Question

Let \( X = \{ f \in C[0,1] : f(0) = 0 = f(1) \} \) with the norm \( \|f\|_\infty = \sup_{0 \leq t \leq 1} |f(t)| \), where \( C[0,1] \) is the space of all real-valued continuous functions on \( [0,1] \).

Let \( Y = C[0,1] \) with the norm \( \|f\|_2 = \left( \int_0^1 |f(t)|^2 \, dt \right)^{\frac{1}{2}} \). Let \( U_X \) and \( U_Y \) be the closed unit balls in \( X \) and \( Y \) centered at the origin, respectively. Consider \( T: X \to \mathbb{R} \) and \( S: Y \to \mathbb{R} \) given by

\[ T(f) = \int_0^1 f(t) \, dt \quad \text{and} \quad S(f) = \int_0^1 f(t) \, dt. \]

Consider the following statements:
S1: \( \sup |T(f)| \) is attained at a point of \( U_X \).
S2: \( \sup |S(f)| \) is attained at a point of \( U_Y \).

Then, which one of the following is correct?

• A. S1 is TRUE and S2 is FALSE
• B. S2 is TRUE and S1 is FALSE
• C. both S1 and S2 are TRUE
• D. neither S1 nor S2 is TRUE

Hint:

When working with suprema in functional spaces, check the properties of the function spaces and the behavior of integrals with respect to the norms to determine where the supremum is attained. The \( L_\infty \) and \( L_2 \) norms may yield different results in terms of where the supremum is reached.

Answer 1

The Correct Option is B

We need to check the two statements \( S1 \) and \( S2 \) and verify if the suprema are attained at a point of the respective unit balls.

Step 1: Analyzing S1:

For \( S1 \), consider the supremum of \( |T(f)| = \left|\int_0^1 f(t) \, dt\right| \). The function \( f(t) = t(1 - t) \) maximizes this integral within the unit ball \( U_X \). However, the supremum of \( |T(f)| \) is not attained at any point of \( U_X \) since the supremum involves a function that is not contained in the closed unit ball of the \( L_\infty \) norm. For functions in \( U_X \), the integral of \( f(t) \) cannot reach its maximum value, and thus, S1 is FALSE.

Step 2: Analyzing S2:

For \( S2 \), consider the supremum of \( |S(f)| = \left|\int_0^1 f(t) \, dt\right| \) with respect to the \( L_2 \) norm. The function \( f(t) = 1 \) maximizes the integral of \( f(t) \) while staying within the unit ball \( U_Y \) in the \( L_2 \) norm. Specifically, we calculate:

\[ \int_0^1 1 \, dt = 1, \quad \text{and} \quad \|f\|_2 = \left( \int_0^1 1^2 \, dt \right)^{\frac{1}{2}} = 1. \]

Therefore, \( |S(f)| \) attains its supremum at \( f(t) = 1 \), which is an element of the unit ball \( U_Y \). Hence, S2 is TRUE.

Thus, the correct answer is:

\[ \boxed{S2 \text{ is TRUE and } S1 \text{ is FALSE.}} \]