Question

Let \( \ell^2_{{Z}} = \{ (x_j)_{j \in {Z} : x_j \in {R} { and } \sum_{j = -\infty}^\infty x_j^2<\infty \} \) endowed with the inner product \[ \langle x, y \rangle = \sum_{j = -\infty}^\infty x_j y_j, \quad x = (x_j)_{j \in {Z}}, \, y = (y_j)_{j \in {Z}}. \] Let \( T : \ell^2_{{Z}} \to \ell^2_{{Z}} \) be given by \( T((x_j)_{j \in {Z}}) = (y_j)_{j \in {Z}} \), where \[ y_j = \frac{x_j + x_{-j}}{2}, \quad j \in {Z}. \] Which of the following is/are correct?}

• A. \( T \) is a compact operator
• B. The operator norm of \( T \) is 1
• C. \( T \) is a self-adjoint operator
• D. \( {Range}(T) \) is closed

Hint:

For operators on \( \ell^2 \), verify norm properties, symmetry, and the structure of the range carefully.

Answer 1

The Correct Option is B

Step 1: Operator norm of \( T \). The operator \( T \) maps elements of \( \ell^2_{{Z}} \) such that \( \|T(x)\| \leq \|x\| \). The norm of \( T \) is 1 since \( T(x) = x \) when \( x_j = x_{-j} \). Step 2: Self-adjoint property. \( T \) is self-adjoint since \( \langle T(x), y \rangle = \langle x, T(y) \rangle \) holds for all \( x, y \in \ell^2_{{Z}} \). Step 3: Range of \( T \). The range of \( T \) is the subspace of symmetric sequences in \( \ell^2_{{Z}} \), which is closed in \( \ell^2_{{Z}} \). Step 4: Compactness. \( T \) is not compact, as it is not a finite-rank operator. Step 5: Conclusion. The correct answers are \( {(2), (3), (4)} \).