Set $ -\Delta: H^2(\mathbb{R}^3) \subseteq L^2(\mathbb{R}^3) \to L^2(\mathbb{R}^3) $. Then $ \mathcal{R}(-\Delta) $ is non-closed?


Sorry if this question is trivial. I am not familiar with theory of linear partial differential operators. This is a result I may use as a "black box".


Thank you in advance!



Non-closed range space of Laplace operators?
The range of the Laplace operator $-\Delta: L^2(\mathbb{R^d}) \supseteq H^2(\mathbb{R^d}) \to L^2(\mathbb{R^d})$ is **not** closed (for any dimension $d \ge 1$).


To see this, one can for instance use the following observations:


* $-\Delta$ has empty point spectrum, so $0$ is not an eigenvalue of $-\Delta$.
* $0$ is a spectral value of $-\Delta$, and the spectrum of $-\Delta$ has empty interior (within the complex plane) since $-\Delta$ is self-adjoint and its spectrum is thus real.


Those (and more) spectral properties of $-\Delta$ are very well-known in PDE theory and in Mathematical Physics; see for instance Theorem 7.17 in "G. Teschl: Mathematical Methods in Quantum Mechanics - With Applications to Schrödinger Operators (2014)".


Now the claim follows from the following general result:


**Proposition.** Let $A: E \supseteq D(A) \to E$ be a closed linear operator on a (complex) Banach space $E$. Assume that a given number $\lambda \in \mathbb{C}$ is not an eigenvalue of $A$, but contained in the topological boundary of the spectrum of $A$. Then $\lambda - A$ has non-closed range.


*Proof.* The operator $\lambda - A$ is an injective and continuous linear operator from $D(A)$ to $E$ (where $D(A)$ is endowed with the graph norm $\|\cdot\|\_{D(A)}$). Hence, if $\lambda - A$ had closed range, it would be a linear homeomorphism from the Banach space $D(A)$ to the range of $\lambda - A$. In particular, $\lambda - A$ would be *bounded below* in the sense that there exists a constant $c > 0$ such that $\|(\lambda - A)x\|\_E \ge c \|x\|\_{D(A)}$ for all $x \in D(A)$.


However, as $\lambda$ is a value in the topological boundary of the spectrum of $A$, it follows that $\lambda$ is an *approximate eigenvalue* of $A$, meaning that there exists a sequence $(x\_n) \subseteq D(A)$, normalized in $E$, such that $(\lambda - A)x\_n \to 0$ in $E$. Note that $\|x\_n\|\_{D(A)} \ge \|x\_n\|\_E = 1$ for all $n$, so we obtain a contradiction to the fact that $\lambda - A$ is bounded below.


**Remark.** The fact that every $\lambda$ in the boundary of the spectrum $\sigma(A)$ is an approximate eigenvalue of $A$ is a simple consequence of standard properties of the resolvent of $A$; see for instance Lemma IV.1.9 in [Engel, Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)].


