4
$\begingroup$

Let $T$ be a densely defined closed unbounded operator on a Hilbert space $H$. A number $\lambda \in C$ is called an approximate eigenvalue of T if there is a sequence ${X_n} \subset D(T)$, with $\|X_n\|=1, n=1,2,... $ s.t. $(T-\lambda I)X_n \rightarrow 0$ as $n \rightarrow \infty$.

Then how to prove that if $T$ is self adjoint then any $\lambda \in \sigma(T)$ is an approximate eigenvalue?

Was thinking about using spectral theorem, first assume there doesn't exist the sequence $X_n$ and try to induce the contradiction.

$\endgroup$
0

1 Answer 1

5
$\begingroup$

If $\lambda$ is not in the approximate spectrum of $T$, then $T-\lambda I$ is bounded below (because if it weren't, then you would be able to find a sequence as in the statement, implying that $\lambda$ is in the approximate spectrum).

What the above shows is that any element in the spectrum of $T$ which is not in the approximate spectrum is in the residual spectrum (i.e. the range of $T-\lambda I$ has nontrivial orthogonal complement). Indeed, if the range of $T-\lambda I$ is dense and $T-\lambda I$ is bounded below, this implies that one can define $(T-\lambda I)^{-1}$ in a dense subspace and extend by continuity. So $T-\lambda I$ would be invertible, contradicting the fact that $\lambda\in\sigma(T)$.

Now looking at this question, we see that if $T$ is selfadjoint it cannot have residual spectrum. So every element in $\sigma(T)$ is an approximate eigenvalue.

$\endgroup$
1
  • $\begingroup$ very nice. Thanks $\endgroup$ Oct 22, 2018 at 22:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .