Question

Let \( \ell^p = \left\{ x = (x_n)_{n \geq 1 : x_n \in {R}, \|x\|_p = \left( \sum_{n=1}^\infty |x_n|^p \right)^{1/p}<\infty \right\} \) for \( p = 1, 2 \). Let \[ c_0 = \{ (x_n)_{n \geq 1} : x_n = 0 { for all but finitely many } n \geq 1 \}. \] For \( x = (x_n)_{n \geq 1} \in c_0 \), define \( f(x) = \sum_{n=1}^\infty \frac{x_n}{\sqrt{n}} \). Consider the following statements:} 1. There exists a continuous linear functional \( F \) on \( (\ell^1, \|\cdot\|_1) \) such that \( F = f { on } c_0 \).
2. There exists a continuous linear functional \( G \) on \( (\ell^2, \|\cdot\|_2) \) such that \( G = f { on } c_0 \).
Which one of the following is correct?

• A. Both I and II are TRUE
• B. I is TRUE and II is FALSE
• C. I is FALSE and II is TRUE
• D. Both I and II are FALSE

Hint:

For functionals on sequence spaces, check summability conditions carefully under the given norms.

Answer 1

The Correct Option is B

Step 1: Verifying Statement I. In \( \ell^1 \), every element of \( c_0 \) is in \( \ell^1 \), and \( f(x) = \sum_{n=1}^\infty \frac{x_n}{\sqrt{n}} \) defines a continuous linear functional due to the absolute summability of the series. Hence, \( F \) exists. Step 2: Verifying Statement II. In \( \ell^2 \), \( \frac{1}{\sqrt{n}} \notin \ell^2 \), as the series \( \sum_{n=1}^\infty \frac{1}{n} \) diverges. Therefore, \( G \) does not exist. Step 3: Conclusion. Statement I is true, but Statement II is false. The correct answer is \( {(2)} \).