I know from Orthogonal sum with non-closed subspace that if an Hilbert space $H$ is the orthogonal sum of linear subspaces $M$ and $N$ then $M$ and $N$ are closed.
I need to prove that orthogonality $M \perp N$ is a necessary hypotesis with an example only knowing $H=M+N$ and $M \cap N={0}$. My idea was using $l^2(\mathbb{N})$ and the subspace $c_0=\{x=(x_1, \ldots,x_N,0,\ldots)\colon N>0,x_1,\ldots,x_N \in \mathbb{F}\}$ that is not closed. I'm having trouble finding $N$ such that $l^2(\mathbb{N})=c_0+N$.
Hilbert space as sum of not closed subspaces
functional-analysishilbert-spacesreal-analysis
Best Answer
Let $f$ be a dis-continuous linear functional on $H=\ell^{2}$. Let $M$ be the kernel of $f$. Let $x_0$ be a vector with $f(x_0)=1$ and let $N$ be the span of $x_0$. Then $H$ is the direct sum of $M$ and $N$ (becasue $x=(x-f(x)x_0)+(f(x)x_0)$ and $M$ is not closed. $M$ and $N$ are not orthogonal by the argument in the link provided by you. However I will avoid that (simple) argument to make this self-contained.
If there exists $x_1$ such that $f(x_1)=0$ but $x_1$ not orthogonal to $x_0$ it would follow that $M$ and $N$ are not orthogonal. Such a point $x_1$ may not exist but we can make following modification in the construction.
Choose $x_1$ such that $x_0$ and $x_0$ are L.I. but not orthogonal. [Can you see why such a thing is possible?]. Let $g$ be a dis-continuous linear functional on $\{x_0,x_1\}^{\perp}$. Define $f(ax_0+bx_1+y)=a+g(y)$ for all scalars $a,b$ and all $y \in \{x_0,x_1\}^{\perp}$. Define $M$ and $N$ as before for this $f$.