# Non-closed range space of Laplace operators?

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!

## 2 Answers

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)].

• 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". – Yidong Luo Sep 12 '19 at 6:51
• @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. – Jochen Glueck Sep 12 '19 at 8:22
• I failed in searching the appropriate materials for the spectral properties shown above. Could you help recommend some materials? – Yidong Luo Sep 12 '19 at 9:49
• @YidongLuo: I added a reference. – Jochen Glueck Sep 12 '19 at 10:18
• @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? – Ivan Nov 10 '20 at 12:32

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.