A New Definition of Derivative

Question

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.

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Original Question:

I've come across "Carathéodory 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 satisfies $$1) \quad f(x)-f(a)=\varphi(x)\cdot(x-a),\forall x\in \mathbb{R};$$ $$2) \quad \text{$\varphi $ is continuous at the point a} ,$$ 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 satisfies $$1) \quad f(x)-f(a)=\varphi(x)\cdot(x-a),\forall x\in \mathbb{R}^n;$$ $$2) \quad \text{$\varphi $ is continuous at the point a} ,$$ 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 $\mathrm{def}\ 3)\Rightarrow \mathrm{def}\ 4).$ I doubt whether $\mathrm{def}\ 4)\Rightarrow \mathrm{def}\ 3)$ 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.)

Answer 1

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.

Answer 2

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 $a$.

Answer 3

the definition 4 is standard definition of derivative that is Frechet derivative. used in lang cartan and Dieudonne. Lang in one of his texts redefines this using tangent function and remarks that being tangent is independent of norm. in his book in differential geometry he gives yet one more definition in topological vector spaces/. amap f : X -> y is differentiable at a point a if there exists a continuous linear transformation A from X to Y and f (x) -f(a) -A(x-a) = S( x-a) and S is tangent at 0 meansS(x-a) where s is tangent at 0 In banach spaces s is tangent at 0 means s(h)/| h| --> 0as h --> 0. where here h = x(x-a) in topologocal vector spaces s( h) is tangent at 0 is defined by him as given a nbd W of 0 in Y there exists a nbd U of 0 in X, such that S( th) is conatined in o(t)W where t is real number and o(t) is areal valued function of o(t) that is o(t)/| t| --> 0. He remarks that in Banach space it reduces to usual tangent definition and this is not clear to me. further intuition is faint for this definition,,. on the other hand definition 3 carathedory definition can be perfectly generalized in top vector space verbatim. but showing 4 imples 3 requires hahan Banach theorem which is unpleasant as even in Banach spaces Calculus otherwise does not use the theorem. Also on its own HBT implies Banach tarski paradox . HBT uses axiom of choice. on its own implies the paradox.Even for separable Banach spaces HBT involves axiom of dependent choice which also has some unpleasant consequences. Recent research in computer science and foundations indicates that uncountable axiom of choice is counterintuitive. A simple prisoners puzzle discussed in Cornell university shows it vividly. So we welcome any modification of caratheodory definition so that in Banach spaces frechet differentiability implies caratheodory. Calculus in Banach spaces is almost identical to calculus in euclidean spaces Also i invite readers to show the tangent in tvs implies and equivalent to tangent in Banach spaces in the next answer i propose a modification of caratheodory definition

Answer 4

Let X,Y be topological spaces we say a map f X --> Y is differentiable at a if f(x) - f(a) = s(x) (x-a) and there exists a continuous linear map A from X to Y with s(a) = A i suggest alternatives for s

1. for each x, s(x) is a continuous linear map defined on one dimensional subspace of X to Y, and as x--->a s(x) tends to s(a)
2. for each x, s(x)( x-a) is a vector y in Y and as x-->a , y ---> a , s(x)( h) ---> A.h where h = x-a
3. we do not require s(x) to be linear but just continuous and condition as in 2. can we now avoid hahn banach theory hhbt does not hold in tvs so direct caratheodory definition though nice in tvs will pit severe indirect restriction on differentiation