Extreme points of a closed unit ball












1












$begingroup$


H is the closed unit ball of $ell_1(mathbb{N})$ Show that the set of extreme points of H is Ex(H)= ${lambda_{e_n} : lambda in mathbb{C }, |lambda| = 1, n ≥ 1 },$ where $(e_n)_{ngeq 1}$ the standard (unit vector) basis for $ell_1(mathbb{N}). $ and show that the closure of the extreme points is equal to $H cap c_e(mathbb{N})$






functional-analysis







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$endgroup$












  • $begingroup$
    what's $lambda_{e_n}$? also, you should show your thoughts on the problem
    $endgroup$
    – mathworker21
    Jan 6 at 13:03












  • $begingroup$
    Vector to the standard (unit vector) basis.
    $endgroup$
    – Anna Schmitz
    Jan 6 at 13:12










  • $begingroup$
    you probably mean $lambda e_n$, not $lambda_{e_n}$. i also dont know what "vector to the standard (unit vector) basis" means. sounds like word vomit
    $endgroup$
    – mathworker21
    Jan 6 at 13:35












  • $begingroup$
    That might be true. I am just super lost
    $endgroup$
    – Anna Schmitz
    Jan 6 at 13:47










  • $begingroup$
    idk, maybe reread the book or find other references online.
    $endgroup$
    – mathworker21
    Jan 6 at 13:53
















1












$begingroup$


H is the closed unit ball of $ell_1(mathbb{N})$ Show that the set of extreme points of H is Ex(H)= ${lambda_{e_n} : lambda in mathbb{C }, |lambda| = 1, n ≥ 1 },$ where $(e_n)_{ngeq 1}$ the standard (unit vector) basis for $ell_1(mathbb{N}). $ and show that the closure of the extreme points is equal to $H cap c_e(mathbb{N})$






functional-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    what's $lambda_{e_n}$? also, you should show your thoughts on the problem
    $endgroup$
    – mathworker21
    Jan 6 at 13:03












  • $begingroup$
    Vector to the standard (unit vector) basis.
    $endgroup$
    – Anna Schmitz
    Jan 6 at 13:12










  • $begingroup$
    you probably mean $lambda e_n$, not $lambda_{e_n}$. i also dont know what "vector to the standard (unit vector) basis" means. sounds like word vomit
    $endgroup$
    – mathworker21
    Jan 6 at 13:35












  • $begingroup$
    That might be true. I am just super lost
    $endgroup$
    – Anna Schmitz
    Jan 6 at 13:47










  • $begingroup$
    idk, maybe reread the book or find other references online.
    $endgroup$
    – mathworker21
    Jan 6 at 13:53














1












1








1





$begingroup$


H is the closed unit ball of $ell_1(mathbb{N})$ Show that the set of extreme points of H is Ex(H)= ${lambda_{e_n} : lambda in mathbb{C }, |lambda| = 1, n ≥ 1 },$ where $(e_n)_{ngeq 1}$ the standard (unit vector) basis for $ell_1(mathbb{N}). $ and show that the closure of the extreme points is equal to $H cap c_e(mathbb{N})$






functional-analysis







share|cite|improve this question











$endgroup$




H is the closed unit ball of $ell_1(mathbb{N})$ Show that the set of extreme points of H is Ex(H)= ${lambda_{e_n} : lambda in mathbb{C }, |lambda| = 1, n ≥ 1 },$ where $(e_n)_{ngeq 1}$ the standard (unit vector) basis for $ell_1(mathbb{N}). $ and show that the closure of the extreme points is equal to $H cap c_e(mathbb{N})$






functional-analysis




functional-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 12:49







Anna Schmitz

















asked Jan 6 at 12:44









Anna SchmitzAnna Schmitz

917




917












  • $begingroup$
    what's $lambda_{e_n}$? also, you should show your thoughts on the problem
    $endgroup$
    – mathworker21
    Jan 6 at 13:03












  • $begingroup$
    Vector to the standard (unit vector) basis.
    $endgroup$
    – Anna Schmitz
    Jan 6 at 13:12










  • $begingroup$
    you probably mean $lambda e_n$, not $lambda_{e_n}$. i also dont know what "vector to the standard (unit vector) basis" means. sounds like word vomit
    $endgroup$
    – mathworker21
    Jan 6 at 13:35












  • $begingroup$
    That might be true. I am just super lost
    $endgroup$
    – Anna Schmitz
    Jan 6 at 13:47










  • $begingroup$
    idk, maybe reread the book or find other references online.
    $endgroup$
    – mathworker21
    Jan 6 at 13:53


















  • $begingroup$
    what's $lambda_{e_n}$? also, you should show your thoughts on the problem
    $endgroup$
    – mathworker21
    Jan 6 at 13:03












  • $begingroup$
    Vector to the standard (unit vector) basis.
    $endgroup$
    – Anna Schmitz
    Jan 6 at 13:12










  • $begingroup$
    you probably mean $lambda e_n$, not $lambda_{e_n}$. i also dont know what "vector to the standard (unit vector) basis" means. sounds like word vomit
    $endgroup$
    – mathworker21
    Jan 6 at 13:35












  • $begingroup$
    That might be true. I am just super lost
    $endgroup$
    – Anna Schmitz
    Jan 6 at 13:47










  • $begingroup$
    idk, maybe reread the book or find other references online.
    $endgroup$
    – mathworker21
    Jan 6 at 13:53
















$begingroup$
what's $lambda_{e_n}$? also, you should show your thoughts on the problem
$endgroup$
– mathworker21
Jan 6 at 13:03






$begingroup$
what's $lambda_{e_n}$? also, you should show your thoughts on the problem
$endgroup$
– mathworker21
Jan 6 at 13:03














$begingroup$
Vector to the standard (unit vector) basis.
$endgroup$
– Anna Schmitz
Jan 6 at 13:12




$begingroup$
Vector to the standard (unit vector) basis.
$endgroup$
– Anna Schmitz
Jan 6 at 13:12












$begingroup$
you probably mean $lambda e_n$, not $lambda_{e_n}$. i also dont know what "vector to the standard (unit vector) basis" means. sounds like word vomit
$endgroup$
– mathworker21
Jan 6 at 13:35






$begingroup$
you probably mean $lambda e_n$, not $lambda_{e_n}$. i also dont know what "vector to the standard (unit vector) basis" means. sounds like word vomit
$endgroup$
– mathworker21
Jan 6 at 13:35














$begingroup$
That might be true. I am just super lost
$endgroup$
– Anna Schmitz
Jan 6 at 13:47




$begingroup$
That might be true. I am just super lost
$endgroup$
– Anna Schmitz
Jan 6 at 13:47












$begingroup$
idk, maybe reread the book or find other references online.
$endgroup$
– mathworker21
Jan 6 at 13:53




$begingroup$
idk, maybe reread the book or find other references online.
$endgroup$
– mathworker21
Jan 6 at 13:53










1 Answer
1






active

oldest

votes


















0












$begingroup$

Hints:



$1). e_n$ is an extreme point for each integer $n$: suppose $0<t<1; x,yin H$. Then, consider the components of $e_n=tx+(1-t)y$ to conclude.



$2). $ Suppose $xin $Ex$ H$ but $xneq lambda e_n$ for some integer $n, |lambda|=1.$ Then, $x$ has a component $x^{(m)}$ such that $|x^{(m)}|<1.$ Consider $y=e_m$ and $z=$ the sequence that is $0$ in the $m-th$ component and $frac{x^{(n)}}{1-x^{(m)}}$ everywhere else. Note that $y$ and $z$ have norm $1$. To conclude, write $x$ as a convex combination of $y$ and $z$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have a question about your second hint: I would write $x=x^{(m)}y+(1-x^{(m)})z$ but no one tells me that $0 < x^{(m)} < 1$
    $endgroup$
    – user289143
    Jan 8 at 16:06












  • $begingroup$
    It follows from the hypothesis: wlog real components to make it easier to see. We have $; |lambda |=1$ and $|x|le 1$, so if $xneq lambda e_n$ then either $x=zeta e_n$ with $zeta<1$ or $x$ has at least two components so at least one if these components must be less than $1$
    $endgroup$
    – Matematleta
    Jan 8 at 16:41












  • $begingroup$
    Sorry, my comment wasn't precise. I agree that $|x^{(m)}|<1$, but it doesn't need to be real. And I can't find any other $0 < alpha <1$ such that $x=alpha y +(1-alpha)z$
    $endgroup$
    – user289143
    Jan 8 at 17:11










  • $begingroup$
    But you can choose $theta$ such that $e^{itheta}x^{(m)}$ is real.
    $endgroup$
    – Matematleta
    Jan 8 at 17:36













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0












$begingroup$

Hints:



$1). e_n$ is an extreme point for each integer $n$: suppose $0<t<1; x,yin H$. Then, consider the components of $e_n=tx+(1-t)y$ to conclude.



$2). $ Suppose $xin $Ex$ H$ but $xneq lambda e_n$ for some integer $n, |lambda|=1.$ Then, $x$ has a component $x^{(m)}$ such that $|x^{(m)}|<1.$ Consider $y=e_m$ and $z=$ the sequence that is $0$ in the $m-th$ component and $frac{x^{(n)}}{1-x^{(m)}}$ everywhere else. Note that $y$ and $z$ have norm $1$. To conclude, write $x$ as a convex combination of $y$ and $z$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have a question about your second hint: I would write $x=x^{(m)}y+(1-x^{(m)})z$ but no one tells me that $0 < x^{(m)} < 1$
    $endgroup$
    – user289143
    Jan 8 at 16:06












  • $begingroup$
    It follows from the hypothesis: wlog real components to make it easier to see. We have $; |lambda |=1$ and $|x|le 1$, so if $xneq lambda e_n$ then either $x=zeta e_n$ with $zeta<1$ or $x$ has at least two components so at least one if these components must be less than $1$
    $endgroup$
    – Matematleta
    Jan 8 at 16:41












  • $begingroup$
    Sorry, my comment wasn't precise. I agree that $|x^{(m)}|<1$, but it doesn't need to be real. And I can't find any other $0 < alpha <1$ such that $x=alpha y +(1-alpha)z$
    $endgroup$
    – user289143
    Jan 8 at 17:11










  • $begingroup$
    But you can choose $theta$ such that $e^{itheta}x^{(m)}$ is real.
    $endgroup$
    – Matematleta
    Jan 8 at 17:36


















0












$begingroup$

Hints:



$1). e_n$ is an extreme point for each integer $n$: suppose $0<t<1; x,yin H$. Then, consider the components of $e_n=tx+(1-t)y$ to conclude.



$2). $ Suppose $xin $Ex$ H$ but $xneq lambda e_n$ for some integer $n, |lambda|=1.$ Then, $x$ has a component $x^{(m)}$ such that $|x^{(m)}|<1.$ Consider $y=e_m$ and $z=$ the sequence that is $0$ in the $m-th$ component and $frac{x^{(n)}}{1-x^{(m)}}$ everywhere else. Note that $y$ and $z$ have norm $1$. To conclude, write $x$ as a convex combination of $y$ and $z$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have a question about your second hint: I would write $x=x^{(m)}y+(1-x^{(m)})z$ but no one tells me that $0 < x^{(m)} < 1$
    $endgroup$
    – user289143
    Jan 8 at 16:06












  • $begingroup$
    It follows from the hypothesis: wlog real components to make it easier to see. We have $; |lambda |=1$ and $|x|le 1$, so if $xneq lambda e_n$ then either $x=zeta e_n$ with $zeta<1$ or $x$ has at least two components so at least one if these components must be less than $1$
    $endgroup$
    – Matematleta
    Jan 8 at 16:41












  • $begingroup$
    Sorry, my comment wasn't precise. I agree that $|x^{(m)}|<1$, but it doesn't need to be real. And I can't find any other $0 < alpha <1$ such that $x=alpha y +(1-alpha)z$
    $endgroup$
    – user289143
    Jan 8 at 17:11










  • $begingroup$
    But you can choose $theta$ such that $e^{itheta}x^{(m)}$ is real.
    $endgroup$
    – Matematleta
    Jan 8 at 17:36
















0












0








0





$begingroup$

Hints:



$1). e_n$ is an extreme point for each integer $n$: suppose $0<t<1; x,yin H$. Then, consider the components of $e_n=tx+(1-t)y$ to conclude.



$2). $ Suppose $xin $Ex$ H$ but $xneq lambda e_n$ for some integer $n, |lambda|=1.$ Then, $x$ has a component $x^{(m)}$ such that $|x^{(m)}|<1.$ Consider $y=e_m$ and $z=$ the sequence that is $0$ in the $m-th$ component and $frac{x^{(n)}}{1-x^{(m)}}$ everywhere else. Note that $y$ and $z$ have norm $1$. To conclude, write $x$ as a convex combination of $y$ and $z$






share|cite|improve this answer









$endgroup$



Hints:



$1). e_n$ is an extreme point for each integer $n$: suppose $0<t<1; x,yin H$. Then, consider the components of $e_n=tx+(1-t)y$ to conclude.



$2). $ Suppose $xin $Ex$ H$ but $xneq lambda e_n$ for some integer $n, |lambda|=1.$ Then, $x$ has a component $x^{(m)}$ such that $|x^{(m)}|<1.$ Consider $y=e_m$ and $z=$ the sequence that is $0$ in the $m-th$ component and $frac{x^{(n)}}{1-x^{(m)}}$ everywhere else. Note that $y$ and $z$ have norm $1$. To conclude, write $x$ as a convex combination of $y$ and $z$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 16:11









MatematletaMatematleta

11.6k2920




11.6k2920












  • $begingroup$
    I have a question about your second hint: I would write $x=x^{(m)}y+(1-x^{(m)})z$ but no one tells me that $0 < x^{(m)} < 1$
    $endgroup$
    – user289143
    Jan 8 at 16:06












  • $begingroup$
    It follows from the hypothesis: wlog real components to make it easier to see. We have $; |lambda |=1$ and $|x|le 1$, so if $xneq lambda e_n$ then either $x=zeta e_n$ with $zeta<1$ or $x$ has at least two components so at least one if these components must be less than $1$
    $endgroup$
    – Matematleta
    Jan 8 at 16:41












  • $begingroup$
    Sorry, my comment wasn't precise. I agree that $|x^{(m)}|<1$, but it doesn't need to be real. And I can't find any other $0 < alpha <1$ such that $x=alpha y +(1-alpha)z$
    $endgroup$
    – user289143
    Jan 8 at 17:11










  • $begingroup$
    But you can choose $theta$ such that $e^{itheta}x^{(m)}$ is real.
    $endgroup$
    – Matematleta
    Jan 8 at 17:36




















  • $begingroup$
    I have a question about your second hint: I would write $x=x^{(m)}y+(1-x^{(m)})z$ but no one tells me that $0 < x^{(m)} < 1$
    $endgroup$
    – user289143
    Jan 8 at 16:06












  • $begingroup$
    It follows from the hypothesis: wlog real components to make it easier to see. We have $; |lambda |=1$ and $|x|le 1$, so if $xneq lambda e_n$ then either $x=zeta e_n$ with $zeta<1$ or $x$ has at least two components so at least one if these components must be less than $1$
    $endgroup$
    – Matematleta
    Jan 8 at 16:41












  • $begingroup$
    Sorry, my comment wasn't precise. I agree that $|x^{(m)}|<1$, but it doesn't need to be real. And I can't find any other $0 < alpha <1$ such that $x=alpha y +(1-alpha)z$
    $endgroup$
    – user289143
    Jan 8 at 17:11










  • $begingroup$
    But you can choose $theta$ such that $e^{itheta}x^{(m)}$ is real.
    $endgroup$
    – Matematleta
    Jan 8 at 17:36


















$begingroup$
I have a question about your second hint: I would write $x=x^{(m)}y+(1-x^{(m)})z$ but no one tells me that $0 < x^{(m)} < 1$
$endgroup$
– user289143
Jan 8 at 16:06






$begingroup$
I have a question about your second hint: I would write $x=x^{(m)}y+(1-x^{(m)})z$ but no one tells me that $0 < x^{(m)} < 1$
$endgroup$
– user289143
Jan 8 at 16:06














$begingroup$
It follows from the hypothesis: wlog real components to make it easier to see. We have $; |lambda |=1$ and $|x|le 1$, so if $xneq lambda e_n$ then either $x=zeta e_n$ with $zeta<1$ or $x$ has at least two components so at least one if these components must be less than $1$
$endgroup$
– Matematleta
Jan 8 at 16:41






$begingroup$
It follows from the hypothesis: wlog real components to make it easier to see. We have $; |lambda |=1$ and $|x|le 1$, so if $xneq lambda e_n$ then either $x=zeta e_n$ with $zeta<1$ or $x$ has at least two components so at least one if these components must be less than $1$
$endgroup$
– Matematleta
Jan 8 at 16:41














$begingroup$
Sorry, my comment wasn't precise. I agree that $|x^{(m)}|<1$, but it doesn't need to be real. And I can't find any other $0 < alpha <1$ such that $x=alpha y +(1-alpha)z$
$endgroup$
– user289143
Jan 8 at 17:11




$begingroup$
Sorry, my comment wasn't precise. I agree that $|x^{(m)}|<1$, but it doesn't need to be real. And I can't find any other $0 < alpha <1$ such that $x=alpha y +(1-alpha)z$
$endgroup$
– user289143
Jan 8 at 17:11












$begingroup$
But you can choose $theta$ such that $e^{itheta}x^{(m)}$ is real.
$endgroup$
– Matematleta
Jan 8 at 17:36






$begingroup$
But you can choose $theta$ such that $e^{itheta}x^{(m)}$ is real.
$endgroup$
– Matematleta
Jan 8 at 17:36




















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