Timeline for Does these commutator estimates bound in $L^{2}$

Current License: CC BY-SA 3.0

10 events

when toggle format	what		by	license	comment
Apr 28, 2012 at 1:37	vote	accept	user23078
Apr 25, 2012 at 12:21	history	edited	user23078	CC BY-SA 3.0	deleted 16 characters in body
Apr 24, 2012 at 15:52	answer	added	Jeff Schenker		timeline score: 2
Apr 24, 2012 at 15:03	comment	added	user23078		oops...you are right.
Apr 24, 2012 at 14:40	comment	added	Jeff Schenker		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.)
Apr 24, 2012 at 7:50	history	edited	user23078	CC BY-SA 3.0	added 94 characters in body
Apr 24, 2012 at 7:13	comment	added	user23078		note that only $\alpha< -\frac{n}{2}$ can assure boundedness.what about $0>\alpha \geq -\frac{n}{2}$ ?
Apr 24, 2012 at 1:12	comment	added	Jeff Schenker		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 $0<\alpha <2$.
Apr 23, 2012 at 20:16	answer	added	Bazin		timeline score: 1
Apr 23, 2012 at 14:08	history	asked	user23078	CC BY-SA 3.0