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Does a function f whose domain is a closed interval have an antiderivative (a function F (whose domain may not be the same as f) that is only differentiable on the domain of f, and the derivative is equal to f on the domain of f)?

For a function to be the antiderivative (with the definition above) of the above function, its domain must include values outside and near the endpoints of the closed interval in order to differentiable, and it also has to be not differentiable for those values.

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  • $\begingroup$ Sorry, what? Why is a constant function not an antiderivative to the $0$ function? $\endgroup$
    – FShrike
    Feb 26, 2022 at 19:13
  • $\begingroup$ The fundamental theorem of calculus' proof actually assumes that the domain is a compact interval $[a,b]$, where $F$ will be continuous, and that $F'$ must exist everywhere in the open subinterval $(a,b)$. In the reverse direction, $f$ is assumed continuous on the compact interval $[a,b]$ (otherwise $f$ might not be integrable...) $\endgroup$
    – FShrike
    Feb 26, 2022 at 19:14
  • $\begingroup$ @FShrike According to that definition of an antiderivative, they are not because derivatives have different domains. $\endgroup$
    – TFR
    Feb 26, 2022 at 20:19
  • $\begingroup$ @FShrike Does a function whose domain is a closed interval have an antiderivative? $\endgroup$
    – TFR
    Feb 26, 2022 at 20:20

1 Answer 1

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There appear to be two issues in question: The distinction between open and closed intervals, and differing domains of $f$ and $F$.

If $a < b$ are real, one says a real-valued function $f$ is differentiable on $[a, b]$ if $f$ extends to a differentiable function on a larger open interval $(\alpha, \beta) \supset [a, b]$.

If it's of interest, we say $f$ is continuously-differentiable if $f$ extends to a continuously-differentiable function on a larger open interval. This is equivalent to saying $f$ is continuous, differentiable in $(a, b)$, and the one-sided limits of the derivative exist at the endpoints.

Particularly, if $f$ is defined on an interval $I$, a function $F$ on $I$ is an antiderivative of $f$ if $F$ is differentiable in the preceding sense and $F' = f$ at each point of $I$. For example, if $f$ is identically zero on some interval $I$, then every constant function $F$ on $I$ is an antiderivative of $f$.

Finally, if it matters, "any function whose domain is one or more open interval has an antiderivative" isn't true, but is true if we assume $f$ is continuous (for example).

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  • $\begingroup$ So no function whose domain is one closed interval is differentiable on that closed interval because it is not defined for an open interval contains the domain? $\endgroup$
    – TFR
    Feb 26, 2022 at 21:37
  • $\begingroup$ Is it possible for a function to be differentiable at any point in a closed interval which is a subset of its domain, and not differentiable at any point outside that closed interval? $\endgroup$
    – TFR
    Feb 26, 2022 at 21:39
  • $\begingroup$ Just to calibrate regarding the first comment, have you heard the term extend, which is used here in a specific mathematical sense? (You've also changed the question, though this answer still addresses the new version: "Yes" a function on a closed bounded interval can have an antiderivative on that interval.) $\endgroup$ Feb 26, 2022 at 23:04
  • $\begingroup$ I was trying to find if there is any agreed upon definition, but it seems like there is no. I am looking for an antiderivative with my definition (see the bracket in my question). If so, could you give some examples? Is there a general one? $\endgroup$
    – TFR
    Feb 26, 2022 at 23:13
  • $\begingroup$ Not sure I understand what seems to be missing. There is a standard of antiderivative for a real-valued function on a real interval (which may or may not be closed and bounded), given in this answer. $\endgroup$ Feb 27, 2022 at 0:46

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