# Connection of orthogonal decomposition of Hilbert space and with eigenspace of linear operator on Hilbert space

functional-analysishilbert-spaceslinear algebrareal-analysisspectral-theory

We know that $$L^2(S^1)$$ is Hilbert space. Also
$$-i\frac{d}{dx}:L^2(S^1)\to L^2(S^1)$$
is linear operator on that. Also $$-i\frac{d}{dx}e^{ikx}=ke^{ikx}.$$

So each $$k$$, $$e^{ikx}$$ is eigenvector.
Define
$$H_k=\{f\in L^2(S^1):-i\frac{d}{dx}f=kf \}$$
Due to eigenspace $$H_k$$ for different $$k$$ are orthogonal.
We also know that $$f\in L^{2}(S^1)$$ is written in the Fourier series.

$$f(x)=\sum_{k\in \mathbb Z}\hat f(k)e^{ikz}$$
That is $$L^2(S^1)=\oplus_{k\in \mathbb Z} H_k$$.

Is such a phenomenon occur for all Hilbert spaces? What would be the choice of operator?

Also,
if we consider $$C^{\infty}(S^1)$$ space, on that we can define operator and that space is dense in $$L^2(S^1)$$. Our operator is self adjoint also closed. Can there is any spectral decomposition theorem for closed self adjoint operator defined on dense set.
I wanted to link this with more general $$L^2(M)$$ Hilbert space.

Any help/hint/reference will be highly appreciated.

First of all, your operator is not well-defined. Its domain should be the Sobolev space $$H^1(S^1)$$. Then it is well-defined and self-adjoint. The point here is that this self-adjoint operator (let's call it $$A$$) has a compact resolvent, i.e., $$(A - \lambda I)^{-1}$$ is a compact operator for any $$\lambda\notin\mathbb R$$. For these operators it is always true that the closed linear span of its eigenspaces coincides with the whole space.