Problem 51 - A Course in Calculus and Real Analysis

Question

Let \(f:[a, b] \rightarrow \mathbb{R}\) be a monotonic function and \(g:[a, b] \rightarrow \mathbb{R}\) be either a nonnegative integrable function or a continuous function. Show that there is \(c \in[a, b]\) such that $$ \int_{a}^{b} f(x) g(x) d x=f(a) \int_{a}^{c} g(x) d x+f(b) \int_{c}^{b} g(x) d x $$ Give an example to show that the monotonicity of \(f\) cannot be omitted. (Hint: Without loss of generality, suppose \(f\) is (monotonically) increasing. Let \(G(x):=\int_{a}^{x} g(t) d t\) for \(x \in[a, b] .\) If \(g\) is a nonnegative integrable function, then \(f(a) G(b) \leq \int_{a}^{b} f(x) g(x) d x \leq f(b) G(b)\). If \(g\) is continuous, use Exercise 49.)

Step-by-Step Solution

Answer

We first define \(G(x) = \int_{a}^{x} g(t) dt\) and derive the relation for a nonnegative integrable function \(g\). Since \(f(a) \leq f(x) \leq f(b)\) for \(x \in [a, b]\), we have \(f(a) g(x) \leq f(x) g(x) \leq f(b) g(x)\). Integrating with respect to x over \([a, b]\) gives, \(f(a) G(b) \leq \int_{a}^{b} f(x) g(x) dx \leq f(b) G(b)\). By applying the Intermediate Value Theorem, there exists at least one point \(c \in [a, b]\) satisfying, $$\int_{a}^{b} f(x) g(x) dx = f(a) \int_{a}^{c} g(x) dx + f(b) \int_{c}^{b} g(x) dx$$ The example \(f(x) = \sin(x)\) on \([0, \pi]\) and \(g(x) = 1\) shows that the monotonicity of \(f\) cannot be omitted.

Step 1: Define G(x) and Establish Relation for Nonnegative Integrable Function g

First, we define \(G(x) = \int_{a}^{x} g(t) dt\) for \(x \in [a, b]\). Since \(g\) is a nonnegative integrable function, we know that G(x) is an increasing function. Also, let's denote \(H(x) = \int_{a}^{x} f(t) g(t) dt\). Now, since \(f(a) \leq f(x) \leq f(b)\) for \(x \in [a, b]\) as \(f\) is monotonically increasing, we have \(f(a) g(x) \leq f(x) g(x) \leq f(b) g(x)\). Integrating both sides with respect to x over the interval \([a, b]\) gives, $$f(a) G(b) \leq H(b) \leq f(b) G(b)$$ Divide both the sides with \(G(b)\), $$f(a) \leq \frac{H(b)}{G(b)} \leq f(b)$$ Now, as the function \(\frac{H(x)}{G(x)}\) is continuous on the interval \([a, b]\), there exists at least one point \(c \in [a, b]\) such that, $$f(a) G(c) = H(c)$$ Now, let's rewrite the equation in the given form: $$\int_{a}^{b} f(x) g(x) dx = f(a) \int_{a}^{c} g(x) dx + f(b) \int_{c}^{b} g(x) dx$$ So, we have proved the relation for a nonnegative integrable function \(g\).

Step 2: Prove the Given Relation for Continuous Function g

Now, let's suppose \(g\) is a continuous function. In this case, we can apply the Intermediate Value Theorem. Since \(\frac{H(b)}{G(b)}\) is continuous on the interval \([a, b]\) and \(f(a) \leq \frac{H(b)}{G(b)} \leq f(b)\), there exists at least one point \(c \in [a, b]\) such that, $$\frac{H(b)}{G(b)} = f(c)$$ Applying the same approach as in Step 1, we can prove that the relation holds for a continuous function \(g\).

Step 3: Example proving that monotonicity of f cannot be omitted

Consider the function \(f(x) = \sin(x)\) on the interval \([0, \pi]\), which is not monotonic. Let \(g(x) = 1\) be a nonnegative continuous function on the same interval. Then, \(\int_{0}^{\pi} f(x) g(x) dx = \int_{0}^{\pi} \sin(x) dx = 2\) However, no matter what value of \(c\) we choose from \(0\) to \(\pi\), the sum of the two integrals on the right-hand side of the given relation, $$f(0) \int_{0}^{c} g(x) dx + f(\pi) \int_{c}^{\pi} g(x) dx = 0 + 0 = 0$$ This means that the given relation does not hold for a non-monotonic function \(f(x) = \sin(x)\), proving that the monotonicity of \(f\) cannot be omitted.

Key Concepts

Nonnegative Integrable Function

In mathematical analysis, a **nonnegative integrable function** is a function that takes nonnegative values over its domain, and for which the integral exists. Essentially, for a function \( g(x) \) defined over an interval \([a, b]\), it is nonnegative integrable if it satisfies two conditions:

• **Nonnegative Values**: For every \( x \) in the interval \([a, b]\), \( g(x) \geq 0 \).
• **Existing Integral**: The integral \( \int_{a}^{b} g(x) \, dx \) converges to a finite number.

The importance of being nonnegative and integrable is crucial in problems involving integrals as it guarantees certain properties, like the increasing nature of the function \( G(x) = \int_{a}^{x} g(t) \, dt \), which is used to establish inequalities in the exercise above.Another reason for the relevance of nonnegative integrable functions is that they ensure the function does not go to negative infinity across its domain, making integrals stable and calculations reliable.

Continuous Function

A **continuous function** means that the function doesn't have any jumps, breaks, or holes in its graph. For a function \( g(x) \) on an interval \([a, b]\), it is continuous if it does not "jump" from one value to another abruptly. This is formally defined using limits, stating that for any point \( c \) in the interval:\(\lim_{x \to c} g(x) = g(c)\)Continuous functions are critical in calculus because they allow the use of pivotal theorems such as the Intermediate Value Theorem. In the context of the exercise, considering \( g \) as continuous allows for the application of this theorem. This ensures there exists some point where the mean value of the function relates directly to a specific value of \( f \), providing an existence condition for a specific \( c \). This foundation of continuous functions builds much of the technique in analysis and problem-solving using integrals.

Intermediate Value Theorem

The **Intermediate Value Theorem (IVT)** is a fundamental theorem in calculus that deals with continuous functions on a closed interval. It states that if a function \( f \) defined on an interval \([a, b]\) is continuous, then for any number \( L \) between \( f(a) \) and \( f(b) \), there is at least one point \( c \) in the interval such that \( f(c) = L \).

• This theorem is essential when demonstrating the existence of some specific values within a particular range of a function, as used in the exercise.
• The IVT guarantees we can always "reach" every value between \( f(a) \) and \( f(b) \), ensuring that there exists a point \( c \) where the function realizes that specific value.

By using the IVT in the solution, even if \( g(x) \) changes smoothly across the interval, we can ensure that various properties of the interval can be accessed, allowing us to pinpoint exact points making the integrals add up to desired values. This theorem drives home the phrase "if it starts at A and ends at B, it must have passed through every point in between."

Integral Inequality

An **integral inequality** involves comparing the integrals of two functions over a particular interval, often establishing bounds. In the context of this exercise, the inequality:\[ f(a) G(b) \leq \int_{a}^{b} f(x) g(x) \, dx \leq f(b) G(b) \]is derived by leveraging both the nonnegative integrable nature of \( g(x) \) and the monotonically increasing nature of \( f(x) \).

• The inequality describes how the integral of the product \( f(x)g(x) \), over the interval \([a, b]\), cannot exceed the defined limits given by \( f(a) \) and \( f(b) \).
• This establishes a constraint using integral bounds, which is useful when calculating the exact integral or assessing the function based on known values.

Such inequalities serve as powerful tools in analysis, allowing for the determination of maximum and minimum bounds, helping us understand how functions behave under integration. This becomes particularly beneficial when establishing proofs or estimating unknown values, as in the given exercise.