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A differentiable function of one real variable is a function whose derivative exists at each point in its domain. Would the derivative still exists if the domain of the function is the null set?

For example, the function

$$ f(x)= \left( \sin \frac{1}{x} \right)^{\cos x} + \arcsin (4^x) + \frac{\sin x}{\ln x} $$ is undefined for any real values of $x$, hence, its domain is $\varnothing$. Are functions like this still differentiable at this condition?

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    $\begingroup$ Are you asking if the empty function is differentiable? $\endgroup$
    – Klaus
    May 24, 2021 at 12:45
  • $\begingroup$ Yes Sir. I am confused because for a function to be differentiable, it has to be continuous. And a function is continuous at some point c of its domain if the limit of f(x), as x approaches c through the domain of f, exists and is equal to f(c). But for the given function we can't choose a value of c since the domain is empty. This implies that the given function is not continuous and hence not differentiable? Am I understanding this correctly? $\endgroup$
    – user50605
    May 24, 2021 at 13:46

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This is really an answer to the comment, but I'll post this as an answer because it is too long for a comment and probably also resolves the question in the OP.

The empty function is continuous by definition. Indeed, according to your definition every $c \in \varnothing$ satisfies the requirement. That may look a little bit odd, but think about it this way: The function is continuous at no point. So the set of points where the function is continuous is the empty set. As the empty set is all you have, the function is continuous everywhere. It's just one of these oddities with the empty set like "all cows on the moon are green" is a true statement.

Similarly, you can argue that the empty function is differentiable.

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  • $\begingroup$ Hi..thank you for your insight. But I am quite confused on $c \in \varnothing$. If there is such $c$ in the empty set, then it can be argued that the empty is no longer empty which is a contradiction. Am I right, sir? $\endgroup$
    – user50605
    May 25, 2021 at 2:03
  • $\begingroup$ @user50605 The phrase "every $c \in \varnothing$" doesn't imply that there is an element in $\varnothing$. It just says that in case there is one, then the assertion holds. Of course, there is none, so you could claim anything. Similarly with the cow example above. Of course, there is no cow on the moon, so you might as well claim anything for cows on the moon. This really goes into the fundamentals of logic and for this I recommend reading the truth table article on Wikipedia, especially the part about logical implication. $\endgroup$
    – Klaus
    May 25, 2021 at 13:16

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