What is $|P|_infty $ or $|P|_p$ for a polynomial?
Clash Royale CLAN TAG#URR8PPP
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Let $(mathbb R[X],|cdot |_p)$
I don't really know what $|P|_p$ stands for a polynomial of $Pin mathbb R[X]$, and neither do I know what is $|P|_infty $
Is it $$|P|_infty =sup_xin [0,1]|P(x)| ?$$
$$|P|_2=sqrt^2dx ?$$
Furthermore, for the $$|P|_p$$, I really have no idea.
functional-analysis norm
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up vote
1
down vote
favorite
Let $(mathbb R[X],|cdot |_p)$
I don't really know what $|P|_p$ stands for a polynomial of $Pin mathbb R[X]$, and neither do I know what is $|P|_infty $
Is it $$|P|_infty =sup_xin [0,1]|P(x)| ?$$
$$|P|_2=sqrt^2dx ?$$
Furthermore, for the $$|P|_p$$, I really have no idea.
functional-analysis norm
4
It depends on the context. Where did you find this ?
â Kolja
Aug 25 at 9:40
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(mathbb R[X],|cdot |_p)$
I don't really know what $|P|_p$ stands for a polynomial of $Pin mathbb R[X]$, and neither do I know what is $|P|_infty $
Is it $$|P|_infty =sup_xin [0,1]|P(x)| ?$$
$$|P|_2=sqrt^2dx ?$$
Furthermore, for the $$|P|_p$$, I really have no idea.
functional-analysis norm
Let $(mathbb R[X],|cdot |_p)$
I don't really know what $|P|_p$ stands for a polynomial of $Pin mathbb R[X]$, and neither do I know what is $|P|_infty $
Is it $$|P|_infty =sup_xin [0,1]|P(x)| ?$$
$$|P|_2=sqrt^2dx ?$$
Furthermore, for the $$|P|_p$$, I really have no idea.
functional-analysis norm
edited Aug 25 at 10:22
dmtri
766317
766317
asked Aug 25 at 9:24
user330587
828310
828310
4
It depends on the context. Where did you find this ?
â Kolja
Aug 25 at 9:40
add a comment |Â
4
It depends on the context. Where did you find this ?
â Kolja
Aug 25 at 9:40
4
4
It depends on the context. Where did you find this ?
â Kolja
Aug 25 at 9:40
It depends on the context. Where did you find this ?
â Kolja
Aug 25 at 9:40
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
It could be $L^p$ norm defined as
$$
||P||_p=sqrt[p]^p dp
$$
Here we consider the space $L^p(S)$, with the normal Lebegue measure, where $Ssubseteq mathbbR$. If $S=[0,1]$ then this is always a finite value, and the $||P||_2$ and the $||P||_infty$ norms are exactly what you wrote. If on the other hand $S=mathbbR$, all of the norms are infinite since polynomials diverge, and in fact polynomials are not elements of $L^p(mathbbR)$. In that case I believe Surb's answer is the more correct one.
add a comment |Â
up vote
1
down vote
In this situation, I suppose that a polynomial $P(x)=a_0+a_1X+...+a_nX^n$ is identified as the vector $(a_0,a_1,...,a_n,0,0,...)$ in $mathbb R^mathbb N $ so I could imagine that $$|P|_p^p=|a_0|^p+cdots |a_n|^p$$
and $$|P|_infty =sup_iinmathbb N|a_i|.$$
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It could be $L^p$ norm defined as
$$
||P||_p=sqrt[p]^p dp
$$
Here we consider the space $L^p(S)$, with the normal Lebegue measure, where $Ssubseteq mathbbR$. If $S=[0,1]$ then this is always a finite value, and the $||P||_2$ and the $||P||_infty$ norms are exactly what you wrote. If on the other hand $S=mathbbR$, all of the norms are infinite since polynomials diverge, and in fact polynomials are not elements of $L^p(mathbbR)$. In that case I believe Surb's answer is the more correct one.
add a comment |Â
up vote
2
down vote
It could be $L^p$ norm defined as
$$
||P||_p=sqrt[p]^p dp
$$
Here we consider the space $L^p(S)$, with the normal Lebegue measure, where $Ssubseteq mathbbR$. If $S=[0,1]$ then this is always a finite value, and the $||P||_2$ and the $||P||_infty$ norms are exactly what you wrote. If on the other hand $S=mathbbR$, all of the norms are infinite since polynomials diverge, and in fact polynomials are not elements of $L^p(mathbbR)$. In that case I believe Surb's answer is the more correct one.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It could be $L^p$ norm defined as
$$
||P||_p=sqrt[p]^p dp
$$
Here we consider the space $L^p(S)$, with the normal Lebegue measure, where $Ssubseteq mathbbR$. If $S=[0,1]$ then this is always a finite value, and the $||P||_2$ and the $||P||_infty$ norms are exactly what you wrote. If on the other hand $S=mathbbR$, all of the norms are infinite since polynomials diverge, and in fact polynomials are not elements of $L^p(mathbbR)$. In that case I believe Surb's answer is the more correct one.
It could be $L^p$ norm defined as
$$
||P||_p=sqrt[p]^p dp
$$
Here we consider the space $L^p(S)$, with the normal Lebegue measure, where $Ssubseteq mathbbR$. If $S=[0,1]$ then this is always a finite value, and the $||P||_2$ and the $||P||_infty$ norms are exactly what you wrote. If on the other hand $S=mathbbR$, all of the norms are infinite since polynomials diverge, and in fact polynomials are not elements of $L^p(mathbbR)$. In that case I believe Surb's answer is the more correct one.
edited Aug 27 at 17:40
answered Aug 25 at 9:39
Kolja
19319
19319
add a comment |Â
add a comment |Â
up vote
1
down vote
In this situation, I suppose that a polynomial $P(x)=a_0+a_1X+...+a_nX^n$ is identified as the vector $(a_0,a_1,...,a_n,0,0,...)$ in $mathbb R^mathbb N $ so I could imagine that $$|P|_p^p=|a_0|^p+cdots |a_n|^p$$
and $$|P|_infty =sup_iinmathbb N|a_i|.$$
add a comment |Â
up vote
1
down vote
In this situation, I suppose that a polynomial $P(x)=a_0+a_1X+...+a_nX^n$ is identified as the vector $(a_0,a_1,...,a_n,0,0,...)$ in $mathbb R^mathbb N $ so I could imagine that $$|P|_p^p=|a_0|^p+cdots |a_n|^p$$
and $$|P|_infty =sup_iinmathbb N|a_i|.$$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In this situation, I suppose that a polynomial $P(x)=a_0+a_1X+...+a_nX^n$ is identified as the vector $(a_0,a_1,...,a_n,0,0,...)$ in $mathbb R^mathbb N $ so I could imagine that $$|P|_p^p=|a_0|^p+cdots |a_n|^p$$
and $$|P|_infty =sup_iinmathbb N|a_i|.$$
In this situation, I suppose that a polynomial $P(x)=a_0+a_1X+...+a_nX^n$ is identified as the vector $(a_0,a_1,...,a_n,0,0,...)$ in $mathbb R^mathbb N $ so I could imagine that $$|P|_p^p=|a_0|^p+cdots |a_n|^p$$
and $$|P|_infty =sup_iinmathbb N|a_i|.$$
edited Aug 25 at 9:57
answered Aug 25 at 9:28
Surb
36.6k84376
36.6k84376
add a comment |Â
add a comment |Â
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4
It depends on the context. Where did you find this ?
â Kolja
Aug 25 at 9:40