I'm struggling to get my head around Hilbert spaces and orthogonality. I've been using this example as a reference point to help deepen my understanding but I still have some questions.
We have the Hilbert space $H = L^2$, whose elements are real valued sequences $x = (ξ_{j} ) = (ξ_{1}, ξ_{2}, . . .)$ such that
$\sum_{j=1}^∞ |ξ_{j}|^2 < ∞$,
with inner product
$\langle$ x, y $\rangle$ $= \sum_{j=1}^∞ ξ_jη_j$ , for every $x, y ∈ L^2$ where $x = (ξ_1, ξ_2, . . .)$ and $y = (η_1, η_2, . . .)$.
Consider now the set $Y ⊂ L^2$ defined in the following way: $y ∈ Y ⇐⇒ y ∈ L^2$ and $∃ N_y ∈ N$ :
$η_j = 0$ for every $j > 2N_y$,
$η_{2i} = 0$ for every $i = 1, 2, . . . , Ny$
(The last part should be a piecewise function but I'm unsure how to create one in MathJax).
The things that I am finding difficult to do of my own accord are: showing that $Y$ is a subspace of $L^2$, showing whether $Y$ is closed or not and proving it, finding $Y^⊥$ and showing that $L^2$ is not the direct sum of $Y$ and $Y^⊥$.
A simpler description of $Y$ is: it consists of sequences $(a_n)$ whose only non -zero entries are the first few odd ones. (This means $a_n=0$ except when $n=2k-1$ with $ k \leq $ some integer). From this description you can see immediately that $Y$ is a subspace. $Y$ is not closed because the sequence $a_n=0$ for all $n$ even and $a_n=\frac 1 n$ for $n$ odd is in the closure of $Y$ but not itself in $Y$. I will let you verify that $Y^{\perp}$ is nothing but sequences whose odd coordinates are all $0$. A sequence which is not in $Y+Y^{\perp}$ is given by $a_n=0$ for $n$ even and $a_n=\frac 1 n$ for all odd $n$.