# Unit Ball of $\mathcal{l}_2$

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?

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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

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 $k\neq j$.

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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

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. 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 $R\sqrt{2}$.

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@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 $d =\sqrt{2}+0.1>\sqrt{2}$ –  Norbert 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
@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
@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