Im looking at an exercise in my book on functional analysis. I did work out a solution but i believe there is still something missing as I have not used all the given conditions (which is always highly suspicious)
The problem is as follow:
Let H be a separable Hilbert space and T a compact operator with infinite spectrum $\sigma(T) = \{0\} \cup \{\lambda_i \mid i \in \mathbb{N}\}$ Assume $V_{\lambda_i} = E_{\lambda_i}$ for all $i \in \mathbb{N}$ and you can write x as: $$x = y_m +z_m$$ according to the separation $$H = \bigcap^m_{i = 1} F_{\lambda_i} \oplus \bigoplus^m_{i=1}E_{\lambda_i}$$ Construct a sequence $(e_k)_k$ of linear independent eigenvectors of T for which $$\overline{<e_k \mid k \in \mathbb{N}>} = H$$ Under which extra condition is this a complete orthonormal system?
My proof idea is the following:
As H is separable there exists a countable orthonomal basis. The Riesz theorem tell us that the spectral values will be eigenvalues and the (generalized) eigenspace will be finite. Therefore, we can find orthonormal eigenvectors are bases for each eigenspace: $$\{e_1,e_2,...,e_n \mid n \in \mathbb{N} \}$$ We can choose them orthonormal as eigenvectors will be orthogonal for the same eigenvalue, and we can scale them to length 1 and still be an eigenvector.
Finding an orthonormal basis for every eigenspace we can construct a countable sequence of orthonormal eigenvectors: $$e = \{e_{\lambda i} \mid \lambda \in \sigma(T)/\{0\},i \in N\}$$ Claim: The closure of the span of this sequence is the whole Hilbert space H $$H = \overline{<e_{\lambda i}>}$$ Proof Claim:
Pick an $x \in H$. We can write it as $$x = y_m + z_m$$ This x is a limit point to $z_m \in <e_{\lambda i}>$ $$\lim_{m \xrightarrow{} \infty } |y_m +z_m - z_m | = \lim_{m \xrightarrow{} \infty } |y_m| = 0$$ Which proves the claim and therefore also that this is a complete orthonormal system.
Is this proof correct or am I missing something? In my proof there is no extra condition for it to be completeness of the orthonormal system. Why is it given that H is separable? also, what do I do exactly with the eigenvalue 0?
You have orthonormality only inside an eigenspace, but two eigenspaces corresponding to different eigenvalues may fail to be orthogonal, see for example this question Are eigen spaces orthogonal?