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Suppose we are given the following: $$\left| \int_{a}^{b} f(x) g(x) \ dx \right| \leq \int_{a}^{b} |f(x)|\cdot |g(x)| \ dx$$

How would we prove this? Does this follow from Cauchy Schwarz? Intuitively this is how I see it: In the LHS we could have a negative area that reduces the positive area. In the RHS the area can only increase because we take the absolute values of the functions first.

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    $\begingroup$ What type of integration are you dealing with? Lebesgue or Riemann? $\endgroup$ – Nick Peterson Jun 25 '13 at 16:28
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    $\begingroup$ @nrpeterson Does it make a difference? $\endgroup$ – Amr Jun 25 '13 at 16:29
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    $\begingroup$ His level and what he knows about integration make a difference in how I would explain a proof of the inequality, yes. $\endgroup$ – Nick Peterson Jun 25 '13 at 16:30
  • $\begingroup$ @nrpeterson: Riemann. $\endgroup$ – fourierguy Jun 25 '13 at 16:33
  • $\begingroup$ Would it be sufficient to expand both sides to their Riemann sum definitions and apply the triangle inequality? $\endgroup$ – A.E Jun 25 '13 at 16:34
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The big idea here is this:

First: it is enough to show that $$ \left\lvert\int_a^b f(x)\,dx\right\rvert\leq\int_a^b\lvert f(x)\rvert dx, $$ since you can replace $f(x)$ by $f(x)\cdot g(x)$ to get the desired result.

Now, notice that $$ -\lvert f(x)\rvert\leq f(x)\leq \lvert f(x)\rvert $$ for all $x$; hence $$ -\int_a^b\lvert f(x)\rvert\,dx\leq \int_a^b f(x)\,dx\leq\int_a^b\lvert f(x)\rvert\,dx. $$ Can you finish it from here?

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    $\begingroup$ Since $|x| \leq a \Longleftrightarrow -a \leq x \leq a$, we have the desired result. $\endgroup$ – fourierguy Jun 25 '13 at 16:36
  • $\begingroup$ That's exactly it. $\endgroup$ – Nick Peterson Jun 25 '13 at 16:36
  • $\begingroup$ Is $a < b$ assumed? $\endgroup$ – user124384 Feb 24 '16 at 4:27
  • $\begingroup$ @user124384 since we are integrating over the interval $[a,b]$, the convention is that $a<b$. $\endgroup$ – Ephraim May 3 '17 at 4:40
  • $\begingroup$ @fourierguy I think, it would be better to use a different letter than "a". It got me confused with the boundary of the integral for a while. $\endgroup$ – 2chromatic Sep 17 '18 at 10:51

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