Orthogonal complement of closed subset in Hilbert space

Question

Consider a subset $K\subset L^2 (\mathbb{R})$ defined by $$ K=\{f\in L^2(\mathbb{R})\mid \forall n\in\mathbb{Z}: \int_{n}^{n+1}f(x)dx=0\} $$

I want to determine the orthogonal complement $K^\bot$ in this Hilbert space. I was already able to verify that $K$ is a closed subset. So if I would somehow be able to suggest how the orthogonal complement should look like by a small calculation, then I can simply verify that this suggestion is correct by verifying that $(K^\bot)^\bot=K$ holds.

I know the calculation should start in the trend of: let $g\in L^2 (\mathbb{R})$ and $f\in K$ s.t. \begin{align} \langle f,g\rangle &=\int_{\mathbb{R}}f(x)\overline{g(x)} dx \\ &= ...,\\ \end{align} which should equal zero, where the term that remains inside the integral might then tell us something about the elements of $K^\bot$. However, it is not clear what manipulative calculation I should perform. Thank you in advance!

Answer 1

$K$ is defined by the relation $$\int_\mathbb{R} f(x)\mathbf{1}_{[n,n+1)}(x)\,dx = 0$$ for all $n \in \mathbb{Z}.$ That looks a lot like we're defining basic elements of $K^\perp.$ In particular, this seems to define $$K^\perp = \{g \in L^2(\mathbb{R}) : \forall n \in \mathbb{Z}, g \text{ is constant on } [n,n+1)\}$$ which can be naturally identified with $\ell^2(\mathbb{Z})$

Answer 2

Let $M$ consist of functions that have a constant value $c_n$ on $(n,n+1)$ ($n\in \mathbb Z$) with $\sum_n |c_n|^{2} <\infty$. Using the fact that convergence in $L^{2}$ implies almost everywhere convergence of some subsequence we can easily see that $M$ is a closed subspace of $L^{2}(\mathbb R)$. $K$ is nothing but $M^{\perp}$. Hence $K^{\perp}=M^{\perp \perp}=M$.