Quotient of two smooth functions is smooth

Solution 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.

Remark:

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 $k$.