I would like to know if there is a proof of the following statement that only requires "elementary means" (i.e. not the spectral theorem for Hilbert spaces):
"Let $H$ be an Hilbert Space and let $K$ be an $\hat X$-invariant closed subspace, where $\hat X$ is a self-adjoint and everywhere-defined operator on $H$. If $\hat X$ admits an orthonormal basis of eigenvectors in $H$ then $\hat X$ also admits an orthonormal basis of eigenvectors in $K$."
The reason behind my question is that I am trying to prove a statement about compatible observables in quantum mechanics and this is the last Lemma I would need.
Let $P_K$ be the orthgonal projection on $K$. The assumption $\hat{X}(K)\subseteq K$ implies $P_K\hat{X}P_K=\hat{X}P_K$. So $$P_K\hat{X}={P_K}^*{\hat{X}}^*=(\hat{X}P_K)^*=(P_K\hat{X}P_K)^*={P_K}^*{\hat{X}}^*{P_K}^*=P_K\hat{X}P_K$$ hence $P_K\hat{X}=\hat{X}P_K$.
Let $\{\lambda_i\}_{i\in I}$ be the distinct eigenvalues of $\hat{X}$, and $P_i$ the orthogonal projection on the eigenspace of $\lambda_i$. Then $P_i P_j=0$ for $i\ne j$.
If $P_i x=x$ then $\hat{X}P_K x=P_K\hat{X} x=P_K\lambda_i x=\lambda_i P_K x$. Hence $P_K(\operatorname{range}(P_i))\subseteq \operatorname{range}(P_i)$. By an argument similar to above, $P_K P_i = P_i P_K$ for all $i\in I$. Moreover if $i\ne j$ then $(P_K P_i)(P_K P_j)=P_K P_i P_j = 0$.
Finally, $$P_K=P_K \mathbf{1}_H=P_K\left(\sum_{i\in I}P_i\right)=\sum_{i\in I}P_K P_i$$ where the sum is in the strong operator topology.