Question

If \(f,g\) are bounded functions defined on \([a,b]\) and let \(P\) be any partition of \([a,b]\), then which of the following is NOT TRUE?

• A. \(L(P,-f)=-U(P,f)\)
• B. \(U(P,-f)=-L(P,f)\)
• C. \(L(P,f+g)\leq L(P,f)+L(P,g)\)
• D. \(U(P,f+g)\leq U(P,f)+U(P,g)\)

Hint:

For lower sums, \(L(P,f+g)\geq L(P,f)+L(P,g)\). For upper sums, \(U(P,f+g)\leq U(P,f)+U(P,g)\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question tests the properties of lower and upper Darboux sums for bounded functions over a partition P of an interval [a,b]. We need to verify the given relationships involving the sums for \(-f\) and \(f+g\).

Step 2: Key Formula or Approach:
Let \(m_i(f) = \inf_{x \in [x_{i-1}, x_i]} f(x)\) and \(M_i(f) = \sup_{x \in [x_{i-1}, x_i]} f(x)\). Then \(L(P,f) = \sum m_i(f) \Delta x_i\) and \(U(P,f) = \sum M_i(f) \Delta x_i\). We use the properties of infimum and supremum: - \(\inf(-f) = -\sup(f)\) and \(\sup(-f) = -\inf(f)\). - \(\sup(f+g) \le \sup(f) + \sup(g)\). - \(\inf(f+g) \ge \inf(f) + \inf(g)\).

Step 3: Detailed Explanation:
Let's analyze each option: - (A) \(L(P, -f) = -U(P, f)\): For any subinterval, \(m_i(-f) = \inf(-f) = -\sup(f) = -M_i(f)\). So, \(L(P, -f) = \sum m_i(-f) \Delta x_i = \sum (-M_i(f)) \Delta x_i = - \sum M_i(f) \Delta x_i = -U(P, f)\). This statement is TRUE. - (B) \(U(P, -f) = -L(P, f)\): For any subinterval, \(M_i(-f) = \sup(-f) = -\inf(f) = -m_i(f)\). So, \(U(P, -f) = \sum M_i(-f) \Delta x_i = \sum (-m_i(f)) \Delta x_i = - \sum m_i(f) \Delta x_i = -L(P, f)\). This statement is TRUE. - (D) \(U(P, f+g) \le U(P, f) + U(P, g)\): For any subinterval, \(M_i(f+g) = \sup(f+g) \le \sup(f) + \sup(g) = M_i(f) + M_i(g)\). So, \(U(P, f+g) = \sum M_i(f+g) \Delta x_i \le \sum (M_i(f) + M_i(g)) \Delta x_i = \sum M_i(f) \Delta x_i + \sum M_i(g) \Delta x_i = U(P, f) + U(P, g)\). This statement is TRUE. - (C) \(L(P, f+g) \le L(P, f) + L(P, g)\): For any subinterval, \(m_i(f+g) = \inf(f+g) \ge \inf(f) + \inf(g) = m_i(f) + m_i(g)\). So, \(L(P, f+g) = \sum m_i(f+g) \Delta x_i \ge \sum (m_i(f) + m_i(g)) \Delta x_i = \sum m_i(f) \Delta x_i + \sum m_i(g) \Delta x_i = L(P, f) + L(P, g)\). The correct inequality is \(L(P, f+g) \ge L(P, f) + L(P, g)\). The statement given in option (C) has the inequality reversed (\(\le\)). Therefore, this statement is NOT TRUE in general.

Step 4: Final Answer:
The statement \(L(P, f+g) \le L(P, f) + L(P, g)\) is not generally true; the correct inequality is \(L(P, f+g) \ge L(P, f) + L(P, g)\). This corresponds to option (C).