In the theory of Pseudo-differential operators,when a symbol $a(x,\xi)\in S^{0}$,then the operator $a(x,D)$ defined by$$a(x,D)u=\int{e^{ix\xi}a(x,\xi)\widehat{u}}d \xi$$ is $L^2$ bounded.$ $ My question is when add what condition (as least as possible) to the symbol $a(x,\xi)$ can assure the operator $a(x,D)$ to be compact on $L^2$ ?Or is there a equivalent condition of this ?I guess some decayed assumption on $a(x,\xi)$ (about $\xi$) is necessary.but I'm not sure.some references about this are also appreciated

Added:There is a equivalent condition of a pseudo-differential operators to be compact.Assume that $g \leq g^{\sigma}$,that g is $\sigma$ temperate.and that m is $\sigma$,g temperate.then the operators $a^{w}(x.D)$with $a\in S(m,g)$are compact (bounded) in $L^{2}$ if and only if $m \to 0 $ at$\infty$ (m is bound).When we let g to be the metric $|dx|^{2}+|d \xi|^{2}/(1+|\xi|^2)$,and $m=(1+|\xi|^{2})^{\frac{\mu}{2}}.$then the class $S(m,g)$ become the usual$S^{\mu}$,Is it implying that when $\mu<0$,then $a^{w}(x.D)$ is compact in $L^{2}$ without further assuming the kernel to be compact supported ?

More precisely, considering the symbol $a(x,\xi)=V(x)(1+|\xi|^2)^{-1}$(it appears in the scattering problem of the schr\ddot{o}dinger operators),now we have the differential condition (rather than the integral condition)$\partial^{\alpha}{V} \leq (1+|x|)^{-\beta-|\alpha|}$,with $|\beta|>\frac{1}{2}$. Then does the operator $a(x,D)$ compact in $L^{2}$ ?

I try to prove it by decomposing both the $\xi-space$ and $x-space$ with a partition of unity as the almost orthogonality method. But it seems without the use of the decay of $x-space$,the compactness wouldn't be obtained(just think about the case $(1-\triangle)^{-1}$.


2 Answers 2


First a simple remark: in the formulation of the question $\mu$ should be replaced by $\mu/2$ to get $S(m,g)=S^\mu_{1,0}$.

Next the "if and only if" is correct but misleading since it is a condition on all the class of operators with symbols in $S(m,g)$. It is of course possible that a given operator is compact without that condition: take for instance the resolvent of the 2D operator $$ -∆_{x,y}+x^2y^2. $$ It is compact in $L^2(\mathbb R^2)$. That example is due to Barry Simon and was extended to a more general setting by Charles Fefferman (Bull. AMS, 1983) in his discussion of Schrödinger equation with polynomial potentials.

  • $\begingroup$ @Bazin:Thanks very much!That mistake has been corrected.your example is very illuminated.I will go for the reference.In general.consider the symbol:$a(x,\xi)=V(x)(1+|\xi|^2)^{\mu}$,where $\partial^{\alpha}{V}\leq (1+|x|)^{m-|\alpha|}$.where $\mu \leq -1$,and m<0.In this case,does the operator $a(x,D)$ be compact,why? $\endgroup$
    – user23078
    May 15, 2012 at 9:15

The following sufficient condition for operators on $\mathbb{R}^n$ is well-known: If the kernel of $a(x,D)$ has compact support and $\mathop{sup}_{x} |a(x,\xi)| \to 0$ as $\xi \to \infty$, then A extends to a compact operator on $L^2$.

A possible reference is M. Shubin, Pseudodifferential operators and spectral theory, Second Edition, Springer.

It follows easily that on a compact manifold a classical pseudodifferential operator is compact if and only if it has negative order.

  • $\begingroup$ thank you very much. I'm considering if the condition that require compact support of the kernel would be replaced by some other conditions of the synbol since it can be checked more conveniently $\endgroup$
    – user23078
    May 15, 2012 at 2:14
  • $\begingroup$ It's easy to check whether a symbol has negative order or not. $\endgroup$
    – Deane Yang
    May 15, 2012 at 5:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.