Existence of Bounded Linear Functional That Maps Elements of Normed Space to Distance from Subspace












1














I'm attempting to prove the following:




Suppose X is a normed space and Y a subspace of X. Let $x ∈ X$ and denote $d$ the distance from $x$ to $Y$; that is, $d = d(x, Y ) = inf_{y∈Y} || x − y ||$. Prove that there exists a bounded linear functional $f : X → R$ such that
$f(x) = d$, $f(y) = 0$, for all $y ∈ Y$ and $||f||_{op} = sup_{x in X, ||x||=1}||fx|| ≤ 1$.




It seems like this can be proven using the Hahn-Banach Theorem, but I'm not sure how to proceed.



Any help is much appreciated!






functional-analysis







share|cite|improve this question


















  • 1




    That is a consequence of Hanh-Banach theorem
    – Guy Fsone
    Dec 13 '17 at 18:59
















1














I'm attempting to prove the following:




Suppose X is a normed space and Y a subspace of X. Let $x ∈ X$ and denote $d$ the distance from $x$ to $Y$; that is, $d = d(x, Y ) = inf_{y∈Y} || x − y ||$. Prove that there exists a bounded linear functional $f : X → R$ such that
$f(x) = d$, $f(y) = 0$, for all $y ∈ Y$ and $||f||_{op} = sup_{x in X, ||x||=1}||fx|| ≤ 1$.




It seems like this can be proven using the Hahn-Banach Theorem, but I'm not sure how to proceed.



Any help is much appreciated!






functional-analysis







share|cite|improve this question


















  • 1




    That is a consequence of Hanh-Banach theorem
    – Guy Fsone
    Dec 13 '17 at 18:59














1












1








1







I'm attempting to prove the following:




Suppose X is a normed space and Y a subspace of X. Let $x ∈ X$ and denote $d$ the distance from $x$ to $Y$; that is, $d = d(x, Y ) = inf_{y∈Y} || x − y ||$. Prove that there exists a bounded linear functional $f : X → R$ such that
$f(x) = d$, $f(y) = 0$, for all $y ∈ Y$ and $||f||_{op} = sup_{x in X, ||x||=1}||fx|| ≤ 1$.




It seems like this can be proven using the Hahn-Banach Theorem, but I'm not sure how to proceed.



Any help is much appreciated!






functional-analysis







share|cite|improve this question













I'm attempting to prove the following:




Suppose X is a normed space and Y a subspace of X. Let $x ∈ X$ and denote $d$ the distance from $x$ to $Y$; that is, $d = d(x, Y ) = inf_{y∈Y} || x − y ||$. Prove that there exists a bounded linear functional $f : X → R$ such that
$f(x) = d$, $f(y) = 0$, for all $y ∈ Y$ and $||f||_{op} = sup_{x in X, ||x||=1}||fx|| ≤ 1$.




It seems like this can be proven using the Hahn-Banach Theorem, but I'm not sure how to proceed.



Any help is much appreciated!






functional-analysis




functional-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '17 at 18:57









OGBerglemir

1799




1799








  • 1




    That is a consequence of Hanh-Banach theorem
    – Guy Fsone
    Dec 13 '17 at 18:59














  • 1




    That is a consequence of Hanh-Banach theorem
    – Guy Fsone
    Dec 13 '17 at 18:59








1




1




That is a consequence of Hanh-Banach theorem
– Guy Fsone
Dec 13 '17 at 18:59




That is a consequence of Hanh-Banach theorem
– Guy Fsone
Dec 13 '17 at 18:59










1 Answer
1






active

oldest

votes


















2














I suppose that $d>0$. Otherwise, the zero functional trivially solves the problem.



We apply Hahn-Banach Theorem on the subspace $mathrm{span}{Y,x}={tx+yin Xmid tin Bbb{R},yin Y}$ and the functional $g:mathrm{Y,span}{x}toBbb{R},g(tx+y)=t||x||=||tx||$ linear with respect to $t$ and $y$. Check that $g(x) = ||x||$. For any unit vector $uinmathrm{span}{Y,x},$ express $u$ as $tx+y$
$$|g(u)| = left|gleft(frac{tx+y}{||tx+y||}right)right|=frac{|t|,||x||}{||tx+y||} = frac{||x||}{||x-(-y/|t|)||}le frac{||x||}{d},$$ so $||g||_{(mathrm{span}{Y,x})'} le ||x||/d$. (The last inequality makes use of the defintion of $d$ as $inflimits_{y in Y} ||x-y||$, $Y$ as a subspace of $X$, and the assumption $d>0$.)



By Hahn-Banach, there exists a bounded functional $f(in X')$ extending $g$ to the whole space $X$: $$||f||_{X'}= ||g||_{(mathrm{span}{Y,x})'} le frac{||x||}{d} text{ and } f(z) = g(z) quadforall, zinmathrm{span}{Y,x}.$$



In particular, $f(x)=||x||$. We scale this function $f$ by a factor of $d/||x||$ to finish the proof.






share|cite|improve this answer



















  • 1




    I'm following up until the $frac{d}{||x||} = 1$ bit. Why is this true? This step is to prove that $g$ is bounded, correct?
    – OGBerglemir
    Dec 13 '17 at 19:47










  • @OGBerglemir I'm going to fix it. You are right.
    – GNUSupporter 8964民主女神 地下教會
    Dec 13 '17 at 19:52






  • 1




    It might be a little late, but I just saw this which is realted to a question I have. I don't understand the last step in the estimate of $|g(u)|$: This is not true for general norms right? For example $|-2|/|-2+1|=2>1$. So how do you get that your term is smaller or equal to 1?
    – mathstackuser
    Nov 27 '18 at 17:49












  • @mathstackuser Thanks for your comment and spotting out this error. I've edited my question to address the problem in the first comment.
    – GNUSupporter 8964民主女神 地下教會
    Nov 28 '18 at 0:44











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1 Answer
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1 Answer
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2














I suppose that $d>0$. Otherwise, the zero functional trivially solves the problem.



We apply Hahn-Banach Theorem on the subspace $mathrm{span}{Y,x}={tx+yin Xmid tin Bbb{R},yin Y}$ and the functional $g:mathrm{Y,span}{x}toBbb{R},g(tx+y)=t||x||=||tx||$ linear with respect to $t$ and $y$. Check that $g(x) = ||x||$. For any unit vector $uinmathrm{span}{Y,x},$ express $u$ as $tx+y$
$$|g(u)| = left|gleft(frac{tx+y}{||tx+y||}right)right|=frac{|t|,||x||}{||tx+y||} = frac{||x||}{||x-(-y/|t|)||}le frac{||x||}{d},$$ so $||g||_{(mathrm{span}{Y,x})'} le ||x||/d$. (The last inequality makes use of the defintion of $d$ as $inflimits_{y in Y} ||x-y||$, $Y$ as a subspace of $X$, and the assumption $d>0$.)



By Hahn-Banach, there exists a bounded functional $f(in X')$ extending $g$ to the whole space $X$: $$||f||_{X'}= ||g||_{(mathrm{span}{Y,x})'} le frac{||x||}{d} text{ and } f(z) = g(z) quadforall, zinmathrm{span}{Y,x}.$$



In particular, $f(x)=||x||$. We scale this function $f$ by a factor of $d/||x||$ to finish the proof.






share|cite|improve this answer



















  • 1




    I'm following up until the $frac{d}{||x||} = 1$ bit. Why is this true? This step is to prove that $g$ is bounded, correct?
    – OGBerglemir
    Dec 13 '17 at 19:47










  • @OGBerglemir I'm going to fix it. You are right.
    – GNUSupporter 8964民主女神 地下教會
    Dec 13 '17 at 19:52






  • 1




    It might be a little late, but I just saw this which is realted to a question I have. I don't understand the last step in the estimate of $|g(u)|$: This is not true for general norms right? For example $|-2|/|-2+1|=2>1$. So how do you get that your term is smaller or equal to 1?
    – mathstackuser
    Nov 27 '18 at 17:49












  • @mathstackuser Thanks for your comment and spotting out this error. I've edited my question to address the problem in the first comment.
    – GNUSupporter 8964民主女神 地下教會
    Nov 28 '18 at 0:44
















2














I suppose that $d>0$. Otherwise, the zero functional trivially solves the problem.



We apply Hahn-Banach Theorem on the subspace $mathrm{span}{Y,x}={tx+yin Xmid tin Bbb{R},yin Y}$ and the functional $g:mathrm{Y,span}{x}toBbb{R},g(tx+y)=t||x||=||tx||$ linear with respect to $t$ and $y$. Check that $g(x) = ||x||$. For any unit vector $uinmathrm{span}{Y,x},$ express $u$ as $tx+y$
$$|g(u)| = left|gleft(frac{tx+y}{||tx+y||}right)right|=frac{|t|,||x||}{||tx+y||} = frac{||x||}{||x-(-y/|t|)||}le frac{||x||}{d},$$ so $||g||_{(mathrm{span}{Y,x})'} le ||x||/d$. (The last inequality makes use of the defintion of $d$ as $inflimits_{y in Y} ||x-y||$, $Y$ as a subspace of $X$, and the assumption $d>0$.)



By Hahn-Banach, there exists a bounded functional $f(in X')$ extending $g$ to the whole space $X$: $$||f||_{X'}= ||g||_{(mathrm{span}{Y,x})'} le frac{||x||}{d} text{ and } f(z) = g(z) quadforall, zinmathrm{span}{Y,x}.$$



In particular, $f(x)=||x||$. We scale this function $f$ by a factor of $d/||x||$ to finish the proof.






share|cite|improve this answer



















  • 1




    I'm following up until the $frac{d}{||x||} = 1$ bit. Why is this true? This step is to prove that $g$ is bounded, correct?
    – OGBerglemir
    Dec 13 '17 at 19:47










  • @OGBerglemir I'm going to fix it. You are right.
    – GNUSupporter 8964民主女神 地下教會
    Dec 13 '17 at 19:52






  • 1




    It might be a little late, but I just saw this which is realted to a question I have. I don't understand the last step in the estimate of $|g(u)|$: This is not true for general norms right? For example $|-2|/|-2+1|=2>1$. So how do you get that your term is smaller or equal to 1?
    – mathstackuser
    Nov 27 '18 at 17:49












  • @mathstackuser Thanks for your comment and spotting out this error. I've edited my question to address the problem in the first comment.
    – GNUSupporter 8964民主女神 地下教會
    Nov 28 '18 at 0:44














2












2








2






I suppose that $d>0$. Otherwise, the zero functional trivially solves the problem.



We apply Hahn-Banach Theorem on the subspace $mathrm{span}{Y,x}={tx+yin Xmid tin Bbb{R},yin Y}$ and the functional $g:mathrm{Y,span}{x}toBbb{R},g(tx+y)=t||x||=||tx||$ linear with respect to $t$ and $y$. Check that $g(x) = ||x||$. For any unit vector $uinmathrm{span}{Y,x},$ express $u$ as $tx+y$
$$|g(u)| = left|gleft(frac{tx+y}{||tx+y||}right)right|=frac{|t|,||x||}{||tx+y||} = frac{||x||}{||x-(-y/|t|)||}le frac{||x||}{d},$$ so $||g||_{(mathrm{span}{Y,x})'} le ||x||/d$. (The last inequality makes use of the defintion of $d$ as $inflimits_{y in Y} ||x-y||$, $Y$ as a subspace of $X$, and the assumption $d>0$.)



By Hahn-Banach, there exists a bounded functional $f(in X')$ extending $g$ to the whole space $X$: $$||f||_{X'}= ||g||_{(mathrm{span}{Y,x})'} le frac{||x||}{d} text{ and } f(z) = g(z) quadforall, zinmathrm{span}{Y,x}.$$



In particular, $f(x)=||x||$. We scale this function $f$ by a factor of $d/||x||$ to finish the proof.






share|cite|improve this answer














I suppose that $d>0$. Otherwise, the zero functional trivially solves the problem.



We apply Hahn-Banach Theorem on the subspace $mathrm{span}{Y,x}={tx+yin Xmid tin Bbb{R},yin Y}$ and the functional $g:mathrm{Y,span}{x}toBbb{R},g(tx+y)=t||x||=||tx||$ linear with respect to $t$ and $y$. Check that $g(x) = ||x||$. For any unit vector $uinmathrm{span}{Y,x},$ express $u$ as $tx+y$
$$|g(u)| = left|gleft(frac{tx+y}{||tx+y||}right)right|=frac{|t|,||x||}{||tx+y||} = frac{||x||}{||x-(-y/|t|)||}le frac{||x||}{d},$$ so $||g||_{(mathrm{span}{Y,x})'} le ||x||/d$. (The last inequality makes use of the defintion of $d$ as $inflimits_{y in Y} ||x-y||$, $Y$ as a subspace of $X$, and the assumption $d>0$.)



By Hahn-Banach, there exists a bounded functional $f(in X')$ extending $g$ to the whole space $X$: $$||f||_{X'}= ||g||_{(mathrm{span}{Y,x})'} le frac{||x||}{d} text{ and } f(z) = g(z) quadforall, zinmathrm{span}{Y,x}.$$



In particular, $f(x)=||x||$. We scale this function $f$ by a factor of $d/||x||$ to finish the proof.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 14:51

























answered Dec 13 '17 at 19:14









GNUSupporter 8964民主女神 地下教會

12.8k72445




12.8k72445








  • 1




    I'm following up until the $frac{d}{||x||} = 1$ bit. Why is this true? This step is to prove that $g$ is bounded, correct?
    – OGBerglemir
    Dec 13 '17 at 19:47










  • @OGBerglemir I'm going to fix it. You are right.
    – GNUSupporter 8964民主女神 地下教會
    Dec 13 '17 at 19:52






  • 1




    It might be a little late, but I just saw this which is realted to a question I have. I don't understand the last step in the estimate of $|g(u)|$: This is not true for general norms right? For example $|-2|/|-2+1|=2>1$. So how do you get that your term is smaller or equal to 1?
    – mathstackuser
    Nov 27 '18 at 17:49












  • @mathstackuser Thanks for your comment and spotting out this error. I've edited my question to address the problem in the first comment.
    – GNUSupporter 8964民主女神 地下教會
    Nov 28 '18 at 0:44














  • 1




    I'm following up until the $frac{d}{||x||} = 1$ bit. Why is this true? This step is to prove that $g$ is bounded, correct?
    – OGBerglemir
    Dec 13 '17 at 19:47










  • @OGBerglemir I'm going to fix it. You are right.
    – GNUSupporter 8964民主女神 地下教會
    Dec 13 '17 at 19:52






  • 1




    It might be a little late, but I just saw this which is realted to a question I have. I don't understand the last step in the estimate of $|g(u)|$: This is not true for general norms right? For example $|-2|/|-2+1|=2>1$. So how do you get that your term is smaller or equal to 1?
    – mathstackuser
    Nov 27 '18 at 17:49












  • @mathstackuser Thanks for your comment and spotting out this error. I've edited my question to address the problem in the first comment.
    – GNUSupporter 8964民主女神 地下教會
    Nov 28 '18 at 0:44








1




1




I'm following up until the $frac{d}{||x||} = 1$ bit. Why is this true? This step is to prove that $g$ is bounded, correct?
– OGBerglemir
Dec 13 '17 at 19:47




I'm following up until the $frac{d}{||x||} = 1$ bit. Why is this true? This step is to prove that $g$ is bounded, correct?
– OGBerglemir
Dec 13 '17 at 19:47












@OGBerglemir I'm going to fix it. You are right.
– GNUSupporter 8964民主女神 地下教會
Dec 13 '17 at 19:52




@OGBerglemir I'm going to fix it. You are right.
– GNUSupporter 8964民主女神 地下教會
Dec 13 '17 at 19:52




1




1




It might be a little late, but I just saw this which is realted to a question I have. I don't understand the last step in the estimate of $|g(u)|$: This is not true for general norms right? For example $|-2|/|-2+1|=2>1$. So how do you get that your term is smaller or equal to 1?
– mathstackuser
Nov 27 '18 at 17:49






It might be a little late, but I just saw this which is realted to a question I have. I don't understand the last step in the estimate of $|g(u)|$: This is not true for general norms right? For example $|-2|/|-2+1|=2>1$. So how do you get that your term is smaller or equal to 1?
– mathstackuser
Nov 27 '18 at 17:49














@mathstackuser Thanks for your comment and spotting out this error. I've edited my question to address the problem in the first comment.
– GNUSupporter 8964民主女神 地下教會
Nov 28 '18 at 0:44




@mathstackuser Thanks for your comment and spotting out this error. I've edited my question to address the problem in the first comment.
– GNUSupporter 8964民主女神 地下教會
Nov 28 '18 at 0:44


















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