91Ó°ÊÓ

Recall that the upper sum \(U(f)\) of a function \(f\) over the interval \([a, b]\) is defined as: \[U(f) = \sum_{i=1}^{n}M_i (x_i - x_{i-1})\] where \(M_i = \sup\{f(x) : x_{i-1} \leq x \leq x_i\}\) represents the maximum value in the \(i\)-th subinterval. Now, we will determine the maximum value of \(f + g\) over the interval \([a, b]\). By definition: \(M_1^{(f+g)} = \sup \{(f + g)(x) : x_0 \leq x \leq x_1\}\) Using properties of supremum, we have: \(M_1^{(f+g)} \leq \sup \{f(x) : x_0 \leq x \leq x_1\} + \sup \{g(x) : x_0 \leq x \leq x_1\}\) \(M_1^{(f+g)} \leq M_1^f + M_1^g\) Now for the upper sum, \(U(f+g) = \sum_{i=1}^{n} M_i^{(f+g)} (x_i - x_{i-1})\), we have: \(U(f+g) \leq \sum_{i=1}^{n} (M_i^f + M_i^g) (x_i - x_{i-1})\) Now, apply the distributive property: \(U(f+g) \leq \sum_{i=1}^{n} M_i^f (x_i - x_{i-1}) + \sum_{i=1}^{n} M_i^g (x_i - x_{i-1})\) \(U(f+g) \leq U(f) + U(g)\) Hence, we have proved that \(U(f+g) \leq U(f) + U(g)\).

Now, let's construct an example that shows that strict inequality can hold in part (a). Let's consider the functions \(f(x) = x\) and \(g(x) = -x\) on the interval \([0, 1]\). The function \(f+g\) is just \(f(x) + g(x) = x-x = 0\). Hence, \(U(f+g) = 0\) for any partition of the interval \([0, 1]\). Now let's compute the upper sum \(U(f)\) and \(U(g)\). We will take a simple partition of the interval, i.e., \(x_0 = 0\) and \(x_1 = 1\). For \(f\), the supremum \(M_1^f = \sup \{f(x) : x_0 \leq x \leq x_1\} = 1\), so \(U(f) = M_1^f(x_1 - x_0) = 1\). For \(g\), the supremum \(M_1^g = \sup \{g(x) : x_0 \leq x \leq x_1\} = 0\), so \(U(g) = M_1^g(x_1 - x_0) = 0\). Thus, \(U(f) + U(g) = 1 + 0 = 1\). In this example, we have \(U(f+g) = 0\) and \(U(f) + U(g) = 1\). Therefore, the strict inequality \(U(f+g) < U(f) + U(g)\) holds.