Commutator of pseudodifferential operator and multiplication operator

Question

Let $\eta:\mathbb R^n\to[0,1]$ be a smooth and compactly supported function and assume $f:\mathbb R^n \to\mathbb R$ is measurable. I want to bound the commutator of the operators $\eta(-i\nabla)$, which is a well-defined bounded and selfadjoint operator by the functional calculus, and $M_f$, which is the multiplication operator given by $f$, acting on $L^2(\mathbb R^n)$.

If $f$ is Lipschitz-continuous, I can prove the commutator is bounded by using properties of the Fourier transform. However, I would like to use weaker assumptions (for example locally Hölder continuous) on $f$, as my assumption on $\eta$ is rather strong. References, proofs and counterexamples are all warmly welcome.

Edit: I should stress that I especially do not have any integrability or boundedness conditions on $f$. Examples, I am thinking of, are usually diverging at infinity, such as $f(x)=|x|$ (which is an example of the Lipschitz case) or, say, $f(x)=x^2$.

Answer 1

So, as a summary:

• it is sufficient to have functions in $L^2 + C^{0,1}$, so at most locally $L^2$ and asymptotically growing linearly.
• If the growth is bigger, then one can find counterexamples (unless perhaps one adds some special conditions on $\eta$).

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$\bullet$ Local regularity conditions. Let me write $\|\cdot\|_\infty$ the operator norm and $\|\cdot\|_2$ the Hilbert-Schmidt norm, and $L^p = L^p(\mathbb R^{n})$.

Let $A = [M_f,\eta(-i\nabla)] = M_f\,\eta(-i\nabla) -\eta(-i\nabla)\, M_f$.

A sufficient condition is to have $f∈ L^2$. In both these cases, even $f(x)\,\eta(-i\nabla)$ and $f(x)\,\eta(-i\nabla)$ are bounded operators, and so one can bound the commutator by the triangle inequality, leading to $$ \|A\|_∞ \leq 2\,\|M_f\,\eta(-i\nabla)\|_∞ $$ where I used the fact that $\eta(-i\nabla)\, M_f = (M_f\,\eta(-i\nabla))^*$ and $M_f\,\eta(-i\nabla)$ have the same operator norm. Let me now write how to get that the right-hand side term is bounded. These are actually special cases of the sometimes called Kato-Seiler-Simon inequality (see e.g. B. Simon, Trace Ideals and Their Applications, Chapter 4). In the case when $f∈ L^2$, then one can compute the Hilbert-Schmidt norm of the operator by taking the $L^2$ norm of its integral kernel $(2\pi)^{-n}\,f(x)\,\widehat{\eta}(\tfrac{y-x}{2\pi})$, and so $$ \|M_f\,\eta(-i\nabla)\|_{\infty} ≤ \|M_f\,\eta(-i\nabla)\|_{2} = (2\pi)^{-n}\,\|f(x)\,\widehat{\eta}(\tfrac{y-x}{2\pi})\|_{L^2(\mathbb R^{2n})} = (2\pi)^{-n/2}\,\|f\|_{L^2}\,\|\eta\|_{L^2} $$

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$\bullet$ New part after edit: growth condition. Let me just write here some ideas of proof that one can make rigorous with additional work.

• Assume $f(x) = |x|^2$. Then $$ A\,\psi(x) = \int\widehat{\eta}(y-x) \, (|x|^2-|y|^2) \,\psi(y)\,\mathrm d y = \int (x-y)\,\widehat{\eta}(y-x) \cdot (x+y) \,\psi(y)\,\mathrm d y $$ Let $\varphi_k(x) = x^k\,\widehat{\eta}(x)$ ($x^1 = x$ and $x^2 = |x|^2$), and $\psi(x) = (1+|x|^2)^{-n/4}$ with $n \in (d,d+1)$ (so that $x\,\psi(x)\notin L^2$). Then $$ \begin{align*} A\,\psi(x) &= \int \varphi_1(x-y) \cdot (x+y) \,\psi(y)\,\mathrm d y \\ &= \int \varphi_2(x-y) \,\psi(y)\,\mathrm d y + 2\int \varphi_1(x-y) \,y\,\psi(y)\,\mathrm d y \\ &= \varphi_2*\psi + 2\,\varphi_1*(x\,\psi(x)) \end{align*} $$ Since $\eta ∈ C^\infty_c$, for any $k\in\mathbb N$, $\varphi_k$ is a Schwartz function (smooth and decaying faster than any power function), and so $\|\varphi_2*\psi\|_{L^2} ≤ C \,\|\psi\|_{L^2}$ $$ \begin{align*} \|A\psi\|_{L^2} &≥ 2\,\|\varphi_1*(x\,\psi(x))\|_{L^2} - \|\varphi_2*\psi\|_{L^2} \\ &≥ 2\,\|\varphi_1*(x\,\psi(x))\|_{L^2} - C\,\|\psi\|_{L^2} \end{align*} $$ To get the behavior of the function $\varphi_1*(x\,\psi(x))$ when $x$ is large, it is not difficult to see that since $\varphi_1$ is decaying quickly for large values of $x$, we can bound the parts of the integral in $\|\varphi_1*(x\,\psi(x))\|_{L^2}$ where $\varphi_1$ is small by something of the form $C'\,\|\psi\|_{L^2}$. If $\eta$ is general, the remaining part will be bounded by below by $C\,\|x\,\psi(x)\|_{L^2}$ which is not bounded in $L^2$.

• I suppose one can do the same with more general growth like $|x|^a$ with $a>1$. What is important is to use the fact that $|x|^a - |y|^a = u(x-y) + v(x,y)$ with $v$ growing like $|y|^{a-1}$ for fixed $x$.

• To be complete, of course you are right that linear growth is allowed. It is sufficient to write $$ \begin{align*} \|A\,\psi\|_{L^2} &= \Big\|\int\widehat{\eta}(y-x) \, (f(x)-f(y)) \,\psi(y)\,\mathrm d y\Big\|_{L^2_x} \\ &\leq \|f\|_{\dot{C}^{0,1}} \,\Big\|\int |\varphi_1(x-y)| \,|\psi(y)|\,\mathrm d y\Big\|_{L^2_x} \\ &\leq \|f\|_{\dot{C}^{0,1}} \,\|\varphi_1\|_{L^1} \,\|\psi\|_{L^2} \end{align*} $$ by Young's inequality for convolution.