Define f : X → K by f(x) = x'(1), x € X. Again, the linear functional f is not continuous. By 6.4, Z(ƒ) = {x € X : x' (1) = 0} cannot be closed. This also follows directly by considering n(t) t-(t/n), t€ [0, 1], and noting that → ₁ in X, Zn € Z(f), but x₁ & Z(f).
Define f : X → K by f(x) = x'(1), x € X. Again, the linear functional f is not continuous. By 6.4, Z(ƒ) = {x € X : x' (1) = 0} cannot be closed. This also follows directly by considering n(t) t-(t/n), t€ [0, 1], and noting that → ₁ in X, Zn € Z(f), but x₁ & Z(f).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Request explain the example marked in red related to corollary 6.4
![6.4 Corollary
Let X and Y be normed spaces and F: X→Y be a linear map such
that the range R(F) of F is finite dimensional. Then F is continuous
if and only if the zero space Z(F) of F is closed in X.
In particular, a linear functional f on X is continuous if and only
if Z(f) is closed in X.
Define f : X→ K by f(x) = x'(1), x € X. Again, the linear
functional f is not continuous. By 6.4, Z(ƒ) = {x € X : x' (1) = 0}
cannot be closed. This also follows directly by considering zn(t)
t-(t/n), t = [0, 1], and noting that zn → x₁ in X, z₁ € Z(ƒ), but
x₁ & Z(f).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2F57d5e3a2-b561-4315-898a-c67fd946c165%2F09ythc_processed.png&w=3840&q=75)
Transcribed Image Text:6.4 Corollary
Let X and Y be normed spaces and F: X→Y be a linear map such
that the range R(F) of F is finite dimensional. Then F is continuous
if and only if the zero space Z(F) of F is closed in X.
In particular, a linear functional f on X is continuous if and only
if Z(f) is closed in X.
Define f : X→ K by f(x) = x'(1), x € X. Again, the linear
functional f is not continuous. By 6.4, Z(ƒ) = {x € X : x' (1) = 0}
cannot be closed. This also follows directly by considering zn(t)
t-(t/n), t = [0, 1], and noting that zn → x₁ in X, z₁ € Z(ƒ), but
x₁ & Z(f).
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