Question

Prove: If \(f\) and \(g\) are locally integrable on \([a, b)\) and the improper integrals \(\int_{a}^{b} f^{2}(x) d x\) and \(\int_{a}^{b} g^{2}(x) d x\) converge, then \(\int_{a}^{b} f(x) g(x) d x\) converges absolutely. HINT: \((f \pm\) \(g)^{2} \geq 0\).

Short answer

Question: Prove that if \(f\) and \(g\) are locally integrable on \([a, b)\) and the improper integrals \(\int_{a}^{b} f^{2}(x) d x\) and \(\int_{a}^{b} g^{2}(x) d x\) converge, then \(\int_{a}^{b} f(x) g(x) d x\) converges absolutely.

Step-by-step solution

Expand the inequality

Given the hint \((f \pm g)^{2} \geq 0\), expanding this, we have: \(f^2(x) \pm 2f(x)g(x) + g^2(x) \geq 0\)

Rearrange and sum inequalities

Rearranging the inequality, we have \(|f(x)g(x)| \leq \frac{1}{2}(f^2(x) + g^2(x))\) Add the inequalities corresponding to both \((f+g)^2 \geq 0\) and \((f-g)^2 \geq 0\) to obtain \(|f(x)g(x)| \leq \frac{1}{2}(f^2(x) + g^2(x))\)

Integrate both sides

Now, we integrate both sides of the inequality with respect to x over the interval \([a, b)\): \(\int_{a}^{b} |f(x)g(x)| d x \leq \frac{1}{2} \int_{a}^{b} (f^{2}(x) + g^{2}(x)) d x\)

Apply given convergence

We are given that the integrals \(\int_{a}^{b} f^{2}(x) d x\) and \(\int_{a}^{b} g^{2}(x) d x\) both converge. Therefore, the sum of these integrals converges as well: \(\int_{a}^{b} (f^{2}(x) + g^{2}(x)) d x\) converges

Show convergence of the absolute integral

Since \(\int_{a}^{b} |f(x)g(x)| d x \leq \frac{1}{2} \int_{a}^{b} (f^{2}(x) + g^{2}(x)) d x\) and the right-hand side of the inequality converges, by the comparison test, it follows that the left-hand side also converges. Hence, we have: \(\int_{a}^{b} |f(x)g(x)| d x\) converges This means that the integral of the product of \(f(x)\) and \(g(x)\) converges absolutely on the interval \([a, b)\).

Key concepts

Locally Integrable Functions

Understanding the concept of locally integrable functions is essential when dealing with improper integrals. A function is locally integrable over an interval \([a, b)\) if it can be integrated over every smaller closed sub-interval within \([a, b)\). This makes it possible to handle functions that may have singularities or are not well-behaved at some finite number of points in the interval. To determine if a function is locally integrable:

• Consider breaking down a larger interval into smaller segments where the function behaves nicely.
• Check that the integral over each smaller segment is finite.

If these smaller segments can sum up to cover the entire interval, then the function is said to be locally integrable. This property is a crucial requirement for the integrals in the exercise since it allows us to apply the necessary convergence tests and manipulations on segments where the functions are well-behaved.

Absolute Convergence of Integrals

In our context, when we say an integral converges absolutely, we mean that the integral of the absolute value of the function has a finite result. This ensures a stronger form of convergence compared to regular convergence, as it does not depend on the oscillations of the function across negative and positive values. To grasp absolute convergence:

• Consider the integral \(\int_{a}^{b} |f(x)g(x)| \, dx\).
• This must satisfy that summing the values of \(|f(x)g(x)|\) over \([a, b)\) results in a finite number.

This concept is pivotal in handling products of functions like \(f(x)g(x)\), where each function's behavior might still allow their combination to result in a convergent integral. Absolute convergence assures us that despite possibly significant oscillations, the net value behaves well enough over the interval.

Comparison Test for Convergence

The comparison test is a handy tool when determining the convergence of integrals. It involves comparing a given integral to another that is known to converge or diverge.Here's how to use it:

• Find a function \(|f(x)g(x)|\) and compare it to an easier functions sum like \(\frac{1}{2}(f^2(x) + g^2(x))\).
• Assess if the integral of this simpler function over \([a, b)\) converges.

If the simpler function's integral converges, and it acts as an upper bound to the integral you are investigating, it implies that the integral of \(|f(x)g(x)|\) converges as well.This method gives a relatively straightforward way to establish the behavior of complex functions by simplifying the problem into simpler, already known cases.