Orthogonal of $K=\{f \in L^2(\mathbb{R}) \mid \forall n \in \mathbb{Z}: \int^{n+1}_n f(x)dx = 0 \}$ in $L^2(\mathbb R)$ Define a subspace $K \subset L^2(\mathbb{R})$ given by $$K=\{f \in L^2(\mathbb{R}) \mid \forall n \in \mathbb{Z}: \int^{n+1}_n f(x)dx = 0 \}.$$
I have already proven that $K$ is closed by using a convergence theorem (is this correct?). I want to find $K^{\perp}$. I'm quite confused on how to find this. I first tried writing down $\langle f,g \rangle$ with $g \in K$ and $f \in L^2(\mathbb{R})$ to find a condition for $f$, but I'm stuck there.
 A: Hint
$K$ is closed
The map
\begin{array}{l|rcl}
\psi_n : & L^2(\mathbb{R}) & \longrightarrow & \mathbb R \\
    & f & \longmapsto & \int_n^{n+1} f(x) \ dx\end{array}
is continuous, hence its kernel is closed.
$$K = \bigcap\limits_{n \in \mathbb Z} \psi_n^{-1}[\{0\}]$$ is closed as an intersection of closed subsets.
About $K^\perp$
You know that on $(0,1)$, $\{\sin 2\pi nx, \cos 2\pi nx \mid n \in \mathbb N\}$ is an orthonormal basis of $L^2((0,1))$. They all belong to $K$, except the constant map.
Hence an element $\phi$ of $K^\perp$ has to be constant on each interval $(n, n+1)$.
You also need to have $\int_{\mathbb R} f^2$ convergent. Finally, $K^\perp$ is the subset of $ L^2(\mathbb{R})$ generated by the indicator functions of the intervals $[n,n+1]$ for $n \in \mathbb Z$.
A: You can convince yourself that the orthogonal space must include the span of the set of indicators of the segments (n,n+1). On the other hand every function in the orthogonal space must be constant on such segments up to a set of measure zero (there are details to complete here), so both inclusions apply.
Edit: K is alredy the orthogonal space to the space of these indicators, so use the fact the orthogonal of the orthogonal is the closure of the span.
