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Let $V$ be the set of all infinite real sequences and let $S$ be the Shift Operator for the set of all infinite sequences $(a_n)_{n \in \mathbb N}$ such that $S((a_n)_{n \in \mathbb N})=(a_{n+1})_{n \in \mathbb N}$.

Define a Subspace $W$ of $V$ such that: $$W = \{(a_n)_{n \in \mathbb N} \in V: a_{n+3} = 2a_{n+2} + a_{n+1} -2a_n\}$$

I need to show that $W$ is $S$-Invariant, that is $S((a_n)_{n \in \mathbb N}) \in W$, however, I am really not sure how to start with this question. I only understood that after the transformation, the first term of the sequence is removed and I will then need to show that the remaining sequence is still in $W$. Intuitively $W$ seems to be $S$-invariant but I am not sure on how to prove this.

Any help and advice will be really appreciated!

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Let $w=(w_n)_{n\in\Bbb{N}}\in W$ be a sequence. Then for all $n\in\Bbb{N}$ you have $$w_{n+3}=2w_{n+2}+w_{n+1}-2w_n.$$ Now consider its shift $u:=S(w)$, and denote it by $u=(u_n)_{n\in\Bbb{N}}$ so that $u_n:=w_{n+1}$ for all $n\in\Bbb{N}$. Simply verify that $$u_{n+3}=2u_{n+2}+u_{n+1}-2u_n,$$ for all $n\in\Bbb{N}$.

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  • $\begingroup$ Many thanks! Understood this better with the definition of the new sequence $u_n$ with $u_n = w_{n+1}$ $\endgroup$ – Derp Nov 16 '18 at 18:31

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