Mixing
Showing the set of polynomials of degree $>n$ is open in the set of polynomials with norm...
$begingroup$
Let $(X,|.|)$ be the normed vector space of polynomials with $|p_0+p_1t+dots+p_nt^n|=max{|p_0|,dots,|p_n|}$ with normal addition and multiplication by a scalar.
I have proved that $U_n={P(t) :deg P>n}$ is dense in $(X,|.|)=$polynomials over $Bbb R$ with the $max|p_i|$ norm:
for any $Pin X$ and $delta>0$ if $deg P>n$ then $Pin U_ncap B(P,delta)neemptyset$
If $deg Ple n$ then $Q:=P+{deltaover2}t^{n+1}$ is in the ball $B(P,delta)$ and in $U_n$ simultaneously so $U_n$ is dense since $P$ and $delta$ were arbitrary
But for proving that $U_n$ is open for any $ninBbb N$ I'm not sure if this is ok:
Let $Pin U_n$ and $d:=deg P$. Let $r:=min{|p_i|:iin{n,dots,d}, p_ine 0}$
If $Qin B(P,r)$ then if $deg Qle n$, $|P-Q|ge|p_{n+1}t^{n+1}+p_{n+2}t^{n+2}+dots+p_dt^d|ge r$
Does that make sense?
vector-spaces metric-spaces normed-spaces
asked Jan 24 at 9:34
John CataldoJohn Cataldo
1,1931316
$endgroup$
add a comment |
$begingroup$
Let $(X,|.|)$ be the normed vector space of polynomials with $|p_0+p_1t+dots+p_nt^n|=max{|p_0|,dots,|p_n|}$ with normal addition and multiplication by a scalar.
I have proved that $U_n={P(t) :deg P>n}$ is dense in $(X,|.|)=$polynomials over $Bbb R$ with the $max|p_i|$ norm:
for any $Pin X$ and $delta>0$ if $deg P>n$ then $Pin U_ncap B(P,delta)neemptyset$
If $deg Ple n$ then $Q:=P+{deltaover2}t^{n+1}$ is in the ball $B(P,delta)$ and in $U_n$ simultaneously so $U_n$ is dense since $P$ and $delta$ were arbitrary
But for proving that $U_n$ is open for any $ninBbb N$ I'm not sure if this is ok:
Let $Pin U_n$ and $d:=deg P$. Let $r:=min{|p_i|:iin{n,dots,d}, p_ine 0}$
If $Qin B(P,r)$ then if $deg Qle n$, $|P-Q|ge|p_{n+1}t^{n+1}+p_{n+2}t^{n+2}+dots+p_dt^d|ge r$
Does that make sense?
vector-spaces metric-spaces normed-spaces
asked Jan 24 at 9:34
John CataldoJohn Cataldo
1,1931316
$endgroup$
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 9:46
1
$begingroup$
I think you mean that $Q = P + fracdelta2 t^{n+1}$ in your proof of density.
$endgroup$
– Mees de Vries
Jan 24 at 9:47
$begingroup$
Yes, I wanted to point that out in my comment , but I forgot.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:02
add a comment |
$begingroup$
Let $(X,|.|)$ be the normed vector space of polynomials with $|p_0+p_1t+dots+p_nt^n|=max{|p_0|,dots,|p_n|}$ with normal addition and multiplication by a scalar.
I have proved that $U_n={P(t) :deg P>n}$ is dense in $(X,|.|)=$polynomials over $Bbb R$ with the $max|p_i|$ norm:
for any $Pin X$ and $delta>0$ if $deg P>n$ then $Pin U_ncap B(P,delta)neemptyset$
If $deg Ple n$ then $Q:=P+{deltaover2}t^{n+1}$ is in the ball $B(P,delta)$ and in $U_n$ simultaneously so $U_n$ is dense since $P$ and $delta$ were arbitrary
But for proving that $U_n$ is open for any $ninBbb N$ I'm not sure if this is ok:
Let $Pin U_n$ and $d:=deg P$. Let $r:=min{|p_i|:iin{n,dots,d}, p_ine 0}$
If $Qin B(P,r)$ then if $deg Qle n$, $|P-Q|ge|p_{n+1}t^{n+1}+p_{n+2}t^{n+2}+dots+p_dt^d|ge r$
Does that make sense?
vector-spaces metric-spaces normed-spaces
asked Jan 24 at 9:34
John CataldoJohn Cataldo
1,1931316
$endgroup$
Let $(X,|.|)$ be the normed vector space of polynomials with $|p_0+p_1t+dots+p_nt^n|=max{|p_0|,dots,|p_n|}$ with normal addition and multiplication by a scalar.
I have proved that $U_n={P(t) :deg P>n}$ is dense in $(X,|.|)=$polynomials over $Bbb R$ with the $max|p_i|$ norm:
for any $Pin X$ and $delta>0$ if $deg P>n$ then $Pin U_ncap B(P,delta)neemptyset$
If $deg Ple n$ then $Q:=P+{deltaover2}t^{n+1}$ is in the ball $B(P,delta)$ and in $U_n$ simultaneously so $U_n$ is dense since $P$ and $delta$ were arbitrary
But for proving that $U_n$ is open for any $ninBbb N$ I'm not sure if this is ok:
Let $Pin U_n$ and $d:=deg P$. Let $r:=min{|p_i|:iin{n,dots,d}, p_ine 0}$
If $Qin B(P,r)$ then if $deg Qle n$, $|P-Q|ge|p_{n+1}t^{n+1}+p_{n+2}t^{n+2}+dots+p_dt^d|ge r$
Does that make sense?
vector-spaces metric-spaces normed-spaces
vector-spaces metric-spaces normed-spaces
asked Jan 24 at 9:34
John CataldoJohn Cataldo
1,1931316
asked Jan 24 at 9:34
John CataldoJohn Cataldo
1,1931316
edited Jan 24 at 9:55
John Cataldo
asked Jan 24 at 9:34
John CataldoJohn Cataldo
1,1931316
asked Jan 24 at 9:34
John CataldoJohn Cataldo
1,1931316
asked Jan 24 at 9:34
John CataldoJohn Cataldo
1,1931316
1,1931316
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 9:46
1
$begingroup$
I think you mean that $Q = P + fracdelta2 t^{n+1}$ in your proof of density.
$endgroup$
– Mees de Vries
Jan 24 at 9:47
$begingroup$
Yes, I wanted to point that out in my comment , but I forgot.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:02
add a comment |
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 9:46
1
$begingroup$
I think you mean that $Q = P + fracdelta2 t^{n+1}$ in your proof of density.
$endgroup$
– Mees de Vries
Jan 24 at 9:47
$begingroup$
Yes, I wanted to point that out in my comment , but I forgot.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:02
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 9:46
$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 9:46
1
1
$begingroup$
I think you mean that $Q = P + fracdelta2 t^{n+1}$ in your proof of density.
$endgroup$
– Mees de Vries
Jan 24 at 9:47
$begingroup$
I think you mean that $Q = P + fracdelta2 t^{n+1}$ in your proof of density.
$endgroup$
– Mees de Vries
Jan 24 at 9:47
$begingroup$
Yes, I wanted to point that out in my comment , but I forgot.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:02
$begingroup$
Yes, I wanted to point that out in my comment , but I forgot.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:02
add a comment |
1 Answer
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$begingroup$
I don't know where does the inequality $lVert P-QrVertgeqslantlVert p_nt^n+p_{n+1}t^{n+1}+dots+p_dt^drVertgeqslant r$ came from (or what it means).
Note that, by the definition $lVertcdotrVert$, if $P(x)=p_Nx^N+p_{N-1}x^{N-1}+cdots+p_0in U_n$, and $Q(x)in Bbigl(P(x),lvert p_Nrvertbigr)$, then $Q(x)in U_n$. Therefore, $U_n$ is an open set.
answered Jan 24 at 9:46
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
$begingroup$
$p_n$ could be zero though The inequality comes from the fact that taking the maximum over a larger set that contains another set is $ge$ than the maximum over that smaller set. But it should probably be $n+1$ instead of $n$
$endgroup$
– John Cataldo
Jan 24 at 9:51
$begingroup$
I meant $p_N$, not $p_n$. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 24 at 9:54
$begingroup$
I see how I did too much with $r$ and $r=|p_N|$ works, thanks
$endgroup$
– John Cataldo
Jan 24 at 9:58
$begingroup$
I think the inequality use by OP is correct. I don't see any mistake in his argument.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:04
$begingroup$
@KaviRamaMurthy I did not understand the OP's argument. Perhaps that it is correct.
$endgroup$
– José Carlos Santos
Jan 24 at 10:07
add a comment |
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$begingroup$
I don't know where does the inequality $lVert P-QrVertgeqslantlVert p_nt^n+p_{n+1}t^{n+1}+dots+p_dt^drVertgeqslant r$ came from (or what it means).
Note that, by the definition $lVertcdotrVert$, if $P(x)=p_Nx^N+p_{N-1}x^{N-1}+cdots+p_0in U_n$, and $Q(x)in Bbigl(P(x),lvert p_Nrvertbigr)$, then $Q(x)in U_n$. Therefore, $U_n$ is an open set.
answered Jan 24 at 9:46
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
$begingroup$
$p_n$ could be zero though The inequality comes from the fact that taking the maximum over a larger set that contains another set is $ge$ than the maximum over that smaller set. But it should probably be $n+1$ instead of $n$
$endgroup$
– John Cataldo
Jan 24 at 9:51
$begingroup$
I meant $p_N$, not $p_n$. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 24 at 9:54
$begingroup$
I see how I did too much with $r$ and $r=|p_N|$ works, thanks
$endgroup$
– John Cataldo
Jan 24 at 9:58
$begingroup$
I think the inequality use by OP is correct. I don't see any mistake in his argument.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:04
$begingroup$
@KaviRamaMurthy I did not understand the OP's argument. Perhaps that it is correct.
$endgroup$
– José Carlos Santos
Jan 24 at 10:07
add a comment |
$begingroup$
I don't know where does the inequality $lVert P-QrVertgeqslantlVert p_nt^n+p_{n+1}t^{n+1}+dots+p_dt^drVertgeqslant r$ came from (or what it means).
Note that, by the definition $lVertcdotrVert$, if $P(x)=p_Nx^N+p_{N-1}x^{N-1}+cdots+p_0in U_n$, and $Q(x)in Bbigl(P(x),lvert p_Nrvertbigr)$, then $Q(x)in U_n$. Therefore, $U_n$ is an open set.
answered Jan 24 at 9:46
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
$begingroup$
$p_n$ could be zero though The inequality comes from the fact that taking the maximum over a larger set that contains another set is $ge$ than the maximum over that smaller set. But it should probably be $n+1$ instead of $n$
$endgroup$
– John Cataldo
Jan 24 at 9:51
$begingroup$
I meant $p_N$, not $p_n$. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 24 at 9:54
$begingroup$
I see how I did too much with $r$ and $r=|p_N|$ works, thanks
$endgroup$
– John Cataldo
Jan 24 at 9:58
$begingroup$
I think the inequality use by OP is correct. I don't see any mistake in his argument.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:04
$begingroup$
@KaviRamaMurthy I did not understand the OP's argument. Perhaps that it is correct.
$endgroup$
– José Carlos Santos
Jan 24 at 10:07
add a comment |
$begingroup$
I don't know where does the inequality $lVert P-QrVertgeqslantlVert p_nt^n+p_{n+1}t^{n+1}+dots+p_dt^drVertgeqslant r$ came from (or what it means).
Note that, by the definition $lVertcdotrVert$, if $P(x)=p_Nx^N+p_{N-1}x^{N-1}+cdots+p_0in U_n$, and $Q(x)in Bbigl(P(x),lvert p_Nrvertbigr)$, then $Q(x)in U_n$. Therefore, $U_n$ is an open set.
answered Jan 24 at 9:46
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
I don't know where does the inequality $lVert P-QrVertgeqslantlVert p_nt^n+p_{n+1}t^{n+1}+dots+p_dt^drVertgeqslant r$ came from (or what it means).
Note that, by the definition $lVertcdotrVert$, if $P(x)=p_Nx^N+p_{N-1}x^{N-1}+cdots+p_0in U_n$, and $Q(x)in Bbigl(P(x),lvert p_Nrvertbigr)$, then $Q(x)in U_n$. Therefore, $U_n$ is an open set.
answered Jan 24 at 9:46
José Carlos SantosJosé Carlos Santos
168k22132236
edited Jan 24 at 9:53
answered Jan 24 at 9:46
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 9:46
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 9:46
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
$begingroup$
$p_n$ could be zero though The inequality comes from the fact that taking the maximum over a larger set that contains another set is $ge$ than the maximum over that smaller set. But it should probably be $n+1$ instead of $n$
$endgroup$
– John Cataldo
Jan 24 at 9:51
$begingroup$
I meant $p_N$, not $p_n$. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 24 at 9:54
$begingroup$
I see how I did too much with $r$ and $r=|p_N|$ works, thanks
$endgroup$
– John Cataldo
Jan 24 at 9:58
$begingroup$
I think the inequality use by OP is correct. I don't see any mistake in his argument.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:04
$begingroup$
@KaviRamaMurthy I did not understand the OP's argument. Perhaps that it is correct.
$endgroup$
– José Carlos Santos
Jan 24 at 10:07
add a comment |
$begingroup$
$p_n$ could be zero though The inequality comes from the fact that taking the maximum over a larger set that contains another set is $ge$ than the maximum over that smaller set. But it should probably be $n+1$ instead of $n$
$endgroup$
– John Cataldo
Jan 24 at 9:51
$begingroup$
I meant $p_N$, not $p_n$. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 24 at 9:54
$begingroup$
I see how I did too much with $r$ and $r=|p_N|$ works, thanks
$endgroup$
– John Cataldo
Jan 24 at 9:58
$begingroup$
I think the inequality use by OP is correct. I don't see any mistake in his argument.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:04
$begingroup$
@KaviRamaMurthy I did not understand the OP's argument. Perhaps that it is correct.
$endgroup$
– José Carlos Santos
Jan 24 at 10:07
$begingroup$
$p_n$ could be zero though The inequality comes from the fact that taking the maximum over a larger set that contains another set is $ge$ than the maximum over that smaller set. But it should probably be $n+1$ instead of $n$
$endgroup$
– John Cataldo
Jan 24 at 9:51
$begingroup$
$p_n$ could be zero though The inequality comes from the fact that taking the maximum over a larger set that contains another set is $ge$ than the maximum over that smaller set. But it should probably be $n+1$ instead of $n$
$endgroup$
– John Cataldo
Jan 24 at 9:51
$begingroup$
I meant $p_N$, not $p_n$. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 24 at 9:54
$begingroup$
I meant $p_N$, not $p_n$. I've edited my answer.
$endgroup$
– José Carlos Santos
Jan 24 at 9:54
$begingroup$
I see how I did too much with $r$ and $r=|p_N|$ works, thanks
$endgroup$
– John Cataldo
Jan 24 at 9:58
$begingroup$
I see how I did too much with $r$ and $r=|p_N|$ works, thanks
$endgroup$
– John Cataldo
Jan 24 at 9:58
$begingroup$
I think the inequality use by OP is correct. I don't see any mistake in his argument.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:04
$begingroup$
I think the inequality use by OP is correct. I don't see any mistake in his argument.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:04
$begingroup$
@KaviRamaMurthy I did not understand the OP's argument. Perhaps that it is correct.
$endgroup$
– José Carlos Santos
Jan 24 at 10:07
$begingroup$
@KaviRamaMurthy I did not understand the OP's argument. Perhaps that it is correct.
$endgroup$
– José Carlos Santos
Jan 24 at 10:07
add a comment |
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$begingroup$
Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 9:46
1
$begingroup$
I think you mean that $Q = P + fracdelta2 t^{n+1}$ in your proof of density.
$endgroup$
– Mees de Vries
Jan 24 at 9:47
$begingroup$
Yes, I wanted to point that out in my comment , but I forgot.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 10:02