Operator's convergence
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Operator's convergence
Let \(\displaystyle{\left(X,||\cdot\||\right)}\) be a \(\displaystyle{\rm{Banach}}\) space and \(\displaystyle{T:X\longrightarrow X}\)
be a continuous linear operator, that is \(\displaystyle{T\in\mathbb{B}\,(X,X)}\), and also
\(\displaystyle{||T||=\sup\,\left\{||T(x)||\in\mathbb{R}: x\in X\,,||x||\leq 1\right\}<1}\) .
Prove that \(\displaystyle{Id_{X}-T\in\mathbb{B}\,(X,X)}\) and \(\displaystyle{\left(Id_{X}-T\right)^{-1}=\sum_{n=0}^{\infty}T^{n}}\), where
\(\displaystyle{T^{0}=Id_{X}\,\,,T^{n}=T\circ T^{n-1}\,,n\geq 1}\) .
be a continuous linear operator, that is \(\displaystyle{T\in\mathbb{B}\,(X,X)}\), and also
\(\displaystyle{||T||=\sup\,\left\{||T(x)||\in\mathbb{R}: x\in X\,,||x||\leq 1\right\}<1}\) .
Prove that \(\displaystyle{Id_{X}-T\in\mathbb{B}\,(X,X)}\) and \(\displaystyle{\left(Id_{X}-T\right)^{-1}=\sum_{n=0}^{\infty}T^{n}}\), where
\(\displaystyle{T^{0}=Id_{X}\,\,,T^{n}=T\circ T^{n-1}\,,n\geq 1}\) .
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- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: Operator's convergence
We give a solution:
1st Note
Since the normed space \(\displaystyle{\left(X,||\cdot||\right)}\) is a \(\displaystyle{\rm{Banach}}\) space, we have that the
normed space \(\displaystyle{\left(\mathbb{B}(X,X),||\cdot||\right)}\) is also a \(\displaystyle{\rm{Banach}}\) space. So, the series
\(\displaystyle{\sum_{n=0}^{\infty}T^{n}}\) in this space converges if, and only if, converges absolutely. This holds since :
\(\displaystyle{||T^{n}||\leq ||T||^{n}\, \;\;\forall\,n\in\mathbb{N}\cup\left\{0\right\}}\) and
\(\displaystyle{\sum_{n=0}^{\infty}||T||^{n}=\dfrac{1}{1-||T||}}\) because \(\displaystyle{||T||<1}\).
Therefore, \(\displaystyle{\sum_{n=0}^{\infty}T^{n}=\lim_{n\to +\infty}\,\sum_{k=0}^{n}T^{k}}\) .
Also, \(\displaystyle{T\neq Id_{X}}\). Indeed, if \(\displaystyle{T=Id_{X}}\), then \(\displaystyle{||T||=1}\), a contradiction, and
\(\displaystyle{||Id_{X}-T||>0}\) .
2nd Note
Since \(\displaystyle{Id_{X}\,,T\in\mathbb{B}(X,X)}\) and \(\displaystyle{\left(\mathbb{B}(X,X),+,\cdot\right)}\) is a vector space, we get \(\displaystyle{Id_{X}-T\in\mathbb{B}(X,X)}\) .
3rd Note
\(\displaystyle{\bullet,n=0}\): \( \displaystyle {\left(Id_{X}-T\right)\circ \sum_{k=0}^{0}T^{k}=\left(Id_{X}-T\right)\circ Id_{X}=Id_{X}-T^{1}=\sum_{k=0}^{0}T^{k}\circ \left(Id_{X}-T\right)}\)
Suppose that\(\displaystyle{\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}=I-T^{n+1}=\sum_{k=0}^{n}T^{k}\circ \left(Id_{X}-T\right)\,\,(\ast)}\) for some \( \displaystyle{n\in\mathbb{N}\cup\left\{0 \right\}} \).
\(\displaystyle{\bullet\,n\to n+1}\) :
\(\displaystyle{\begin{aligned} \left(Id_{X}-T\right)\circ \sum_{k=0}^{n+1}T^{k}&=\left(Id_{X}-T\right)\circ\left(\sum_{k=0}^{n}T^{k}+T^{n+1}\right)\\&=\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}+\left(Id_{X}-T\right)\circ T^{n+1}\\&\stackrel{(\ast)}{=}\left(I-T^{n+1}\right)+Id_{X}\circ T^{n+1}-T^{n+2}\\&=I-T^{n+2}\end{aligned}}\)
Similarly, \(\displaystyle{\sum_{k=0}^{n+1}T^{k}\circ \left(Id_{X}-T\right)=I-T^{n+2}}\). So, by induction we have that
\(\displaystyle{\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}=I-T^{n+1}=\sum_{k=0}^{n}T^{k}\circ \left(Id_{X}-T\right)\,\,(I)\,\,,\forall\,n\in\mathbb{N}\cup\left\{0\right\}}\).
4th Note
Let \(\displaystyle{\epsilon>0}\). Since \(\displaystyle{||T||^{n+1}\to 0\,\,,n\to +\infty}\), there is \(\displaystyle{n_{0}\in\mathbb{N}}\) such that
\(\displaystyle{\left|||T||^{n+1}-0\right|=||T||^{n+1}<\epsilon}\) for every \(\displaystyle{n\geq n_{0}}\) .
So, for every \(\displaystyle{n\geq n_{0}}\) we have that
\(\displaystyle{||Id_{X}-T^{n+1}-Id_{X}||=||T^{n+1}||\leq ||T||^{n+1}<\epsilon}\), so: \(\displaystyle{Id_{X}-T^{n+1}\to Id_{X}}\).
5th Note
Let \(\displaystyle{\epsilon>0}\). For the positive number \(\displaystyle{\dfrac{\epsilon}{||Id_{X}-T||}}\), there is \(\displaystyle{n_{0}\in\mathbb{N}}\)
such that\(\displaystyle{||\sum_{k=0}^{n}T^{k}-\sum_{n=0}^{\infty}T^{n}||<\dfrac{\epsilon}{||Id_{X}-T||}}\) for every \(\displaystyle{n\geq n_{0}}\).
Therefore, for every \(\displaystyle{n\geq n_{0}}\) we get:
\(\displaystyle{\begin{aligned} ||\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}-\left(Id_{X}-T\right)\circ \sum_{n=0}^{\infty}T^{n}||&=||\left(Id_{X}-T\right)\circ \left(\sum_{k=0}^{n}T^{k}-\sum_{n=0}^{\infty}T^{n}\right)||\\&=||Id_{X}-T||\cdot ||\sum_{k=0}^{n}T^{k}-\sum_{n=0}^{\infty}T^{n}||<||Id_{X}-T||\cdot \dfrac{\epsilon}{||Id_{X}-T|}\\&=\epsilon\end{aligned} }\)
and thus \(\displaystyle{\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}\to \left(Id_{X}-T\right)\circ \sum_{n=0}^{\infty}T^{n}}\).
Similarly, \(\displaystyle{\sum_{k=0}^{n}T^{k}\circ \left(Id_{X}-T\right)\to \left(Id_{X}-T\right)\circ \sum_{n=0}^{\infty}T^{n}}\).
6th Note
Taking limits at the equation \(\displaystyle{(I)}\), we get :
\(\displaystyle{\left(Id_{X}-T\right)\circ \sum_{n=0}^{\infty}T^{n}=Id_{X}=\sum_{n=0}^{\infty}T^{n}\circ \left(Id_{X}-T\right)}\), which means that:
\(\displaystyle{\left(Id_{X}-T\right)^{-1}=\sum_{n=0}^{\infty}T^{n}}\) .
Remark
The triplet \(\displaystyle{\left(\mathbb{B}(X,X),+,\circ\right)}\) is an associative ring with unity \(\displaystyle{1=Id_{X}}\).
1st Note
Since the normed space \(\displaystyle{\left(X,||\cdot||\right)}\) is a \(\displaystyle{\rm{Banach}}\) space, we have that the
normed space \(\displaystyle{\left(\mathbb{B}(X,X),||\cdot||\right)}\) is also a \(\displaystyle{\rm{Banach}}\) space. So, the series
\(\displaystyle{\sum_{n=0}^{\infty}T^{n}}\) in this space converges if, and only if, converges absolutely. This holds since :
\(\displaystyle{||T^{n}||\leq ||T||^{n}\, \;\;\forall\,n\in\mathbb{N}\cup\left\{0\right\}}\) and
\(\displaystyle{\sum_{n=0}^{\infty}||T||^{n}=\dfrac{1}{1-||T||}}\) because \(\displaystyle{||T||<1}\).
Therefore, \(\displaystyle{\sum_{n=0}^{\infty}T^{n}=\lim_{n\to +\infty}\,\sum_{k=0}^{n}T^{k}}\) .
Also, \(\displaystyle{T\neq Id_{X}}\). Indeed, if \(\displaystyle{T=Id_{X}}\), then \(\displaystyle{||T||=1}\), a contradiction, and
\(\displaystyle{||Id_{X}-T||>0}\) .
2nd Note
Since \(\displaystyle{Id_{X}\,,T\in\mathbb{B}(X,X)}\) and \(\displaystyle{\left(\mathbb{B}(X,X),+,\cdot\right)}\) is a vector space, we get \(\displaystyle{Id_{X}-T\in\mathbb{B}(X,X)}\) .
3rd Note
\(\displaystyle{\bullet,n=0}\): \( \displaystyle {\left(Id_{X}-T\right)\circ \sum_{k=0}^{0}T^{k}=\left(Id_{X}-T\right)\circ Id_{X}=Id_{X}-T^{1}=\sum_{k=0}^{0}T^{k}\circ \left(Id_{X}-T\right)}\)
Suppose that\(\displaystyle{\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}=I-T^{n+1}=\sum_{k=0}^{n}T^{k}\circ \left(Id_{X}-T\right)\,\,(\ast)}\) for some \( \displaystyle{n\in\mathbb{N}\cup\left\{0 \right\}} \).
\(\displaystyle{\bullet\,n\to n+1}\) :
\(\displaystyle{\begin{aligned} \left(Id_{X}-T\right)\circ \sum_{k=0}^{n+1}T^{k}&=\left(Id_{X}-T\right)\circ\left(\sum_{k=0}^{n}T^{k}+T^{n+1}\right)\\&=\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}+\left(Id_{X}-T\right)\circ T^{n+1}\\&\stackrel{(\ast)}{=}\left(I-T^{n+1}\right)+Id_{X}\circ T^{n+1}-T^{n+2}\\&=I-T^{n+2}\end{aligned}}\)
Similarly, \(\displaystyle{\sum_{k=0}^{n+1}T^{k}\circ \left(Id_{X}-T\right)=I-T^{n+2}}\). So, by induction we have that
\(\displaystyle{\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}=I-T^{n+1}=\sum_{k=0}^{n}T^{k}\circ \left(Id_{X}-T\right)\,\,(I)\,\,,\forall\,n\in\mathbb{N}\cup\left\{0\right\}}\).
4th Note
Let \(\displaystyle{\epsilon>0}\). Since \(\displaystyle{||T||^{n+1}\to 0\,\,,n\to +\infty}\), there is \(\displaystyle{n_{0}\in\mathbb{N}}\) such that
\(\displaystyle{\left|||T||^{n+1}-0\right|=||T||^{n+1}<\epsilon}\) for every \(\displaystyle{n\geq n_{0}}\) .
So, for every \(\displaystyle{n\geq n_{0}}\) we have that
\(\displaystyle{||Id_{X}-T^{n+1}-Id_{X}||=||T^{n+1}||\leq ||T||^{n+1}<\epsilon}\), so: \(\displaystyle{Id_{X}-T^{n+1}\to Id_{X}}\).
5th Note
Let \(\displaystyle{\epsilon>0}\). For the positive number \(\displaystyle{\dfrac{\epsilon}{||Id_{X}-T||}}\), there is \(\displaystyle{n_{0}\in\mathbb{N}}\)
such that\(\displaystyle{||\sum_{k=0}^{n}T^{k}-\sum_{n=0}^{\infty}T^{n}||<\dfrac{\epsilon}{||Id_{X}-T||}}\) for every \(\displaystyle{n\geq n_{0}}\).
Therefore, for every \(\displaystyle{n\geq n_{0}}\) we get:
\(\displaystyle{\begin{aligned} ||\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}-\left(Id_{X}-T\right)\circ \sum_{n=0}^{\infty}T^{n}||&=||\left(Id_{X}-T\right)\circ \left(\sum_{k=0}^{n}T^{k}-\sum_{n=0}^{\infty}T^{n}\right)||\\&=||Id_{X}-T||\cdot ||\sum_{k=0}^{n}T^{k}-\sum_{n=0}^{\infty}T^{n}||<||Id_{X}-T||\cdot \dfrac{\epsilon}{||Id_{X}-T|}\\&=\epsilon\end{aligned} }\)
and thus \(\displaystyle{\left(Id_{X}-T\right)\circ \sum_{k=0}^{n}T^{k}\to \left(Id_{X}-T\right)\circ \sum_{n=0}^{\infty}T^{n}}\).
Similarly, \(\displaystyle{\sum_{k=0}^{n}T^{k}\circ \left(Id_{X}-T\right)\to \left(Id_{X}-T\right)\circ \sum_{n=0}^{\infty}T^{n}}\).
6th Note
Taking limits at the equation \(\displaystyle{(I)}\), we get :
\(\displaystyle{\left(Id_{X}-T\right)\circ \sum_{n=0}^{\infty}T^{n}=Id_{X}=\sum_{n=0}^{\infty}T^{n}\circ \left(Id_{X}-T\right)}\), which means that:
\(\displaystyle{\left(Id_{X}-T\right)^{-1}=\sum_{n=0}^{\infty}T^{n}}\) .
Remark
The triplet \(\displaystyle{\left(\mathbb{B}(X,X),+,\circ\right)}\) is an associative ring with unity \(\displaystyle{1=Id_{X}}\).
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