As a non functional analyst, I stumbled over the following question:


Given a self-adjoint Operator $T:D(T) \subset H \rightarrow H.$ Assume we know that $T$ has some eigenvalue $\lambda$ which is isolated in the spectrum of $T$. The eigenvalue may be non-degenerate.


We do have an approximating sequence of eigenfunctions for $(T-\lambda),$ i.e. $(T-\lambda)x\_n \rightarrow 0$ for $x\_n \in D(T)$ normalized.


Now consider a closed restriction $S \subset T.$ Assuming $x\_n \in D(S)$ and $(S-\lambda)x\_n \rightarrow 0$ as well. 


Does this imply that there is an eigenfunction to $T$ with eigenvalue $\lambda$ that is also in $D(S)$?



Spectrum of self-adjoint operator
Since $\lambda$ is an isolated point in the spectrum of $T$, the distance from $x\_n$ to the eigenspace corresponding to $\lambda$ tends to zero. Since we may assume that this eigenvalue is non-degenerate, we may force $(x\_n)$ to converge to some eigenvector $x$, multiplying by appropriate scalars if necessary.


Now we have that $\lim x\_n = x$ and $ \lim (S-\lambda)x\_n=0$, i.e. $\lim Sx\_n=\lambda x$. Since $S$ is closed, we conclude that $x \in D(S)$ and $Sx=\lambda x$, exactly as required.


