \\ & < \sum^{\infty }_{i=0} \biggl( \frac{ \sum_{j=2^{i}-1}^{2^{i+1}-2}\epsilon \vert (e_{n_{0}})_{j} \vert }{2^{i}} \biggr)^{p_{i}} =\varrho (\epsilon e_{n_{0}}), \end{aligned} $$
which is a contradiction. Hence, \(D=\phi \) so that \(|\beta _{k}|\geq \epsilon \), for all \(k\in \ker (\beta )^{c}\). This proves the theorem. □
Theorem 4.8
Let \(\beta \in \mathbb{C}^{\mathbb{N}}\)and \((C(p) )_{\varrho }\)be a pre-quasi-Banach (sss), where \(\varrho (x)= \sum^{\infty }_{i=0} ( \frac{\sum_{j=2^{i}-1}^{2^{i+1}-2}|x_{j}|}{2^{i}} )^{p_{i}}\)for all \(x\in C(p)\). There exist \(a>0\)and \(A>0\)such that \(a<\beta _{n}<A\); for all \(n\in \mathbb{N}\)if and only if \(T_{\beta }\in L( (C(p) )_{\varrho })\)is invertible.
Proof
Let the condition be true. Define \(\eta \in \mathbb{C}^{\mathbb{N}}\) by \(\eta _{n}=\frac{1}{\beta _{n}}\). Then by Theorem 4.2, \(T_{\beta }\) and \(T_{\eta }\) are bounded linear operators. Also \(T_{\beta }.T_{\eta }=T_{\eta }.T_{\beta }=I\). Hence, \(T_{\beta }\) is the inverse of \(T_{\eta }\). Conversely, let \(T_{\beta }\) be invertible. Then \(\operatorname{range}(T_{ \beta })= ( (C(p) )_{\varrho } )_{\mathbb{N}}\). Therefore, \(\operatorname{range}(T_{ \beta })\) is closed. Hence, by Theorem 4.7, there exists \(a>0\) such that \(|\beta _{n}|\geq a\), for all \(n\in \ker (\beta )^{c}\). Now \(\ker (\beta )=\phi \); otherwise \(\beta _{n_{0}}=0\), for some \(n_{0}\in \mathbb{N}\), in which case \(e_{n_{0}}\in \ker (T_{\beta })\) which is a contradiction, since \(\ker (T_{\beta })\) is trivial. Hence, \(|\beta _{n}|\geq a\), for all \(n\in \mathbb{N}\). Since \(T_{\beta }\) is bounded, so by Theorem 4.2, there exists \(A>0\) such that \(|\beta _{n}|\leq A\), for all \(n\in \mathbb{N}\). Thus, we have proved that \(a\leq |\beta _{n}|\leq A\), for all \(n\in \mathbb{N}\). □
Theorem 4.9
Let \((C(p) )_{\varrho }\)be a pre-quasi-Banach (sss), where \(\varrho (x)=\sum^{\infty }_{i=0} ( \frac{\sum_{j=2^{i}-1}^{2^{i+1}-2}|x_{j}|}{2^{i}} )^{p_{i}}\)for all \(x\in C(p)\)and \(T_{\beta }\in L ( (C(p) )_{\varrho } )\). Then \(T_{\beta }\)is a Fredholm operator if and only if:
-
(i)
\(\ker (\beta )\)is a finite subset of \(\mathbb{N}\).
-
(ii)
\(|\beta _{n}|\geq \epsilon \), for all \(n\in \ker (\beta )^{c}\).
Proof
Let \(T_{\beta }\) be Fredholm. If \(\ker (\beta )\) is an infinite subset of \(\mathbb{N}\), then \(e_{n}\in \ker (T_{\beta })\), for all \(n\in \ker (\beta )\). But \(e_{n}\)’s are linearly independent, which shows that \(\ker (T_{\beta })\) is an infinite dimensional which is a contradiction. Hence, \(\ker (\beta )\) must be a finite subset of \(\mathbb{N}\). The condition (ii) comes from Theorem 4.7. Conversely, If the conditions (i) and (ii) are true, then we prove that \(T_{\beta }\) is Fredholm. By Theorem 4.7, the condition (ii) implies that \(T_{\beta }\) has closed range. The condition (i) implies that \(\ker (T_{\beta })\) and \((\operatorname{range}(T_{\beta }))^{c}\) are finite dimensional. This proves that \(T_{\beta }\) is Fredholm. □