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Let $f:\mathbb{R}^n \to \mathbb{R}^m$ be a function. We write $f=\left(f_1,\cdots,f_m\right)$, where $f_i:\mathbb{R}^n \to \mathbb{R}$.

Original Question : The problem from which I was motivated to ask this question is the following : Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be given by $f\left(x,y\right)=\left(\cos{x}+\cos{y},\,\sin{x}+\sin{y}\right)$. Is it true that $f \in C^1\left(\mathbb{R}^2\right)?$

The answer is obviously yes! But to establish that, if I understand it correctly, I have to do a hell lot of work. I have to compute jacobian matrix, check continuity of the entries, establish differentiability of $f$ and then check continuity of the total derivative. I was wondering if there's any easy formulation to verify $C^1$-ness.

Question : I want to know whether or not the following statement holds in general : $f \in C^1\left(\mathbb{R}^n\right)$ if and only if $f_i \in C^1\left(\mathbb{R}^n\right)$ for every $i=1,\cdots,n$. If yes, why? If not, is there any such (sufficient) condition at all?

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  • $\begingroup$ Yes, your equivalence is true. This follows quite easily from the definition (the Jacobian of $f$ is the Jacobians of $f_1,\dots, f_m$ stacked on top of each other). $\endgroup$ – PhoemueX Nov 18 '18 at 8:52
  • $\begingroup$ @PhoemueX Thanks for the comment! So, can I assume that we only need the entries of the Jacobian $\frac{\partial f_i}{\partial x_j}$ to be $C^1$? (It's probably a very basic question, but I need a confirmation) For example, in the case of the function $f:\mathbb{R}^2 \to \mathbb{R}^2$ mentioned above, I get a nice Jacobian with nice $4$ entries which are all $C^1$. Is it enough to say that $f$ is $C^1$? $\endgroup$ – user405743 Nov 18 '18 at 9:33
  • $\begingroup$ Yes, all you need is that all partial derivatives $\partial f_i /\partial x_j$ exist and are continuous. That this suffices is one of the theorems of multivariable calculus. $\endgroup$ – PhoemueX Nov 18 '18 at 15:38
  • $\begingroup$ @PhoemueX: These comments together look like an answer; would you mind posting it as one so we can get this off the Unanswered list? $\endgroup$ – aleph_two Dec 20 '18 at 4:45
  • $\begingroup$ Dragon, you may also be interested in this: math.stackexchange.com/questions/1380650/… $\endgroup$ – aleph_two Dec 20 '18 at 4:46

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