1

How could I find a non-closed linear subspace $X$ of $l^{2}$ , such that $l^{2} \ne X + X^{\perp} ?$

Martin Sleziak
  • 53,687
MathComputC
  • 11
  • 3
    Let $X = c_{00}$ be the subspace of all finite sequences (i.e. those which have a finite number of non-zero elements). Show that $X^{\perp}$ is zero, hence $X$ is an example of such a subspace. – bakula Oct 23 '15 at 09:54
  • 1
    You can take any dense subspace $X$ different from $\ell^2$.. – Jochen Oct 23 '15 at 10:29
  • You can take any non-closed subspace $X$, since $\ell^2 = \bar X + X^\perp$, which is a direct sum. – gerw Oct 23 '15 at 10:40

1 Answers1

2

Recall the following two facts:

(1) For any closed subspace $X$ of $\ell^2$, we have $X \oplus X^\bot = \ell^2$.

(2) For any subspace $X$ of $\ell^2$, we have $X^\bot = \bar X^\bot$.

To prove (1), suppose $x \in \ell^2$ is given, as $X$ is a closed subspace, there is an orthogonal projection $P$ with range $X$, then $x = Px + (x- Px)$ is a partition of $x$ into elements of $X$ and $X^\bot$. For (2) note that $\bar X^\bot \subseteq X^\bot$ follows from the antimonotonicity of $(-)^\bot$, and if $x \in X^\bot$ and $y \in \bar X$ are given, choose a sequence $y_n \to y$ with $y_n \in X$. Then $\left<y,x\right> = \lim_n \left<y_n, x\right> = 0$. Hence $x \in \bar X^\bot$.

So if $X$ is any non-closed subspace of $\ell^2$, we have $$ X \oplus X^\bot \subsetneq \bar X \oplus X^\bot = \bar X \oplus \bar X^\bot = \ell^2 $$

martini
  • 84,101