What is $|P|_infty $ or $|P|_p$ for a polynomial?

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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.







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  • 4




    It depends on the context. Where did you find this ?
    – Kolja
    Aug 25 at 9:40














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.







share|cite|improve this question


















  • 4




    It depends on the context. Where did you find this ?
    – Kolja
    Aug 25 at 9:40












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.







share|cite|improve this question














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.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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 ?
    – Kolja
    Aug 25 at 9:40












  • 4




    It depends on the context. Where did you find this ?
    – Kolja
    Aug 25 at 9:40







4




4




It depends on the context. Where did you find this ?
– Kolja
Aug 25 at 9:40




It depends on the context. Where did you find this ?
– Kolja
Aug 25 at 9:40










2 Answers
2






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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.






share|cite|improve this answer





























    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 $$|P|_infty =sup_iinmathbb N|a_i|.$$






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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      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.






      share|cite|improve this answer


























        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.






        share|cite|improve this answer
























          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 27 at 17:40

























          answered Aug 25 at 9:39









          Kolja

          19319




          19319




















              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 $$|P|_infty =sup_iinmathbb N|a_i|.$$






              share|cite|improve this answer


























                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 $$|P|_infty =sup_iinmathbb N|a_i|.$$






                share|cite|improve this answer
























                  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 $$|P|_infty =sup_iinmathbb N|a_i|.$$






                  share|cite|improve this answer














                  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 $$|P|_infty =sup_iinmathbb N|a_i|.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 25 at 9:57

























                  answered Aug 25 at 9:28









                  Surb

                  36.6k84376




                  36.6k84376



























                       

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