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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,I Believe generally,how to show that $[(1-\triangle)^{\frac{\alpha}{2}},\langle x \rangle^{\alpha}]$ ($\alpha>0$)is $\alpha \leq1$)is $L^2$ bounded.and how to prove them?Does bounded? it maintain 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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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,I Believe that $[(1-\triangle)^{\frac{\alpha}{2}},\langle x \rangle^{\alpha}]$ ($\alpha>0$)is $L^2$ bounded.and how to prove them?Does it maintain 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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# Does these commutator estimates bound in $L^{2}$

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,I Believe that $[(1-\triangle)^{\frac{\alpha}{2}},\langle x \rangle^{\alpha}]$ ($\alpha>0$)is $L^2$ bounded.and how to prove them?Does it maintain true when $\alpha<0$