Question

Let C[0,1] denote the set of all real valued continuous functions defined on [0,1] and \(||f||_\infty = \sup\{|f(x)| : x \in [0,1]\}\) for all \(f \in C[0,1]\). Let
\[ X = \{ f \in C[0,1] : f(0) = f(1) = 0 \}. \]
Define \(F : (C[0,1], ||.||_\infty) \to \mathbb{R}\) by \(F(f) = \int_0^1 f(t)dt\) for all \(f \in C[0,1]\).
Denote \(S_X = \{f \in X : ||f||_\infty = 1\}\).
Then the set \(\{f \in X : F(f) = ||F||\} \cap S_X\) has

• A. NO element
• B. exactly one element
• C. exactly two elements
• D. an infinite number of elements

Hint:

When checking if a functional on a space of continuous functions attains its norm, first find the norm (usually via an inequality and a sequence approaching the bound). Then, analyze the condition for equality in the inequality. Often, this equality condition forces the function to have a property (like being constant) that disqualifies it from being in the original space.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The question asks about the existence of a function that attains the norm of a linear functional. We are given the space \(X\) (continuous functions on [0,1] that are zero at the endpoints), a linear functional \(F\) (the definite integral), and the unit sphere \(S_X\) in this space. The set in question is the set of functions on the unit sphere for which the functional \(F\) achieves its maximum possible value, i.e., its norm.
Step 2: Key Formula or Approach:
1. Calculate the norm of the functional \(F\) when restricted to the subspace \(X\). The norm is given by \(||F|| = \sup_{f \in S_X} |F(f)|\).
2. Determine if there exists a function \(f^* \in S_X\) such that \(F(f^*) = ||F||\). This function is called a norm-attaining function.
Step 3: Detailed Calculation:
Step 3.1: Calculate the norm of F on X.
For any \(f \in S_X\), we have \(||f||_\infty = 1\), which means \(|f(t)| \le 1\) for all \(t \in [0,1]\).
The value of the functional is \(F(f) = \int_0^1 f(t)dt\).
Taking the absolute value:
\[ |F(f)| = \left| \int_0^1 f(t)dt \right| \le \int_0^1 |f(t)|dt \le \int_0^1 ||f||_\infty dt = \int_0^1 1 dt = 1 \] This shows that \(||F|| \le 1\).
To show that \(||F|| = 1\), we must find a sequence of functions \(f_n \in S_X\) such that \(F(f_n) \to 1\). Consider a sequence of "trapezoidal" or "tent" functions \(f_n \in S_X\). For \(n \ge 2\), define \(f_n(t)\) as:
- Linearly increasing from 0 to 1 on \([0, 1/n]\).
- Constant at 1 on \([1/n, 1 - 1/n]\).
- Linearly decreasing from 1 to 0 on \([1 - 1/n, 1]\).
Each \(f_n\) is continuous, \(f_n(0) = f_n(1) = 0\), and \(||f_n||_\infty = 1\), so \(f_n \in S_X\).
The integral is the area under the curve:
\[ F(f_n) = \int_0^1 f_n(t)dt = \frac{1}{2}\left(\frac{1}{n}\right)(1) + \left(1-\frac{2}{n}\right)(1) + \frac{1}{2}\left(\frac{1}{n}\right)(1) = 1 - \frac{1}{n} \] As \(n \to \infty\), \(F(f_n) \to 1\). Since we found a sequence of values approaching 1, the supremum is indeed 1. Thus, \(||F|| = 1\).
Step 3.2: Check for a norm-attaining function.
The question asks for the elements in the set \(\{f \in S_X : F(f) = ||F||\}\). We need to see if there is any function \(f^* \in S_X\) such that \(F(f^*) = 1\). Let's assume such a function \(f^*\) exists.
We have \(F(f^*) = \int_0^1 f^*(t)dt = 1\).
We also know from \(f^* \in S_X\) that \(f^*(t) \le ||f^*||_\infty = 1\) for all \(t \in [0,1]\).
So we have \(\int_0^1 (1 - f^*(t))dt = 0\).
The integrand \(g(t) = 1 - f^*(t)\) is non-negative because \(f^*(t) \le 1\).
Since \(f^*\) is continuous, \(g(t)\) is also continuous.
If the integral of a non-negative continuous function is zero, the function must be identically zero.
Therefore, \(1 - f^*(t) = 0\) for all \(t \in [0,1]\), which implies \(f^*(t) = 1\) for all \(t \in [0,1]\).
However, this constant function \(f^*(t) = 1\) does not belong to the space \(X\), because it does not satisfy the condition \(f^*(0) = 0\) and \(f^*(1) = 0\).
This is a contradiction. Our assumption that such a function \(f^*\) exists in \(X\) must be false.
Therefore, the set of norm-attaining functions is empty.
Step 4: Final Answer:
The set has NO element.
Step 5: Why This is Correct:
The norm of the integral functional on this space is 1. For a continuous function in the unit ball to have an integral of 1, it must be the constant function f(t)=1. However, this function is not in the specified subspace X because it does not equal zero at the endpoints. Therefore, the supremum is approached but never attained by any function in the set.