Product of $\mathcal{C}^k$ multivariable functions is $\mathcal{C}^k$

112 Views Asked by At

Let $A$ be an open set of $\mathbb{R}^n$ and $f,g:A \rightarrow \mathbb{R}$. Prove that if $f,g \in \mathcal{C}^k(A, \mathbb{R})$ then $f \cdot g \in \mathcal{C}^k(A, \mathbb{R})$. (Hint: prove it by induction over $k$).

Case $k=1$:

So , following the hint let's suppose $f,g \in \mathcal{C}^1(A, \mathbb{R})$. This means that $D_i f(x),D_i g(x)$ are continuous $\forall \ i\in \{1,...,n\}$ and $\forall \ x\in A$.

We have to prove that $f \cdot g \in \mathcal{C}^1(A, \mathbb{R})$, this is, $D_i (f \cdot g )(x)$ is continuous $\forall \ i\in \{1,...,n\}$ and $\forall \ x\in A$.

If $D_i (f \cdot g )(x) = D_i f (x) D_i g (x)$ then we would be done, because the product of two continuous functions is continuous, but I don't know if this holds for partial derivatives. So my question is:

  • Does $D_i (f \cdot g )(x) = D_i f (x) D_i g (x)$ hold for $f$ and $g$ continuously differentiable?

  • If not. how should I continue?