Show continuously uniform function

  • Thread starter Thread starter aeronautical
  • Start date Start date
  • Tags Tags
    Function Uniform
Click For Summary
SUMMARY

The discussion centers on demonstrating that the function f defined as f(x) = inf_{y∈F} d(x,y) is continuously uniform across the metric space (X,d). Participants highlight the relationship between Lipschitz continuity and uniform continuity, asserting that proving Lipschitz continuity with a constant C=1 is a stronger approach. The solution involves leveraging the triangle inequality and sequences from the set F to establish the desired properties of the function f.

PREREQUISITES
  • Understanding of metric spaces and the properties of distance functions.
  • Familiarity with Lipschitz continuity and its implications in analysis.
  • Knowledge of the triangle inequality in the context of metric spaces.
  • Experience with sequences and limits in mathematical analysis.
NEXT STEPS
  • Study the concept of Lipschitz continuity in depth, focusing on its applications in metric spaces.
  • Explore the properties of infimum and supremum in real analysis.
  • Review the triangle inequality and its generalizations in metric spaces.
  • Examine exercises from "Topology and Geometry" by G. Bredon to reinforce understanding of continuity concepts.
USEFUL FOR

Mathematicians, students of analysis, and anyone interested in the properties of functions in metric spaces, particularly those studying continuity and distance functions.

aeronautical
Messages
33
Reaction score
0
Let F be a non-empty subset of a metric room (X,d) and define the function f: X→R through f(x)= inf_{y∈F} d(x,y)= inf {d(x,y):y ∈ F}. Please show that f is continuously uniform in the entire X.

Please note that f(x)= inf_{y∈F} d(x,y) means that y∈F is written below the inf...it is hard to show it on one line, so I chose to show as a subscript...Can anyone tell me how I should begin this problem?
 
Physics news on Phys.org
Strange coincidence, I've been trying to solve the same problem! Well, not exactly the same... I have been trying to show that the map [itex]x\mapsto d(x, F)[/itex] is Lipschitz continuous, meaning that there exists a constant C such that |d(x,F)-d(y,F)|≤Cd(x,y) for all x,y in X. Obviously Lipschitz continuity is stronger than uniform continuity since given ε>0, it suffices to take δ=ε/C. As it turns out, there is an exercise in the book Topology and Geometry of G. Bredon that asks the reader to show that the map [itex]x\mapsto d(x, F)[/itex] is (merely!) continuous. And a hint is provided in suggesting to show that this map is actually Lipschitz of constant C=1. Meaning that, somewhat surprisingly, the easiest way to solve your problem and mine is probably to demonstrate the stronger inequality |d(x,F)-d(y,F)|≤d(x,y) for all x,y in X. So let's try to show this.
 
The key for me was to consider first the very special case in which F consists of only two points: F={p,q}. In this scenario, the problem is not so overwhelming psychologically and after solving it, I realized that its solutions admits an immediate generalization to the case of arbitrary F.

The general solution I found is as follows. Tell me what you think.

First, note that we can assume without loss of generality that d(y,F)≤d(x,F) [if not, interchange the labels of x and y...]. In this case, |d(x,F)-d(y,F)|=d(x,F)-d(y,F) and so we must prove that d(x,F)≤d(x,y)+d(y,F) [i.e., we must prove a kind of triangle inequality in which sets are allowed to occupy one of the variable position].

Next, consider {e_n}, {f_n} two sequences of points in F such that d(x,F)=lim d(x,e_n) and d(y,F)=lim d(y,f_n) and further assume that d(x,e_n)≤d(x,f_n) for all n in N.

Then, we have d(x,y)+d(y,f_n) ≥ d(x,f_n) ≥ d(x,e_n) for all n in N and passing to the limit yields the desired conclusion.
 

Similar threads

  • · Replies 40 ·
2
Replies
40
Views
5K
Replies
26
Views
3K
  • · Replies 27 ·
Replies
27
Views
2K
Replies
2
Views
1K
  • · Replies 39 ·
2
Replies
39
Views
6K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K