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Nov 11, 2020 at 0:55 comment added Jochen Glueck @Ivan: No. Every finite dimensional matrix shows that the implication "$\Rightarrow$" is false. To see that the implication "$\Leftarrow$" also fails, just use a direct sum of the operator $-\Delta$ from the question and an injective operator whose spectrum is the closed unit disk.
Nov 10, 2020 at 12:32 comment added Ivan @JochenGlueck Is it in general true that a continuous operator has a closed range if, and only if, 0 is not an eigenvalue and does not lie on the boundary of the spectrum?
Sep 12, 2019 at 10:33 vote accept Yidong Luo
Sep 12, 2019 at 10:18 comment added Jochen Glueck @YidongLuo: I added a reference.
Sep 12, 2019 at 10:15 history edited Jochen Glueck CC BY-SA 4.0
Added a reference to the spectral properties of the Laplace operator.
Sep 12, 2019 at 9:49 comment added Yidong Luo I failed in searching the appropriate materials for the spectral properties shown above. Could you help recommend some materials?
Sep 12, 2019 at 8:22 comment added Jochen Glueck @YidongLuo: The topological boundary of the spectrum is no special concept, but just the boundary of the spectrum with respect to the usual topology in $\mathbb{C}$. I added a reference to the fact that every point in the boundary of the spectrum is an approximate eigenvalue (and I slightly changed the properties of the sequence $(x_n)$ in the above proof as to be consistent with this reference). The spectral properties of the Laplace operator on $\mathbb{R}^d$ are standard and can be probably be found in many books or manuscripts about PDE or matematical physics.
Sep 12, 2019 at 8:15 history edited Jochen Glueck CC BY-SA 4.0
added 388 characters in body
Sep 12, 2019 at 6:51 comment added Yidong Luo This is a really clear answer! However, I still need some appropriate references since I am not familiar with the details of spectrum of $ -\Delta $ and concepts: "topological boundary of spectrum", " approximate eigenvalue".
Sep 12, 2019 at 6:05 history answered Jochen Glueck CC BY-SA 4.0