1. Consider the map π1 ∶ FN Ð→ F given byπ1((an)) = a1.(a) Show that π1 is linear.(b) Prove that the restriction of π1 to ℓ∞ or to ℓp for p ⩾ 1 is continuous and surjective
Question
- Consider the map π1 ∶ FN Ð→ F given byπ1((an)) = a1.(a) Show that π1 is linear.(b) Prove that the restriction of π1 to ℓ∞ or to ℓp for p ⩾ 1 is continuous and surjective
Solution
(a) To show that π1 is linear, we need to show that it satisfies two properties: additivity and homogeneity.
Additivity: π1((a_n) + (b_n)) = π1((a_n + b_n)) = a1 + b1 = π1((a_n)) + π1((b_n))
Homogeneity: π1(c*(a_n)) = π1((ca_n)) = ca1 = c*π1((a_n))
Therefore, π1 is linear.
(b) To prove that the restriction of π1 to ℓ∞ or to ℓp for p ⩾ 1 is continuous and surjective, we need to show that for every sequence (a_n) in ℓ∞ or ℓp and every ε > 0, there exists a δ > 0 such that if ||(b_n) - (a_n)|| < δ, then |π1((b_n)) - π1((a_n))| < ε (continuity), and for every a in F, there exists a sequence (a_n) in ℓ∞ or ℓp such that π1((a_n)) = a (surjectivity).
Continuity: Let (a_n) be a sequence in ℓ∞ or ℓp, and let ε > 0. Choose δ = ε. If ||(b_n) - (a_n)|| < δ, then |π1((b_n)) - π1((a_n))| = |b1 - a1| ≤ ||(b_n) - (a_n)|| < ε.
Surjectivity: Let a be in F. Choose the sequence (a_n) in ℓ∞ or ℓp where a1 = a and an = 0 for n ≠ 1. Then π1((a_n)) = a1 = a.
Therefore, the restriction of π1 to ℓ∞ or to ℓp for p ⩾ 1 is continuous and surjective.
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