Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $B(\mathcal{l}_2) :=\{x \in \mathcal{l}_2 : \|x \| \leq 1 \}$ and $S(\mathcal{l}_2) :=\{x \in \mathcal{l}_2 : \|x \| = 1 \}$ be the unit ball and the unit sphere of $\mathcal{l}_2$, respectively.

I'm trying to show that $B(\mathcal{l}_2)$ contains an infinite set $A$ such that, for every $x,y \in A$ with $x \neq y$, we have $\|x -y \| > \sqrt{2}$.

My intuition tells me that such a set $A$ should lie inside $S(\mathcal{l}_2)$. Next, I've tried to use a point $z$ such that $d(z,S(\mathcal{l}_2)) = \sqrt{2}$ to define $A$. I know that for such a $z$ there exists $w \in S(\mathcal{l}_2)$ such that $\|z - w \| = \sqrt{2}$, and for every $v \in S(\mathcal{l}_2),v \ne w$, we must have $\|z - v \| > \sqrt{2}$.

So, right now I'm confused about how to go ahead with the definition of $A$ in a way that the distance between any two distinct points in $A$ is greater that $\sqrt{2}$. I get the feeling that I should be able to use what I've described in the previous paragraph, but I don't see how.

Could somebody point me in the right way?

share|improve this question
    
Notice that your result implies that $B(l_2)$ is not compact (or that $l_2$ is not locally compact). In fact, every locally compact normed vector space is finite dimensional (and every finite dimensional normed vector space is locally compact). –  Seirios Nov 5 '12 at 8:44
    
@Seirios: I think the OP knows that. The point of the question is that in a Hilbert space an orthonormal set will have its points at distance $\sqrt2$; can points be found that are farther away from each other? –  Martin Argerami Nov 5 '12 at 14:55

2 Answers 2

up vote 5 down vote accepted

Let $(e_1,e_2,\ldots)$ be an orthonormal subset of $\ell^2$.

Let $(a_1,a_2,a_3,\ldots)$ be a sequence of real numbers in $(0,1)$ to be specified later.

Define the sequence $(v_1,v_2,\ldots)$ in the unit ball as follows:

$\begin{align*} v_1&=e_1\\ v_2&=-a_1e_1+\sqrt{1-a_1^2}e_2\\ v_3&=-a_2(e_1+e_2)+\sqrt{1-2a_2^2}e_3\\ v_4&=-a_3(e_1+e_2+e_3)+\sqrt{1-3a_3^2}e_4\\ &\vdots\\ v_{k+1}&=-a_k(e_1+\cdots+e_k)+\sqrt{1-ka_k^2}e_{k+1} \end{align*}$

If you work out the distances and simplify, you find that a sufficient condition for this example to work is that $a_k<\dfrac{1}{\sqrt{k+k^2}}$ for all $k$. (I used Mathematica to help.) E.g., if $a_k=\dfrac{1}{2k}$ for all $k$, then $\|v_k-v_j\|>\sqrt{2}$ for all $k\neq j$.

share|improve this answer
    
What a beautiful, clean, and clever argument! This is just amazing, Jonas. Thank you so much for sharing it!! –  ragrigg Nov 6 '12 at 0:46
    
@Jonas Meyer Does your construction say anything about the optimality of $\sqrt{2}$? For example, can you choose your $(a_k)$ to get. say, $\sqrt{3}$. My intuition, which is normally not very good, tells me that $\sqrt{2}$ is optimal. I am curious about this question, but I am having trouble with Norbert's argument bellow that says the distance can be arbitrarily close to $2$. –  Theo Nov 6 '12 at 15:16
    
@Theo: It doesn't say anything about optimality of $\sqrt 2$ in general, but for any example that has this particular form, $\sqrt 2$ is optimal: $a_k\to 0$ is necessary, and this implies that the distances get arbitrarily close to $\sqrt 2$. I haven't understood Norbert's argument, but I haven't seriously studied it either. Clearly for $2$ vectors in the ball we can have distance $2$, but already for $3$ the bound is not clear to me. What I would do to try to better understand (given the time to do so) is look at the problem of how big the pairwise distances among $3$ unit vectors can be. –  Jonas Meyer Nov 6 '12 at 16:23
    
Such vectors can be expressed in terms of orthonormal vectors $(e_1,e_2,e_3)$ as $(e_1,a_1 e_1+a_2 e_2, b_1e_1+b_2 e_2 +b_3 e_3)$. If say the pairwise distances are greater than $\sqrt 3$ this would imply that $\Re (a_1)<-\frac12$ and $\Re(b_1)<-\frac12$ and it starts to appear difficult for the latter two vectors to be far enough apart...(but this isn't thought through). –  Jonas Meyer Nov 6 '12 at 16:32
    
@Theo: I still don't know the whole story, but for something concrete if inelegant, if $v_1,v_2,v_3$ are unit vectors in an inner product space, not all of the pairwise distances can be greater than $\sqrt{3.5}$. Let $(e_1,e_2,e_3)$ be orthonormal such that $v_1=e_1$, $v_2=a_1 e_1+a_2e_2$, $v_3=b_1e_1+b_2e_2+b_3e_3$. Suppose $\|v_1-v_2\|^2>3.5$ and $\|v_1-v_3\|^2>3.5$. Then $\Re a_1<-\frac34$ and $\Re b_1<-\frac34$, and these inequalities imply that $\|v_2-v_3\|^2<\frac{43}{16}<3.5$. –  Jonas Meyer Nov 6 '12 at 19:38

Let $d\in(0,2R)$. We will construct by induction

  • a family of affine subsets $\{P_n\subset H:n\in\mathbb{N}\}$

  • two sequences of vectors $\{e_n\in P_n:n\in\mathbb{N}\}$, $\{c_n\in P_n:n\in\mathbb{N}\}$

  • a sequence of reals $\{r_n:n\in\mathbb{N}\}\subset \mathbb{R}_+$,

such that

$ S (c_n,r_n)\cap P_n= S (0,R)\cap P_n\tag{1}$

$P_n\subset P_{n-1}\tag{2}$

$e_n\in S (c_n,r_n)\cap P_n\tag{3}$

$\Vert x-e_{n-1} \Vert=d\tag{4}$

for all $n\in\mathbb{N}$ and $x\in P_n\cap S (c_n,r_n)$.

Properties $(1)-(4)$ will guarantee that $\Vert e_k-e_l\Vert=d$ for $k\neq l$. Indeed, without loss of generality we assume $l>k$. Conditions $(1)$ and $(2)$ gives $ S (c_n,r_n)\cap P_n\subset S (c_{n-1},r_{n-1})\cap P_{n-1}$. Hence using $(3)$ we get $e_l\in S (c_l,r_l)\cap P_l\subset S (c_n,r_n)\cap P_n$. Then by $(4)$ we conclude that $\Vert e_k-e_l\Vert=d$. Therefore we may take $A:=\{e_n:n\in\mathbb{N}\}$.

Let $c_1=0$, $r_1=R$ and $P_1=H$, then choose arbitrary $e_1\in S (c_1,r_1)\cap P_1$. Assume we have already constructed sequences $\{c_1,\ldots,c_n\}\subset H$, $\{e_1,\ldots,e_n\}\in H$ and $\{r_1,\ldots,r_n\}\subset\mathbb{R}_+$ satisfying $(1)$,$(2)$ and $(3)$. Consider affine function $f_n(x)=\langle x-c_n, e_n-c_n\rangle$ and define $$ P_{n+1}=\{x\in P_n: f_n(x)=r_n^2-d^2/2\} $$ $$ c_{n+1}=c_n+\left(1-\frac{d^2}{2r_n^2}\right)(e_n-c_n) $$ $$ r_{n+1}=d\sqrt{1-\frac{d^2}{4r_n^2}} $$ Later you'll see why, but now just see this picture enter image description here

From definition of $P_{n+1}$ we see $P_{n+1}\subset P_n$, so condition $(2)$ satisfied.

For arbitrary $x\in S (c_{n+1}, r_{n+1})\cap P_{n+1}$, we have $x\in P_{n+1}\subset P_n\subset\ldots\subset P_1=S (0,R)$, so $x\in S (0,R)\cap P_{n+1}$. This gives inclusion $ S (c_{n+1}, r_{n+1})\cap P_{n+1}\subset S (0,R)\cap P_{n+1}$. Now let $x\in S (0,R)\cap P_{n+1}$, then $\Vert x\Vert=R$ and $f_n(x)=r_n^2-d^2/2$. Moreover since $x\in P_{n+1}\subset P_n$nd $x\in S (0,R)$ we have $x\in S (0,R)\cap P_n$. Recall that $ S (0,R)\cap P_n= S (c_n,r_n)\cap P_n$, so $x\in S(c_n,r_n)$ and $\Vert x-c_n\Vert^2=r_n^2$. Since $e_n\in S(c_n,r_n)\cap P_n$, then $\Vert e_n-c_n\Vert^2=r_n^2$. It is remains to recall definitions of $c_{n+1}$ and $r_{n+1}$ to get $$ \Vert x-c_{n+1}\Vert^2= \Vert x-c_n\Vert^2+\Vert c_{n+1}-c_n\Vert^2-2\langle x-c_n, c_{n+1}-c_n\rangle= $$ $$ \Vert x-c_n\Vert^2+\left(1-\frac{d^2}{2r_n^2}\right)^2\Vert e_n-c_n\Vert^2-2\left(1-\frac{d^2}{2r_n^2}\right)\langle x-c_n, e_n-c_n\rangle= $$ $$ \Vert x-c_n\Vert^2+\left(1-\frac{d^2}{2r_n^2}\right)^2\Vert e_n-c_n\Vert^2-2\left(1-\frac{d^2}{2r_n^2}\right)f_n(x)= $$ $$ r_n^2+\left(1-\frac{d^2}{2r_n^2}\right)^2 r_n^2-2 \left(1-\frac{d^2}{2r_n^2}\right)\left(r_n^2-\frac{d^2}{2}\right)=d^2\left(1-\frac{d^2}{4r_n^2}\right)=r_{n+1}^2 $$ i.e. $x\in S(c_{n+1},r_{n+1})$. Also we know $x\in P_{n+1}$, so $x\in S(c_{n+1},r_{n+1})\cap P_{n+1}$. Since $x\in S (0,R)\cap P_{n+1}$ is arbitrary we get inclusion $S (0,R)\cap P_{n+1}\subset S(c_{n+1},r_{n+1})\cap P_{n+1}$. Both inclusions gives $S (0,R)\cap P_{n+1}= S(c_{n+1},r_{n+1})\cap P_{n+1}$, hence condition $(1)$ is satisfied.

As the consequence $ S (c_{n+1},r_{n+1})\cap P_{n+1}= S(0,R)\cap P_{n+1}\subset S(0,R)\cap P_n=S(c_n,r_n)\cap P_n $.

Now take arbitrary $x\in \ S (c_{n+1},r_{n+1})\cap P_{n+1}$. Since $x\in P_{n+1}$ then $f_n(x)=r_n^2-d^2/2$. Since $x\in S(c_{n+1},r_{n+1})\cap P_n\subset S(c_n,r_n)\cap P_n$, then $\Vert x-c_n\Vert=r_n^2$. Since $e_n\in S(c_n,r_n)\cap P_n$, then $\Vert e_n-c_n\Vert^2=r_n^2$. Thus $$ \Vert x-e_n\Vert^2=\Vert x-c_n\Vert^2+\Vert e_n-c_n\Vert-2\langle x-c_n, e_n-c_n\rangle=r_n^2+r_n^2-2f_n(x)=d^2 $$ Hence for all $x\in S (c_{n+1},r_{n+1})\cap P_{n+1}\subset P_n$ condition $(4)$ holds.

Now take arbitrary $x\in S (c_{n+1}, r_{n+1})\cap P_{n+1}$ and set $e_{n+1}\in x$. By the choice condition $(3)$ holds.

Since all conditions are satisfied we constructed by induction the desired sequences. But in fact this induction may breaks down if at some step $r_n$ becomes a complex number. Thus we need to study when the recurrence defined by $$ r_{n+1}=d\sqrt{1-\frac{d^2}{4r_n^2}} $$ and $r_1=R$ will exist. Denote $x_n=r_n/d$, then we get recurrence $$ x_{n+1}=\sqrt{1-\frac{1}{4x^2}} $$ with $x_1=R/d$. Since $d\in(0,2R)$, then $x_1\in(1/2,+\infty)$. Now take a look a this graph. enter image description here Here we make plots of functions $x$ and $\sqrt{1-\frac{1}{4x^2}}$. We see that they touches at the point $x=2^{-1/2}$

We see that this recurrence is infinite iff $x\geq 2^{-1/2}$, otherwise this is finite. In terms of $d$, this means that our recurrence well defined iff $d\leq R\sqrt{2}$.

Unfortunately we conclude that this method doesn't provide a way to get an infinite set of elements with pairwise distance between elements equal to $d>R\sqrt{2}$. It seems to mee that there is no such sequence. But anyway one can slightly modify formula for $r_{n+1}$ and $c_{n+1}$ to get the sequence of elements with pairwise distance not equalt to the same value but still greater than $R\sqrt{2}$.

share|improve this answer
    
@Norbert: Thanks for your argument; it is certainly really clever. However, as Christopher says, it shows that the pairwise distance of elements of $A$ is exactly $d$, and not greater than $d$. I've been working on this for a long time now... and still without any success. Without a doubt, a really nice problem. As usual, I'll keep working. –  ragrigg Nov 5 '12 at 15:59
    
@ragrigg I don't get what is the problem - just set $d =\sqrt{2}+0.1>\sqrt{2}$ –  userNaN Nov 5 '12 at 16:02
    
@Norbert: I apologize for my silly comment. I'm really tired. I have some questions: 1) $A$ does not lie on $S(0,R)$, right? 2) What's the geometric intuition behind your definition of $P_{n+1}$? Why the required value of $f_n$? –  ragrigg Nov 5 '12 at 20:15
2  
@Norbert I am having trouble understanding some details. Why is the distance from $e_{n+1}$ to $e_j$ with $j<n$ also $d$? Your argument seem to work in finite dimension too, right? In $3$ dimensions you would be able to find $3$ vectors with this property. However, how can you pick $3$ vectors on $S_2$ such that the distance between them is "almost" equal to the diameter (as you claim $d$ can be chosen)? –  Theo Nov 5 '12 at 22:26
1  
@Norbert I feel dumb, but I still don't see how it follows from (2). Your last claim, that $d$ can be arbitrarily close to the diameter seems highly suspicious to me. To say that all vectors $e_n$ are equally distanced $d$ is equivalent to saying the inner product is the same. For the distance to be arbitrarily close to $2R$, the inner product should be arbitrarily close to $-1$. Now, if you take $x, y, z$ such that $<x,y>=<z,y>=-1+\varepsilon$, it can be shown that $<x,z>>1-\delta(\varepsilon)$, which in turn implies $||x-z||$ is necessarily small, so it cannot be close to $2R$. –  Theo Nov 6 '12 at 15:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.