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Is the fractional Laplacian on $\mathbb{R}$ of trace class or not?

I don't know the basis of $\mathrm{L}^{2}(\mathbb{R})$ for applying the definition of a trace class operator. Thank you very much.

John Gowers
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rachid
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  • Actually there are many ONBs for $L^2(\mathbb{R})$. However, do you know the eigenvalues of the Laplacian? Aren't the eigenvalues of the fractional Laplacian (not sure if this is uniquely defined) just powers of the eigenvalues? – Dirk Oct 01 '13 at 12:09
  • Where the domain is unbounded I think that some orthogonal polynoms like these of Hermite are ONB of $L^{2}(\mathbb{R})$ but I don't know their eigenvalues. – rachid Oct 01 '13 at 13:03

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If it is, then so is $\Phi_s:=\mathcal{F}^{-1}\Delta^s\mathcal{F}$, where $\mathcal{F}$ is the Fourier transform. This operator is multiplication by $\lvert\xi\rvert^{2s}$, and that is not even bounded (unless $s=0$). For $s>0$, consider $u_n=\chi_{[n,n+1]}$: Then $\lVert u_n\rVert_2=1$, but $$\lVert \Phi_s u_n\rVert_2^2=\int_n^{n+1} \xi^{4s}\,d\xi\to\infty\qquad(n\to\infty),$$ showing that $\Phi_s$ is unbounded, and therefore not a trace class operator.

Harald Hanche-Olsen
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  • Unfortunatly no, please if you can explain to me more, thank you very much. – rachid Oct 01 '13 at 12:17
  • I added quite a bit of explanation. – Harald Hanche-Olsen Oct 01 '13 at 12:26
  • Thank you very much Professor, please I wish to know if there exists a $\delta\in (0,1)$ such that this $A^{\delta}$ is trace class. Excuse me if my question is trivial. – rachid Oct 01 '13 at 12:52
  • where $َA$ is the fractional laplacian. – rachid Oct 01 '13 at 13:18
  • Well, that is a somewhat different question, but the answer is still no. First, my $\Delta$ is the ordinary Laplacian, and $\Delta^s$ is then a fractional Laplacian. So with $A=\Delta^s$ then $A^\delta=\Delta^{s\delta}$, which is still a fractional Laplacian. So just replacing $s$ by $s\delta$ in my argument shows that $A^\delta$ is not bounded, hence not trace class. – Harald Hanche-Olsen Oct 01 '13 at 14:26
  • Thank you Professor, in fact I would know what is the case for $A^{\delta}$ when $\delta\in(-1,0)$ instead of $(0,1)$. Excuse me Professor. –  Oct 01 '13 at 19:48
  • To get the same reslt for negative $\delta$, just replace $u_n$ by $\sqrt{n(n+1)}\chi_{[1/(n+1),1/n]}$ (or something like that). – Harald Hanche-Olsen Oct 02 '13 at 06:05
  • For $u_{n}=\sqrt{n(n+1)}$, we have $|\mathcal{F}\Delta^{\delta s}\mathcal{F}^{-1}u_{n}|_{2}^{2} = \frac{1}{1+4\delta s}\left( \frac{1}{(n+1)^{1+4\delta s}} - \frac{1}{(n)^{1+4\delta s}}\right)$, and we can choose $-1/2 < \delta < -1/8$ and $1 < s \leq 2$, and in this case this norm doesn't tend to infinity. In fact that the result will be true and I wish know how I can confirm the result. Thank you very much for your help. – rachid Oct 02 '13 at 12:15
  • In fact that "I hope" that the result will be true – rachid Oct 02 '13 at 12:17