This is a follow-up question to my previous question.

Definition: A function $f: D \to \mathbb{R}$ where $D \subset \mathbb{R}$ is differentiable at a limit point $z \in D$ if

$\lim \limits_{h \to 0} \frac{f(z+h)-f(z)}{h}, h \neq 0, z+h \in D$ exists. We then say $f'(z)$ is the derivative of $f$ at $x$.

Note that the conditions $h \neq 0, z+h \in D$ are required to make sure the difference quotient is defined and $z$ needs to be a limit point of $D$ such it is possibe to find a sequence that approaches $z$ other than the constant sequence (for which the difference quotient is undefined). If this was not the case the limit would not be unique.

It is also easy to show from this definiton that this is equal to the existence of $c=f'(z) \in \mathbb{R}$ such that $\forall h:x+h \in D$:

$f(z+h)=f(z)+ch+r_{z}(h)$ with $\lim \limits_{h \to 0} \frac{r_{z}(h)}{h}=0$.

Note that $f'(z)h$ is a linear map defined for all $h:z+h \in D$. Again the conditions $h \neq 0, z+h \in D$ are required for the expression to make sense and $z$ must be a limit point for $c$ to be unqiue.

The limit definition is also easy to generalize to vector-valued functions, but does not make any sense for functions of more than one variable. So the idea is to define differentiability in terms of existence of a linear map.

Definition: Let $U \subset \mathbb{R^{n}}$ be open. A function $f: U \to \mathbb{R^{m}}$ is differentiable at $z \in U$ if there exists a linear map $L: \mathbb{R^{n}} \to \mathbb{R^{m}}$ such that $\forall h:x+h \in U$

$f(z+h)=f(z)+L_{z}(h)+r_{z}(h)$ with $\lim \limits_{h \to 0} \frac{r_{z}(h)}{\|h\|}=0$.

This definition is taken from some lecture notes on real analysis and the linear map $L$ can be thought of as the best approximation of the change in the value of the function $f$ near $z$.

Why we come up with the idea of differentiability as linear apprximability is clear to me. However, what I really don't understand is why we require $U$ to be open. Of course, we need to be able to approach a point in order for it to be differentiable, but I feel like the condition is stronger than necessary. In particular, we did not need it for the case of functions of one variable. Can't we again use the same conditons as in the one variable case?

Am I missing something? Any hints are appreciated. Thanks!

Edit: After thinking about it again something else came to my mind. If we do not require $U$ to be open, then it might be the case that $z$ is a limit point, but we can only approach it from some but not all directions. We then run into problems when we want to link the partial derivatives and the total differential since it might happen that some partial derivatives don't exist even though $f$ is differentiable.

  • $\begingroup$ The requirement that $h\neq0$ is nonsense; it is not a free variable in the expression $\lim \limits_{h \to 0} \frac{f(z+h)-f(z)}{h}$. Similarly $z+h\in D$ is nonsense. $\endgroup$ – Servaes Feb 16 '20 at 9:42
  • $\begingroup$ @Servaes Sorry, but what you have written is nonsense. Of course it makes sense to restrict the set from which a variable should be chosen in expressions for limits. $\endgroup$ – Thomas Feb 16 '20 at 10:03
  • $\begingroup$ @Thomas The 'restrictions' that $h\neq0$ and $z+h\in D$ are part of the definition of the limit itself. $\endgroup$ – Servaes Feb 16 '20 at 10:24
  • $\begingroup$ Not every limit is a quotient. I agree that when considering a limit it is clear that $h \neq 0$ by definition, but you can also consider a limit for a point that is not in the domain of a function as long as it is a limit point. This is exactly the difference between limits and continuity. $\endgroup$ – DerivativesGuy Feb 16 '20 at 10:33
  • $\begingroup$ @Servaes it may be redundant, but not false, in particular not 'nonsense'. $\endgroup$ – Thomas Feb 16 '20 at 10:34

If $U$ is an open set, we are sure it is "full dimensional", i.e. we are sure that every point can be approached from any direction. As you noted, this is important in the definition of differentiability, since otherwise even functions like $f(x,y)=\frac{xy}{x^2+y^2};f(0,0)=0$ could be seen as differentiable, taking an appropriate closed set $U$ in the definition, while as a function defined on $\mathbb{R}^2$ is not even continuous.

Let me add that the requirement for the domain to be open is quite transversal in multi-dimensional analysis. In fact, even once we have defined the differential, a function is usually said to be differentiable on $U$ if $U$ is open (or if $U$ is contained in an open set on which $f$ is differentiable).

This requirement is incredibly important in a lot of situations: for example, one needs it (together with some kind of connection hypotesis) to generalize to $\mathbb{R}^n$ the theorem $D(f)=0\rightarrow f(x)=\text{cost}$. If we drop it, the theorem is no longer true, as proved by H. Whitney in "a function not constant on a connected set of critical points" (see here)*.

*Actually, as shown in the article, the theorem can be extended with some restrictions to the set of critical points of $D(f)$, but they are quite technical.

  • $\begingroup$ @DerivativesGuy Inside the answer there's an example: on $U=\{y=0\}$ the function $\frac{xy}{x^2+y^2}$ is continuous and differentiable, while on $\mathbb{R}^2$ is not even continuous $\endgroup$ – Caffeine Mar 28 '20 at 22:43
  • $\begingroup$ Thanks. Yeah, it is probably a really bad behavior if a function's continuity and differentiability depends on its domain. Since then properties can change if one restricts the domain in applications for some reason. $\endgroup$ – DerivativesGuy Mar 29 '20 at 7:00

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