If $fh$ is smooth and $h$ Lipschitz, what can be said about $f$? If the product of two functions is smooth, then how quickly must one function decay when the other is non-smooth? Suppose we have two functions $f,h$ on $\mathbb{R}$ such that:

*

*$h$ is Lipschitz continuous

*$f$ is smooth (i.e. $C^\infty$), and

*$fh$ is a smooth function.

What can we conclude about $f$ from this requirement? Presumably $f$ must go to zero 'sufficiently fast' at any point where $h$ is non-smooth.
For example, we can compute that a.e. $(fh)' = f'h + fh'$ and since $f'h$ is continuous, $h' = ((fh)' - f'h)/f$ is continuous in any neighborhood where $f$ is non-zero. Therefore $f$ is zero anywhere $h$ is not $C^1$. Must $f^{(k)}$ be zero anywhere $h$ is not locally smooth? Must $f^{(k)}h$ be smooth? Must $f^{(k)}h^{(\ell)}$ be well-defined (i.e. extend to be 0 at the set where $h$ is not $C^\ell$) and smooth?
In my application, $h$ is also smooth in an open, dense set, which may give some context though I doubt it helps with these.
 A: $\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Here are answers to your three questions (the latter two of them partial).
Answer 1: Yes, for any real $a$ and any $k\in\{0,1,\dots\}$,
\begin{equation*}
    \text{if $h$ does not have all the derivatives at $a$, then $f^{(k)}(a)=0$.}\tag{1} 
\end{equation*}
Indeed, say that $h$ is bad at $a$ (or, equivalently, that $a$ is a bad point of $h$0 if $h$ does not have all the derivatives at $a$.
Take any $a\in\R$. Suppose that $h$ is bad at $a$. Take then the smallest $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$, if such a $k$ exists. Let
\begin{equation*}
    g:=fh
\end{equation*}
and
\begin{equation*}
    F(x):=\frac{f(x)}{(x-a)^k}, \quad G(x):=\frac{g(x)}{(x-a)^k}
\end{equation*}
for real $x\ne a$, with $F(a):=f^{(k)}(a)/k!$ and $G(a):=g^{(k)}(a)/k!$. By a Taylor formula, if $k\ge1$, then for all real $x$
\begin{equation*}
    F(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}f^{(k)}(a+(x-a)s)\,ds
\end{equation*}
and hence for all nonnegative integers $l$
\begin{equation*}
    F^{(l)}(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}s^lf^{(k+l)}(a+(x-a)s)\,ds,
\end{equation*}
with the similar formulas for $G(x)$ and $G^{(l)}(x)$.
So, if $k\ge1$, then $F$ and $G$ are smooth, $F(a)\ne0$, and hence $F\ne0$ and $h=G/F$ on a neighborhood $V$ of $a$ (the equality $h(a)=G(a)/F(a)$ follows by continuity); the same conclusions obviously hold for $k=0$.
So, $h$ is smooth on $V$, which contradicts the assumption that $h$ is bad at $a$. So, as claimed, there is no $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$.
We have actually proved more than (1):
\begin{equation*}
    \text{if $a$ is in the closure of the set of all bad points of $h$, } \\ 
    \text{then $f^{(k)}(a)=0$ for all $k\in\{0,1,\dots\}$.} 
\end{equation*}

Answer 2 (partial):
Now it follows that $g_k:=f^{(k)}h$ must be differentiable. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g_k$. If $h$ does not have all the derivatives at $a$, then, by (1), $f'(x)=o(|x-a|)$ as $x\to a$, so that $g_k(x)=o(|x-a|)$ as $x\to a$, so that $g_k'(a)=0$.

Answer 3 (partial):
Finally, let $g_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz.
By (1), $f^{(k)}(x)=o(|x-a|)$ as $x\to a$, so that $g_{k,1}(x)=o(|x-a|)$ as $x\to a$, and hence $g_{k,1}'(a)=0$. So, $g_{k,1}$ is differentiable.
