Let $E:= \{(x_n) \in l^{2}(\mathbb N ) \mid x_{2k}=x_{2k+1}\}$

How can I show that $E$ is a closed subvector space of $l^{2}(\mathbb N )$ ?

I tried to write $E$ as the kernel of a continuous linear form but I ended with $T:l^{2}(\mathbb N) \rightarrow \mathbb R : (x_n) \rightarrow \sum_{n=0}^\infty |x_{2k}-x_{2k+1}|$

which is not linear.

  • $\begingroup$ just use sequential defn of closed $\endgroup$ – mathworker21 Jun 16 at 14:59

You didn't quantify whether it is for all $k\in\mathbb{N}$ or for a single $k\in\mathbb{N}$.

For each $k\in\mathbb{N}$, note that $(x_n)\in\ell^2(\mathbb{N})\mapsto x_{2k}-x_{2k+1}\in\mathbb{F}$ is continuous and linear. So its kernel is a closed subspace of $\ell^2(\mathbb{N})$.

If your $E$ is $\{(x_n)\in\ell^2(\mathbb{N})\mid x_{2k}=x_{2k+1}\quad\forall k\in\mathbb{N}\}$, then we take intersection of closed subspace, so it is also a closed subspace.


I assume you can show it is a subspace. As to why it is closed define a sequence $(x^k)_{k=1}^\infty\subseteq E$ which converges to $x$. We have to show that $x\in E$. By definition of convergence in $l^2$ we have $\sum_{n=1}^\infty |x^k_n-x_n|^2\to 0$ when $k\to\infty$. This of course implies that for each $n\in\mathbb{N}$ we have $|x^k_n-x_n|\to 0$. That means we have convergence in every coordinate. Can you finish from here using the fact that the vectors in the sequence are in $E$?


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