[I asked this question first on MSE but there was no activity.](https://math.stackexchange.com/questions/2004494/estimating-pointwise-multiplication-conjugated-by-a-fourier-multiplier)


Let $m(D)$ be a Fourier multiplier and $f$ a known function. I'm trying to estimate the operator
$$Tu=m^{-1}(D)(f(x)m(D)u)$$
in say $H^s$. I can do it by estimating each operator in succession but I'd like to find something tighter using the fact that the multipliers are inverses of each other. In particular I'm interested in the case where $m(D)$ is like differentiation for the low frequencies and blows up the large frequencies exponentially. Since $m^{-1}$ obviously does the opposite, I'd hope that these two effects partially cancel out. 


Edit: the particular problem I'm interested in is on $\mathbb R$ with $m(\xi)=|\xi|^2e^{|\xi|}$ and $f(x)=1/x$.



Estimating pointwise multiplication conjugated by a Fourier multiplier
Let's just do some naive things. 


1. Your $m(D)$ is not bounded on any $H^s$, due to the exponential growth. In fact, your $m(D)$ is not a bounded map from $H^{s\_1} \to H^{s\_2}$ for any pair $(s\_1,s\_2)$. (I assume you mean the $L^2$ Sobolev space by $H^s$.) [This is mostly a comment on your assertion that you want to try to estimate the operator piece by piece.]
2. Let's suppose you are willing to work with $u$ with compact Fourier support. Then the convolution 
$$ \hat{f} \* (m\hat{u}) $$
is defined as a tempered distribution. If you are lucky so that $\int m \hat{u} ~\mathrm{d}\xi = 0$ this distribution has compact support (in fact the same support as $\hat{u}$); but that's not going to be the case generally. In this case $\widehat{Tu}$ is exponentially decaying, but has non-compact support. So you see even looking at really high regularity objects there is a loss in regularity for the operator $\widehat{Tu}$, so there's really no reason to expect that $T$ is bounded on $H^s$.
3. In addition to the problems at $\infty$, there is also the problem at $0$. If $m\hat{u}$ has any odd portion, $\hat{f}\*(m\hat{u})(0) \neq 0$, and then $Tu$ cannot be in $H^s$ for any $s$ due to the singular behavior at zero in frequency space.


