MATH 202 B - Problem Set 11
نویسنده
چکیده
in fact we have equality. We have {is, s ∈ S} is a family of linear operators bounded on X∗, since for all `, sups∈S ix(`) = sups∈S `(s) which is finite by assumption. Therefore by the Uniform Boundedness Principle (X ∗ is a Banach space), {is, s ∈ S} are uniformly bounded, i.e. there exists M such that for all s ∈ S, ‖is‖X∗∗ ≤ M , i.e. ‖s‖X ≤M . (11.3) Let X,Y be Banach spaces. Let L ∈ B(X,Y ) be surjective. Let (yn) be a convergent sequence in Y . Show that there exists M <∞ and a convergent sequence (xn) in X such that L(xn) = yn for all n. proof Let y be the limit of (yn), and let x ∈ X such that L(x) = y (L is surjective). Since L is surjective, by the open mapping theorem, L(BX(0, 1)) is an open subset of Y , and since it contains 0 = L(0), there exists r > 0 such that BY (0, r) ⊂ L(BX(0, 1))
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MATH 202 B - Problem Set 10
(10.1) Let k ∈ N, and let X = C k ([0, 1] d) to be the set of all functions f ∈ C 0 ([0, 1] d) such that for all α ∈ ∆ k , f (α) = ∂f ∂x α 1 1 ...∂x α d d exists and is continuous on (0, 1) d , and extends continuously on [0, 1] d. Here
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