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According to the basic rules of symbolic caculus,$[a(x,D),x_{j}]=-ia^{j}[x,D]$.So we have $[(1-\triangle)^{\frac{1}{2}},x_i]=\partial_i(1-\triangle)^{-\frac{1}{2}}$ which is $L^2$ bounded. It's also true that the $[(1-\triangle)^{\frac{1}{2}},\langle x \rangle]$ is $L^2$ bounded.Can we write the explicit expression of this commutator? More generally,how to show that $[(1-\triangle)^{\frac{\alpha}{2}},\langle x \rangle^{\alpha}]$ ($\alpha \leq1$)is $L^2$ bounded? it is obviously true when $\alpha<0$ ?

what if we use $(-\triangle)^{\alpha}$ instead which the symbol of it is not smooth at 0?

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  • $\begingroup$ I assume that $\langle x\rangle =\sqrt{1+x^2}$. If so, here are two simple observations: 1) For $\alpha \le 0$ the operators $(1-\Delta)^{\frac{\alpha}{2}}$ and $\langle x\rangle^\alpha$ are each individually bounded on $L^2$ and so their commutator is bonded, with norm no larger than $2$. 2) For even positive $\alpha$, the commutator is an unbounded partial differential operator. For example $$ [(1-\Delta),\langle x \rangle^2]= -4 [x\cdot \nabla +\nabla\cdot x].$$ Thus seems likely that the commutator is unbounded for $\alpha \ge 2$. Not sure about $0<\alpha <2$. $\endgroup$
    – Jeff Schenker
    Apr 24, 2012 at 1:12
  • $\begingroup$ note that only $\alpha< -\frac{n}{2}$ can assure boundedness.what about $0>\alpha \geq -\frac{n}{2}$ ? $\endgroup$
    – user23078
    Apr 24, 2012 at 7:13
  • $\begingroup$ No. The multiplication operator $f \mapsto \langle x \rangle^\alpha f$ is a bounded operator on $L^2$ with norm one once $\alpha \le 0$. Indeed, $$ \int |\langle x \rangle^\alpha f(x)|^2 dx \le \int |f(x)|^2 dx$$ since the function $\langle x \rangle^\alpha \le 1$. (The criterion $\alpha < -\frac{n}{2}$ indicates when the functions $\langle x \rangle^\alpha$ is itself in $L^2$, but that is not relevant to deciding if the multiplication operator is bounded.) $\endgroup$
    – Jeff Schenker
    Apr 24, 2012 at 14:40
  • $\begingroup$ oops...you are right. $\endgroup$
    – user23078
    Apr 24, 2012 at 15:03

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Here is a technique that shows the commutator is bounded for $\alpha <1$. First note, as I observed in my comment, it is a bounded operator for $\alpha \le 0$. Now note that if $0<\eta<1$ and $A$ is a strictly positive operator then we have the identity $$ A^\eta = c_\eta \int_0^\infty t^\eta \left (\frac{1}{t}- \frac{1}{t+A} \right ) dt $$ where $\frac{1}{c_\eta} =\int_0^\infty t^\eta \left [\frac{1}{t}- \frac{1}{t+1} \right ]dt$ and the integral is to be understood in the strong sense (apply both sides to a vector in a dense core for the domain of $A^\eta$.) It follows that if $A$ and $B$ are two such operators then $$ [A^\eta,B^\eta]=c_\eta^2 \int_0^\infty\int_0^\infty t^\eta s^\eta \left [ \frac{1}{t+A},\frac{1}{s+B} \right]ds dt. $$ The commutator of resolvents can be computed, for $A=1-\Delta$ and $B=1+x^2$ to give $$\left [\frac{1}{t+1-\Delta},\frac{1}{s+1+x^2} \right ] = -\frac{1}{t+1 -\Delta} \left [ 1-\Delta,\frac{1}{s+1 +x^2} \right ] \frac{1}{t+1 -\Delta} $$ $$= - 2 \frac{1}{t+1-\Delta} \left ( \frac{x}{(s+1+x^2)^2}\cdot \nabla+\nabla \cdot\frac{x}{(s+1+x^2)^2} \right ) \frac{1}{t+1-\Delta}.$$ The operator norm of the result is bounded by $$ 4 \frac{1}{(t+1)^{\frac{3}{2}}} \frac{1}{(s+1)^{\frac{3}{2}}}.$$ (To see this note that $\sup_x |x|/\sqrt{s+1 +x^2}= 1$. A similar computation on the Fourier side gives $\|\nabla/\sqrt{t+1-\Delta}\|=1$.) Plugging this into the integral representation gives $$\left \| \left [ (1-\Delta)^\frac{\alpha}{2}, (1+x^2)^\frac{\alpha}{2} \right ] \right \| \le \left (2 c_\eta \int_0^\infty \frac{t^\frac{\alpha}{2}}{(t+1)^{3/2}} dt \right )^2$$ which is finite if $0 <\alpha <1$.

Clearly this argument misses something since it doesn't give the boundary case $\alpha=1$, however I feel that a modification of this argument will show the commutator to be unbounded once $1<\alpha <2$ but I don't see the details. For $\alpha \ge 2$ I believe the commutator is unbounded and one can certainly show this, as I mentioned in my comment above, if $\alpha$ is an even natural number since the result is a partial differential operator.

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Take a pseudodifferential operator with a symbol $p\in S^1_{1,0}$ and $f$ a Lipschitz-continuous function. Then the commutator $$ [p(x,D),f] $$ is bounded on $L^2$. When $f$ is $C^\infty$ with bounded derivatives, the principal symbol of this commutator is the Poisson bracket -i{$p,f$}, i.e. in that case $$ c_0(x,\xi)=\frac{1}{i}\sum_{1\le j\le n}\frac{\partial p}{\partial \xi_j}\cdot \frac{\partial f}{\partial x_j},\quad c_0\in S^0_{1,0}. $$ In that case, $ [p(x,D),f]=c_0(x,D)+ r(x,D),\quad r\in S^{-1}_{1,0}. $

Bazin.

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