Non-closed range space of Laplace operators?

Question

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!

Answer 1

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)].

Answer 2

After comprehension of the answer by Jochen Glueck, we give an answer with our background.

Use the following observation:

• $ 0 $ is a spectral value of $ -\Delta $, and further since there exist no point spectrum and residual spectrum of $ -\Delta $, we know $ 0 $ locates in the continuous spectrum of $ -\Delta $.

This yields that \begin{equation*} \mathcal{N}(-\Delta) = \{ 0 \}, \ \ \overline{\mathcal{R}(-\Delta)} = L^2(\mathbb{R}^3) \\ (-\Delta)^{-1} \ \textrm{unbounded}. \end{equation*}

Now assume that $ \mathcal{R}(-\Delta) $ is closed, by

Proposition: If $ A \in \mathcal{C}(X,Y) $, then \begin{equation} A|_{\mathcal{C}(A)} \ \textrm{has a bounded inverse} \Longleftrightarrow \mathcal{R}(A) \ \textrm{closed}, \ \textrm{where} \ \mathcal{C}(A) := \mathcal{D}(A)\cap \mathcal{N}(A)^\perp \end{equation} (See Chapter 9.3, 2 (M) in "A. Israel, T. Greville, Generalized Inverses Theory and Applications. Second Edition, Springer-Verlag, New York, 2003.")

we have $ (-\Delta)^{-1} $ bounded. This is a contradiction.