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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 $C^\infty$?

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    $\begingroup$ Yes it is. It has continuous partial derivatives s of al orders. $\endgroup$ – Kavi Rama Murthy Apr 8 at 5:38
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    $\begingroup$ Hint: Note that the partial derivatives distribute independently to each component $\endgroup$ – Pink Panther Apr 8 at 5:41
  • $\begingroup$ 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. $\endgroup$ – reuns Apr 8 at 6:44
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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 $C^\infty$.

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  • $\begingroup$ Please, consider typing out the answer using MathJax instead of an image. $\endgroup$ – Aloizio Macedo Apr 14 at 16:28

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