Let W be a subspace of $L^2[0, 2\pi]$ of all functions that $\hat{f}(n)=0$ for all $n \leq 0 $ (where $\hat{f}(n)=\frac{1}{2 \pi }\int_\limits{0}^{2\pi}f(t)e^{-int}dt$ denotes n-th Fourier coefficient of $f$). What is orthogonal complement of W and what form does orthogonal projection onto W take?
Now, I know that $f \in L^2[0, 2\pi]$ can take form of $f(x) = \sum_\limits{n=-\infty}^{\infty} \hat{f}(n)e^{inx}$ and ${e^{izx}}_{z \in \mathbb{Z}}$ is complete orthogonal set in $L^2[0,2\pi]$.
If W contains those functions with all Fourier coefficients $n \leq 0$ equal to 0, then its orthogonal complement would be functions where $\hat{f}(n) = 0$ for all $n>0$. Is that correct?
Now, the projection. We know that $$ L^2[0, 2\pi]\ni f(x) = \sum_\limits{n=-\infty}^{\infty} \hat{f}(n)e^{inx} = \sum_\limits{n=-\infty}^{0} \hat{f}(n)e^{inx} + \sum_\limits{n=1}^{\infty} \hat{f}(n)e^{inx} $$ Does it mean that projection onto W will be just: $$proj(f) = \sum_\limits{n=-\infty}^{0} \hat{f}(n)e^{inx}$$
If so, why? I have seen similar example from $L^2[\mathbb{R}]$ with set V being all f such that $f = 0$ on $(0, \infty)$, its orthogonal complement was all functions $f = 0$ on $(-\infty, 0)$ and the projection onto V was defined as $proj(f) = f \mathbb{1}_{(0, \infty)}$. So I kinda guessed it will be the same idea applied here.
I don't see clearly why a projection of any function $g$ from $L^2[0, 2 \pi]$ (not necesarilly with all coefficients $n \leq 0$ of $g$ equal to 0) will be projected onto $\sum_\limits{n=-\infty}^{0} \hat{g}(n)e^{inx}$. Any help and explanation appreciated.
In general $(M^{\perp})^{\perp} =M$ for any closed subspace of a Hilbert space. If $(e_n)$ is an orthonormal basis for a Hilbert space $H$ then the orthogonal complement of $\{e_i: i \in A\}$ is given by $\{\sum_{i \notin A} a_i e_i: \sum |a_i|^{2} <\infty\}$. And the orthogonal complement of this space is exactly the closed subspace spanned by $\{e_i: i \in A\}$.