Problem 9 - A Primer of Real Analysis

Question

Show that if \(a, b \in \mathbb{R}\) with \(a

Step-by-Step Solution

Answer

The weighted average of \( a \) and \( b \) is between \( a \) and \( b \) when \( a < b \) and \( 0 < \lambda < 1 \).

Step 1: Understand the Concept of a Weighted Average

The expression \( \lambda a + (1 - \lambda) b \) is a convex combination (weighted average) of \( a \) and \( b \). Since \( 0 < \lambda < 1 \), the coefficients \( \lambda \) and \( 1 - \lambda \) are both positive and sum to 1.

Step 2: Analyze the Condition \( a < b \)

We are given that \( a < b \), which means \( a \) is less than \( b \). Our goal is to show that the weighted average \( \lambda a + (1 - \lambda) b \) also falls between \( a \) and \( b \).

Step 3: Evaluate \( \lambda a + (1-\lambda) a < \lambda a + (1-\lambda)b \)

Consider \( \lambda a + (1-\lambda) a \). This simplifies to \( a \), since \( \lambda + (1-\lambda) = 1 \). We want to show \( a < \lambda a + (1-\lambda) b \). Since \( b > a \), \((1-\lambda)b > (1-\lambda)a \) for any \(0 < \lambda < 1\). Thus, \(a < \lambda a + (1-\lambda) b\).

Step 4: Evaluate \( \lambda a + (1-\lambda)b < \lambda b + (1-\lambda) b \)

We want to show that the other inequality, \( \lambda a + (1-\lambda)b < b \), holds. Notice this is equivalent to showing that \( \lambda a + (1-\lambda)b < \lambda b + (1-\lambda)b \), which simplifies to \( \lambda a + (1-\lambda)b < b \). Again, since \( a < b \), \( \lambda a < \lambda b \), ensuring the inequality holds.

Step 5: Conclude the Proof

Since both \( a < \lambda a + (1-\lambda) b\) and \( \lambda a + (1-\lambda) b < b \) are true, we have shown that \( a < \lambda a + (1-\lambda) b < b \). Hence, the statement is proven for \( \lambda \in (0, 1) \).

Key Concepts

Convex Combination

A convex combination is essentially a way to blend or combine two numbers, like turning two different colors of paint into a new shade. Mathematically, when you have real numbers \(a\) and \(b\), a convex combination looks like \( \lambda a + (1-\lambda) b \). The key is in the coefficients, \(\lambda\) and \(1-\lambda\), which are both non-negative and sum up to one.
This ensures that the new number \( \lambda a + (1-\lambda) b \) is a "mix" of \(a\) and \(b\). When \( \lambda \) is between 0 and 1, the combination always lands somewhere between \(a\) and \(b\).

• If \(\lambda\) is closer to 0, the combination leans towards \(b\).
• If \(\lambda\) is closer to 1, it favors \(a\).

This property is very useful in proving inequalities because it helps us show that \( \lambda a + (1-\lambda) b \) lies between \(a\) and \(b\), as shown in the original exercise.

Weighted Average

In simple terms, a weighted average is a "fair" means of averaging numbers, especially when some numbers are more important than others. Imagine you have two test scores, and one matters more because it's a bigger part of your overall grade. A weighted average accounts for that difference.
For any two numbers \(a\) and \(b\), the expression \( \lambda a + (1-\lambda) b \) acts as their weighted average. The weights, \( \lambda \) and \(1-\lambda\), define how much influence each number has on the final result. In this specific problem, \(\lambda\) falls between 0 and 1.

• \(\lambda\) determines the influence of \(a\).
• \(1-\lambda\) determines the influence of \(b\).

Since these weights sum to one, the formula keeps everything balanced, ensuring the average stays within the limits set by \(a\) and \(b\). Thus, the weighted average naturally falls between \(a\) and \(b\), which is crucial for understanding the inequalities in the exercise.

Inequalities

Inequalities are expressions about the size relationship between numbers, like saying one number is greater or less than another. They are often represented using symbols such as \(<\), \(>\), \(\leq\), or \(\geq\). In mathematics, proving inequalities means showing that these relationships are always true under certain conditions.
In the given exercise, the inequality we need to prove is \( a < \lambda a + (1-\lambda) b < b \). This involves two smaller inequalities:

• First, \( a < \lambda a + (1-\lambda) b \), meaning the weighted average must be greater than \(a\).
• Second, \( \lambda a + (1-\lambda) b < b \), indicating it must be less than \(b\).

By combining these, we neatly position the weighted average between \(a\) and \(b\). Thus, inequalities help us describe precise mathematical boundaries and relationships between numbers, like meeting a goalpost in a target range. By rigorously proving these inequalities through logical steps, as we did in the solution, we confirm their validity in ensuring such balance is always maintained.