Let $I$ be an interval and $g$ be a continuous function $g\colon I\to\mathbb{R}$. Suppose that $f\colon I\to\mathbb{R}$ is a nonzero function and $f$ has antiderivative. Show that the product $fg$ has antiderivative.

I tried to use the fact that $fg=\frac{\left(f+g\right)^{2}-f^{2}-g^{2}}{2}$ and since $g$ is continuous then $g^{2}$ is also continuous and therefore it has antiderivative. Also $f+g$ has antiderivative, but I don't know if $f^{2}$ and $\left(f+g\right)^{2}$ have antiderivative. I couldn't prove that the square of these functions has antiderivative and also I don't know if this statement is false. Could you give me some suggestion?


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    $\begingroup$ If you're trying to show that $fg$ has an elementary antiderivative, it doesn't. Counterexample: $f(x)=1,$ and $g(x)=e^{-x^2}.$ If you're trying to show that $fg$ is integrable, then please change your wording, as that's quite different. $\endgroup$ – Adrian Keister Jun 28 '19 at 18:53
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    $\begingroup$ @AdrianKeister, the OP's terminology seems OK to me. See, e.g., en.wikipedia.org/wiki/Antiderivative $\endgroup$ – Barry Cipra Jun 28 '19 at 18:56
  • $\begingroup$ @AdrianKeister No, I only have to show that $fg$ has antiderivative, but not that is elementary. $\endgroup$ – oioa Jun 28 '19 at 19:07
  • $\begingroup$ The condition that $f$ is not zero must be relevant, compare math.stackexchange.com/q/2439669/42969. $\endgroup$ – Martin R Jun 28 '19 at 19:09
  • $\begingroup$ @MartinR May be that's important. Even so I don't know how can I prove that the product has antiderivative. Is there a sufficient condition to prove that a function has antiderivative? Since there are counterexamples that show that the function could not have an elementary antiderivative. $\endgroup$ – oioa Jun 28 '19 at 19:30

Short answer: If $F$ is the antiderivative of $f$, then: $$\int g \cdot f = \int g\circ \mathbb{1}_{I} \cdot f = \int g\circ F^{-1}\circ F \cdot f = \left(\int g\circ F^{-1}\right)\circ F$$

Long answer:

Let $F$ is the antiderivative of $f$. Obviously $F$ is continuous.

By Darboux's theorem, $f$ has the intermediate value property.

As $f(x)\ne 0$, it follows that either $f>0$ or $f<0$ (otherwise by intermediate value property $f$ would take the value zero).

It follows that $F$ is strictly monotone, therefore injective.

Let $J$ be the image of $F$. By intermediate value property, $J$ is an interval. Moreover, $F:I\to J$ is continuous and injective, so it has a continuous inverse $F^{-1}:J\to I$

Let $h:J\to \mathbb{R}$ with $h=g\circ F^{-1}$. $h$ is continuous as composition of two continuous functions, thus it has an antiderivative $H:J\to \mathbb{R}$

Then $H\circ F:I\to \mathbb{R}$ is differentiable, as composition of two differentiable functions, and by the chain rule: $$(H\circ F)'=H'\circ F\cdot F'=H'\circ F\cdot f=g\circ F^{-1}\circ F\cdot f=g\circ\mathbb{1}_I\cdot f=g\cdot f$$

That's all, folks!


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