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(f, g) = ||fg|l, Vf, 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(f, g) = ||fg|l, Vf, g € C[0, 1] Our aim is to prove that < C[0, 1], p > is a metric space.
Chapter3: Functions
Section3.3: Rates Of Change And Behavior Of Graphs
Problem 2SE: If a functionfis increasing on (a,b) and decreasing on (b,c) , then what can be said about the local...
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![Given the family C[0, 1] of continuous functions ƒ : [0, 1] → R, and the norm
||f||| E=
sup |f(x)].
x= [0,1]
Let us define a distance function
p(f, g) = ||fg||, Vf, g € C[0, 1]
Our aim is to prove that < C[0, 1], p > is a metric space.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd74c38e2-bc89-4af1-ad7f-05db36328edb%2Fdf787184-2c6d-4c02-a86c-cd6ef8a2c1f8%2Fp6ik5pb_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||| E=
sup |f(x)].
x= [0,1]
Let us define a distance function
p(f, g) = ||fg||, Vf, g € C[0, 1]
Our aim is to prove that < C[0, 1], p > is a metric space.
![The metric function p also satisfies the triangle inequality, because
p(f, g) = ||ƒ — g|| = sup |ƒ(x) — g(x)]
x= [0,1]
= sup {f(x) − h(x) + h(x) − g(x)]}
x=[0,1]
= sup {[f(x) − h(x)| + |h(x) − g(x)]}
x= [0,1]
≤ sup f(x) - h(x)| + sup |h(x) — g(x)|
x= [0,1]
x=[0,1]
= || ƒ − h|| + ||h − g|| = p(f, h) + p(h, g), f, g, h = C[0, 1].
-
O True
False](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd74c38e2-bc89-4af1-ad7f-05db36328edb%2Fdf787184-2c6d-4c02-a86c-cd6ef8a2c1f8%2Fs4gun6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The metric function p also satisfies the triangle inequality, because
p(f, g) = ||ƒ — g|| = sup |ƒ(x) — g(x)]
x= [0,1]
= sup {f(x) − h(x) + h(x) − g(x)]}
x=[0,1]
= sup {[f(x) − h(x)| + |h(x) − g(x)]}
x= [0,1]
≤ sup f(x) - h(x)| + sup |h(x) — g(x)|
x= [0,1]
x=[0,1]
= || ƒ − h|| + ||h − g|| = p(f, h) + p(h, g), f, g, h = C[0, 1].
-
O True
False
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