# Show that Y is a closed subspace of l2

This might be a straight forward problem but I wouldn't ask if I knew how to continue. Apologies in advance, I am not sure how to use the mathematical formatting.

We are currently busy with inner product spaces and I'm struggling to prove that: $Y = \{x|x=(\xi_{j}) \in l^{2}, \xi_{2n}=0, n \in \mathbb{N}\}$ is a closed subspace of $l^{2}$.

From this we need to find orthogonal complement of Y which I should manage to do. I just need to show the subspace is closed.

workings:

$l^{2} = \{\xi_{1},\xi_{2},\xi_{3},\xi_{4},...\}$, $\sum_{i=1}^{\infty} |\xi_{i}|^{2}$ finite therefore convergent series.

$Y = \{\xi_{1},0,\xi_{3},0,\xi_{5},...\}$

Y is a proper subspace of $l^{2}$ the separable infinite dimensional Hilbert space. Clearly the sum |of the elements of Y|^2 converge however it will be slower convergence than $l^{2}$'s elements. Am I missing something elementary? I dont know how to put it together.

Also if $x_{n} \in Y$ and $x_{n} \to x$ implies $x \in Y$ then Y will be closed. How do I write out the formal proof?

Thank you!

-

If $x_n\to x$ (in the $\ell^2$ norm) strongly, in particular it converges weakly, i.e. $$x_n\to x \text{ (weak) iff }\langle x_n, y \rangle \to \langle x, y \rangle, \;\forall \; y\in \ell^2.$$ Then you have $$\langle x_n, e_{2k} \rangle=0 \to \langle x, e_{2k} \rangle=0,$$ where $e_{2k}=(0,\dots,1,\dots,0)$ and the $1$ is the $2k$th entry. Thus, every $2k$th entry of $x$ vanishes. This implies that $x\in Y$