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I Must functions really have interval domains for derivatives?

  1. Jul 26, 2017 #1
    Nearly every analysis reference I come across defines the derivative for functions on an open interval ##f:(a, b) \rightarrow \mathbb{R}##. I understand that, in constructing the definition of ##f## being differentiable on a point ##c##, we of course want it to first be a point it's domain, so require ##c \in dom(f)##. In order to make sense of the limit ##\lim_{x \to c} \frac{f(x) - f(c)}{x - c}##, we must make sure that the expression is a function, on an appropriate domain. It is easy to see that it indeed is, and has domain ##dom{f} - \{c\}##. Thus, we can form the difference quotient function ##\phi: dom(f) - \{c\} \rightarrow \mathbb{R}##, defined by, for ##x \in dom(f) - \{c\}##, ##\phi(x) = \frac{f(x) - f(c)}{x - c}##. So then we would have that ##\lim_{x \to c} \phi(x) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}## as intended. Now, for us to be able to even consider the limit, we would then need ##c## to be a limit point of ##dom{f} - \{c\}##, which we could satisfy by letting ##c## be a limit point of ##dom{f}##. Now, I am aware that one can define differentiation for a function on closed intervals if one considers one-sided derivatives, or one could also go to general open sets which wouldn't cause much more issue since each point in an open set is locally contained in an interval. However, given my flow of reasoning above, should it be correct, wouldn't this mean that one could define differentiation of a point in a domain only for which the point is a limit point? I know that if it's in an interval to begin with, then it is a limit point, but this seems like an arbitrary restriction. Thanks in advance for any response.
     
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  3. Jul 26, 2017 #2

    andrewkirk

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    I can't see any reason why one could not do what you say. If the domain of a function is everywhere dense in the reals, as the rationals and irrationals both are, I suspect a generalised concept of differentiation could be defined, along the lines you suggest. More broadly, generalised differentiation could be defined at any points of the domain that are limit points of the reals. For instance, if the domain is
    $$\{b/n\ :\ b\in\{-1,0,1\},\ n\in\mathbb Z,n>0\}$$
    then a function on that domain may be differentiable at 0, but nowhere else.

    I'd guess the reason analysis texts use the more restrictive definition is that some of the theorems about derivatives and differentiability may no longer apply under a more general approach, making it a less useful concept.
     
  4. Jul 26, 2017 #3

    fresh_42

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    So far so good, except that I cannot see why you correctly said ##c \in \operatorname{dom}(f)## has to be required and then remove of point from the domain. If ##f(c)## isn't defined, ##\left.\frac{d}{dx}\right|_{x=c} f(x)## doesn't make any sense.
    This is not true. You have to be able to approach ##c##. The interval ##[c,c]## is closed and you cannot define a derivative here.
    This is the definition of open sets. It guarantees that a two-sided approach to ##c## is available. That's the only reason.
    How? E.g. how would you define ##\left.\frac{d}{dx}\right|_{x=0} |x|\,##?
    It's not arbitrary. It reflects the fact, that differentiability is a local phenomenon, and local means open set.
     
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