Given the family C[0, 1] of continuous functions ƒ [0, 1] → R, and the norm ||f|| = sup |f(x)]. x= [0,1] Let us define a distance function p(ƒ, g) = ||ƒ — g||, f, g € C[0, 1] - Our aim is to prove that < C[0, 1], p > is a metric space.
Given the family C[0, 1] of continuous functions ƒ [0, 1] → R, and the norm ||f|| = sup |f(x)]. x= [0,1] Let us define a distance function p(ƒ, g) = ||ƒ — g||, f, g € C[0, 1] - Our aim is to prove that < C[0, 1], p > is a metric space.
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
![Given the family C[0, 1] of continuous functions
ƒ [0, 1] → R, and the norm
||f|| :=
sup [ƒ(x)].
x= [0,1]
Let us define a distance function
p(ƒ, g) = ||ƒ — g||, \ƒ,g € C[0, 1]
Our aim is to prove that < C[0, 1], p > is a metric space.
By the Extreme Value Theorem, ||ƒ|| is
finite for every continuous function
fe C[0, 1]. The distance function,
p(f, g) = ||ƒ - g|| ≥ 0 for all
f, g € C[0, 1]. For f = g on domain
[0, 1], then p(f, g) = 0. This proves
that p is positive definite, a property
that every metric function satisfies.
O True
O False
Rationale:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd74c38e2-bc89-4af1-ad7f-05db36328edb%2Fbc0d5030-b7a7-4808-a52f-887fb134e3b1%2Fenrr8rp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given the family C[0, 1] of continuous functions
ƒ [0, 1] → R, and the norm
||f|| :=
sup [ƒ(x)].
x= [0,1]
Let us define a distance function
p(ƒ, g) = ||ƒ — g||, \ƒ,g € C[0, 1]
Our aim is to prove that < C[0, 1], p > is a metric space.
By the Extreme Value Theorem, ||ƒ|| is
finite for every continuous function
fe C[0, 1]. The distance function,
p(f, g) = ||ƒ - g|| ≥ 0 for all
f, g € C[0, 1]. For f = g on domain
[0, 1], then p(f, g) = 0. This proves
that p is positive definite, a property
that every metric function satisfies.
O True
O False
Rationale:
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