Linear and bounded operator.
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Linear and bounded operator.
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 \(\displaystyle{||T||=1}\) .
\(\displaystyle{a=\left(a_{n}\right)_{n\in\mathbb{N}}\mapsto T(a):=\left(0,a_1,a_2,...,a_n,a_{n+1},...\right)}\) .
Prove that the function \(\displaystyle{T}\) is linear, bounded and \(\displaystyle{||T||=1}\) .
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Re: Linear and bounded operator.
Linearity is trivial. For boundedness we can do better and observe that $$||T(a)|| = ||a||$$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 \(\displaystyle{||T||=1}\) .
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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