Discussion
Let
\[ \begin{aligned} \ell^{2} = \{&(x_{1},x_{2},x_{3},\ldots) : x_{n} \in \mathbb{R} \text{ for all } n \in \mathbb{N} \\ &\text{and } \sum_{n=1}^{\infty}x_{n}^{2} \lt \infty\}. \end{aligned} \]
For a sequence \((x_{1},x_{2},x_{3},\ldots) \in \ell^{2}\), define
\[ \begin{aligned} \|(x_{1},x_{2},x_{3},\ldots)\|_{2} \\ = \left(\sum_{n=1}^{\infty}x_{n}^{2}\right)^{1/2}. \end{aligned} \]
Let \(S : (\ell^{2}, \|\cdot\|_{2}) \to (\ell^{2}, \|\cdot\|_{2})\) and \(T : (\ell^{2}, \|\cdot\|_{2}) \to (\ell^{2}, \|\cdot\|_{2})\) be defined by
\(S(x_{1},x_{2},x_{3},\ldots) = (y_{1},y_{2},y_{3},\ldots)\), where
\[ y_{n} = \begin{cases} 0, & n = 1 \\ x_{n-1}, & n \ge 2 \end{cases} \]
\(T(x_{1},x_{2},x_{3},\ldots) = (y_{1},y_{2},y_{3},\ldots)\), where
\[ y_{n} = \begin{cases} 0, & n \text{ is odd} \\ x_{n}, & n \text{ is even} \end{cases} \]
Then
Answer: D
NEITHER S NOR T is a compact linear map.
A compact operator maps the bounded sequence of unit vectors \(e_{n}\) to a sequence with a convergent subsequence.
S is the right shift: \(Se_{n} = e_{n+1}\), and \(\|e_{n+1} - e_{m+1}\| = \sqrt{2}\) for \(n \ne m\), so no subsequence converges. S is an isometry on an infinite-dimensional space, so it is not compact.
T is the projection onto the even coordinates: \(Te_{2n} = e_{2n}\), again a sequence with mutual distances \(\sqrt{2}\). T is the identity on an infinite-dimensional subspace, so it is not compact either.
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