Question

Let \( \{ e_n : n = 1, 2, 3, \dots \} \) be an orthonormal basis of a complex Hilbert space \( H \). Consider the following statements: P: There exists a bounded linear functional \( f: H \to \mathbb{C} \) such that \( f(e_n) = \frac{1}{n} \) for \( n = 1, 2, 3, \dots \)
Q: There exists a bounded linear functional \( g: H \to \mathbb{C} \) such that \( g(e_n) = \frac{1}{\sqrt{n}} \) for \( n = 1, 2, 3, \dots \)

• A. both P and Q are TRUE
• B. P is TRUE and Q is FALSE
• C. P is FALSE and Q is TRUE
• D. both P and Q are FALSE

Hint:

In Hilbert spaces, a sequence \( \{ a_n \} \) can define a bounded linear functional if and only if the sequence is square-summable.

Answer 1

The Correct Option is B

Step 1: Statement P Analysis.
In a Hilbert space, the existence of a bounded linear functional \( f \) such that \( f(e_n) = \frac{1}{n} \) for all \( n \) is possible. This follows from the fact that the sequence \( \{ \frac{1}{n} \} \) is square-summable, and a functional can be defined in terms of the inner product with the elements of the orthonormal basis.

Step 2: Statement Q Analysis.
The sequence \( \{ \frac{1}{\sqrt{n}} \} \) is not square-summable, and thus no bounded linear functional \( g \) can exist such that \( g(e_n) = \frac{1}{\sqrt{n}} \). This violates the condition for defining a bounded linear functional in a Hilbert space.

Step 3: Conclusion.
Therefore, statement P is true, but statement Q is false.

Final Answer: \[ \boxed{P \text{ is TRUE and } Q \text{ is FALSE}} \]