Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$












4












$begingroup$


Can you check if my proof is correct?




Let $X$ be a normed linear space. Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$




Let $fin X^{*}$, then $f$ is a bounded linear functional. Assume that $x,in X$ such that begin{align}f(x)=f(y)&iff f(x)-f(y)=0, \& iff f(x-y)=0, ;text{since};f ;text{is a linear functional};\& iff x-yin ker f ={0}\& iff x=yend{align}






linear-algebra functional-analysis analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:25










  • $begingroup$
    @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:27












  • $begingroup$
    @Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:29












  • $begingroup$
    LOL, What if $X=Bbb R^n$? What's the kernel then?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:29






  • 1




    $begingroup$
    Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:31
















4












$begingroup$


Can you check if my proof is correct?




Let $X$ be a normed linear space. Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$




Let $fin X^{*}$, then $f$ is a bounded linear functional. Assume that $x,in X$ such that begin{align}f(x)=f(y)&iff f(x)-f(y)=0, \& iff f(x-y)=0, ;text{since};f ;text{is a linear functional};\& iff x-yin ker f ={0}\& iff x=yend{align}






linear-algebra functional-analysis analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:25










  • $begingroup$
    @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:27












  • $begingroup$
    @Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:29












  • $begingroup$
    LOL, What if $X=Bbb R^n$? What's the kernel then?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:29






  • 1




    $begingroup$
    Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:31














4












4








4


2



$begingroup$


Can you check if my proof is correct?




Let $X$ be a normed linear space. Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$




Let $fin X^{*}$, then $f$ is a bounded linear functional. Assume that $x,in X$ such that begin{align}f(x)=f(y)&iff f(x)-f(y)=0, \& iff f(x-y)=0, ;text{since};f ;text{is a linear functional};\& iff x-yin ker f ={0}\& iff x=yend{align}






linear-algebra functional-analysis analysis







share|cite|improve this question











$endgroup$




Can you check if my proof is correct?




Let $X$ be a normed linear space. Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$




Let $fin X^{*}$, then $f$ is a bounded linear functional. Assume that $x,in X$ such that begin{align}f(x)=f(y)&iff f(x)-f(y)=0, \& iff f(x-y)=0, ;text{since};f ;text{is a linear functional};\& iff x-yin ker f ={0}\& iff x=yend{align}






linear-algebra functional-analysis analysis




linear-algebra functional-analysis analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 19:26







Omojola Micheal

















asked Jan 10 at 19:23









Omojola MichealOmojola Micheal

1,831324




1,831324








  • 1




    $begingroup$
    Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:25










  • $begingroup$
    @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:27












  • $begingroup$
    @Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:29












  • $begingroup$
    LOL, What if $X=Bbb R^n$? What's the kernel then?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:29






  • 1




    $begingroup$
    Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:31














  • 1




    $begingroup$
    Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:25










  • $begingroup$
    @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:27












  • $begingroup$
    @Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:29












  • $begingroup$
    LOL, What if $X=Bbb R^n$? What's the kernel then?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:29






  • 1




    $begingroup$
    Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
    $endgroup$
    – Ted Shifrin
    Jan 10 at 19:31








1




1




$begingroup$
Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
$endgroup$
– Ted Shifrin
Jan 10 at 19:25




$begingroup$
Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $ker f = {0}$ when $fcolon XtoBbb C$ (or $Bbb R$)?
$endgroup$
– Ted Shifrin
Jan 10 at 19:25












$begingroup$
@Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
$endgroup$
– Omojola Micheal
Jan 10 at 19:27






$begingroup$
@Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals
$endgroup$
– Omojola Micheal
Jan 10 at 19:27














$begingroup$
@Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
$endgroup$
– Omojola Micheal
Jan 10 at 19:29






$begingroup$
@Ted Shifrin: However, I am not certain if $ker f={0}$ when $f:Xto Bbb{R}.$ What do you think?
$endgroup$
– Omojola Micheal
Jan 10 at 19:29














$begingroup$
LOL, What if $X=Bbb R^n$? What's the kernel then?
$endgroup$
– Ted Shifrin
Jan 10 at 19:29




$begingroup$
LOL, What if $X=Bbb R^n$? What's the kernel then?
$endgroup$
– Ted Shifrin
Jan 10 at 19:29




1




1




$begingroup$
Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
$endgroup$
– Ted Shifrin
Jan 10 at 19:31




$begingroup$
Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $fin X^*$. Try this in $Bbb R^n$: Suppose $xcdot v = 0$ for all $vinBbb R^n$. Why must $x=0$?
$endgroup$
– Ted Shifrin
Jan 10 at 19:31










2 Answers
2






active

oldest

votes


















4












$begingroup$

The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10



















2












$begingroup$

Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10
















4












$begingroup$

The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10














4












4








4





$begingroup$

The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).






share|cite|improve this answer









$endgroup$



The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $fin X^*$, then $ker f={0}$, which is not true in general (think of the zero functional).



A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-yneq 0$, then there exists some $fin X^*$ such that $f(x-y)=|x-y|neq 0$, so that $f(x)neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 19:36









triple_sectriple_sec

15.8k21851




15.8k21851












  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10


















  • $begingroup$
    Thanks for your quick response. I'll read through (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:38










  • $begingroup$
    I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:48












  • $begingroup$
    The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
    $endgroup$
    – triple_sec
    Jan 10 at 20:05












  • $begingroup$
    Yes, that's true!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 20:10
















$begingroup$
Thanks for your quick response. I'll read through (+1)
$endgroup$
– Omojola Micheal
Jan 10 at 19:38




$begingroup$
Thanks for your quick response. I'll read through (+1)
$endgroup$
– Omojola Micheal
Jan 10 at 19:38












$begingroup$
I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
$endgroup$
– Omojola Micheal
Jan 10 at 19:48






$begingroup$
I can't access the book. Is there anyway of getting it? Or can you share with me on my email?
$endgroup$
– Omojola Micheal
Jan 10 at 19:48














$begingroup$
The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
$endgroup$
– triple_sec
Jan 10 at 20:05






$begingroup$
The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout.
$endgroup$
– triple_sec
Jan 10 at 20:05














$begingroup$
Yes, that's true!
$endgroup$
– Omojola Micheal
Jan 10 at 20:10




$begingroup$
Yes, that's true!
$endgroup$
– Omojola Micheal
Jan 10 at 20:10











2












$begingroup$

Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37
















2












$begingroup$

Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37














2












2








2





$begingroup$

Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}






share|cite|improve this answer









$endgroup$



Your proof is not correct, because, unles $dim Xleqslant1$, $ker f$ cannot possibly be ${0}$.



Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$begin{array}{rccc}gcolon&Z&longrightarrow&mathbb R\&lambda z&mapsto&lambda.end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $fin X^*$. Butbegin{align}f(x)-f(y)&=f(x-y)\&=f(z)\&=g(z)\&=1\&neq0.end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 19:34









José Carlos SantosJosé Carlos Santos

156k22125227




156k22125227












  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37


















  • $begingroup$
    Thanks for your quick response. (+1)
    $endgroup$
    – Omojola Micheal
    Jan 10 at 19:37
















$begingroup$
Thanks for your quick response. (+1)
$endgroup$
– Omojola Micheal
Jan 10 at 19:37




$begingroup$
Thanks for your quick response. (+1)
$endgroup$
– Omojola Micheal
Jan 10 at 19:37


















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Mario Kart Wii

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
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The Binding of Isaac: Rebirth/Afterbirth

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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