Question
Download Solution PDFLet β2 = {(π₯1, π₯2, π₯3, … ) βΆ π₯π ∈ β for all π ∈ β and \(\rm \Sigma_{n=1}^\infty x_n^2<\infty \}\)
For a sequence (π₯1, π₯2, π₯3, … ) ∈ β2 , define β(π₯1, π₯2, π₯3, … )β2 = \(\rm (\Sigma_{n=1}^\infty x_n^2)^{\frac{1}{2}}\)
Let π βΆ (β2 , β⋅β2) → (β2 , β⋅β2) and π βΆ (β2 , β⋅β2) → (β2 , β⋅β2) be defined by
π(π₯1, π₯2, π₯3, … ) = (π¦1, π¦2, π¦3, … ), where π¦π = \(\rm \left\{\begin{matrix}0,&n=1\\\ x_{n-1},& n\ge2\end{matrix}\right.\)
π(π₯1, π₯2, π₯3, … ) = (π¦1, π¦2, π¦3, … ), where π¦π = \(\rm \left\{\begin{matrix}0,&n\ is\ odd\\\ x_{n},& n\ is\ even\end{matrix}\right.\)
Then
- π is a compact linear map and π is NOT a compact linear map
- π is NOT a compact linear map and π is a compact linear map
- both π and π are compact linear maps
- NEITHER π NOR π is a compact linear map
Answer (Detailed Solution Below)
Option 4 : NEITHER π NOR π is a compact linear map
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