Question

Let \(\ell^2 = \{(x_1, x_2, x_3, \dots) : x_n \in \mathbb{R} \text{ for all } n \in \mathbb{N} \text{ and } \sum_{n=1}^\infty x_n^2 < \infty\}\).
For a sequence \((x_1, x_2, x_3, \dots) \in \ell^2\), define \[ ||(x_1, x_2, x_3, \dots)||_2 = \left(\sum_{n=1}^\infty x_n^2\right)^{1/2}. \]
Let \(S: (\ell^2, ||.||_2) \to (\ell^2, ||.||_2)\) and \(T: (\ell^2, ||.||_2) \to (\ell^2, ||.||_2)\) be defined by
\[ S(x_1, x_2, x_3, \dots) = (y_1, y_2, y_3, \dots), \text{ where } y_n = \begin{cases} 0, & n=1 \\ x_{n-1}, & n \ge 2 \end{cases} \]
\[ T(x_1, x_2, x_3, \dots) = (y_1, y_2, y_3, \dots), \text{ where } y_n = \begin{cases} 0, & n \text{ is odd} \\ x_n, & n \text{ is even} \end{cases} \]
Then

• A. S is a compact linear map and T is NOT a compact linear map
• B. S is NOT a compact linear map and T is a compact linear map
• C. both S and T are compact linear maps
• D. NEITHER S NOR T is a compact linear map

Hint:

A useful test for compactness: an operator on an infinite-dimensional Hilbert space is NOT compact if it has an infinite number of eigenvalues bounded away from zero. S has no eigenvalues. T has eigenvalues 0 and 1, with 1 having infinite multiplicity. The infinite multiplicity of a non-zero eigenvalue implies non-compactness. Another quick check: if an operator maps an infinite orthonormal sequence to another infinite orthonormal sequence (or one nearly so), it is not compact.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A linear map (or operator) \(A: H \to H\) on a Hilbert space \(H\) is compact if it maps bounded sets into precompact sets (sets whose closure is compact). In an infinite-dimensional space like \(\ell^2\), this means the image of the unit ball under A has a compact closure. A key property is that a compact operator must map weakly convergent sequences to strongly convergent sequences. Also, an operator is compact if and only if it is the limit of a sequence of finite-rank operators.
Step 3: Detailed Explanation:
Let's analyze the operators S and T.
\(S\) is the right-shift operator: \(S(x_1, x_2, ......) = (0, x_1, x_2, ......)\).
\(T\) is a projection-like operator: \(T(x_1, x_2, x_3, x_4 ......) = (0, x_2, 0, x_4, ......)\).
Analysis of Operator S (Right-shift):
Let's consider the standard orthonormal basis of \(\ell^2\), \(\{e_n\}\), where \(e_n\) has a 1 in the nth position and 0s elsewhere. This is a bounded set since \(||e_n||_2 = 1\) for all n.
Let's see where S maps this set: \(S(e_n) = e_{n+1}\).
The image set is \(\{e_2, e_3, e_4, ......\}\).
To check if this set is precompact, we check if the sequence \(\{S(e_n)\}_{n=1}^\infty = \{e_{n+1}\}_{n=1}^\infty\) has a convergent subsequence. Let's compute the distance between any two distinct elements in the image sequence:
\[ ||S(e_n) - S(e_m)||_2 = ||e_{n+1} - e_{m+1}||_2 = \sqrt{1^2 + (-1)^2} = \sqrt{2} \quad (\text{for } n \neq m) \] Since the distance between any two distinct points in the sequence is a constant \(\sqrt{2}\), no subsequence can be a Cauchy sequence. Therefore, no subsequence can converge.
This means the image of the bounded set \(\{e_n\}\) is not precompact.
Thus, S is NOT a compact linear map.
Analysis of Operator T:
Let's use the same sequence of basis vectors, \(\{e_n\}\).
\(T(e_n)\) will be \(e_n\) if n is even, and the zero vector if n is odd.
Consider the subsequence of even-indexed basis vectors: \(\{e_{2k}\}_{k=1}^\infty\). This is a bounded set.
The image under T is \(T(e_{2k}) = e_{2k}\).
The image sequence is \(\{e_2, e_4, e_6, ......\}\).
Let's compute the distance between distinct elements in this image sequence: \[ ||T(e_{2k}) - T(e_{2j})||_2 = ||e_{2k} - e_{2j}||_2 = \sqrt{2} \quad (\text{for } k \neq j) \] Again, no subsequence can be Cauchy, so the sequence has no convergent subsequence. The image of the bounded set \(\{e_{2k}\}\) is not precompact. Thus, T is NOT a compact linear map.
Step 4: Final Answer:
NEITHER S NOR T is a compact linear map.
Step 5: Why This is Correct:
For both operators S and T, we found a bounded sequence (a subset of the orthonormal basis) whose image is not precompact. The image sequence consists of vectors that are all a fixed distance \(\sqrt{2}\) apart, which prevents any subsequence from converging. This property directly shows that neither operator is compact.