Consider a subspace $K$ of $L^2(\mathbb{R})$ defined by
$$ K =\{f \in L^2(\mathbb{R})\mid \forall n\in\mathbb{Z}:\int_n^{n+1}f(x)\, dx = 0\}$$
I now want to determine the orthogonal complement $K^\perp$. I thought it would be most easily done by proving that $K$ is a closed subspace in $L^2(\mathbb{R})$ such that $L^2(\mathbb{R})=K\oplus K^\perp$.
Proving the fact that $K$ is close can be easily done by using a convergent sequence in $K.$ However, I'm not sure how to proceed to determine the complement $K^\perp$. Normally when I want to determine the orthogonal complement, I consider $f\in L^2(\mathbb{R})$ and $g\in K$, and look at the inner product $\langle f,g\rangle$. By manipulating this expression, you obtain an idea about how $K^\perp$ should look like. But I'm not sure if this is the right way to proceed in this situation and if so, how I should write the inner product to see how $K^\perp$ should look like.
Some hints are welcome! Thank you in advance.
Best Answer
In the comments there already is written something close to full answer. Consider functions $\alpha_n = \chi_{[n,n+1]}$.
At first observe that $K = \{f \in L_2(\mathbb{R}): (f,\alpha_n) = 0\; \forall n\in \mathbb{Z}\}$. From here it follows that $K = L^\perp$ where $L = \operatorname{span} \{\alpha_n:\; n \in \mathbb{Z}\}$. And therefore $K^{\perp} = L^{\perp \perp} = \overline{L}$.
Now we can write down the exact form of $\overline{L}$. In $L$ (a prehilbert space) there is an orthonormal (Hamel) basis $\alpha_n$. So, $\alpha_n$ form a Hilbert orthonormal basis for $\overline{L}$. Therefore $$\overline{L} = \left\{\sum_{n \in \mathbb{Z}} b_n \alpha_n: \sum_{n \in \mathbb{Z}} |b_n|^2 < \infty\right\} = \left\{\sum_{n \in \mathbb{Z}} b_n \chi_{[n,n+1]}: \sum_{n \in \mathbb{Z}} |b_n|^2 < \infty\right\}.$$