I'm interested in finding an analogue of a derivative that applies to functions which are defined more general subsets of $\mathbb{R}^n$ than open subsets. In particularly, I'm looking at functions defined on the non-negative orthant of $\mathbb{R}^n$.

I've been thinking that one could use tangent cones to such an end. The question is organised as follows, first are the standard definitions of a tangent cone and a differentiable function, then comes my candidate extension of differentiability and finally are the questions.

Thank you very much in advanced (even if for just having a read!).

EDIT: Does anyone think that this would be an appropriate (or not) question for MathOverflow?

Tangent Cones: Let $X\subseteq\mathbb{R}^n$ and $x\in X$. Then the tangent cone to $X$ at $x$, $T_X(x)$ is defined as the closure of the cone formed by all half-lines emanating from $x$ and intersecting $X$ in at least one point $y\in X$ distinct from $x$. Formally

$$T_X(x)=\{0\}\cup\left\{y:y\neq0,\exists (x_k)_{k\in\mathbb{N}}\subseteq X,\quad x_k\neq x\quad \forall k,\quad \frac{x_k-x}{||x_k-x||}\rightarrow\frac{y}{||y||}\right\}.$$

Differentiable function: Suppose $E$ is an open set in $\mathbb{R}^n$, $f$ is a function that maps $E$ into $\mathbb{R}^m$, and $x\in E$. If there exists a linear transformation $A$ from $\mathbb{R}^n$ to $\mathbb{R}^m$ such that

$$\lim_{h\rightarrow 0}\frac{|f(x+h)-f(x)-A(h)|}{|h|}=0,$$

where $|\cdot|$ denotes any p-norm, then we say that $f$ is differentiable at $x$, and we write


Tentative extension of "differentiability": Suppose $X\subseteq\mathbb{R}^n$, $x\in X$ and $f:X\rightarrow\mathbb{R}^m$. We say that $f$ is differentiable at $x$ if there exists a "pseudo-linear" transformation $\tilde{A}$ from $T_X(x)$ to $\mathbb{R}^m$ such that for any sequence

$$(h_k)_{k\in\mathbb{N}}\subset T_X(x)$$

that satisfies $h_k\rightarrow 0$ as $k\rightarrow\infty$


then we say that $f$ is differentiable at $x$, and we write


By $\tilde{A}$ being pseudo-linear I mean that for any $a,b\in\mathbb{R}$ and $x,y\in X$ such that $ax+by\in X$


Note that, because $x\in int(X)$ implies that $T_X(x)=\mathbb{R}^n$, the above definition coincides with the usual one if $X$ is open. My questions then are:

  1. Does the above notion of a differentiable function and its derivative already exist? If so what is it called? Or is there a more general notion for which the above is a special case?
  2. If 1., does there exist an analogue of the chain rule that applies to it?
  3. Similarly, if $\tilde{A}$ is a continuous, is there an easy way to compute the $\tilde{A}$ in the standard basis of $\mathbb{R}^n$ (much in the same way we use the partial derivatives to compute the derivative of a continuously differentiable functions)?
  4. Similarly, can one extend a differentiable (in the sense above) function on $X$ to a differentiable (in the usual sense) function on $\mathbb{R}^n$?
  5. Is there any reason why any or all the above could not be answered affirmatively (for example, something that does not make sense in the derivative)?

Any answers or references that might contain them would be greatly appreciated. If it helps, please add any extra conditions on $X$ that are satisfied by the orthant (closed, convex, closure of an open set, ...etc).

  • $\begingroup$ Have you ever heard of a Monge cone? en.wikipedia.org/wiki/Monge_cone $\endgroup$ – Ron Gordon Apr 18 '13 at 21:13
  • $\begingroup$ Hi, sorry its taken me this long to reply - I've been attempting to find out what a Monge cone is. However, my lack of familiarity with PDEs and notions of differential geometry is getting in the way. Would you mind elaborating a bit or recommending a reference (apart from the wikipedia entry) or a list of topics I should look into that would help me understand the relation between Monge cones and my question? By the way, thank you for replying. $\endgroup$ – jkn Apr 19 '13 at 14:24
  • $\begingroup$ Yes: mathoverflow.net/questions/77175/… $\endgroup$ – user117644 Jan 4 '16 at 7:13

Derivatives of set-valued mappings are naturally defined via different types of tangent cones. In particular, your definition is very close to the definition of the contingent derivative of a set-valued map. You can read about it in the book "Set-valued analysis" by J.-P. Aubin and H.Frankowska.

  • $\begingroup$ Thank you, I've had a quick look and the definitions are indeed very close. However, they assume that the functions map set to and from normed spaces, which the orthant is not (I'll have to check how many of the results remain by relaxing the definition). So, I'm going to wait a little bit to see if anything else crops up. Otherwise, I'll accept this and give you bounty. $\endgroup$ – jkn Apr 26 '13 at 18:03
  • 1
    $\begingroup$ Set-valued mappings, unlike single valued functions, can have the empty set as their values. Therefore you can consider a function $f$ as a set-valued map and define $f(x) = \emptyset$ outside the ortant. See the example before proposition 5.1.2 in "Set-valued analysis". $\endgroup$ – Leon Shutikoff Apr 26 '13 at 19:11

I came across the following paper by accident this morning while looking through some of my things. The paper appears to be a close fit to what you're asking about. I've included the introductory remarks that make up the first page of the paper.

José M. Bayod and J. M. Olazábal, Points of uniqueness of differentials, International Journal of Mathematical Education in Science and Technology 20 #3 (1989), 361-363.

In advanced calculus textbooks, the differentiability of functions of several variables is usually defined at points interior to the domain of definition of the function. The obvious reason for this is that on interior points you can easily ensure uniqueness of the differential (when it exists, of course).

On the other hand, for functions of one real variable it is not uncommon to consider differentiability on points that are not interior to the domain of the function – most notably on endpoints of an interval, where it is known as one-sided differentiability. Some texts find it convenient to consider functions defined on general (i.e., open or not) intervals when dealing with functions of one single variable, and consider only functions defined on open sets when dealing with functions of several variables (see, for example, [1], [2], [3]).

Even [4] {this is Flett's 1980 book Differential Analysis}, where a quite general theory can be found, studies differentiability of functions of one variable at non-isolated points of its domain (the most general possible meaningful approach, since at isolated points you can only obtain trivialities), but restricts itself to interior points when studying several variables functions. The same is done in other popular textbooks, such as [5].

In fact, whenever differentiability is needed at a point that is in the boundary of the domain of the function (for functions of several variables), the usual way to define the differential is via the differentiability of some extension of the original function to a neighbourhood of that point. In more specialized treatises, one can find the differential at an adherent point defined only over the tangent cone at that point (see [6]) {this is Federer's 1966 book Geometric Measure Theory}; the reason for that is the theorem below. Both our statement and our proof are readily accessible to students of a rigorous course in advanced calculus, and we feel that bright students could benefit from learning the geometrical implications of the statement (see the remarks at the end of this paper).

Our purpose is to prove that the points of any given subset $A$ of ${\mathbb R}^n$ where the usual definition of differentiability of a function must necessarily provide uniqueness of the differential, are exactly those points of $A$ where the tangent cone of $A$ spans ${\mathbb R}^{n}.$ The result we give may be a part of mathematical folklore, and is implicit in works like [6], but we have not been able to trace it in texts for students.

In order to keep the exposition as simple as possible, we only consider real functions of a finite number of real variables. For vector valued functions the changes are obvious; however, finite dimensionality is essential in the second part of our proof.

  • $\begingroup$ Thank you very much for posting this paper - it's insightful and the references discussed in it sound promising (in particular, Federer's book). $\endgroup$ – jkn Jun 11 '13 at 21:37

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