Let $f:\mathbb R\to \mathbb R$ be a $C^\infty$-smooth function. Suppose that $f^{(k)}(0)=0$ for $k=0,\dots,n-1$. Prove that the function $g(x)=f(x)/x^n$ extends to a $C^\infty$-smooth function on $\mathbb R$.

Comment: by l'Hôpital's rule, $g$ has a finite limit at $0$, namely $f^{(n)}(0)/n!$. So, it extends to a continuous function on $\mathbb R$. However, I do not see any elementary way to show that $g$ is $C^\infty$-smooth. (One could chop up the Fourier transform of $g$ and thus reduce the problem to analytic functions, as one does in the proof of the Malgrange preparation theorem. But this looks like an overkill.)

Related posts:


1 Answer 1


Applying Taylor's expansion with integral form of the remainder to $f$ at $0$, and noting that $f^{(k)}(0)=0$ for $k=0,\dots,n-1$, we have:

$$f(x)=\frac{1}{(n-1)!}\int_0^x(x-t)^{n-1} f^{(n)}(t)dt,\quad \forall x\in\Bbb R.\tag{1}$$ Substituting $t=sx$ into $(1)$, it follows that $$f(x)=\frac{x^n}{(n-1)!}\int_0^1(1-s)^{n-1} f^{(n)}(sx)ds,\quad \forall x\in\Bbb R.\tag{2}$$

As a result, $g$ can be expressed as follows: $$g(x)=\int_0^1h(s,x)ds,\quad \forall x\in\Bbb R, \tag{3}$$ where $$h(s,x)=\frac{1}{(n-1)!}(1-s)^{n-1} f^{(n)}(sx),\quad \forall (s,x)\in\Bbb R^2.$$

Evidently $h$ is a $C^\infty$ function on $\Bbb R^2$, so we can interchange the order of $k$-th differentiation w.r.t. $x$ and integration w.r.t. $s$ in $(3)$ freely for every $k\ge 1$. That is to say, $g$ is a $C^\infty$ function.


  1. Equation $(1)$ can be easily proved by using integration by parts repeatedly.
  2. For the validity of differentiation under the integral sign, one may refer to this page.
  3. A slightly different approach to the problem is using induction on $n$ to reduce the problem to the $n=1$ case. More precisely, by induction on $n$, it suffices to prove that $f_1(x)=\frac{f(x)}{x}$ can extend to a $C^\infty$ function on $\Bbb R$ and $f_1^{(k)}(0)=0$, $k=0,\dots, n-2$. The proof of smoothness of $f_1$ is similar, i.e. using the expression $f_1(x)=\int_0^1f'(sx)ds$. $f_1^{(k)}(0)=0~(0\le k\le n-2)$ follows from applying Leibniz rule to $f(x)=xf_1(x)$ and induction on $k$.
  • $\begingroup$ wrong taylor formula... It should be $f(x)=\frac{1}{n!}\int_0^x(x-t)^{n} f^{(n+1)}(t)dt $ $\endgroup$ Oct 29, 2013 at 12:14
  • $\begingroup$ @GabrielR.: Thank you. By the way, yours formula is not quite accurate in this situation, because we don't know $f^{(n)}(0)=0$. $\endgroup$ Oct 29, 2013 at 12:20
  • $\begingroup$ Still wrong. It is either $f(x)= \frac{x^{n}f^{(n)}(0)}{n!} + \frac{1}{n!}\int_0^x(x-t)^{n} f^{(n+1)}(t)dt $ or $f(x)= \frac{1}{(n-1)!} \int_0^x(x-t)^{n-1} f^{(n)}(t)dt$ $\endgroup$ Oct 29, 2013 at 12:23
  • 1
    $\begingroup$ @GabrielR.: Yes. Sorry about that. It was just some typo on indexes. $\endgroup$ Oct 29, 2013 at 12:34
  • $\begingroup$ Nice combination of the integral form of the remainder and a change of variables. :) $\endgroup$ Oct 29, 2013 at 23:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.