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Assume there is no eigenvalues of $T$. Let $\lambda$ is an approximate eigenvalue but not eigenvalue of a self-adjoint operator $T$. That means there exist sequences $\{x_{n}\}$ unit vectors in $\mathcal{H}$ such that $\|(T-\lambda I)x_{n}\| \to 0$. Can there always exist $y_{n}$ orthogonal to $x_{n}$ i.e $\langle y_{n},x_{n}\rangle=0$., such that $\|(T-\lambda I)y_{n}\| \to 0$, with $\|y_{n}\|=1$.

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    $\begingroup$ If $\lambda\neq 0$ then these will imply $|\lambda|\|x_{n}-y_{n}\| \to 0$, that says $\lambda\sqrt{2} \to 0$, a contradiction. $\endgroup$ – mathlover Feb 11 at 10:07
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You already noted that it is not possible if $\lambda \ne 0$. I believe the answer is yes in the case $\lambda = 0$.

Assume $Tx_n \to 0$ where $\|x_n\| = 1$.

Notice that $|\langle x_{n+1}, x_n\rangle| \le 1$ and it is equal to $1$ if and only if there exists $q_n \in \mathbb{C}$, $|q_n| = 1$ such that $x_{n+1} = q_nx_n$.

Let $y_1$ be any vector such that $y_1 \perp x_1$ and $\|y_1\| = 1$. Define inductively $$y_n = \begin{cases} \frac{x_{n+1}-\langle x_n, x_{n+1}\rangle x_n}{\sqrt{2\left(1-|\langle x_n, x_{n+1}\rangle|^2\right)}} &\text{if } |\langle x_{n+1}, x_n\rangle| < 1\\ y_{n-1} &\text{if } |\langle x_{n+1}, x_n\rangle| = 1 \end{cases}$$

Then $y_n \perp x_n$, $\|y_n\| = 1$, and $Ty_n \to 0$ since it is not possible that $y_n = y_1$ for all $n \in \mathbb{N}$. In that case we would have $q_n(Tx_1) = Tx_n \to 0$ so $x_1 = 0$ since $|q_n| = 1$. But this is a contradiction with $\|x_1\| = 1$.

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The question was modified after I posted my answer. The requirement that $T$ has no eigen values has been added.

On $\ell^{2}$ let $Te_n=\frac 1 n e_n$ where $(e_n)$ is the usual basis. Then $(T-0I)e_n \to 0$ and $(T-0I)e_{n+1} \to 0$; of course $\langle e_n, e_{n+1} \rangle =0$ for all $n$. Also, $0$ is not an eigen value and $T$ is self adjoint.

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  • $\begingroup$ But if $\sigma_{eig}(T)=\emptyset$ then?? $\endgroup$ – mathlover Feb 11 at 9:17
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    $\begingroup$ If $T(a_n)=0 a_n$ then $\frac 1 n a_n =0$ for all $n$, This implies $a_n=0$ for all $n$. Hence $0$ is not an eigen value. $\endgroup$ – Kavi Rama Murthy Feb 11 at 9:21
  • $\begingroup$ @mathlover I have modified my answer. $\endgroup$ – Kavi Rama Murthy Feb 11 at 9:30
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    $\begingroup$ Yes, ok I see. Does it always exist? $\endgroup$ – mathlover Feb 11 at 9:32
  • $\begingroup$ @mathlover If you want t make major changes to a question you should ask a new question. If that is not possible, you should at least display the edited part of the question. $\endgroup$ – Kavi Rama Murthy Feb 11 at 9:48

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