Show that $x_0 in overline{langle M rangle } $ if and only if $f(x_0) = 0, ; forall f in X^* : f|_M = 0$.












1












$begingroup$


Exercise :




Let $X$ be a normed space and $M subset X$. Show that an element $x_0 in X$ belongs to the set $overline{langle M rangle}$ if and only if $f(x_0) = 0$ for all $f in X^*$ such that $f|_M = 0$.




Attempt :



First of all, clarifying that $overline{langle M rangle}$ denotes the closure of the linear hull of $m in M$, thus it is the set :



$$overline{langle M rangle} = overline{{lambda cdot m | m in M , lambda in mathbb R}}$$



I have the intuition that what we want to prove essentialy is that $langle M rangle$ is dense over $X$, but I'm not sure if that's correct.



Other than that, I can't seem to find any other way around it, so I would really appreciate any tips or elaborations.






real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem







share|cite|improve this question









$endgroup$












  • $begingroup$
    The definition you state about linear hull is wrong by the way.
    $endgroup$
    – Shashi
    Dec 17 '18 at 17:22
















1












$begingroup$


Exercise :




Let $X$ be a normed space and $M subset X$. Show that an element $x_0 in X$ belongs to the set $overline{langle M rangle}$ if and only if $f(x_0) = 0$ for all $f in X^*$ such that $f|_M = 0$.




Attempt :



First of all, clarifying that $overline{langle M rangle}$ denotes the closure of the linear hull of $m in M$, thus it is the set :



$$overline{langle M rangle} = overline{{lambda cdot m | m in M , lambda in mathbb R}}$$



I have the intuition that what we want to prove essentialy is that $langle M rangle$ is dense over $X$, but I'm not sure if that's correct.



Other than that, I can't seem to find any other way around it, so I would really appreciate any tips or elaborations.






real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem







share|cite|improve this question









$endgroup$












  • $begingroup$
    The definition you state about linear hull is wrong by the way.
    $endgroup$
    – Shashi
    Dec 17 '18 at 17:22














1












1








1





$begingroup$


Exercise :




Let $X$ be a normed space and $M subset X$. Show that an element $x_0 in X$ belongs to the set $overline{langle M rangle}$ if and only if $f(x_0) = 0$ for all $f in X^*$ such that $f|_M = 0$.




Attempt :



First of all, clarifying that $overline{langle M rangle}$ denotes the closure of the linear hull of $m in M$, thus it is the set :



$$overline{langle M rangle} = overline{{lambda cdot m | m in M , lambda in mathbb R}}$$



I have the intuition that what we want to prove essentialy is that $langle M rangle$ is dense over $X$, but I'm not sure if that's correct.



Other than that, I can't seem to find any other way around it, so I would really appreciate any tips or elaborations.






real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem







share|cite|improve this question









$endgroup$




Exercise :




Let $X$ be a normed space and $M subset X$. Show that an element $x_0 in X$ belongs to the set $overline{langle M rangle}$ if and only if $f(x_0) = 0$ for all $f in X^*$ such that $f|_M = 0$.




Attempt :



First of all, clarifying that $overline{langle M rangle}$ denotes the closure of the linear hull of $m in M$, thus it is the set :



$$overline{langle M rangle} = overline{{lambda cdot m | m in M , lambda in mathbb R}}$$



I have the intuition that what we want to prove essentialy is that $langle M rangle$ is dense over $X$, but I'm not sure if that's correct.



Other than that, I can't seem to find any other way around it, so I would really appreciate any tips or elaborations.






real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem




real-analysis functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 16:16









RebellosRebellos

14.6k31247




14.6k31247












  • $begingroup$
    The definition you state about linear hull is wrong by the way.
    $endgroup$
    – Shashi
    Dec 17 '18 at 17:22


















  • $begingroup$
    The definition you state about linear hull is wrong by the way.
    $endgroup$
    – Shashi
    Dec 17 '18 at 17:22
















$begingroup$
The definition you state about linear hull is wrong by the way.
$endgroup$
– Shashi
Dec 17 '18 at 17:22




$begingroup$
The definition you state about linear hull is wrong by the way.
$endgroup$
– Shashi
Dec 17 '18 at 17:22










1 Answer
1






active

oldest

votes


















2












$begingroup$

$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19











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1 Answer
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1 Answer
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2












$begingroup$

$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19
















2












$begingroup$

$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19














2












2








2





$begingroup$

$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.






share|cite|improve this answer











$endgroup$



$newcommand{span}{operatorname{span}}$No, the exercise is not about proving that $span M$ is dense. The set $M$ could be a singleton for instance. A sketch of a proof could be as follows:



"$Rightarrow$": Let $xin overline{span M}$, then there is a sequence $x_nin span M$ such that $x_nto x$. Moreover if $f|_M=0$ for some $fin X^*$ then what can you say about $f(x)$? Use the continuity of $f$.



"$Leftarrow$": Assume that $f(x)=0$ for all $fin X^*$ such that $f|_M=0$. Assume that $xnotin overline{span M}$ then there is some ball $B(x,delta)$ such that $B(x,delta)capoverline{span M}=emptyset$. In particular for every $zin span M$ we have $|x-z|geq delta$.



We will make use of this. First notice that each $zin span(Mcup x)$ can be written as follows
begin{align}
z=lambda x+sum_{i=1}^n mu_i m_i
end{align}
for some $lambda$, $mu_iinmathbb K$ and $m_iin M$. We define the operator $g:span(Mcup x)tomathbb K$ as follows:
begin{align}
g(z)=lambda
end{align}
It is easy to see that it is linear. Moreover $g|_M=0$ and $g(x)=1$ by definition. It is also bounded, since (assuming $lambdaneq 0$)
begin{align}
frac{|g(z)|}{|z|}=frac{|lambda|}{|lambda x + sum_{i}mu_i m_i |}=frac{1}{|x+sum_{i}frac{mu_i}{lambda} m_i|}leq frac 1 delta
end{align}
This bound trivially holds if $lambda=0$.
Hahn Banach now says that there is an extension of $g$, call it $fin X^*$ satisfying $f|_{span(Mcup {x})}=g$. This $f$ satisfies $f(x)=1$ while $f|_M=0$ a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 22:06

























answered Dec 17 '18 at 16:49









ShashiShashi

7,1981628




7,1981628












  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19


















  • $begingroup$
    Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 17:59










  • $begingroup$
    @Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:00










  • $begingroup$
    I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
    $endgroup$
    – Rebellos
    Dec 17 '18 at 18:04










  • $begingroup$
    @Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:06






  • 1




    $begingroup$
    @Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
    $endgroup$
    – Shashi
    Dec 17 '18 at 18:19
















$begingroup$
Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
$endgroup$
– Rebellos
Dec 17 '18 at 17:59




$begingroup$
Hi and thanks a lot for your input. Well, for the first, if $f|_M = 0 implies f(x) = 0 Leftrightarrow f(lim_{n to infty} x_n) = 0 Leftrightarrow lim_{n to infty} f(x_n) = 0$ ? Where does that lead ?
$endgroup$
– Rebellos
Dec 17 '18 at 17:59












$begingroup$
@Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
$endgroup$
– Shashi
Dec 17 '18 at 18:00




$begingroup$
@Rebellos Don't you want to show that $f(x)=0$? That is what you have in your comment? I don't understand what you mean.
$endgroup$
– Shashi
Dec 17 '18 at 18:00












$begingroup$
I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
$endgroup$
– Rebellos
Dec 17 '18 at 18:04




$begingroup$
I want to show that $f(x) =0$ with $x in M$. What we have is $x in overline{text{span}M}$. I am probably missing something thought.
$endgroup$
– Rebellos
Dec 17 '18 at 18:04












$begingroup$
@Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
$endgroup$
– Shashi
Dec 17 '18 at 18:06




$begingroup$
@Rebellos Oh I had to clarify $text{ span } M$ is just $langle Mrangle$. Does it make sense now?
$endgroup$
– Shashi
Dec 17 '18 at 18:06




1




1




$begingroup$
@Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
$endgroup$
– Shashi
Dec 17 '18 at 18:19




$begingroup$
@Rebellos I'm happy as long as you understand it! Let me know if you come up with something unclear!
$endgroup$
– Shashi
Dec 17 '18 at 18:19


















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Ellipse (mathématiques)

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Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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