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I know that this result is quite elementary, but nevertheless, I don't think that the result is trivial. So, let $f,g \in C^1 (\mathbb R)$ with $f, g, f', g' \in L^2 (\mathbb R)$.

Why is it true that $\displaystyle\int_{\mathbb R} f'g\,d\lambda = - \int_{\mathbb R} fg'\,d\lambda\;?$

Thanks for your help!

StMan
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    If this is supposed to be integration by parts, I think you're forgetting a term... – Arthur Dec 12 '19 at 13:44
  • @Arthur: In $L^2,$ with finite integrals for the functions and their derivatives on the entire space, the boundary term would have to vanish. – Adrian Keister Dec 12 '19 at 14:10
  • @AdrianKeister So this is about definite integrals over $\Bbb R$, not about indefinite integrals? I should've thought about that when we were talking about $L^2$. – Arthur Dec 12 '19 at 14:16
  • Is it true that L^1 functions vanish on infinity? I don't think so. –  Dec 12 '19 at 14:40
  • @user661541: It's $L^2,$ not $L^1,$ and we have the functions and their derivatives all in $L^2.$ I think that's sufficient to get vanishing at infinity. – Adrian Keister Dec 12 '19 at 17:42
  • @Arthur: Well, it's not clear, I'll give you that! It certainly looks like the integrals are being computed in the Lebesgue sense w.r.t. the measure $\lambda.$ I suppose you could define the indefinite integral, but the definite integral over all of $\mathbb{R}$ would seem to fit better, here. The OP should definitely clarify that. – Adrian Keister Dec 12 '19 at 17:43
  • @StMan: Do you mean the definite integral over all $\mathbb{R}?$ If so, please edit your question accordingly. – Adrian Keister Dec 12 '19 at 17:46
  • I edited the question. @AdrianKeister – StMan Dec 12 '19 at 18:15

1 Answers1

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You are right that this is usually treated "in the literature" (i.e., in textbooks and papers) as trivial, although it is not quite that.

First, note (by Cauchy Schwarz) that $F := f \cdot g \in L^1$ is continuously differentiable with $F'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x) \in L^1$. It is well-known (see for instance here Absolute continuity of the Lebesgue integral) that this ($F'$ being an integrable function) implies for $\epsilon > 0$ that there is $\delta > 0$ satisfying $\int_A |F'(x)|dx < \epsilon$ as soon as $\lambda(A) < \delta$ (here, $\lambda$ is the Lebesgue measure). Therefore, if $|x-y| < \delta$ and (without loss of generality) $x \leq y$, then $|F(x) - F(y)| \leq \int_x^y |F'(t)| dt \leq \epsilon$.

In other words, $F$ is uniformly continuous. Since $F$ is also integrable, this implies (see here: Limit at infinity of a uniformly continuous integrable function) that $F$ vanishes at infinity, that is, $F(x) \to 0$ as $x \to \infty$ or $x \to -\infty$.

Now, we can apply the fundamental theorem of calculus to deduce $$ \int_{\Bbb{R}} F'(t) dt = \lim_n \int_{-n}^n F'(t) d t = \lim_n \big[ F(n) - F(-n) \big] = 0. $$ If you recall that $F'(t) = f'(t) g(t) + f(t) g'(t)$, this implies the claim.

PhoemueX
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