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I am looking for an example such that in a pre-Hilbert space $H$ we have for a subspace $U$ that

(i) $\bar{U} \oplus U^\perp \neq H$

(ii) $ \bar{U} \neq U^{\perp \perp}$ Since finite and closed subspaces will probably not do it, I am not good at inventing nice examples. Does anybody here have a few at hand?

1 Answers1

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Let $H \subset \ell^2$ the space of sequences with only finitely many nonzero terms. Let $\xi \in \ell^2$ any element with infinitely many nonzero terms.

Then consider $U = H \cap (\mathbb{K}\cdot \xi)^\perp$.

Since it's homework, I leave the details to you.

$U$ is the intersection of a closed subspace of $\ell^2$ with $H$, so it's closed in $H$. Let $k$ be an index with $\xi_k \neq 0$. For all $n \neq k$, $U$ contains $\overline{\xi_k}\cdot e_n - \overline{\xi_n}\cdot e_k$, so if $v\in U^\perp$, then

$$0 = \langle v,\overline{\xi_k}e_n - \overline{\xi_n}e_k\rangle = \xi_k v_n - \xi_n v_k,$$

i.e. $v_n = \frac{\xi_n}{\xi_k}\cdot v_k$ for all $n\neq k$. If $v_k = 0$, it follows that $v = 0$, and if we had $v_k \neq 0$, then $v = \frac{v_k}{\xi_k}\cdot \xi \notin H$, hence $U^\perp = \{0\}$ in $H$. Since $U \neq H$ (we have $e_k \in H\setminus U$), it follows that

  • $\overline{U} \oplus U^\perp = \overline{U} = U \neq H$, and
  • $(U^\perp)^\perp = \{0\}^\perp = H \supsetneq \overline{U} = U$.
Daniel Fischer
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  • well, honestly I see that $U:={x \in H; \langle x,\xi \rangle =0 }$ is a closed subspace, since it is a nullspace, but I don't see what it's complement should be and I don't see what will be missing in it's orthogonal complement. (As I don't know what the orthogonal complement will be, I cannot say anything about the complement of the complement). So, could you tell me a few more things about your example? –  Jan 23 '14 at 06:26
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    Since $\xi \notin H$, you have $U^\perp = {0}$ in $H$. – Daniel Fischer Jan 23 '14 at 09:21