Orthogonal complements in arbitrary normed spaces












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1


0



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













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edited Dec 8 '18 at 12:04









amWhy

1




1










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


















  • $begingroup$
    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










1 Answer
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$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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