7
$\begingroup$

Update 2018/4/18: I've found a book in which the definition 5) is discussed. See Topology, Calculus and Approximation by Vilmos Komornik, published by Springer-Verlag , page 98, Lemma 4.1.


Original Question:

I've come across "Caratheodory Derivative" in my textbook, which is,

Definition 1) Let $f:\mathbb{R}\to \mathbb{R},\quad t\mapsto f(t)$ be a function, $a\in \mathbb{R}.$ Then if there exists a map $\varphi:\mathbb{R}\to \mathbb{R}, \quad t\mapsto \varphi(t)$, which satisfys $$1) \quad f(x)-f(a)=\varphi(x)\cdot(x-a),\forall x\in \mathbb{R};$$ $$2) \quad \text{$\varphi $ is continuous at the point a} ,$$ then we call $\varphi(a)$ the derivative of $f$ at point $a.$

And compared with the traditional definition of derivative:

Definition 2) Let $f:\mathbb{R}\to \mathbb{R},\quad t\mapsto f(t)$ be a function, $a\in \mathbb{R}.$ Then if the limit $$\lim_{x\to a}{f(x)-f(a)\over{x-a}}$$ exists, then the value of this limit is called the derivative of $f$ at point $a$.

I can prove that (it's not difficult) these two definitions above are equivalent to each other.But when I look at the high-dimensional condition,things get complicated.

Definition 3) Let $f:\mathbb{R}^n\to \mathbb{R}^m,\quad t\mapsto f(t)$ be a multivariate function, $a\in \mathbb{R}^n,$ Then if there exists a map $\varphi:\mathbb{R}\to M_{m\times n}(\mathbb{R}),\quad t\mapsto \varphi(t)$, which satisfys $$1) \quad f(x)-f(a)=\varphi(x)\cdot(x-a),\forall x\in \mathbb{R}^n;$$ $$2) \quad \text{$\varphi $ is continuous at the point a} ,$$ then we call $\varphi(a)$ the derivative of $f$ at point $a.$

And consider the traditional definition of derivative

Definition 4) Let $f:\mathbb{R}^n\to \mathbb{R}^m,\quad t\mapsto f(t)$ be a multivariate function, $a\in \mathbb{R}^n.$ Then if there exists a matrix $A\in M_{m\times n}(\mathbb{R}),$ such that $$\lim_{x\to a}{||f(x)-f(a)-A\cdot (x-a)||\over{||x-a||}}=0,$$ then matrix $A$ is called the derivative of $f$ at point $a$.

Question: I expect that definition 3) is equivalent to definition 4), but I can only prove that $def\ 3)\Rightarrow def\ 4).$ I doubt whether $def\ 4)\Rightarrow def\ 3)$ is correct. Any help is appreciated.

P.S. Now I am able to do some generalization to definition 3).

Definition 5) Let $E,F$ be two Banach spaces, $a\in E.$ $\mathcal{L}(E;F)$ be the set of continuous linear map $E\to F,$ then consider function $f:E\to F, \quad t\mapsto f(t),$ then if there exists a map $\varphi:E\to \mathcal{L}(E;F), \ t\mapsto \varphi(t),$ such that$$1) \quad f(x)-f(a)=(\varphi(x))(x-a),\forall x\in E;$$ $$2) \quad \text{$\varphi $ is continuous at the point a} ,$$ then we call $\varphi(a)$ the derivative of $f$ at point $a.$

Using Hahn-Banach theorem, we can see this definition is also equivalent to the classic definition of derivative on Banach space.

P.P.S: A more general condition is,

Definition 6) Let $E,F$ be two additive groups, and $\mathcal{T}$ be a topology over $E,$ $\mathcal{T'}$ be a topology over $\mathcal{L}(E;F)$, $a\in E.$ Here $\mathcal{L}(E;F)$ is the set of continuous linear map $E\to F.$ Consider function $f:E\to F, \quad t\mapsto f(t),$ then if there exists a map $\varphi:(E,\mathcal{T})\to (\mathcal{L}(E;F),\mathcal{T'}), \ t\mapsto \varphi(t),$ such that$$1) \quad f(x)-f(a)=(\varphi(x))(x-a),\forall x\in E;$$ $$2) \quad \text{$\varphi $ is continuous at the point a} ,$$ then we call $\varphi(a)$ a derivative of $f$ at point $a,$ with respect to topology $\mathcal{T}$ and topology $\mathcal{T'}.$ (Under this condition the derivative may not be unique.)

$\endgroup$
  • 1
    $\begingroup$ Is 4) really widely used? It requires the use of the Euclidean norm, which isn't very desirable. We want to be able to do calculus on spaces that don't have a norm defined on them, or for which the norm isn't Euclidean. For example, we want to be able to use coordinates that are not cartesian. $\endgroup$ – Ben Crowell Jun 13 '17 at 13:25
  • $\begingroup$ @ Ben Crowell: Right, I just take it for example (we're learning multivariate calculus now, so I'm familiar with this definition.) $\endgroup$ – painday Jun 13 '17 at 13:29
  • 1
    $\begingroup$ @BenCrowell (4) is the standard definition. Note that in a finite-dimensional space, any two norms are equivalent (i.e., their ratio is bounded above and bounded away from zero), so the definition of the derivative is actually independent of the norm chosen. $\endgroup$ – user49640 Jun 13 '17 at 13:34
  • 1
    $\begingroup$ This definition is discussed in the finite-dimensional case in: Ernesto Acosta G., and Cesar Delgado G. "Frechet vs. Caratheodory." The American Mathematical Monthly 101, no. 4 (1994): 332-38. doi:10.2307/2975625 It also seems to have been extended to more general spaces in a 1995 master's thesis by R.C. Tovar, La derivada de Caratheodory en espacios vectoriales pseudotopológicos, reference to which I found on this page: scienti.colciencias.gov.co:8081/cvlac/visualizador/… However, I have no idea how to get a master's thesis from Colombia. $\endgroup$ – user49640 Jun 14 '17 at 4:28
  • $\begingroup$ @user49640: Wow, that's surprising! I'll read the first paper right now. $\endgroup$ – painday Jun 14 '17 at 5:53
4
$\begingroup$

After translating and subtracting a linear function from $f$, we can assume that $a = 0$, that $f(0) = 0$ and that $A = 0$. So we're assuming that $f(x) = \varepsilon(x)||x||$ for some vector-valued function $\varepsilon(x)$ with $\varepsilon(x) \to 0$ as $x \to 0$. We must show that there is a matrix-valued function $\varphi(x)$ with $\varphi(x) \to 0$ as $x \to 0$ and $f(x) = \varphi(x) \cdot x$.

To achieve this, for all $x \ne 0$ we define $\varphi(x) \cdot h = \langle \frac{x}{||x||},h \rangle \varepsilon(x)$. We have $||\varphi(x)|| = ||\varepsilon(x)||$, where by $||\varphi(x)||$ I mean the operator-norm of $\varphi(x)$, so it is clear that $\varphi(x)$ satisfies our requirements.

$\endgroup$
  • $\begingroup$ Great! It's an amazingly creative method, and the application of inner product is excellent and really clever! $\endgroup$ – painday Jun 13 '17 at 15:33
  • $\begingroup$ You can use this method in an arbitrary normed vector space, even an infinite-dimensional one, but you need to replace the use of the inner product by an appeal to the Hahn-Banach theorem. $\endgroup$ – user49640 Jun 13 '17 at 15:35
  • $\begingroup$ Thanks a lot, and with your help now I can avoid the annoying fraction in the definition of derivative ! I dislike the fraction appearing in a limit... :) $\endgroup$ – painday Jun 13 '17 at 15:39
  • $\begingroup$ In practice, I do this. I write $f(x) = f(a) + df(x) \cdot (x-a) + \varepsilon(x) ||x - a||$ with $\varepsilon(x) \to 0$ as $x \to a$. So there are no fractions there. I don't think I had ever seen form (3) before doing this problem. Obviously $\varphi(x)$ is underspecified, so that presents a problem. With the other form, you know exactly what $A$ has to be if you know your function's partial derivatives. $\endgroup$ – user49640 Jun 13 '17 at 15:42
  • $\begingroup$ Right, and I have established many theorems to talk about this problem. In fact , I've proved that as long as all the partial derivatives exist, then $\varphi(a)$ is uniquely determined. And actually as long as $u_1,\cdots,u_n$ is a basis of $\mathbb{R}^n$, and all the direction derivative $ D_{u_i} f_j $ exist, then $\varphi(a)$ is also uniquely determined by them. $\endgroup$ – painday Jun 13 '17 at 15:48
2
$\begingroup$

Define the mapping $\psi$ as

$$\psi(x) = {f(x)-f(a) - A(x-a)\over |x-a|^2} (x-a)\cdot$$

then if $x\ne a$ you have that $x\ne a$ you have that $f(x)-f(a) - A(x-a) = \psi(x) (x-a)$. And you have that

$$||\psi(x)|| = {||f(x)-f(a) - A(x-a)||\over||x-a||}$$

So you have that $||\psi(x)||\to 0$ as $x\to a$. Now we have

$$f(x)-f(a) = A(x-a) + \psi(x) (x-a) = (A-\psi(x))(x-a)$$

Now we have that since $||\psi(x)||\to 0$ that $\varphi(x) = A-\psi(x)$ is continuous at $a$.

$\endgroup$
  • $\begingroup$ @painday It seem there were a lapse in my reasoning. However I think that can be fixed as we can chose $\psi$ in such a way that equality holds as we don't require $\psi(x) u$ to be non-zero if $u$ is perpendicular to $x-a$. $\endgroup$ – skyking Jun 13 '17 at 13:48
  • $\begingroup$ But it's quite difficult to choose such a mapping, and I highly suspect there are some counter-examples for some certain functions... $\endgroup$ – painday Jun 13 '17 at 13:52
  • $\begingroup$ @ChristianBlatter No, but that wasn't the expression, it was $\psi(x)(x-a)$ that is apply the mapping $\psi(x)$ to $(x-a)$. $\endgroup$ – skyking Jun 13 '17 at 15:47
  • $\begingroup$ @painday I've updated the answer with a concrete definition of $\psi(x)$ with those properties. $\endgroup$ – skyking Jun 13 '17 at 15:48
  • $\begingroup$ It's correct, and this method is similar to @user 49640 's. Inner product is so useful! $\endgroup$ – painday Jun 13 '17 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.