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Hausdorff distance between balls

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data

    Hello. Let be n integer and let's considere $$\mathbb{R}^{n} = E$$.
    Let be the distance $$d : (x; y) \in \mathbb{R}^{2} \rightarrow ||x - y||_{2}$$.
    We wright $$\forall p \in [1; \infty], B_{p}$$ the unit closed ball for $$||.||_{p}$$.
    Forall compacts A, B of $$E$$, we define $$d'(A; B) = sup_{x \in \mathbb{R}^{n}} |inf_{y \in A}d(x; y) - inf_{y \in B}d(x; y)|$$.
    My goal is to show tha forall $$p_{0} \in [1; +\infty]$$, $$d'(B_{p_{0}}, B_{p}) \rightarrow_{p \rightarrow p_{0}} 0$$.



    2. Relevant equations
    $$d'(A; B) = sup_{x \in \mathbb{R}^{n}} |inf_{y \in A}d(x; y) - inf_{y \in B}d(x; y)|$$


    I also simplify this distance by showing $$d'(A; B) = max (sup_{x \in A} d(x; B), sup_{x \in B}d(x; A))$$.



    3. The attempt at a solution

    At the moment I just think about the case where n = 2 : the sup is reach I think
    is reach on the $$y = x$$ space.
    I also ask my self about the inequality between norm.


    What do you think please?

    Thank you in advance and have a nice afternoon:oldbiggrin:.

     
  2. jcsd
  3. Jan 27, 2016 #2

    fresh_42

    Staff: Mentor

    What happens to the distance, if you apply it to the balls in question?
     
  4. Jan 28, 2016 #3
    We can suppose for exemple that ##p \leq p_{0}##, so ##B_{p} \subset B_{p_{0}}## so
    ##d(B_p, B_{p_0}) = sup_{x \in B_{p_0}} d(x, B_p)##
     
    Last edited by a moderator: Jan 28, 2016
  5. Jan 28, 2016 #4
    Now what could be intresting is to explicite this sup. And at dimension 2, the sup is reach as I think(I think we could show it.). on the line define by y = x.
     
  6. Jan 28, 2016 #5

    fresh_42

    Staff: Mentor

    I will write ##q < p## because for ##p = q## is nothing to do: ##d(B_p,B_q) = d(B_p, B_p) = 0##.
    It's faster to type with less indices.
    Then ##B_q ⊂ B_p##. I will also use the original definition of ##d## since you have not shown the simplification you stated.
    Further I drop the ' from the ##d##. It doesn't make sense to have it. Your original ##d## is nothing else but ##||x-y||##.

    First I would draw a two dimensional picture of the situation.

    For the proof it does not become easier in the case ##n=2##. Therefore we can stay in the general case.

    Next: Can you exclude that a possible ##x## from the supremum is not within ##B_p##? How?
    (I'm not sure whether you need it, but it is a good exercise to use your drawing and to visualize what has to be done.)
     
  7. Jan 28, 2016 #6
    Hello and thanks : the supremum $$sup_{x \in B_{p}}d(x, B_{q})$$ is the distance you wright and this sup is reach on a certain x of $$B_{p}$$(because of its compacity.).
    Don't worry I visualisz the situation at dimension 2.
    It's just I try to stay general.
     
  8. Jan 28, 2016 #7

    fresh_42

    Staff: Mentor

    Correction to #5: Of course I meant, that the points within ##B_p## cannot be those ##x## in the supremum (double negation error). Why?
    Hint: From this on you can tell which point ##y## of ##B_p## has to be taken for the infimum to a given ##x##.
     
  9. Jan 28, 2016 #8
    You mean the sup :
    is reach on a point in $$B_{p}$$.
    Of course it is : $$B_{p}$$ is compact and the application $$d(., A)$$ is continious. So it reach is sup.
     
  10. Jan 28, 2016 #9
    Forall A subset of E.
     
  11. Jan 28, 2016 #10

    fresh_42

    Staff: Mentor

    I have difficulties to follow you.
    I mean:
    1) ##d(B_q,B_p) = \sup_{x\in ℝ^n} | ## some function of ## x|## is what you have to examine.
    2) Start to consider a single given ##x##.
    3) In order to build the supremum as required, can this ##x## be an element of ##B_p## ?
     
  12. Jan 28, 2016 #11
    1)
    d(Bq,Bp)=supx∈Rn|d(Bq,Bp)=supx∈ℝn|d(B_q,B_p) = \sup_{x\in ℝ^n} | some function of x|
    The some fonction is $$x \in B_{p} \rightarrow d(x, B_{q}) \in \mathbb{R}^{+}$$.
    2) Now I have to find forall x in $$B_{p}$$ I have to find $$d(x, B_{q})$$.
    3)I don't understand you're question.
    My first goal was to find my sup.
     
  13. Jan 28, 2016 #12

    fresh_42

    Staff: Mentor

    Step by step.
    Let us write ##f(x) = \inf_{y \in B_q} ||x-y|| - \inf_{y \in B_p} ||x-y||## and ## d(B_q,B_p) = \sup_{x\in ℝ^n} | f(x)|##.
    It is only for we will have less to type, if we introduce ##f(x)##. That's all. And it is the definition of distance here. I do not take the other formula, since you have not proven it. First I want you to understand what has to be done.

    Why?
    I mean those ##x \in ℝ^n## among which we are searching for the supremum.
    Of all possible ##x## I asked
    The answer is no, but why?
    What is the value of ##f(x)## in case ##x \in B_q##?
    What is the value of ##f(x)## in case ##x \in B_p##?
    Remember, ##B_q ⊂ B_p##.
     
  14. Jan 29, 2016 #13
    Hello first I'm sorry(I'm french so sorry if I don't undesrtand you.). so you want to solve it by using :
    OK.

    If : $$
    x \in B_q$$
    we have $$f(x) = \inf_{y \in B_p} ||x-y||$$. Same by exchanging p and q.
    OK.
     
  15. Jan 29, 2016 #14
    OK so : now you want to find where the sup of $$f$$ is reach if it's reach(remmber f is define on $$\mathbb{R}^{n}$$ which is not compact.).
    OK.
     
  16. Jan 29, 2016 #15

    fresh_42

    Staff: Mentor

    Je sais ("some fonction").
    If ##x## is in ##B_q## then it is in both balls, because ##B_q ⊂ B_p## and ##f(x) = 0##.
    If ##x ∈ B_p / B_q## then $$f(x) = \inf_{y \in B_q} ||x-y||$$.
    Hint: we can give an upper bound for ##|f(x)|##. Which? What does that mean for the supremum over all ##x∈ℝ^n##?
     
  17. Jan 29, 2016 #16
    I aggry with what you wright it's juste I think it was uselesse to said it. And abbs would be better since we considere a distance.
    I put the abs since my first trade.
     
  18. Jan 29, 2016 #17
    I'm sorry to follow you step it's just I don't see where you want to go.
     
  19. Jan 29, 2016 #18

    fresh_42

    Staff: Mentor

    Me, too. I defined ##d= sup |f(x)|##. If you pull the abs into the definition of ##f## or not doesn't make a difference.
     
  20. Jan 29, 2016 #19
    The things is it not permitt me to calculate he sup which was my original problem.
     
  21. Jan 29, 2016 #20

    fresh_42

    Staff: Mentor

    I want to show that the points ##x## in the supremum function cannot be inside the balls.
    If it is so, I know they are outside and I don't have to think about special cases anymore.
    Then I will take a single point ##x## (outside of ##B_p##) and compute its minimal distances to ##B_q## and ##B_p## plus the points on the balls where the straight of minimal distance goes through. The difference of these points will answer the question about convergence.

    (The problem is: If you draw a picture you will see, that your ##x## for the supremum are at infinite distance. The proof is to show why this does not matter.)
     
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