Orthogonal complements in arbitrary normed spaces
$begingroup$
I'm struggling to follow a certain proof in Luenberger's book "Optimization by Vector Space Methods". If anyone could help shed some light on what I'm missing, I would certainly appreciate it. Let me start with a couple of definitions.
Definition 1 Let $S$ be a subset of a normed linear space $X$. The orthogonal
complement of $S$, denoted $S^perp$, consists of all elements $x^* in X^*$ ($X^*$ denotes the continuous dual space) orthogonal to every vector in $S$.
Definition 2 Given a subset $U$ of the dual space $X^*$, we define the orthogonal complement of $U$ in $X$ as the set $^perp U subset X$ consisting of all elements in $X$ orthogonal to every vector in $U$.
Definition 3 The vectors $x in X$ and $x^* in X^*$ are said to be orthogonal if $langle x, x^*rangle =0$.
$langle cdot,cdot rangle$ denotes the dual pairing.
Theorem Let $M$ be a closed subspace of a normed space $X$. Then
$^perp[M^perp] = M$.
Proof It is clear that $M subset{}^perp[M^perp]$. To prove the converse, let $x notin M$. On the subspace $[x + M]$ generated by $x$ and $M$, define the linear functional $f(alpha x + m) = alpha $ for $m in M$. Then
begin{equation}
|,f|=sup_{min M}frac{f(x+m)}{|x+m|}=frac{1}{inf_{m}|x+m|}
end{equation}
and since $M$ is closed, $|,f| < infty$. Thus by the Hahn-Banach theorem, we
can extend $f$ to an $x^* in X^*$. Since $f$ vanishes on M, we have $x^* in M^perp$ But also $langle x, x^*rangle = 1$ and thus $x notin{}^perp[M^perp]$.
Comments I'm struggling to understand the necessity of the subspace being closed. Would bounded not suffice?
My understanding of the proof is as follows. Using the definition of the norm of an $f$ in the dual space and positivity we get the first equality in the displayed equation. Noticing that $alpha=1$ and interchanging $sup$ and $inf$ by taking the $inf$ into the denominator gives the second equality. I would have thought from here we would be able to guarantee finiteness of the norm. We then use the Hahn-Banach Theorem to extend $f$ to $x^* in X^*$. Then using $alpha=0$ implies $f=0$ on $M$. Thus $f(x_1)=langle x^*,x_1rangle=0$ where $x_1 in M$ so they are orthogonal. However, $alpha=1$ implies $f(x)=1=langle x^*,xrangle$ hence $x notin{}^perp[M^perp]$.
functional-analysis vector-spaces orthogonality
edited Dec 8 '18 at 12:04
amWhy
1
asked Aug 19 '18 at 14:33
markmark
7531712
$endgroup$
add a comment |
$begingroup$
I'm struggling to follow a certain proof in Luenberger's book "Optimization by Vector Space Methods". If anyone could help shed some light on what I'm missing, I would certainly appreciate it. Let me start with a couple of definitions.
Definition 1 Let $S$ be a subset of a normed linear space $X$. The orthogonal
complement of $S$, denoted $S^perp$, consists of all elements $x^* in X^*$ ($X^*$ denotes the continuous dual space) orthogonal to every vector in $S$.
Definition 2 Given a subset $U$ of the dual space $X^*$, we define the orthogonal complement of $U$ in $X$ as the set $^perp U subset X$ consisting of all elements in $X$ orthogonal to every vector in $U$.
Definition 3 The vectors $x in X$ and $x^* in X^*$ are said to be orthogonal if $langle x, x^*rangle =0$.
$langle cdot,cdot rangle$ denotes the dual pairing.
Theorem Let $M$ be a closed subspace of a normed space $X$. Then
$^perp[M^perp] = M$.
Proof It is clear that $M subset{}^perp[M^perp]$. To prove the converse, let $x notin M$. On the subspace $[x + M]$ generated by $x$ and $M$, define the linear functional $f(alpha x + m) = alpha $ for $m in M$. Then
begin{equation}
|,f|=sup_{min M}frac{f(x+m)}{|x+m|}=frac{1}{inf_{m}|x+m|}
end{equation}
and since $M$ is closed, $|,f| < infty$. Thus by the Hahn-Banach theorem, we
can extend $f$ to an $x^* in X^*$. Since $f$ vanishes on M, we have $x^* in M^perp$ But also $langle x, x^*rangle = 1$ and thus $x notin{}^perp[M^perp]$.
Comments I'm struggling to understand the necessity of the subspace being closed. Would bounded not suffice?
My understanding of the proof is as follows. Using the definition of the norm of an $f$ in the dual space and positivity we get the first equality in the displayed equation. Noticing that $alpha=1$ and interchanging $sup$ and $inf$ by taking the $inf$ into the denominator gives the second equality. I would have thought from here we would be able to guarantee finiteness of the norm. We then use the Hahn-Banach Theorem to extend $f$ to $x^* in X^*$. Then using $alpha=0$ implies $f=0$ on $M$. Thus $f(x_1)=langle x^*,x_1rangle=0$ where $x_1 in M$ so they are orthogonal. However, $alpha=1$ implies $f(x)=1=langle x^*,xrangle$ hence $x notin{}^perp[M^perp]$.
functional-analysis vector-spaces orthogonality
edited Dec 8 '18 at 12:04
amWhy
1
asked Aug 19 '18 at 14:33
markmark
7531712
$endgroup$
$begingroup$
Which subspace of $X$ is bounded?
$endgroup$
– Paul Frost
Aug 19 '18 at 14:39
$begingroup$
Check out my edit to see how to prevent a left superscript from being attached to the symbol to its left.
$endgroup$
– joriki
Aug 19 '18 at 14:53
add a comment |
$begingroup$
I'm struggling to follow a certain proof in Luenberger's book "Optimization by Vector Space Methods". If anyone could help shed some light on what I'm missing, I would certainly appreciate it. Let me start with a couple of definitions.
Definition 1 Let $S$ be a subset of a normed linear space $X$. The orthogonal
complement of $S$, denoted $S^perp$, consists of all elements $x^* in X^*$ ($X^*$ denotes the continuous dual space) orthogonal to every vector in $S$.
Definition 2 Given a subset $U$ of the dual space $X^*$, we define the orthogonal complement of $U$ in $X$ as the set $^perp U subset X$ consisting of all elements in $X$ orthogonal to every vector in $U$.
Definition 3 The vectors $x in X$ and $x^* in X^*$ are said to be orthogonal if $langle x, x^*rangle =0$.
$langle cdot,cdot rangle$ denotes the dual pairing.
Theorem Let $M$ be a closed subspace of a normed space $X$. Then
$^perp[M^perp] = M$.
Proof It is clear that $M subset{}^perp[M^perp]$. To prove the converse, let $x notin M$. On the subspace $[x + M]$ generated by $x$ and $M$, define the linear functional $f(alpha x + m) = alpha $ for $m in M$. Then
begin{equation}
|,f|=sup_{min M}frac{f(x+m)}{|x+m|}=frac{1}{inf_{m}|x+m|}
end{equation}
and since $M$ is closed, $|,f| < infty$. Thus by the Hahn-Banach theorem, we
can extend $f$ to an $x^* in X^*$. Since $f$ vanishes on M, we have $x^* in M^perp$ But also $langle x, x^*rangle = 1$ and thus $x notin{}^perp[M^perp]$.
Comments I'm struggling to understand the necessity of the subspace being closed. Would bounded not suffice?
My understanding of the proof is as follows. Using the definition of the norm of an $f$ in the dual space and positivity we get the first equality in the displayed equation. Noticing that $alpha=1$ and interchanging $sup$ and $inf$ by taking the $inf$ into the denominator gives the second equality. I would have thought from here we would be able to guarantee finiteness of the norm. We then use the Hahn-Banach Theorem to extend $f$ to $x^* in X^*$. Then using $alpha=0$ implies $f=0$ on $M$. Thus $f(x_1)=langle x^*,x_1rangle=0$ where $x_1 in M$ so they are orthogonal. However, $alpha=1$ implies $f(x)=1=langle x^*,xrangle$ hence $x notin{}^perp[M^perp]$.
functional-analysis vector-spaces orthogonality
edited Dec 8 '18 at 12:04
amWhy
1
asked Aug 19 '18 at 14:33
markmark
7531712
$endgroup$
I'm struggling to follow a certain proof in Luenberger's book "Optimization by Vector Space Methods". If anyone could help shed some light on what I'm missing, I would certainly appreciate it. Let me start with a couple of definitions.
Definition 1 Let $S$ be a subset of a normed linear space $X$. The orthogonal
complement of $S$, denoted $S^perp$, consists of all elements $x^* in X^*$ ($X^*$ denotes the continuous dual space) orthogonal to every vector in $S$.
Definition 2 Given a subset $U$ of the dual space $X^*$, we define the orthogonal complement of $U$ in $X$ as the set $^perp U subset X$ consisting of all elements in $X$ orthogonal to every vector in $U$.
Definition 3 The vectors $x in X$ and $x^* in X^*$ are said to be orthogonal if $langle x, x^*rangle =0$.
$langle cdot,cdot rangle$ denotes the dual pairing.
Theorem Let $M$ be a closed subspace of a normed space $X$. Then
$^perp[M^perp] = M$.
Proof It is clear that $M subset{}^perp[M^perp]$. To prove the converse, let $x notin M$. On the subspace $[x + M]$ generated by $x$ and $M$, define the linear functional $f(alpha x + m) = alpha $ for $m in M$. Then
begin{equation}
|,f|=sup_{min M}frac{f(x+m)}{|x+m|}=frac{1}{inf_{m}|x+m|}
end{equation}
and since $M$ is closed, $|,f| < infty$. Thus by the Hahn-Banach theorem, we
can extend $f$ to an $x^* in X^*$. Since $f$ vanishes on M, we have $x^* in M^perp$ But also $langle x, x^*rangle = 1$ and thus $x notin{}^perp[M^perp]$.
Comments I'm struggling to understand the necessity of the subspace being closed. Would bounded not suffice?
My understanding of the proof is as follows. Using the definition of the norm of an $f$ in the dual space and positivity we get the first equality in the displayed equation. Noticing that $alpha=1$ and interchanging $sup$ and $inf$ by taking the $inf$ into the denominator gives the second equality. I would have thought from here we would be able to guarantee finiteness of the norm. We then use the Hahn-Banach Theorem to extend $f$ to $x^* in X^*$. Then using $alpha=0$ implies $f=0$ on $M$. Thus $f(x_1)=langle x^*,x_1rangle=0$ where $x_1 in M$ so they are orthogonal. However, $alpha=1$ implies $f(x)=1=langle x^*,xrangle$ hence $x notin{}^perp[M^perp]$.
functional-analysis vector-spaces orthogonality
functional-analysis vector-spaces orthogonality
edited Dec 8 '18 at 12:04
amWhy
1
asked Aug 19 '18 at 14:33
markmark
7531712
edited Dec 8 '18 at 12:04
amWhy
1
asked Aug 19 '18 at 14:33
markmark
7531712
edited Dec 8 '18 at 12:04
amWhy
1
edited Dec 8 '18 at 12:04
amWhy
1
edited Dec 8 '18 at 12:04
amWhy
1
1
asked Aug 19 '18 at 14:33
markmark
7531712
asked Aug 19 '18 at 14:33
markmark
7531712
asked Aug 19 '18 at 14:33
markmark
7531712
7531712
$begingroup$
Which subspace of $X$ is bounded?
$endgroup$
– Paul Frost
Aug 19 '18 at 14:39
$begingroup$
Check out my edit to see how to prevent a left superscript from being attached to the symbol to its left.
$endgroup$
– joriki
Aug 19 '18 at 14:53
add a comment |
$begingroup$
Which subspace of $X$ is bounded?
$endgroup$
– Paul Frost
Aug 19 '18 at 14:39
$begingroup$
Check out my edit to see how to prevent a left superscript from being attached to the symbol to its left.
$endgroup$
– joriki
Aug 19 '18 at 14:53
$begingroup$
Which subspace of $X$ is bounded?
$endgroup$
– Paul Frost
Aug 19 '18 at 14:39
$begingroup$
Which subspace of $X$ is bounded?
$endgroup$
– Paul Frost
Aug 19 '18 at 14:39
$begingroup$
Check out my edit to see how to prevent a left superscript from being attached to the symbol to its left.
$endgroup$
– joriki
Aug 19 '18 at 14:53
$begingroup$
Check out my edit to see how to prevent a left superscript from being attached to the symbol to its left.
$endgroup$
– joriki
Aug 19 '18 at 14:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $M$ is not closed, then for each $x in overline{M} backslash M$ there is a sequence $(m_n)$ in $M$ such that $lVert x - m_n rVert to 0$ as $n to infty$. But then $inf_m lVert x + m rVert = 0$ since $-m_n in M$.
Note that no linear subspace except the trivial one (having $0$ as its only element) is bounded.
answered Aug 19 '18 at 16:14
Paul FrostPaul Frost
10.2k3933
$endgroup$
$begingroup$
I see. Sorry. I realise now that that was a somewhat silly thing not to notice. Thanks.
$endgroup$
– mark
Aug 19 '18 at 21:28
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2887791%2forthogonal-complements-in-arbitrary-normed-spaces%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $M$ is not closed, then for each $x in overline{M} backslash M$ there is a sequence $(m_n)$ in $M$ such that $lVert x - m_n rVert to 0$ as $n to infty$. But then $inf_m lVert x + m rVert = 0$ since $-m_n in M$.
Note that no linear subspace except the trivial one (having $0$ as its only element) is bounded.
answered Aug 19 '18 at 16:14
Paul FrostPaul Frost
10.2k3933
$endgroup$
$begingroup$
I see. Sorry. I realise now that that was a somewhat silly thing not to notice. Thanks.
$endgroup$
– mark
Aug 19 '18 at 21:28
add a comment |
$begingroup$
If $M$ is not closed, then for each $x in overline{M} backslash M$ there is a sequence $(m_n)$ in $M$ such that $lVert x - m_n rVert to 0$ as $n to infty$. But then $inf_m lVert x + m rVert = 0$ since $-m_n in M$.
Note that no linear subspace except the trivial one (having $0$ as its only element) is bounded.
answered Aug 19 '18 at 16:14
Paul FrostPaul Frost
10.2k3933
$endgroup$
$begingroup$
I see. Sorry. I realise now that that was a somewhat silly thing not to notice. Thanks.
$endgroup$
– mark
Aug 19 '18 at 21:28
add a comment |
$begingroup$
If $M$ is not closed, then for each $x in overline{M} backslash M$ there is a sequence $(m_n)$ in $M$ such that $lVert x - m_n rVert to 0$ as $n to infty$. But then $inf_m lVert x + m rVert = 0$ since $-m_n in M$.
Note that no linear subspace except the trivial one (having $0$ as its only element) is bounded.
answered Aug 19 '18 at 16:14
Paul FrostPaul Frost
10.2k3933
$endgroup$
If $M$ is not closed, then for each $x in overline{M} backslash M$ there is a sequence $(m_n)$ in $M$ such that $lVert x - m_n rVert to 0$ as $n to infty$. But then $inf_m lVert x + m rVert = 0$ since $-m_n in M$.
Note that no linear subspace except the trivial one (having $0$ as its only element) is bounded.
answered Aug 19 '18 at 16:14
Paul FrostPaul Frost
10.2k3933
answered Aug 19 '18 at 16:14
Paul FrostPaul Frost
10.2k3933
answered Aug 19 '18 at 16:14
Paul FrostPaul Frost
10.2k3933
answered Aug 19 '18 at 16:14
Paul FrostPaul Frost
10.2k3933
10.2k3933
$begingroup$
I see. Sorry. I realise now that that was a somewhat silly thing not to notice. Thanks.
$endgroup$
– mark
Aug 19 '18 at 21:28
add a comment |
$begingroup$
I see. Sorry. I realise now that that was a somewhat silly thing not to notice. Thanks.
$endgroup$
– mark
Aug 19 '18 at 21:28
$begingroup$
I see. Sorry. I realise now that that was a somewhat silly thing not to notice. Thanks.
$endgroup$
– mark
Aug 19 '18 at 21:28
$begingroup$
I see. Sorry. I realise now that that was a somewhat silly thing not to notice. Thanks.
$endgroup$
– mark
Aug 19 '18 at 21:28
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2887791%2forthogonal-complements-in-arbitrary-normed-spaces%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Which subspace of $X$ is bounded?
$endgroup$
– Paul Frost
Aug 19 '18 at 14:39
$begingroup$
Check out my edit to see how to prevent a left superscript from being attached to the symbol to its left.
$endgroup$
– joriki
Aug 19 '18 at 14:53