Let $f,g: \mathbb{R} \to \mathbb{R}$ smooth functions on $\mathbb {R}$, prove that $fg$ and $\frac{f}{g}$ are smooth functions on $\mathbb{R}$, where $g \not\equiv 0$.
Logically, they are smooth because if $f$ and $g$ are smooth on $\mathbb{R}$ that means they are $n$ times differentiable and each derivative is derivative. Therefore, from calculus, $f.g$ and $\frac{f}{g}$ are also $n$ times differentiable and each derivative is continuous. However, I am not sure about the formal proof.
My attempt:
- Assuming that $f$ and $g$ are smooth on $\mathbb{R}$, we want to prove that $f.g$ is smooth. In order to do so, we should prove that $f.g$ is $n$ times differentiable and each derivative is continuous.
That’s true in the case of $n=1$. Let’s assume that when $f$ and $g$ are $n-1$ times differentiable then $f .g$ is $n-1$ times differentiable.
Now, we have
$(f. g)’=f’.g + g’. f$
$f,\ g,\ f’ \text{and} \ g’$ are $n-1$ times differentiable, so by the induction step $(f.g)’$ is $n-1$ times differentiable. Then $f.g$ is $n$ times differentiable and all of the derivatives are continuous, as desired.
- We know that $\frac{f}{g}=f .\frac{1}{g}$, so by (1), if we prove that $\frac{1}{g}$ is $n$ times differentiable on $\mathbb{R} - \{g ^{-1}(0)\}$ whenever $g$ is $n$ times differentiable on $\mathbb{R}$, we find that $\frac{f}{g}$ is smooth.
Assuming that if $g$ is $n-1$ times differentiable on $\mathbb{R}$, then $\frac{1}{g}$ is $n-1$ times differentiable on $\mathbb{R} - \{g^{-1}(0)\}$, we obtain
$(\frac{1}{g})’= \frac{-g’}{g^2}$ is $n-1$ times differentiable on $\mathbb{R} - \{g^{-1}(0)\}$ because it is a quotient of two $n-1$ times differentiable functions.
As a result, $\frac{f}{g}$ is smooth.
Is my answer correct? Thank you in advance.