Problems 1.1 | Question 8 | 𝑑(𝑥,𝑦)=∫|𝑥(𝑡)−𝑦(𝑡)|𝑑𝑡 | Metric Space Ch 1 | Functional Analysis Kreyszig
Problems 1.1 | Question 8 | 𝑑(𝑥,𝑦)=∫|𝑥(𝑡)−𝑦(𝑡)|𝑑𝑡 | Metric Space Chapter 01 | Functional Analysis Kreyszig
𝑑(𝑥,𝑦)=∫|𝑥(𝑡)−𝑦(𝑡)|𝑑𝑡 | Question No 8 | Problems 1.1 | Metric Space | Chapter 01 | Problems 1.1 | Introductory Functional Analysis with Applications | Erwin Kreyszig
𝑑(𝑥,𝑦)=∫|𝑥(𝑡)−𝑦(𝑡)|𝑑𝑡
Book Name : Introductory Functional Analysis with Applications
By : Erwin Kreyszig
Chapter Number : 01
Chapter Name : Metric Space
Lecture Number : 10
By (Name) : Awais Rasool
Exercise Number : 1.1
Problems Number: 1.1
Question Number : 08
Part Number : 0
Example Number : 03
Awais Rasool Shah
Topics Name :
Chapter 1. Metric Spaces
1.1 Metric Space
1.2 Further Examples of Metric Spaces
1.3 Open Set, Closed Set, Neighborhood
1.4 Convergence, Cauchy Sequence, Completeness
1.5 Examples. Completeness Proofs
1.6 Completion of Metric Spaces
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Definition (Metric space , Metric).
Consider a non-empty set 𝑥 and a function 𝑑 : 𝑥∗𝑥 𝑅^+ "∪{0}".
This function 𝑑 is called metric on 𝑥 if following conditions are holds:
1) 𝑑(𝑥,𝑦)≥0
2) 𝑑(𝑥,𝑦)=0 𝑖𝑓𝑓 𝑥=𝑦
3) 𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) (Symmetry)
4) 𝑑(𝑥,𝑦)≤𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)
The set X with d is called metric Space and written as (X, 𝑑).
Show that another metric d on the set X. If [𝑎,𝑏]×[𝑎,𝑏] 𝑅.
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|𝑥(𝑡)−𝑦(𝑡)|〗 𝑑𝑡
Sol:
𝑑(𝑥,𝑦)≥0
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|𝑥(𝑡)−𝑦(𝑡)|𝑑𝑡〗≥0
If 𝑥(𝑡) & 𝑦(𝑡)∈𝑅 and the subtraction of a two real number is also real number and the absolute value of any real number is always non-negative real number , so this property is satisfied.
𝑑(𝑥,𝑦)=0 𝑖𝑓𝑓 𝑥=𝑦
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|𝑥(𝑡)−𝑦(𝑡)|〗 𝑑𝑡=0
"⟺" |𝑥(𝑡)−𝑦(𝑡)|=0
"⟺" 𝑥(𝑡)−𝑦(𝑡)=0
"⟺" 𝑥(𝑡)=𝑦(𝑡)
"⟺" 𝑥=𝑦
𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) (Symmetry)
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|𝑥(𝑡)−𝑦(𝑡)|𝑑𝑡〗
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|(−1)(−𝑥(𝑡)+𝑦(𝑡))|𝑑𝑡〗
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|(−1)(−𝑥(𝑡)+𝑦(𝑡))|𝑑𝑡〗
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|−1||𝑦(𝑡)−𝑥(𝑡))|𝑑𝑡〗
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|𝑦(𝑡)−𝑥(𝑡))|𝑑𝑡〗
𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥)
𝑑(𝑥,𝑦)≤𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|𝑥(𝑡)−𝑦(𝑡)|𝑑𝑡〗
𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|𝑥(𝑡)−𝑧(𝑡)+𝑧(𝑡)−𝑦(𝑡)|𝑑𝑡〗
Using triangle inequality |𝑎+𝑏|≤|𝑎|+|𝑏|
𝑑(𝑥,𝑦)≤∫1_𝑎^𝑏▒〖|𝑥(𝑡)−𝑧(𝑡)|+|𝑧(𝑡)−𝑦(𝑡)|𝑑𝑡〗
𝑑(𝑥,𝑦)≤∫1_𝑎^𝑏▒〖|𝑥(𝑡)−𝑧(𝑡)|𝑑𝑡+∫_𝑎^𝑏▒〖|𝑧(𝑡)−𝑦(𝑡)|𝑑𝑡〗〗
𝑑(𝑥,𝑦)≤𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)
Since the function 𝑑(𝑥,𝑦)=∫1_𝑎^𝑏▒〖|𝑥(𝑡)−𝑦(𝑡)|〗 𝑑𝑡 is satisfies the four properties of a metric space.