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Let $F: R^n \to R^m $ be a vector valued function with components $F = (f_1 ,f_2 ,..., f_m).$ What is the most proper and universal accepted definition for the function $F$ be twice differentiable at the point $x$? Does that simply mean $f_i$ be twice-differentiable at point $x$ for all $i = 1,2,...,m$. I know this is the case when we are talking about a function being $C^2$ (see What does it mean for a vector function to be twice differentiable? ).

Or does it mean $F$ be differentiable around $x$ and the derivative of $F$ which is a matrix valued function, denoted by $\nabla F : U \to R^m \times R^n $ be differentiable (in some sense) at the point $x$.

Any reference about that would be appreciated .

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    $\begingroup$ I think you can define it as just each coordinate being twice continuously differentiable. But this is equivalent to thinking about it as one object, i.e. differentiating a 1-form. Take a look at this: math.stackexchange.com/questions/925322/… $\endgroup$ – zoidberg Dec 11 '18 at 2:41
  • $\begingroup$ Yes, I'm pretty sure that the dimension of the codomain doesn't complicate things - the two definitions ($f_i$ being twice-differentiable vs $\nabla F : U \to \mathbb R^{m\times n}$ being Fréchet differentiable) should coincide. $\endgroup$ – Anthony Carapetis Dec 11 '18 at 3:28
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The derivative of $F$ at $x \in \mathbb{R}^n$ is a linear map $dF(x) : \mathbb{R}^n \to \mathbb{R}^m$. It is characterized as the best linear approximation of $F$ of $x$. See for example my answer to Higher dimensional total derivative.

If $F$ is differentiable in an open neighborhood $U$ if $x$, we obtain a map $dF : U \to \mathcal{L} = \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m)$ = vector space of all linear maps $\mathbb{R}^n \to \mathbb{R}^m$. This is a finite-dimensional vector space which can be identified with $\mathbb{R}^{n\cdot m}$. We call $F$ twice differentiable at $x$ is $dF$ is differentiable at $x$.

Note that it essential that $dF(x')$ exists in a neighborhood of $x$.

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