Linear and bounded operator.

Functional Analysis
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Papapetros Vaggelis
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Joined: Mon Nov 09, 2015 1:52 pm

Linear and bounded operator.

#1

Post by Papapetros Vaggelis »

We define \(\displaystyle{T:\mathbb{l}_{2}(\mathbb{N})\longrightarrow \mathbb{l}_{2}(\mathbb{N})}\) by

\(\displaystyle{a=\left(a_{n}\right)_{n\in\mathbb{N}}\mapsto T(a):=\left(0,a_1,a_2,...,a_n,a_{n+1},...\right)}\) .

Prove that the function \(\displaystyle{T}\) is linear, bounded and \(\displaystyle{||T||=1}\) .
Michael Lambrou
Posts: 1
Joined: Tue Nov 10, 2015 5:41 pm

Re: Linear and bounded operator.

#2

Post by Michael Lambrou »

Papapetros Vaggelis wrote:We define \(\displaystyle{T:\mathbb{l}_{2}(\mathbb{N})\longrightarrow \mathbb{l}_{2}(\mathbb{N})}\) by

\(\displaystyle{a=\left(a_{n}\right)_{n\in\mathbb{N}}\mapsto T(a):=\left(0,a_1,a_2,...,a_n,a_{n+1},...\right)}\) .

Prove that the function \(\displaystyle{T}\) is linear, bounded and \(\displaystyle{||T||=1}\) .
Linearity is trivial. For boundedness we can do better and observe that $$||T(a)|| = ||a||$$
as both equal to $$\sqrt {\sum {a_k^2} }, $$ from which the result follows trivially.

Let me add that this operator is called the forward shift and is VERY well studied.
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