Question

Let \( \ell^1 = \{ x = (x(1), x(2), \dots, x(n), \dots) : \sum_{n=1}^{\infty} |x(n)| < \infty \} \) be the sequence space equipped with the norm \( \|x\| = \sum_{n=1}^{\infty} |x(n)| \). Consider the subspace \[ X = \left\{ x \in \ell^1 : \sum_{n=1}^{\infty} |x(n)| < \infty \right\}, \] and the linear transformation \( T: X \to \ell^1 \) given by \( (Tx)(n) = n x(n) \text{  for  } n = 1, 2, 3, \dots. \) Then:

• A. T is closed but NOT bounded
• B. T is bounded
• C. T is neither closed nor bounded
• D. \( T^{-1} \) exists and is an open map

Hint:

An operator is bounded if the norm of the transformed sequence is bounded by a constant multiple of the norm of the original sequence. In this case, the multiplication by \( n \) makes the operator unbounded.

Answer 1

The Correct Option is A

Step 1: Boundedness of \( T \).
The operator \( T \) multiplies each component of the sequence by its index \( n \). This leads to an unbounded operator, because the sequence \( \{n x(n)\} \) grows too quickly for many sequences in \( \ell^1 \), making \( T \) unbounded.

Step 2: Closedness of \( T \).
Even though \( T \) is unbounded, it is closed in the sense that the image of a convergent sequence under \( T \) is also convergent. However, because it is not bounded, it does not satisfy the conditions for being a continuous operator.

Step 3: Conclusion.
Thus, \( T \) is closed but not bounded, so the correct answer is (A).

Final Answer: \[ \boxed{(A) \text{ T is closed but NOT bounded}} \]