The existence of a vector arbitrarily separated from a close subspace
$begingroup$
Lax Contains a proof of the following claim:
Claim: Let $Y$ be a closed, proper subspace of the normed linear space $X$. Then there is a vector $z$ in $X$ of length 1, $$|z| = 1$$ satisfying $$|z-y| >1/2$$ for all $y$ in $Y$.
Proof: Since $Y$ is a proper subspace of $X$, some point $x$ of $X$ does not
belong to $Y$. Since $Y$ is closed, $x$ has a positive distance to
$Y$: $$inf_{y in Y} |x-y| = d > 0$$ There is then a $y_0 in Y$ such that $$|x - y_0| < 2d quad (*)$$
Denote $z' = x - y_0$; we can then write $(*)$ as $|z'| < 2d$. It follows from definition of $d$ above that $$|z' - y | geq d quad (**)$$ for all $y$ in $Y$. Setting $z = frac{z'}{|z'|}$ gives the vector and combining $(*)$ and $(**)$ gives the condition.
I'm failing to see how to put together $(*)$ and $(**)$ to finish the proof. Thoughts?
functional-analysis convex-analysis
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
$endgroup$
add a comment |
$begingroup$
Lax Contains a proof of the following claim:
Claim: Let $Y$ be a closed, proper subspace of the normed linear space $X$. Then there is a vector $z$ in $X$ of length 1, $$|z| = 1$$ satisfying $$|z-y| >1/2$$ for all $y$ in $Y$.
Proof: Since $Y$ is a proper subspace of $X$, some point $x$ of $X$ does not
belong to $Y$. Since $Y$ is closed, $x$ has a positive distance to
$Y$: $$inf_{y in Y} |x-y| = d > 0$$ There is then a $y_0 in Y$ such that $$|x - y_0| < 2d quad (*)$$
Denote $z' = x - y_0$; we can then write $(*)$ as $|z'| < 2d$. It follows from definition of $d$ above that $$|z' - y | geq d quad (**)$$ for all $y$ in $Y$. Setting $z = frac{z'}{|z'|}$ gives the vector and combining $(*)$ and $(**)$ gives the condition.
I'm failing to see how to put together $(*)$ and $(**)$ to finish the proof. Thoughts?
functional-analysis convex-analysis
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
$endgroup$
add a comment |
$begingroup$
Lax Contains a proof of the following claim:
Claim: Let $Y$ be a closed, proper subspace of the normed linear space $X$. Then there is a vector $z$ in $X$ of length 1, $$|z| = 1$$ satisfying $$|z-y| >1/2$$ for all $y$ in $Y$.
Proof: Since $Y$ is a proper subspace of $X$, some point $x$ of $X$ does not
belong to $Y$. Since $Y$ is closed, $x$ has a positive distance to
$Y$: $$inf_{y in Y} |x-y| = d > 0$$ There is then a $y_0 in Y$ such that $$|x - y_0| < 2d quad (*)$$
Denote $z' = x - y_0$; we can then write $(*)$ as $|z'| < 2d$. It follows from definition of $d$ above that $$|z' - y | geq d quad (**)$$ for all $y$ in $Y$. Setting $z = frac{z'}{|z'|}$ gives the vector and combining $(*)$ and $(**)$ gives the condition.
I'm failing to see how to put together $(*)$ and $(**)$ to finish the proof. Thoughts?
functional-analysis convex-analysis
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
$endgroup$
Lax Contains a proof of the following claim:
Claim: Let $Y$ be a closed, proper subspace of the normed linear space $X$. Then there is a vector $z$ in $X$ of length 1, $$|z| = 1$$ satisfying $$|z-y| >1/2$$ for all $y$ in $Y$.
Proof: Since $Y$ is a proper subspace of $X$, some point $x$ of $X$ does not
belong to $Y$. Since $Y$ is closed, $x$ has a positive distance to
$Y$: $$inf_{y in Y} |x-y| = d > 0$$ There is then a $y_0 in Y$ such that $$|x - y_0| < 2d quad (*)$$
Denote $z' = x - y_0$; we can then write $(*)$ as $|z'| < 2d$. It follows from definition of $d$ above that $$|z' - y | geq d quad (**)$$ for all $y$ in $Y$. Setting $z = frac{z'}{|z'|}$ gives the vector and combining $(*)$ and $(**)$ gives the condition.
I'm failing to see how to put together $(*)$ and $(**)$ to finish the proof. Thoughts?
functional-analysis convex-analysis
functional-analysis convex-analysis
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
1,242917
add a comment |
add a comment |
1 Answer
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$begingroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
add a comment |
$begingroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
add a comment |
$begingroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.6k11446
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.6k11446
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.6k11446
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.6k11446
59.6k11446
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
add a comment |
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
add a comment |
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