# 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 Im struggling to prove that: Y = {x|x=(Ej) element of l2 ("little el two"), E2n=0, n element of Natural numbers} is a closed subspace of l2.

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:

l2 = {E1,E2,E3,E4,...} sum |Ei|^2 finite for 1 <= i <= infinity therefore convergent series.

Y = {E1,0,E3,0,E5,...}

Y is a proper subspace of l2 the separable infinite dimensional Hilbert space. Clearly the sum |of the elements of Y|^2 converge however it will be slower convergence than l2's elements. Am I missing something elementary? I dont know how to put it together.

Also if xn element of Y and xn-->x implies x element of 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 $x\in Y$