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.


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.

• I assume in your first two paragraphs you are looking at the situation when $f(a)=0$, $f^{(k)}(a)\not= 0$ (so "neighborhood" should presumably be "punctured neighborhood"). But then what do you mean by $hf/f$ ? 2 days ago
• By the way, $f(x)\not= 0$ for $x\not= a$ close to $a$ in this situation is also immediate from a Taylor expansion. 2 days ago
• I'm not sure I follow the argument in the first two paragraphs. In the third one, $g_k(x) = o(|x-a|)$ shows that $g_k$ has derivative $0$ at $a$ but is that sufficient to show that that it is continuously differentiable? I believe there are counter-examples. 2 days ago
• I have redone the answer. 2 days ago
• This is very nice. I think you've actually shown a slightly stronger result for Part 1 by defining 'bad' to instead mean that h is not locally smooth. Then every point either has h locally smooth or $f^{(k)} = 0$ for all $k$. Then we can get part 2 up to saying $f'h$ is $C^1$. You've shown it's differentiable and at points where $h$ is bad, the derivative is 0. At all other points, $(f'h)' = f''h + f'h'$, which goes to zero at bad points since $h$ and $h'$ are locally bounded. This also shows $fh'$ extends to a $C^1$ function $(fh)' - f'h$. yesterday