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Consider $w,g\in L^2(\mathbb{R}^n)$ such that $\lambda w-\Delta w=g$. Show that for all $\lambda>0, \lambda\in \rho(\Delta)$. That is, $\sigma(\Delta)=(-\infty,0]$. Where $\sigma(\Delta)$ is the spectrum of $\Delta$.

In my answer, I have used Fourier transform by as in the following:

$\hat{\lambda w}-\hat{\Delta w}=\hat{g}$ which led me to $\hat{w}=\hat{g}/(\lambda+\left|\xi\right|^2)$. Thus for $g \in L^2$, $(\lambda I - \Delta)^{-1} g = \left((\lambda +|\xi|^2)^{-1} \hat{g}\right)^\vee$

I followed the solution on this link:

[What is spectrum for Laplacian in $\mathbb{R}^n$?

From which I was able to prove that $\sigma(\Delta)\subseteq(-\infty,0]$.

However, I have found it hard to elementarily show that $(-\infty,0]\subseteq \sigma(\Delta)$, in order to establish the equality. Even the proof on the link is complicated.

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The resolvent of $\Delta$ is $(\Delta-\lambda I)^{-1}$, which is defined at least for $\lambda\notin\mathbb{R}$, and can be found with the Fourier transform to be $$ (\Delta-\lambda I)^{-1}f = \frac{1}{(2\pi)^{n/2}}\int\frac{1}{-|\xi|^2-\lambda}\hat{f}(\xi)d\xi,\;\;\; \lambda\in\mathbb{C}\setminus\mathbb{R}. $$ From this you can also see that resolvent set includes $\mathbb{C}\setminus(-\infty,0]$. So $\sigma(\Delta)\subseteq(-\infty,0]$.

To show that $(-\infty,0]\subseteq\sigma(\Delta)$, let $r > 0$. For any $0 < \epsilon < r$, let $f_{r,\epsilon}$ be defined so that $$ \hat{f_{r,\epsilon}}(\xi) = \chi_{[r-\epsilon,r+\epsilon]}(|\xi|). $$ Then $\|f_{r,\epsilon}\|=\|\hat{f_{r,\epsilon}}\|=\|\chi_{[r-\epsilon,r+\epsilon]}\|$, and $$ \|(\Delta+rI)f_{r,\epsilon}\|=\|(|\xi|-r)\chi_{[r-\epsilon,r+\epsilon]}(|\xi|)\| \le \epsilon\|\chi_{[r-\epsilon,r+\epsilon]}\|=\epsilon\|f_{r,\epsilon}\|. $$ So $(\Delta +rI)$ cannot have a bounded inverse for any $r > 0$. So $(-\infty,0]=\sigma(\Delta)$.

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  • $\begingroup$ @Sulyman : I'm curious why you unchecked my answer. $\endgroup$ – DisintegratingByParts Jul 23 '18 at 23:27

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