4
$\begingroup$

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: $$Y = \{x\mid x=(\xi_j) \in \ell^2, \xi_{2n}=0, n \in \mathbb{N}\}$$ is a closed subspace of $\ell^2$.

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:

$\ell^2 = \{\xi_1,\xi_2,\xi_3,\xi_4,\ldots\}$, $\sum_{i=1}^\infty |\xi_i|^2$ finite therefore convergent series.

$$Y = \{\xi_1,0,\xi_3,0,\xi_5,\ldots\}$$

$Y$ is a proper subspace of $\ell^2$ the separable infinite dimensional Hilbert space. Clearly the sum |of the elements of Y|^2 converges however it will be slower convergence than $\ell^2$'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!

$\endgroup$
5
$\begingroup$

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 $x\in Y$

$\endgroup$
2
  • $\begingroup$ You are welcome :-) $\endgroup$
    – guacho
    May 23 '13 at 8:02
  • $\begingroup$ Simple and elegant. $\endgroup$
    – user459879
    Dec 13 '20 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.