# What does it mean for a vector function to be twice differentiable at a point?

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 .

• 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/… – zoidberg Dec 11 '18 at 2:41
• 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. – Anthony Carapetis Dec 11 '18 at 3:28

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 $$x$$.