# Integral Inequality Absolute Value: $\left| \int_{a}^{b} f(x) g(x) \ dx \right| \leq \int_{a}^{b} |f(x)|\cdot |g(x)| \ dx$

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.

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

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?
• Since $|x| \leq a \Longleftrightarrow -a \leq x \leq a$, we have the desired result. – fourierguy Jun 25 '13 at 16:36
• Is $a < b$ assumed? – user124384 Feb 24 '16 at 4:27
• @user124384 since we are integrating over the interval $[a,b]$, the convention is that $a<b$. – Ephraim May 3 '17 at 4:40