Commutator operator into convolution operator

Question

Consider two functions $f,g\in L^2(\mathbb{R})$ and $N>1$ a dyadic number. Denote by $\mathbb{P}_N$ the frequency proyector on frequencies of size $N$, that is, the multiplier operator associated with the symbol $\phi_N(\xi)$, with $\phi_N$ a standard bump function supported on the interval $\left[\tfrac12N,2N\right]$. I was reading a book and the author says that by "direct computations" one could verify that $$ \left[\mathbb{P}_N^2\partial_x,g\right]f=\int K(x,y) f(y)dy $$ where $[\cdot,\cdot]$ is the commutator and $$ K(x,y)=icN^2\int e^{iN(x-y)\mu}\mu\phi^2(\mu)\big(g(y)-g(x)\big)d\mu. $$ Although it actually looks like something I should be able to verify, after many tries I haven't been able to come up with anything even remotely close to that expression. I was wondering if anyone has an idea how to get there. What I've been trying so far is to rewrite the commutator as the inverse fourier transform of the fourier transform, and then expand, however I don't see how go back and get $K(x,y)$.

Answer 1

Here we use the convention that $$\hat f(\xi)=\int_\mathbb R f(x)e^{-ix\xi}\,dx,\qquad \check f(x)=\frac1{2\pi}\int_\mathbb R f(\xi)e^{ix\xi}\,d\xi.$$ Then $\widehat{f\ast g}=\hat f\hat g, (f\ast g)^\vee=2\pi \check f \check g, \widehat{fg}=\frac1{2\pi}\hat f\ast\hat g, (fg)^\vee=\check f\ast \check g$.

According to your context, we have $\phi_N(\xi)=\phi(\xi/N)$. Let $\psi(\xi)=\xi\phi^2(\xi)$ and $\psi_N(\xi)=N\psi(\xi/N)$, then $\psi_N(\xi)=\xi\phi_N^2(\xi)$. By definition, \begin{align*} ([\mathbb{P}_N^2\partial_x,g]f)^{\hat\ }(\xi)&=(\mathbb{P}_N^2\partial_x(gf))^{\hat\ }(\xi)-(g\mathbb{P}_N^2\partial_xf)^{\hat\ }(\xi)\\ &=\phi_N^2(\xi)i\xi(\hat f\ast \hat g)(\xi)-i\left(\hat g\ast (\psi_N\hat f)\right)(\xi)\\ &=i\psi_N(\xi)(\hat f\ast \hat g)(\xi)-i\left(\hat g\ast (\psi_N\hat f)\right)(\xi). \end{align*} We have \begin{align*} \left(\psi_N(\hat f\ast\hat g)\right)^\vee(x)=2\pi\left(\check\psi_N\ast(fg)\right)(x)=2\pi\int_\mathbb R\check\psi_N(x-y)f(y)g(y)\,dy. \end{align*} It follows from the definition of $\psi_N$ that \begin{align*} \check\psi_N(x)=\frac1{2\pi}\int_\mathbb{R}\psi_N(\xi)e^{ix\xi}\,d\xi=\frac1{2\pi}\int_\mathbb{R}\xi\phi^2\left(\frac\xi N\right)e^{ix\xi}\,d\xi=\frac{N^2}{2\pi}\int_\mathbb{R}\mu\phi^2(\mu)e^{iNx\mu}\,d\mu, \end{align*} hence $$\left(\psi_N(\hat f\ast\hat g)\right)^\vee(x)=\int_\mathbb{R}\left(N^2\int_\mathbb{R}\mu\phi^2(\mu)e^{iN(x-y)\mu}g(y)\,d\mu\right)f(y)\,dy.$$ We also have \begin{align*} \left(\hat g\ast (\psi_N\hat f)\right)^\vee(x)&=2\pi g(x)\left(\check\psi_N\ast f\right)(x)=2\pi g(x)\int_\mathbb R\check\psi_N(x-y)f(y)\,dy\\ &=\int_\mathbb{R}\left(N^2\int_\mathbb{R}\mu\phi^2(\mu)e^{iN(x-y)\mu}g(x)\,d\mu\right)f(y)\,dy. \end{align*} As a result, $[\mathbb{P}_N^2\partial_x,g]f=\int_\mathbb R K(x,y)f(y)\,dy$ with $$K(x,y)=iN^2\int_\mathbb{R}\mu\phi^2(\mu)e^{iN(x-y)\mu}(g(y)-g(x))\,d\mu.$$