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So, I know for a function $f :\mathbb{R}^m\to\mathbb{R}^n$, $f$ is differentiable on $U\subset\mathbb{R}^m$ iff for every point $a\in U$ there exists a linear transformation, call it $Df(a)$ such that $$ \lim_{\vec{h}\to 0}\frac{\|f(a + \vec{h}) - f(a) - Df(a)\vec{h}\|}{\|\vec{h}\|} = 0 $$ You can determine from this that if $f$ is differentiable then all the partial derivatives of $f$ exist on $U$ and $Df(a)$ is the Jacobian of $f$. We write $f\in C^1(U)$ to say it is once-differentiable on $U$.

So, what exactly does it mean to say $f\in C^k(U)$ for $U\subset\mathbb{R}^m$, and $k>1$? My only thought is that it requires all the partial derivatives of order $k$ to exist, but this is not enough information for the $C^1$ case since not only must the partials exist, but the above limit with $Df(a)$ equal to the Jacobian must also be zero (which are not necessarily always simultaneously true). My second thought was that, at least for the $k = 2$, the Hessian of $f$ must exist, and some limit involving the Hessian must go to zero, but I can't figure out what it is.

There really is, as far as I can tell, no reasonable idea of the "second-derivative" of a vector function exists. The "derivative" of a vector function, as I understand it, is $Df(a)$, which is a function that maps a vector $a$ to a matrix. So if we were to take the derivative of this, the "second" derivative, what kind of object would it be? Would it be a tensor?

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  • $\begingroup$ If all partial derivatives of $f:\mathbb{R^m}\longrightarrow\mathbb{R}$ exist and are continuous in some neighborhood $U$, then $f$ is $C^1$ on $U$. One can use this notion to extend the idea of differentiability classes for vector functions and for values other than $k=1$ $\endgroup$ Nov 24, 2017 at 16:33
  • $\begingroup$ @Fimpellizieri I've only ever seen the theorem that $f$ is differentiable (that is, there exists a linear operator such that the limit I wrote goes to zero) iff all the partial derivatives exist and the limit I wrote, with $Df(a)$ replaced by the Jacobian, goes to zero. This doesn't require any notion of continuity of the partial derivatives. How do you use continuity? $\endgroup$ Nov 24, 2017 at 16:39
  • $\begingroup$ I have posted a proof. $\endgroup$ Nov 24, 2017 at 18:24

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Well, that's pretty simple if you take the abstract point of view: as you mention it, the differential of a map $f\colon U\subset \mathbf R^m\longrightarrow \mathbf R^n$ at a point $a\in\mathbf R^m$ is a linear map map $Df(a)\in \mathcal L(\mathbf R^m,\mathbf R^n)$ which is tangent to $f$ at $a$.

Now, if $f$ is differentiable at every point of $U$, you define a (first order) differential map \begin{align} Df\colon U\subset\mathbf R^m & \longrightarrow \mathcal L(\mathbf R^m,\mathbf R^n),\\ a & \longmapsto Df(a). \end{align} This differential map $Df$ may in turn be differentiable at a point $a\in U$. You then obtain a second order differential $D^2f(a)$, which is a linear map in $\;\mathcal L\bigl(\mathbf R^m,\mathcal L(\mathbf R^m,\mathbf R^n)\bigr)$, tangent to $Df$ at $a$.

As we have a canonical isomorphism $\;\mathcal L\bigl(\mathbf R^m,\mathcal L(\mathbf R^m,\mathbf R^n)\bigr)\simeq\mathcal L^2(\mathbf R^m,\mathbf R^n)$ (the set of bilinear maps from $\mathbf R^m$ to $\mathbf R^n$, we identify $D^2f(a)$ with the corresponding bilinear map, which is represented by the Jacobian matrix once a basis has been chosen.

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  • $\begingroup$ Ok, so my idea of it being a "tensor" was somewhat in the right direction. Can I ask, does this notion of higher order derivatives of vector functions lend itself easily to other generalizations of facts of higher order derivatives of scalar valued functions? For example, how is your second derivative related to the partial derivatives of $f$? What norm do you use in the limit definition of $D^2f(a)$, the Frobenius norm for matrices? If a function is $k$-differentiable in the sense you give, does it have an $k$th order approximation by Taylor polynomials? $\endgroup$ Nov 24, 2017 at 16:58
  • $\begingroup$ The questions I asked above are more-or-less rhetorical (especially the one on Taylor polynomials, but I would be interested in the relationship to partial derivatives if you have time), I really just want to know if your $D^2f(a)$ is the "right" notion of the second derivative to work with generalizations of theorems of calculus on $\mathbb{R}$. $\endgroup$ Nov 24, 2017 at 17:00
  • $\begingroup$ The norm in use is the usual norm for linear maps (which does not depend on any matrix representation), and, yes, you obtain a Taylor's formula. My reference would be Henri Cartan's Differential Calculus. For your information, Henri Cartan was one of French mathematicians who founded the Bourbaki group, towards 1935. $\endgroup$
    – Bernard
    Nov 24, 2017 at 17:10
  • $\begingroup$ This book? amazon.ca/Differential-Calculus-Henri-Cartan/dp/0901665053/… $\endgroup$ Nov 24, 2017 at 17:14
  • $\begingroup$ I guess so – I used the French version when I was a student. $\endgroup$
    – Bernard
    Nov 24, 2017 at 17:16
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Thoerem: Let $U\subset \mathbb{R}^m$ be open and let $f:U\longrightarrow \mathbb{R}$ be such that, for each $i\in\{1,\dots,m\}$, the partial derivative $f_i=\frac\partial{\partial x_i}f$ exists and is continuous throughout $U$. Then $f\in C^1(U)$.

Proof: For each $x\in U$, $i\in\{1,\dots,m\}$ and $h_i\in\mathbb{R}$ such that $x+h_ie_i \in U$, we have that

$$f(x+h_ie_i)=f(x)+h_if_i(x)+h_i\epsilon_{i}(h_i;x), \tag{1}$$

where $\epsilon_{i}$ satisfies $\lim_{t\to0}\epsilon_{i}(t;x)=0$ for all $x$. This limit condition implies that

$$\lim_{h_i\to0}\frac{f(x+h_ie_i)-f(x)}{h_i}-f_i(x)=\lim_{h_i\to0}\epsilon_i(h_i;x)=0$$

and hence that $f_i(x)=\lim_{h_i\to0}\frac{f(x+h_ie_i)-f(x)}{h_i}$. Notice that because $U$ is open, there is always a neighborhood of $0$ in $\mathbb{R}$ of valid $h_i$.

The value of $\epsilon_i$ when $h_i=0$ does not change the validity of equation $(1)$, so we set $\epsilon_i(0,x)=0$ for all $x$. It then follows that $\epsilon_i(\cdot\,; x)$ is continuous in $h_i$ for all $x$, and that $\epsilon_i(0\,;\, \cdot)$ is continuous in $x$.
Moreover, it follows from the continuity of $f_i$ that $\epsilon_{i}(h_i\,;\,\cdot)$ is continuous in $x$ for all $h_i\neq 0$, and hence for all $h_i$.

Thus, $\epsilon_{i}$ is continuous and $\epsilon_{i}(0;x)=0$ for all $x$.

Now, let $a\in U$ and $h=(h_1,\dots,h_m)\in\mathbb{R}^m$ be such that $a+h\in U$. Then:

\begin{align} f(a+h)-f(a) =f\left(a+\sum_{i=1}^mh_ie_i\right)-f(a) =\sum_{k=1}^m\,f\left(a+\sum_{i=1}^{k}h_ie_i\right)-f\left(a+\sum_{i=1}^{k-1}h_ie_i\right), \end{align}

where the sum $\sum_{i=1}^{-1}$ is taken to be the empty sum, $0$. We may then rewrite it as

\begin{align} f(a+h)-f(a)&=\sum_{k=1}^m\,f\left(\left(a+\sum_{i=1}^{k-1}h_ie_i\right)+h_ke_k\right)-f\left(a+\sum_{i=1}^{k-1}h_ie_i\right)\\ &=\sum_{k=1}^m\,h_k\cdot f_k\left(a+\sum_{i=1}^{k-1}h_ie_i\right) + h_k\epsilon_{k}\left(h_k;a+\sum_{i=1}^{k-1}h_ie_i\right) \end{align}

Consider $\nabla f:U\longrightarrow \mathbb{R}^m$ the gradient of $f$, that is, $\nabla f(x)=(f_1(x),\dots,f_m(x))$. We define $Df(a)h=\langle\nabla f(a), h\rangle=\sum_{i=1}^m\,h_if_i(a)$. Then:

\begin{align} f(a+h)-f(a)-Df(a)h &=\sum_{k=1}^m\,h_k\cdot \left(f_k\left(a+\sum_{i=1}^{k-1}h_ie_i\right) -f_k(a)\right) + h_k\epsilon_{k}\left(h_k;a+\sum_{i=1}^{k-1}h_ie_i\right) \end{align}

Now, as $h\to 0$, we have that

$$f_k\left(a+\sum_{i=1}^{k-1}h_ie_i\right) -f_k(a)\longrightarrow 0$$

because the $f_k$ are continuous. Moreover, $|h_k|\leq \lVert h\rVert$, so $\frac{h_k}{\lVert h \rVert}$ remains bounded as $h\to 0$ and hence $\frac{h_k}{\lVert h \rVert}\cdot \left(f_k\left(a+\sum_{i=1}^{k-1}h_ie_i\right) -f_k(a)\right)$ vanishes as $h\to 0$.

Finally, in a similar fashion we have that $\epsilon_{k}\left(h_k;a+\sum_{i=1}^{k-1}h_ie_i\right) \to 0$ as $h\to0$, because $\epsilon_{k}$ is continuous and it vanishes whenever the first argument is $0$. Since $\frac{h_k}{\lVert h \rVert}$ remains bounded as $h\to 0$, once again the product $\frac{h_k}{\lVert h \rVert}\cdot \epsilon_{k}\left(h_k;a+\sum_{i=1}^{k-1}h_ie_i\right)$ vanishes as $h\to 0$.

It follows that the limit $\lim_{h\to0}\frac{f(a+h)-f(a)-Df(a)h}{\lVert h \rVert}$ exists and is $0$, that is, $f$ is differentiable at $a$ and $Df(a)=\langle\nabla f(a),\cdot\rangle$.

Since $a$ was arbitrary, we have that $Df$ is given by $x\mapsto \langle \nabla f(x),\cdot\rangle$. Finally, because the $f_i$ are continuous, $Df$ itself is continuous, which concludes the proof. $\square$

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