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How can I prove that the spectrum of the Laplace operator $$\Delta: H^2(\mathbb{R}^N)\subset L^2(\mathbb{R}^N)\rightarrow L^2(\mathbb{R}^N)$$ is $\sigma(\Delta)=[-\infty,0]$?

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2 Answers 2

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We determine the resolvent set of $(\Delta, H^2(\mathbb{R}^n))$. By definition, the complex number $\lambda$ belongs to the resolvent set if and only if the following equation (named resolvent equation) $$\tag{1}\Delta f -\lambda f = g$$ has a unique solution $f\in H^2$ for any fixed $g\in L^2$. The spectrum is the complementary of the resolvent set and its elements are called spectral values.

[EDIT: Summary. Here's two different approaches. The first uses the Fourier transform. The second computes explicitly a set of (generalized) eigenfunctions.]


Method 1. Taking the Fourier transform termwise we see that (1) is equivalent to $$(-\lvert\xi\rvert^2 - \lambda)\hat{f}=\hat{g}, $$ which has the unique formal solution $$\tag{2}\hat{f}=-\frac{\hat{g}}{\lambda+ \lvert\xi\rvert^2}.$$ Equation (2) determines a solution $f\in H^2$ of the resolvent equation (1) if and only if $f\in H^2$, that is $$\tag{3}\begin{array}{ccc} \lambda \text{ is in the resolvent set }&\iff& (1+\lvert\xi\rvert^2 )\hat{f} = -\frac{1+\lvert\xi\rvert^2}{\lambda+ \lvert\xi\rvert^2}\hat{g}\in L^2.\end{array}$$ When $\lambda\notin (-\infty, 0]$ the function $\frac{1+\lvert\xi\rvert^2}{\lambda + \lvert\xi\rvert^2}$ is bounded, so condition (3) is satisfied. Therefore $\lambda$ belongs to the resolvent set.

On the other hand, when $\lambda\in (-\infty, 0]$, the function $\frac{1+\lvert\xi\rvert^2}{\lambda + \lvert\xi\rvert^2}$ has a singularity at $\lvert \xi \rvert= -\sqrt{\lambda}$. If $\lambda \ne 0$ then we write $$ \frac{1+\lvert \xi\rvert^2}{\lambda +\lvert \xi\rvert^2}=\frac{1+\lvert \xi\rvert^2}{\sqrt{-\lambda}+\lvert \xi\rvert}\frac{1}{\sqrt{-\lambda}-\lvert \xi\rvert}.$$ We take a function $\hat{g}$ that is equal to $\lvert\sqrt{-\lambda}-\lvert\xi\rvert^2\rvert^{-\frac{N-1}{2}}$ in a neighborhood of the annulus $\lvert\xi\rvert=\sqrt{-\lambda}$ and vanishes elsewhere. This function is in $L^2$ but its product against $\frac{1+\lvert\xi\rvert^2}{\lambda +\lvert\xi\rvert^2}$ is not, because of the strong singularity around $\lvert\xi\rvert=\sqrt{-\lambda}$. Therefore condition (3) is not satisfied and $\lambda$ is not in the resolvent. The proof that $\lambda=0$ is not in the resolvent is completely analogous. (Alternatively, one can recall that the spectrum is always a closed set, so if the open half-line $(-\infty, 0)$ is contained in the spectrum, the value $\{0\}$ must be in the spectrum as well).


Method 2 (Added in response to Neal's comment below). Here's an alternative computation of the spectrum that determines a system of approximate eigenfunctions (in Weyl's sense).

Definition. Let $(A, D(A))$ be a closed operator on a Hilbert space $H$. We say that $f_k\in D(A)$ is a sequence of approximate eigenfunctions of $A$ relative to $\lambda\in \mathbb{C}$ if the following two conditions are satisfied:

  1. The sequence is normalized, meaning that $\lVert f_k\rVert =1$ for all indices $k$.
  2. It holds that: $\lVert Af_k-\lambda f_k\rVert \to 0$ as $k\to \infty$.

    If $\lambda$ admits a sequence of approximate eigenfunctions then it is a spectral value. The converse holds if $\lambda$ is on the boundary of the spectrum. (For more information see Teschl, Mathematical methods in quantum mechanics, Lemma 2.16 pag.76).

In the case at hand the eigenfunction equation $$ \Delta f-\lambda f=0 $$

is formally solved by the plane wave profiles $e^{i\sqrt{-\lambda}\xi_0\cdot x}$, where $\xi_0$ is a unit vector. However, those profiles have not a finite $L^2$ norm. To circumvent this difficulty we fix a smooth bump function $\Phi$, normalized so that $\lVert \Phi\rVert_2=1$, and we define \begin{align*} f^0_k(x)&=\frac{1}{k^{N/2}}\Phi\left(\frac{x}{k}\right); \\ f^{\lambda}_k(x)&=e^{i\sqrt{-\lambda}\xi_0\cdot x} f^0_k(x). \end{align*} Here $\xi_0\in \mathbb{S}^{N-1}$ is arbitrarily chosen.

Claim. For any fixed $\lambda\in (-\infty, 0]$ the sequence $f^\lambda_k$ consists of approximate eigenfunctions of $(\Delta, H^2)$ relative to the spectral value $\lambda$.

Proof. We consider the case $\lambda=0$ first. It is obvious that the sequence $f^0_k$ belongs to $H^2$ and that it is normalized. We have \begin{equation*} \lVert \Delta f^0_k\rVert_2=\frac{1}{k^2}\left\lVert \frac{1}{k^{N/2}} \Delta\Phi\left(\frac{x}{k}\right) \right\rVert_2 =\frac{C}{k^2}\to 0. \end{equation*} So $f^0_k$ is a sequence of approximate eigenfunctions. We note also that \begin{equation*} \left\lVert \nabla f^0_k\right\rVert_2=\frac{1}{k}\left\lVert \frac{1}{k^{N/2}} \nabla\Phi\left(\frac{x}{k}\right) \right\rVert_2\to 0; \end{equation*} this will be useful later. Now we consider the sequence $f^\lambda_k$ for a $\lambda\in (-\infty, 0)$. It is again obvious that $f^\lambda_k$ is normalized and that it belongs to $H^2$. Using the Leibniz formula for the Laplacian (that is, $\Delta(fg)=\Delta f g + 2 \nabla f\cdot\nabla g + f \Delta g$), we compute \begin{equation*} \begin{split} \lVert \Delta f^\lambda_k -\lambda f^\lambda_k \rVert_2 &=\left\lVert 2\nabla \left( e^{i\sqrt{-\lambda} \xi_0\cdot x}\right)\cdot \nabla f^0_k + e^{i\sqrt{-\lambda}\xi_0\cdot x} \Delta f^0_k \right\rVert_2 \\ &\le2\sqrt{-\lambda}\lVert \xi_0\cdot \nabla f^0_k\rVert_2 + \lVert \Delta f^0_k\rVert_2 \\ &\le 2\sqrt{-\lambda} \lVert \nabla f^0_k\rVert_2 +\lVert \Delta f^0_k\rVert_2 \to 0. \end{split} \end{equation*} This proves our claim.

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  • $\begingroup$ Do you know of any constructive proofs, e.g., constructing a sequence of $\psi_k\in H^2(\mathbb{R}^n)$ satisfying Weyl's criterion? $\endgroup$
    – Neal
    May 11, 2014 at 17:32
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    $\begingroup$ Actually, the "eigenfunctions" of $\Delta$ are the plane wave profiles $f_\lambda(x)=e^{i\sqrt{-\lambda}\xi_0\cdot x}$, where $\xi_0\in\mathbb{R}^n$ is a unit vector. At Fourier side those correspond roughly to Dirac deltas concentrated at the point $\sqrt{-\lambda} \xi_0$. So a sequence $f_k\in H^2$ such that $\hat{f}_k$ is a bump function that concentrates more and more at $\sqrt{-\lambda} \xi_0$ should do the trick here. I have not checked the details though $\endgroup$ May 11, 2014 at 17:58
  • $\begingroup$ Thank you! That makes perfect sense. $\endgroup$
    – Neal
    May 11, 2014 at 20:39
  • $\begingroup$ I am adding a commentary to this answer to account for your observation. $\endgroup$ May 11, 2014 at 20:45
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    $\begingroup$ Quite clear! Thank you for the alternative approach. It looks like the Fourier transform of your argument should be what you sketched in your comment. $\endgroup$
    – Neal
    May 11, 2014 at 23:52
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You can also prove that the operator is sectorial. Recall that an operator is sectorial if

there exists $a\in \mathbb{R}$, $\phi\in (0, \frac{\pi}{2})$ and $M>0$ such that the spectrum contain the sector $S=\{ \lambda\in \mathbb{C}; \phi\leq |arg(\lambda-a)|\leq \pi; \pi\neq a \}$ and $|(\lambda I-A)^{-1}|\leq \frac{M}{|\lambda-a|}$, for each $\lambda\in S$.

You can see that, if the operator is sectorial, then the spectrum contain $(-\infty, a)$. You can find in Dlotko's book that, if an operator is symmetric and bounded below, then its sectorial and you win that a must be positive by this result.

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