Operator's convergence

Functional Analysis
Post Reply
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Operator's convergence

#1

Post by Papapetros Vaggelis »

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


\(\displaystyle{T^{0}=Id_{X}\,\,,T^{n}=T\circ T^{n-1}\,,n\geq 1}\) .
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Operator's convergence

#2

Post by Papapetros Vaggelis »

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 \(\displaystyle{1=Id_{X}}\).
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 6 guests