Is the limit of approximate eigenvectors an eigenvector? Suppose $\mathcal{H}$ is a Hilbert space $H:\mathcal{H}\to\mathcal{H}$ is a self-adjoint operator. If $\lambda\in\sigma(H)$, then we have a sequence of vectors $\{\phi_n\}\subset\mathcal{H}$ with unit norm, $\|\phi_n\|=1$, such that
$$\lim_{n\to\infty}\|(H-\lambda)\phi_n\| = 0\, .$$
If we suppose that $\phi_n$ converges to $\phi\in\mathcal{H}$ (i.e. $\phi_n\to\phi$) will $\phi$ be an eigenvector of $H$?
This is true if $H$ is bounded as
$$\begin{align}\|(H-\lambda)\phi \| &\leq \|(H-\lambda)(\phi-\phi_n)\| +\| (H-\lambda)\phi_n\|\\
&\leq \|H-\lambda\|\|\phi-\phi_n\| + \|(H-\lambda)\phi_n\|\to 0 
\end{align}\, ,$$
by the assumption that $H$ (and hence $H-\lambda$) is bounded and that $\phi_n$ is a sequence of approximate eigenvectors.

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*If it is not true for unbounded $H$, can someone provide an example?

*Are there extra criteria one can impose on unbounded $H$ to make the statement true?

 A: If $H$ is self-adjoint, then $\sigma(H) \subseteq \Bbb{R}$, so we have $\lambda \in \Bbb{R}$, which means $H-\lambda I$ is also self-adjoint. (You can prove this by first observing that $H-\lambda I$ is symmetric, then check the domains of $H^*=H$ and $(H-\lambda I)^*$.)
Thus, $H-\lambda I$ is closed. As a consequence, $(\phi_n \rightarrow \phi) \; \wedge \; ((H-\lambda I)\phi_n \rightarrow 0) \implies (H-\lambda I)\phi = 0$, which means $\phi$ is an eigenvector of $H$ w.r.t. $\lambda$, as long as $\phi \ne 0$.
In fact, the sum of a closed and a continuous linear operator (denote by $A,T$ respectively) is always closed: we have both $$\lVert x \rVert + \lVert Ax \rVert = \lVert x \rVert + \lVert (A+T)x-Tx \rVert \le \lVert x \rVert + \lVert (A+T)x \rVert + \lVert Tx \rVert \le (1 + \lVert T \rVert)(\lVert x \rVert + \lVert (A+T)x \rVert)$$ and $$\lVert x \rVert + \lVert (A+T)x \rVert=\lVert x \rVert + \lVert Ax+Tx \rVert \le \lVert x \rVert + \lVert Ax \rVert + \lVert Tx \rVert \le (1 + \lVert T \rVert)(\lVert x \rVert + \lVert Ax \rVert)$$ which means the graph norms w.r.t $A$ and $A+T$ are equivalent on $D(A)=D(A+T) \subseteq H$, so their closedness are also equivalent.
