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If not all continuous functions are differentiable, so how is it that all continuous functions have anti-derivatives?

José Carlos Santos
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Sayed Abdullah Qutb
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  • Because $g(x)=\int_0^x f(t),dt$ is differentiable and $g'(x)=f(x)$. –  Dec 10 '17 at 08:48
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    Not all natural numbers can be divided by 2, how is it that all can be multiplied by 2? –  Dec 10 '17 at 08:49
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    Let's say very it openly: continuous functions possess anti-derivatives and the whole gamut of elementary calculus textbooks has failed miserably to explain why it is so. The proof is not easy and fundamental theorem of calculus is just one part (the easier one) of the story here. The more difficult part deals with the integrability of continuous functions. – Paramanand Singh Dec 10 '17 at 09:43
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    Also @ProfessorVector: both facts from number theory are almost trivial to prove. The case here is that it is difficult to show that continous functions possess anti-derivatives. I suppose your analogy is more to illustrate that such contrasts should not be hard to believe. – Paramanand Singh Dec 10 '17 at 09:46
  • @Paramanand Singh That's relative, I find that proof very simple, almost trivial. And it explains very well why it works: because a continuous function doesn't change much in a small interval. So upper and lower integrals don't differ much, and the mean value you get from the difference quotient of the integral function doesn't differ much from the function value. –  Dec 10 '17 at 10:08
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    @ProfessorVector : if it were that trivial most calculus books would have proved it. It is trivial only when you are already aware of it. The fact that continuous functions are integrable is a typical example of "theorems whose proofs are beyond the scope of the introductory books" but somehow their statements are in their scope. – Paramanand Singh Dec 10 '17 at 10:51

1 Answers1

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Let $I\subset\mathbb R$ be an interval with more than one point. If $f\colon I\longrightarrow\mathbb R$ is a continuous function, the existence of an anti-derivative of $f$ can be proved as follows: take $a\in I$ and define$$\begin{array}{rccc}F\colon&I&\longrightarrow&\mathbb R\\&x&\mapsto&\displaystyle\int_a^xf(t)\,\mathrm dt.\end{array}$$Then $F$ is a primitive of $f$, by the Fundamental Theorem of Calculus.

You seem to find it strange that every continuous has an anti-derivative while not all continuous functions are differentiable, but it's up to you to explain what's strange about it.

José Carlos Santos
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