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Mathematical Solutions and Examples

Exercise 1: Continuity and Bounded Variation

Function: \( g(x) = \begin{cases} 0 & \text{if } x = 0 \\ x \sin(1/x) & \text{if } 0 < x \leq 1 \end{cases} \) on \([0, 1]\)

Continuity:

• For \( x > 0 \), \( g(x) = x \sin(1/x) \). The function \( x \) is continuous, and \( \sin(1/x) \) is continuous on \( (0, 1] \) (since \( 1/x \) is continuous and \( \sin \) is continuous). Their product is continuous on \( (0, 1] \).
• At \( x = 0 \): \( \lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x \sin(1/x) \). Since \( |\sin(1/x)| \leq 1 \), \( -|x| \leq x \sin(1/x) \leq |x| \), and \( \lim_{x \to 0^+} |x| = 0 \). By the squeeze theorem, \( \lim_{x \to 0^+} g(x) = 0 = g(0) \).
• Thus, \( g(x) \) is continuous on \([0, 1]\).

Bounded Variation:

• The total variation \( V_0^1(g) = \sup \sum_{i=1}^n |g(x_i) - g(x_{i-1})| \) over all partitions \( \{x_0, x_1, \ldots, x_n\} \) of \([0, 1]\) must be finite for \( g \) to have bounded variation.
• Consider the sequence \( x_n = 1/(2n\pi + \pi/2) \), where \( \sin(1/x_n) = 1 \), so \( g(x_n) = x_n \).
• Then \( x_{n+1} = 1/(2(n+1)\pi + \pi/2) \), where \( \sin(1/x_{n+1}) = -1 \), so \( g(x_{n+1}) = -x_{n+1} \).
• The variation over \( [x_{n+1}, x_n] \) includes \( |g(x_n) - g(x_{n+1})| = |x_n - (-x_{n+1})| = x_n + x_{n+1} \).
• As \( n \to \infty \), \( x_n \approx 1/(n\pi) \), and \( \sum_{n=1}^\infty (x_n + x_{n+1}) \approx \sum 2/(n\pi) \), which diverges (harmonic series behavior).
• Thus, the total variation is unbounded, so \( g \) is not of bounded variation.

Exercise 2

Part a: Uniform Continuity

• Let \( g \) be continuous on the closed interval \([a, b]\). By the extreme value theorem, \( g \) is bounded and attains its maximum and minimum.
• Since \([a, b]\) is compact, the Heine-Cantor theorem states that \( g \) is uniformly continuous: for every \( \epsilon > 0 \), there exists \( \delta > 0 \) such that for all \( x, y \in [a, b] \), \( |x - y| < \delta \) implies \( |g(x) - g(y)| < \epsilon \).

Part b: Limit of Sum of Squared Differences

• Since \( g \) is uniformly continuous, for \( \epsilon > 0 \), there exists \( \delta > 0 \) such that \( |x - y| < \delta \) implies \( |g(x) - g(y)| < \sqrt{\epsilon / (b - a)} \).
• For a partition \( \Pi \) with \( \|\Pi\| = \max_{j} (x_{j+1} - x_j) < \delta \), \( |g(x_{j+1}) - g(x_j)| < \sqrt{\epsilon / (b - a)} \).
• Then, \( \sum_{j=0}^{n-1} [g(x_{j+1}) - g(x_j)]^2 < \sum_{j=0}^{n-1} (\sqrt{\epsilon / (b - a)})^2 = n \cdot \epsilon / (b - a) \).
• Since \( \sum_{j=0}^{n-1} (x_{j+1} - x_j) = b - a \), and as \( \|\Pi\| \to 0 \), \( \max_j |g(x_{j+1}) - g(x_j)| \to 0 \) (by uniform continuity), the sum approaches 0.

Additional Example 4: Continuity and Bounded Variation

Function: \( m(x) = |x - 1/2| \) on \([0, 1]\)

Continuity:

• \( m(x) = |x - 1/2| \) is a composition of the continuous function \( x - 1/2 \) and the absolute value function, both continuous everywhere. Thus, \( m(x) \) is continuous on \([0, 1]\).

Bounded Variation:

• The total variation \( V_0^1(m) = \sup \sum |m(x_i) - m(x_{i-1})| \).
• \( m(x) \) increases from 0 to 1/2 at \( x = 1/2 \), then decreases to 0 at \( x = 1 \), so the variation is \( 1/2 + 1/2 = 1 \), which is finite.
• Thus, \( m(x) \) is of bounded variation.

Additional Example 5: Uniform Continuity and Sum of Squares

Function: \( p(x) = \sin(x) \) on \([0, \pi]\)

Uniform Continuity:

• \( p(x) = \sin(x) \) is continuous on \([0, \pi]\) (closed and bounded), so by the Heine-Cantor theorem, it is uniformly continuous.

Limit of Sum:

• For partition \( \Pi \) with \( \|\Pi\| \to 0 \), \( \sum_{j=0}^{n-1} [p(x_{j+1}) - p(x_j)]^2 \).
• Since \( p'(x) = \cos(x) \) is bounded by 1, \( |p(x_{j+1}) - p(x_j)| \leq (x_{j+1} - x_j) \).
• Then, \( \sum (x_{j+1} - x_j)^2 \leq \|\Pi\| \cdot \pi \to 0 \) as \( \|\Pi\| \to 0 \).
• Thus, the limit is 0.

These examples reinforce the concepts with varied functions, ensuring a deeper understanding.