Hausdorff distance between balls

[Calabi]

Homework Statement

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 $$p_{0} \in [1; +\infty]$$, $$d'(B_{p_{0}}, B_{p}) \rightarrow_{p \rightarrow p_{0}} 0$$.

Homework Equations

$$d'(A; B) = sup_{x \in \mathbb{R}^{n}} |inf_{y \in A}d(x; y) - inf_{y \in B}d(x; y)|$$[/B]

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

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.

 

[fresh_42]

What happens to the distance, if you apply it to the balls in question?

 

[Calabi]

We can suppose for exemple that $p \leq p_{0}$, so $B_{p} \subset B_{p_{0}}$ so
$d(B_p, B_{p_0}) = sup_{x \in B_{p_0}} d(x, B_p)$

 

[Calabi]

Now what could be interesting 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.

 

[fresh_42]

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.)

 

[Calabi]

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.

 

[fresh_42]

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 $x$.

 

[Calabi]

You mean the sup :

supx∈Bpd(x,Bq)

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.

 

[Calabi]

Forall A subset of E.

 

[fresh_42]

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 $B_p$ ?

 

[Calabi]

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.

 

[fresh_42]

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 $$x \in B_{p} \rightarrow d(x, B_{q}) \in \mathbb{R}^{+}$$.

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.

2) Now I have to find forall x in $$B_{p}$$ I have to find $$d(x, B_{q})$$.

Why?

Start to consider a single given $x$.

I mean those $x \in ℝ^n$ among which we are searching for the supremum.
Of all possible $x$ I asked

... can this $x$ be an element of $B_p$ ?

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, $B_q ⊂ B_p$.

 

[Calabi]

Hello first I'm sorry(I'm french so sorry if I don't undesrtand you.). so you want to solve it by using :

f(x)=infy∈Bq||x−y||−infy∈Bp||x−y||f(x)=infy∈Bq||x−y||−infy∈Bp||x−y||f(x) = \inf_{y \in B_q} ||x-y|| - \inf_{y \in B_p} ||x-y|| and d(Bq,Bp)=supx∈Rn|f(x)

OK.

If : $$
x \in B_q$$
we have $$f(x) = \inf_{y \in B_p} ||x-y||$$. Same by exchanging p and q.
OK.

 

[Calabi]

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.

 

[fresh_42]

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.

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 $x∈ℝ^n$?

 

[Calabi]

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.

 

[Calabi]

I'm sorry to follow you step it's just I don't see where you want to go.

 

[fresh_42]

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.

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.

 

[Calabi]

The things is it not permitt me to calculate he sup which was my original problem.

 

[fresh_42]

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.)