Given any continuous function f(x), does there always exist a function whose integral is f(x) ?

For instance, is there a function whose integral is Weierstrass function


No, any integral function of an $L^1$ function is absolutely continuous, while there exist continuous functions that aren't absolutely continuous. For example, the Weierstrass function, being not differentiable in any point, isn't of bounded variation (any function of bounded variation is differentiable almost everywhere) and then not absolutely continuous (because absolute continuity implies bounded variation). So the Weierstrass function is a continuous function that is not an integral function of any $L^1$ function.


If, when you write “integral”, what you mean is “primitive”, then the answer is negative, of course. Because if $F$ is a primitive of $f$ then, by definition, $F'=f$. In particular, $F$ is differentiable. But the Weierstrass function isn't.

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    $\begingroup$ But there can exist f such that F is not differentiable. For example, if f(x)=sign(1-x) for x in [0,2], then F is a triangular function which is not differentiable. Here f need not be continuous. $\endgroup$ – Vinay Sipani Jun 13 '18 at 8:01
  • $\begingroup$ But you wrote in your question that $f$ is continuous. $\endgroup$ – José Carlos Santos Jun 13 '18 at 8:04
  • $\begingroup$ Sorry, in my previous comment and in the answer, the roles of F and f are interchanged. f is continuous but F need not be continuous. I wanted to know, is there F such that integral (F) is f=weierstrass function. $\endgroup$ – Vinay Sipani Jun 13 '18 at 8:15
  • $\begingroup$ What do you by integral ($F$)? $\endgroup$ – José Carlos Santos Jun 13 '18 at 8:16
  • $\begingroup$ By integral(F), I mean integrating F in a suitable interval that has same domain as $f$. $\endgroup$ – Vinay Sipani Jun 13 '18 at 8:24

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