I found the following claim in page 3 of
- Weinstein, Alan, A symbol class for some Schrödinger equations on ${\mathbb{R}}^ n$, Am. J. Math. 107, 1-21 (1985). ZBL0574.35023:
Operators whose symbols have bounded partial derivatives can move or even create singularities. For instance $e^{d/dx}$ is a translation operator with symbol $e^{i\xi}$, while $\sum_{j=0}^\infty 2^{-j}e^{j d/dx}$ can act on a $C^\infty$ function in $\mathcal S'$ and produce a function which is not smooth.
(emphasis mine). I can't see how this is true. Lets call $T=\sum_{j=0}^\infty 2^{-j}e^{j d/dx}$. Obviously for a smooth function without the $\mathcal S'$ requirement, you can just take a function that grows too fast like $f(x)= 4^x$, then $Tf$ is not even finite. But attempts at an explicit counterexample in the relevant class either don't belong to $\mathcal S'$(Weierstrass function) or end up smooth, since $2^{-j}$ decays geometrically.
If $Tf$ is not smooth, then $\chi Tf$ is not smooth for some cutoff $\chi\in C^\infty_c$. So we should be able to see this in the Fourier transform as slow decay. So in this line of attack, the game is to find $f$ so that $$ F(\xi):=\sum_{j=0}^\infty 2^{-j} (\hat\chi * e^{ij\bullet}\hat f)(\xi)$$ has slow decay at infinity. But why should this be possible?
For $f(x)=\sin2^x$ function $$ F(x)=\sum_{j=1}^\infty 2^{-j}\sin2^{x+j} $$ is bounded and continuous (and therefore belongs to $\mathcal S'$), but not smooth.