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.

## Best Answer

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.