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Let $x_n$ be a sequence in a Hilbert space such that $\left\Vert x_n \right\Vert=1$ and $ \langle x_n,\ x_m \rangle =0 $, for all $n \neq m$.

Let $ K= \{ x_n/ n : n \in \mathbb{N} \} \cup \{0\} $.

I need to show that $K$ is compact, $\operatorname{co}(K)$ is bounded, but not closed and finally find all the extreme points of $ \overline{\operatorname{co}(K)} $ .

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$co(M)$ is the convex hull of $M$? –  user20266 Jun 24 '12 at 9:06
    
Yes, that's right. –  user34378 Jun 24 '12 at 9:10
1  
Are you sure you stated this correctly? Is, possibly, $||x_n||=1/n$? Or are you supposed to show weak compactness? –  user20266 Jun 24 '12 at 9:10
    
Oh... I've made a mistake. Let me fix it! –  user34378 Jun 24 '12 at 9:11
1  
Yes. Any subsequence of a convergent sequence converges. –  user20266 Jun 24 '12 at 9:22

2 Answers 2

We can assume without loss of generality that $\mathcal H$ is the separable Hilbert space with $x_n$ as a basis (because all the objects considered lie within the complete span of $x_n$).

$K$ is compact because it is sequentially compact -- any infinite sequence in $K$ tends to $0\in K$.

$co(K)$ is bounded, because the norm of a convex combination of any points is no greater than the largest of their norms (by triangle inequality), and thus because $K$ is bounded, so is $co(K)$.

$co(K)$ is not closed, because for example $\sum x_n2^{-n}/n$ lies in the closure, but is not a convex combination of $x_n/n$ (because any convex combination of $x_n$ is eventually orthogonal to $x_n$). In fact, $cl(co(K))$ is the set of infinite convex combinations of $x_n/n$.

The extreme points of $K'=cl(co(K))$ are precisely the elements of $K$:

  • First, we can see that $K$ are extreme points of $K'$, beacause for any nontrivial convex combination $x$ of elements of $K'$ we have for some $n$ that $\langle x\vert x_n\rangle\in (0,1/n)$, so $x$ is not any of $K$
  • Suppose $x=\sum \alpha_nx_n$ is an extreme point of $K'$.
  • Then $\sum n\alpha_n\leq 1$ and $0\leq n\alpha_n\leq 1$.
  • If there for some $n\neq m$ we have $\alpha_n,\alpha_m\neq 0$, then choose $\varepsilon>0$ smaller than $n\alpha_n$ and $m\alpha_m$. Put $y_1=x-x_n\varepsilon/n+x_m\varepsilon/m$, $y_2=x+x_n\varepsilon/n-x_m\varepsilon/m$. Then clearly $x=(y_1+y_2)/2$, and because of the way we have chosen $\varepsilon$, we have $y_1,y_2\in K'$, so we have a contradiction.
  • Therefore it must be that at most one of $\alpha_n$ is nonzero. If all are zero, then we're done. If there is $\alpha_n\neq 0$, then it is easy to see that $\alpha_n=1/n$ and we're done.
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I've wasted so much time for my answer and it turns out the same as yours! –  userNaN Jun 24 '12 at 13:27
    
@Norbert: I did not write mine quickly, either. ;) Though it is quite a bit more compact, if I may say so. ;) –  tomasz Jun 24 '12 at 13:30
1  
It may be worth mentioning the following facts (all due to Milman, I believe): 1. A closed set $K$ in a Banach space is compact if and only if there is a sequence $x_n \to 0$ such that $K \subset \mathop{\overline{\rm co} \{x_n\}}$. 2. The closed convex hull of a compact set in a Banach space is compact. 3. If $K = \mathop{\overline{\rm co}} A$ then the extremal points of $K$ are contained in the closure of $A$. –  t.b. Jun 24 '12 at 19:07

1. Proof that $K$ is a compact. Consider arbitrary infinite set $\{y_k:k\in\mathbb{N}\}\subset K$, then it has the form $y_k=x_{n_k}/n_k$ for some sequence $\{n_k:k\in\mathbb{N}\}$. Since $\{y_k:k\in\mathbb{N}\}$ is infinite, then does $\{n_k:k\in\mathbb{N}\}$, hence there exist subsequence $\{n_{k_l}:l\in\mathbb{N}\}$ tending to infinity. Now note that $$ \lim\limits_{l\to\infty}\Vert y_{k_l}\Vert= \lim\limits_{l\to\infty}\Vert n_{k_l}^{-1}x_{n_{k_l}}\Vert\leq \lim\limits_{l\to\infty}n_{k_l}^{-1}\left(\limsup\limits_{l\to\infty}\Vert x_{n_{k_l}}\Vert\right)=0 $$ This means that $\{y_k:k\in\mathbb{N}\}$ have limit convergent subsequence $\{y_{k_l}:l\in\mathbb{N}\}$. Since $\{y_k:k\in\mathbb{N}\}\subset K$, then $K$ is a compact.

2. Proof that $co(K)$ is bounded. Consider arbitrary $x\in co(K)$, then $$ x=\sum\limits_{k=1}^{m-1}\lambda_k n_k^{-1}x_{n_k}+\left(1-\sum\limits_{k=1}^{m-1}\lambda_{k}\right)0 $$ for some real numbers $\{\lambda_k\in[0,1]:k\in\{1,\ldots,m\}\}$ such that $\sum_{k=1}^m\lambda_k\leq 1$ and some indexes $\{n_k\in\mathbb{N}:k\in\{1,\ldots,m-1\}\}$. In this case $$ \Vert x\Vert=\left\Vert\sum\limits_{k=1}^{m-1}\lambda_k x_{n_k}\right\Vert\leq \sum\limits_{k=1}^{m-1}|\lambda_k| \Vert n_k^{-1}x_{n_k}\Vert\leq \sum\limits_{k=1}^{m-1}|\lambda_k| n_k^{-1}\leq \sum\limits_{k=1}^{m-1} \lambda_k\leq 1 $$ Since $x\in co(K)$ is arbitrary, then $co(K)$ is bounded.

3. Proof that $co(K)$ is not closed. Note that the series $$ \sum\limits_{n=1}^\infty\Vert 2^{-n} n^{-1} x_n\Vert\leq\sum\limits_{n=1}^\infty 2^{-n}=1 $$ is convergent and we are working in a Hilbert space, hence we have well defined vector $$ \hat{x}=\sum\limits_{n=1}^\infty 2^{-n} n^{-1} x_n $$ From definition of $co(K)$ we see that each element of $co(K)$ is finite linear combination of elemnts of $K$, hence $\hat{x}\notin co(K)$. But $x\in\overline{co(K)}$. Indeed for each $\varepsilon>0$ consider natural number $m=\lceil\log_2 \varepsilon^{-1}\rceil+1$ and the vector $$ x_\varepsilon=\sum\limits_{n=1}^{m-1} 2^{-n}n^{-1}x_n+\left(1-\sum\limits_{n=1}^{m-1} 2^{-n}\right) 0\in co(K) $$ Now we see that $$ \Vert \hat{x}-x_\varepsilon\Vert=\left\Vert\sum\limits_{n=m+1}^\infty 2^{-n}n^{-1}x_n\right\Vert\leq \sum\limits_{n=m+1}^\infty 2^{-n}n^{-1}\Vert x_n\Vert\leq \sum\limits_{n=m+1}^\infty 2^{-n}=2^{-m}<\varepsilon $$ Since $\varepsilon>0$ is arbitrary then $\hat{x}\in\overline{co(K)}$, though $\hat{x}\notin co(K)$. Hence $co(K)$ is not closed.

4. Description of elements of $\overline{co(K)}$. Since $x\in\overline{co(K)}\subset\overline{\mathrm{span}\{x_n:n\in\mathbb{N}\}}$ and the system $\{x_n:n\in\mathbb{N}\}$ is orthonormal, then $x=\sum_{n=1}^\infty\mu_n x_n$ for some real numbers $\{\mu_n\in\mathbb{R}:n\in\mathbb{N}\}$. But for convenience we will consider the following representation $$ x=\sum_{n=1}^\infty\lambda_n n^{-1} x_n $$ for some real numbers $\{\lambda_n\in\mathbb{R}:n\in\mathbb{N}\}$. We claim that for all $n\in\mathbb{N}$ we have $\lambda_n\in[0,1]$ and $\sum_{n=1}^\infty\lambda_n\in(0,1)$. Fix arbitrary $\varepsilon>0$. Since $x\in\overline{co(K)}$, then there exist real numbers $\{\lambda_n^{\varepsilon}\in[0,1]:n\in\{1,\ldots,m^{\varepsilon}\}\}$ such that $$ \left\Vert x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n\right\Vert\leq\varepsilon,\qquad \sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n^{\varepsilon}\in[0,1] $$ Then from orthogonality for all $k\in\mathbb{N}$ we get $$ |\lambda_k-\lambda_k^{\varepsilon}|= \left|\langle x, k x_k\rangle-\langle \sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n, k x_k\rangle\right|= k\left|\langle x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n, x_k\rangle\right|\leq $$ $$ k\left\Vert x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n \right\Vert\Vert x_k\Vert\leq k\varepsilon $$ Using the fact $\lambda_k^{\varepsilon}\in[0,1]$ we see that $$ -k\varepsilon<\lambda_k^{\varepsilon}-k\varepsilon<\lambda_k<\lambda_k^{\varepsilon}+k\varepsilon<1+k\varepsilon $$ Since $\varepsilon>0$ is arbitrary we conclude $\lambda_k\in[0,1]$. Since $k\in\mathbb{N}$ is also arbitrary we have $\{\lambda_n:n\in\mathbb{N}\}\subset[0,1]$.

Similarly, for all $k\in\mathbb{N}$ from orthogonality we get $$ \left|\sum\limits_{n=1}^k\lambda_k - \sum\limits_{n=1}^k\lambda_n^{\varepsilon}\right|= \left|\langle x, \sum\limits_{l=1}^k l x_l\rangle-\langle \sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n, \sum\limits_{l=1}^k l x_l\rangle\right|= \left|\langle x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n, \sum\limits_{l=1}^k l x_l\rangle\right|\leq $$ $$ \left\Vert x-\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n n^{-1} x_n\right\Vert \left\Vert \sum\limits_{l=1}^{k} l x_l\right\Vert\leq \varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2} $$ Using the fact that $0\leq\sum_{n=1}^{m^{\varepsilon}}\lambda_n^{\varepsilon}\leq 1$ we see that $$ -\varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2}<\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n^{\varepsilon}-\varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2}<\sum\limits_{n=1}^k\lambda_k<\sum\limits_{n=1}^{m^{\varepsilon}}\lambda_n^{\varepsilon}+\varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2}<1+\varepsilon\left(\sum\limits_{l=1}^k l^2\right)^{1/2} $$ Since $\varepsilon>0$ is arbitrary we conclude $\sum_{n=1}^k\lambda_k\in[0,1]$. Since $k\in\mathbb{N}$ is also arbitrary we have $\sum_{n=1}^\infty\lambda_k\in[0,1]$. Thus we proved that each element $x\in \overline{co(K)}$ has the form $$ x=\sum_{n=1}^\infty\lambda_n n^{-1} x_n $$ for some real numbers $\{\lambda_n\in[0,1]:n\in\mathbb{N}\}$ satisfying $\sum_{n=1}^\infty\lambda_k\in[0,1]$.

5. Description of extreme points of $\overline{co(K)}$. Fix some $x\in \overline{co(K)}$ and consider its representation given above.

Case 1. Consider case when $\sum_{n=1}^\infty\lambda_n\in(0,1)$, then there exist some $k\in\mathbb{N}$ such that $\lambda_k\in(0,1)$. Define $$ \alpha=\min\left(\frac{1}{2},\left(\sum\limits_{n=1}^\infty\lambda_n\right)^{-1}-1\right) \qquad \delta_n= \begin{cases} \min(\lambda_n,1-\lambda_n) &\text{ if }\quad \lambda_n\in(0,1)\\ 0 &\text{ if }\quad \lambda_n\in\{0,1\} \end{cases} $$ $$ x^{(1)}=\sum\limits_{n=1}^\infty\lambda_n^{(1)} n^{-1} x_n\quad\text{ where }\quad\lambda_n^{(1)}=\lambda_n+\alpha\delta_n $$ $$ x^{(2)}=\sum\limits_{n=1}^\infty\lambda_n^{(2)} n^{-1} x_n\quad\text{ where }\quad\lambda_n^{(2)}=\lambda_n-\alpha\delta_n $$ Now we check that $x^{(1)}$ and $x^{(2)}$ are in $\overline{co(K)}$. Lets go $$ \lambda_n^{(1)}=\lambda_n+\alpha\delta_n>\lambda_n\geq 0 $$ $$ \lambda_n^{(1)}=\lambda_n+\alpha\delta_n\leq\lambda_n+\frac{1}{2}\delta_n\leq\lambda_n+\delta_n\leq\lambda_n+1-\lambda_n=1 $$ $$ \lambda_n^{(2)}=\lambda_n-\alpha\delta_n\geq\lambda_n-\frac{1}{2}\delta_n\geq\lambda_n-\frac{1}{2}\lambda_n=\frac{1}{2}\lambda_n\geq 0 $$ $$ \lambda_n^{(2)}=\lambda_n-\alpha\delta_n\leq\lambda_n\leq 1 $$ $$ \sum\limits_{n=1}^\infty\lambda_n^{(1)}=\sum\limits_{n=1}^\infty(\lambda_n+\alpha\delta_n)= \sum\limits_{n=1}^\infty\lambda_n+\alpha\sum\limits_{n=1}^\infty\delta_n\leq \sum\limits_{n=1}^\infty\lambda_n+\alpha\sum\limits_{n=1}^\infty\lambda_n= (\alpha+1)\sum\limits_{n=1}^\infty\lambda\leq 1 $$ $$ \sum\limits_{n=1}^\infty\lambda_n^{(2)}=\sum\limits_{n=1}^\infty(\lambda_n-\alpha\delta_n)\leq \sum\limits_{n=1}^\infty\lambda_n\leq 1 $$ Since all necessary conditions are satisfied $x^{(1)}$, $x^{(2)}\in \overline{co(K)}$. This elements are distinct. Indeed, from orthogonality we have $$ \Vert x^{(1)}-x^{(2)}\Vert^2=\left\Vert \sum\limits_{n=1}^\infty 2\alpha\delta_n x_n\right\Vert^2= 4\alpha^2\delta_n^2\sum\limits_{n=1}^\infty\Vert x_n\Vert^2> 4\alpha^2\delta_k^2\Vert x_k\Vert^2>0 $$ hence $x^{(1)}\neq x^{(2)}$. And finally we see that $$ x=\sum\limits_{n=1}^\infty \lambda_n n^{-1} x_n= \frac{1}{2}\sum\limits_{n=1}^\infty \lambda_n^{(1)} n^{-1} x_n+ \frac{1}{2}\sum\limits_{n=1}^\infty \lambda_n^{(2)} n^{-1} x_n=\frac{x^{(1)}+x^{(2)}}{2}. $$ This means that $x$ can be middle of some non-trivial segment $[x^{(2)},x^{(2)}]$ with ends in $\overline{co(K)}$, so $x$ is not an extreme point.

Case 2. Consider case, when $\sum_{n=1}^\infty\lambda_n=0$, i.e. $x=0$. Assume we are given representation $x=t x^{(1)}+(1-t)x^{(2)}$ with $t\in(0,1)$ and $$ x^{(1)}=\sum\limits_{n=1}^\infty\lambda_n^{(1)} n^{-1} x_n\qquad x^{(2)}=\sum\limits_{n=1}^\infty\lambda_n^{(2)} n^{-1} x_n $$ where $0\leq\sum_{n=1}^\infty\lambda_n^{(1)}\leq 1$ and $0\leq\sum_{n=1}^\infty\lambda_n^{(2)}\leq 1$. Then from orthogonality for all $k\in\mathbb{N}$ we obtain $$ t\lambda_r^{(1)}+(1-t)\lambda_r^{(2)}=t\langle x^{(1)},x_r\rangle+(1-t)\langle x^{(1)},x_r\rangle=\langle tx^{(1)}+(1-t)x^{(2)},x_r\rangle=\langle x, x_r\rangle=0 $$ Because $t\in(0,1)$ and $\lambda_n^{(1)},\lambda_n^{(2)}\geq 0$ for all $n\in\mathbb{N}$ we conclude that the last equality is possible iff $\lambda_r^{(1)},\lambda_r^{(2)}=0$ for all $r\in\mathbb{N}$. As the consequence $x^{(1)}=x^{(2)}=0$. This means that $x=0$ is an extreme point.

Case 3. Now the third case, when $\sum_{n=1}^\infty\lambda_n=1$, i. e. $\lambda_n=\delta_{n,k}$ for some $k\in\mathbb{N}$. In this case $x=\sum_{n=1}^\infty\lambda_n n^{-1} x_n=k^{-1} x_k$. Again assume we are given representation $x=t x^{(1)}+(1-t)x^{(2)}$ with $t\in(0,1)$ and $$ x^{(1)}=\sum\limits_{n=1}^\infty\lambda_n^{(1)} n^{-1} x_n\qquad x^{(2)}=\sum\limits_{n=1}^\infty\lambda_n^{(2)} n^{-1} x_n $$ where $0\leq\sum_{n=1}^\infty\lambda_n^{(1)}\leq 1$ and $0\leq\sum_{n=1}^\infty\lambda_n^{(2)}\leq 1$. Then from orthogonality for all $k\in\mathbb{N}$ we obtain $$ t\lambda_r^{(1)}+(1-t)\lambda_r^{(2)}=t\langle x^{(1)},x_r\rangle+(1-t)\langle x^{(1)},x_r\rangle=\langle tx^{(1)}+(1-t)x^{(2)},x_r\rangle=\langle x, x_r\rangle=\delta_{r,k} $$ Because $t\in(0,1)$ and $\lambda_n^{(1)},\lambda_n^{(2)}\geq 0$ for all $n\in\mathbb{N}$ we conclude that the last equality is possible iff $\lambda_r^{(1)}=\lambda_r^{(2)}=\delta_{r,k}$ and as the consequence $x^{(1)}=x^{(2)}=k^{-1}x_k$. This means that $x=x_k$ is an extreme point.

Whew, all cases are considered and we conclude that extreme points of $\overline{co(K)}$ are $0$ and $\{n^{-1} x_n:n\in\mathbb{N}\}$, i.e. a set $K$.

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I did not check all the calculations, but it seems that it is indeed essentially the same solution as mine, if a bit too elaborate for my taste. Still, I have been criticized for being too brief, so it may be just me. +1 :) –  tomasz Jun 24 '12 at 13:40
    
Also, the unproved fact is not hard to rigorously show. Suppose that some $\lambda_n\notin [0,1]$ or $\sum\lambda_n>1$, then we cannot have a sequence of elements satisfying those converging to $\sum\lambda_nx_n$ (the representation of any $x$ as such a sum comes from the fact that $x_n$ is an orthonormal sequence, wlog basis) –  tomasz Jun 24 '12 at 13:42
    
thanks for the idea, later I'll write a proof of this fact –  userNaN Jun 24 '12 at 13:48

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