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.


$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$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.

  • 1
    $\begingroup$ 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 $hf/f$ ? $\endgroup$ 2 days ago
  • $\begingroup$ By the way, $f(x)\not= 0$ for $x\not= a$ close to $a$ in this situation is also immediate from a Taylor expansion. $\endgroup$ 2 days ago
  • $\begingroup$ 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. $\endgroup$
    – user32157
    2 days ago
  • 1
    $\begingroup$ I have redone the answer. $\endgroup$ 2 days ago
  • $\begingroup$ 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 $(fh)' - f'h$. $\endgroup$
    – user32157

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