How to show: $\exists$ a subspace $L$ of a Hilbert space $H$ such that $H\neq L\oplus L^{\perp}$.

Question

How to show: $\exists$ a subspace $L$ of a Hilbert space $H$ such that $H\neq L\oplus L^{\perp}$.

Let consider $H=l_2$ where $l_2=\lbrace x=(x_n)^\infty_1: \sum^\infty_1 |x_n|^2<\infty \rbrace $

and if we choose $L$ s.t. $L=\mathrm{span}(e_{2k})^\infty_{k=1}$, then we cannot conclude that $H\neq L\oplus L^{\perp}$.

Are we on right track? Could you please help?

Answer 1

The $L$ suggested in the OP or any closed subspace won't work: we can use orthogonal projection in order to show that $L\oplus L^\perp=H$.

However, if $L$ is not closed, the equality $L\oplus L^\perp=H$ may not hold. For example, consider $\ell^2(\mathbb Z)$ and $$L:=\operatorname{span}(e_k,k\geqslant 0).$$ Then $L^\perp=\{(x_k)_{k\in\mathbb Z}\in\ell^2(\mathbb Z),x_k=0 \mbox{ for each }k\geqslant 0\}$, hence $L\oplus L^\perp=\{(x_k)\in\ell^2(\mathbb Z), \mbox{ for some }n, x_k=0 \mbox{ whenever }k\geqslant n\}$.

Alternatively, as kahen suggest, each dense subspace $V$ of $H$ may be written as $L\oplus L^{\perp}$. for some $L$ (actually take $L=V$).