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

To solve the problem, we need to determine whether the supremum values of \( |T(f)| \) and \( |S(f)| \) are attained at points in their respective closed unit balls, \( U_X \) and \( U_Y \).

Step 1: Understanding the Function Spaces

• X = \{ f \in C[0,1] : f(0) = 0 = f(1) \} is the space of all continuous functions that equal zero at the endpoints of the interval \([0,1]\), with the norm \|f\|_\infty = \sup_{0 \leq t \leq 1} |f(t)|.
• Y = C[0,1] is the space of all continuous functions on \([0,1]\) with the norm \|f\|_2 = \left( \int_0^1 |f(t)|^2 \, dt \right)^{\frac{1}{2}}, the \(L^2\) norm.

Step 2: Evaluating \( T: X \to \mathbb{R} \)

The operator \( T(f) = \int_0^1 f(t) \, dt \) is evaluated over the unit ball in \( X \). Consider:

• The norm \|f\|_\infty \leq 1, which means the maximum absolute value of \( f(t) \) over \([0,1]\) is 1.
• Since f(0) = 0 = f(1), it limits the form of \( f \).
• By the intermediate value property, the integral may not reach its maximum over \( U_X \) because the function must remain continuous and zero at ends.

Therefore, the supremum of \( |T(f)| \) is not generally attained at a point of \( U_X \) because the constraints limit \( f(t) \)'s ability to reach extreme values without exceeding the norm bound.

Step 3: Evaluating \( S: Y \to \mathbb{R} \)

For \( S(f) = \int_0^1 f(t) \, dt \) in the context of \( Y \), we have:

• The \(L^2\) norm \|f\|_2 \leq 1 implies constraints on the average energy or magnitude, not the pointwise values.
• Choosing f(t) = 1 demonstrates feasibility since it is continuous, maintains the \(L^2\) norm of 1 and achieves an integral value of 1, attaining the supremum for \( |S(f)| \).

Therefore, the supremum of \( |S(f)| \) is attained at a point of \( U_Y \) because the norm constraint in \( Y \) allows enough flexibility for constant functions to reach the maximal integral value.

Conclusion:

• Statement S1: \( \sup |T(f)| \) is not attained at a point of \( U_X \), so S1 is FALSE.
• Statement S2: \( \sup |S(f)| \) is attained at a point of \( U_Y \), so S2 is TRUE.

Therefore, the correct answer is: S2 is TRUE and S1 is FALSE.