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.

  • 1
    $\begingroup$ What type of integration are you dealing with? Lebesgue or Riemann? $\endgroup$ – Nick Peterson Jun 25 '13 at 16:28
  • 1
    $\begingroup$ @nrpeterson Does it make a difference? $\endgroup$ – Amr Jun 25 '13 at 16:29
  • 6
    $\begingroup$ 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

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?

  • 4
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.