# Is the Cartesian product of $C^\infty$ functions a $C^\infty$ function?

Define the Cartesian product of two functions $$f:\mathbb{R}^a\to\mathbb{R}^b$$ and $$g:\mathbb{R}^c\to\mathbb{R}^d$$ as $$(f\times g)(x,y)=(f(x),g(y)).$$ If the function $$f$$ and $$g$$ are $$C^\infty$$, is the function $$f\times g$$ also $$C^\infty$$?

• Yes it is. It has continuous partial derivatives s of al orders. – Kavi Rama Murthy Apr 8 at 5:38
• Hint: Note that the partial derivatives distribute independently to each component – Pink Panther Apr 8 at 5:41
• The case $h(x,y) = f(x)g(y)$ is supposedly obvious. In general the key step is to express the derivative of $t\mapsto h(tv,tw)$ in term of the partial derivatives, then the result follows easily. – reuns Apr 8 at 6:44

The following argument shows that $$f\times g$$ is $$C^1$$ if $$f$$ and $$g$$ are both $$C^1$$, and then induction and be applied to prove that it is $$C^\infty$$ if they are $$C^\infty$$.