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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!

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1 Answer 1

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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 $x\in Y$

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  • $\begingroup$ You are welcome :-) $\endgroup$
    – guacho
    Commented May 23, 2013 at 8:02
  • $\begingroup$ Simple and elegant. $\endgroup$
    – user459879
    Commented Dec 13, 2020 at 21:40

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