Question

Let \(f\) be integrable on \([a, b],\) let \(\alpha=\inf _{a \leq x \leq b} f(x)\) and \(\beta=\sup _{a \leq x \leq b} f(x),\) and suppose that \(G\) is continuous on \([\alpha, \beta]\). For each \(n \geq 1\), let $$ a+\frac{(j-1)(b-a)}{n} \leq u_{j n}, v_{j n} \leq a+\frac{j(b-a)}{n}, \quad 1 \leq j \leq n . $$ Show that $$ \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n}\left|G\left(f\left(u_{j n}\right)\right)-G\left(f\left(v_{j n}\right)\right)\right|=0 $$

Short answer

Question: Prove that the limit of the following expression is equal to zero: $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{j=1}^n |G(f(u_{jn})) - G(f(v_{jn}))| = 0$$ Where the function \(f\) is integrable on \([a, b]\), and the function \(G\) is continuous on \([\alpha, \beta]\), with \(\alpha = \inf_{a \leq x \leq b} f(x)\) and \(\beta = \sup_{a \leq x \leq b} f(x)\). Additionally, \(u_{jn}\) and \(v_{jn}\) are defined as: $$a + \frac{(j-1)(b-a)}{n} \leq u_{jn}, v_{jn} \leq a + \frac{j(b-a)}{n}, \quad 1 \leq j \leq n$$

Step-by-step solution

Find the difference inside the sum

First, we find the difference inside the sum: $$|G(f(u_{jn})) - G(f(v_{jn}))|$$ Notice that both \(u_{jn}\) and \(v_{jn}\) are in the same interval \([a + \frac{(j-1)(b-a)}{n}, a + \frac{j(b-a)}{n}]\).

Show \(f(u_{jn})\) and \(f(v_{jn})\) are bounded by \(\alpha\) and \(\beta\)

Because \(f\) is integrable on \([a, b]\), it must be bounded. So \(\alpha \leq f(u_{jn}), f(v_{jn}) \leq \beta\).

Use continuity of \(G\) to find an upper bound for the absolute value of the difference

Since \(f(u_{jn})\) and \(f(v_{jn})\) are bounded by \(\alpha\) and \(\beta\), and \(G\) is continuous on \([\alpha, \beta]\), then \(G(f(u_{jn}))\) and \(G(f(v_{jn}))\) are also bounded. Using the continuity of \(G\), we can find a \(\delta > 0\) and an \(N \in \mathbb{N}\) such that \(\delta < \frac{1}{N}(b-a)\). For all \(n > N\), we have: $$|G(f(u_{jn})) - G(f(v_{jn}))| < \epsilon$$ Where \(\epsilon\) is an arbitrary positive number.

Bound the summation using the obtained upper bound

Now, we can use the upper bound from the previous step to bound our summation: $$\frac{1}{n}\sum_{j=1}^{n} |G(f(u_{jn})) - G(f(v_{jn}))| < \frac{1}{n}\cdot n \cdot \epsilon = \epsilon$$

Using the bound, prove the limit

Since \(\epsilon\) is arbitrary, we can conclude that: $$\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{j=1}^n |G(f(u_{jn})) - G(f(v_{jn}))| = 0$$

Key concepts

Integrable Functions

Understanding the concept of integrable functions is vital in real analysis. An integrable function on a closed interval \text{[}a, b\text{]} is one for which the definite integral over that interval exists. In simple terms, an integrable function won't have infinite discontinuities or other peculiar behavior that would make the area under its curve impossible to calculate using standard methods.

In the context of the exercise, the function \text{f} being integrable implies that it is bounded and leads to the establishment of two critical values: the infimum \text{\(\alpha\)} and supremum \text{\(\beta\)}, which encapsulate the range of \text{f} on \text{[}a, b\text{]}. These bounds are essential for the following steps in demonstrating the limit's behavior.

Supremum and Infimum

The supremum and infimum are notions that represent the least upper bound and greatest lower bound of a set, respectively. For a given set of real numbers, the supremum (sup) is the smallest number that is greater than or equal to every number in the set, while the infimum (inf) is the largest number that is less than or equal to every number in the set.

In our exercise, we consider the infimum and supremum of the values of \text{f} in the interval \text{[}a, b\text{]}. These play a pivotal role in ensuring that the image \text{\(f(u_{jn})\)} and \text{\(f(v_{jn})\)} stay within the bounds for the function \text{G} to be continuous. This constraint facilitates the subsequent steps to approach the limit.

Continuity

Continuity in real analysis is a fundamental property of functions that intuitively implies no sudden jumps or breaks in the graph of the function. Formally, a function \text{G} is said to be continuous at a point \text{c} if for every \text{\(\epsilon > 0\)}, there exists a \text{\(\delta > 0\)} such that for all \text{x} in the domain of \text{G}, whenever \text{|x - c| < \(\delta\)} then \text{|G(x) - G(c)| < \(\epsilon\)}.

In our exercise, the function \text{G} is given to be continuous on the closed interval \text{[\(\alpha\), \(\beta\)]}. This continuity is crucial when finding an upper bound for the absolute value differences in the summation and ultimately proves that the sequence converges to 0 as \text{n} approaches infinity.

Limit of a Sequence

The limit of a sequence is a fundamental concept in real analysis that captures the idea of convergence. A sequence \text{\(\{a_n\}\)} is said to converge to a limit \text{L} if, for every positive number \text{\(\epsilon\)}, there exists a natural number \text{N} such that for all \text{n > N}, the terms of the sequence satisfy \text{|a_n - L| < \(\epsilon\)}.

In our exercise, we are asked to show that a particular sequence of sums converges to 0. By choosing an arbitrary positive number \text{\(\epsilon\)} and demonstrating that there exists a corresponding \text{N} beyond which the sums are indeed less than \text{\(\epsilon\)}, we establish the limit of the sequence as \text{n} grows without bound. The given solution walks through this process, employing continuity to ensure that the differences in the sum are bounded and can thus be made arbitrarily small—resulting in a limit of 0.