Show some theorem concerning of the uniform convergence on compacts of a sequence of polynomials of order...












4












$begingroup$


Let $k,sin mathbb N$,



let $x_0,x_1,...,x_s$ be given pairweise different real numbers,



let $m_0,m_1,...,m_s$ be given nonnegative integers such that $sum_{i=0}^s m_i=k$,



and let
$$
P_n(x)=a_{n0}+a_{n1}x+...+a_{n,k-1}x^{k-1} (textrm{ for } nin mathbb N, xin mathbb R)
$$
be a sequence of real polynomials of order $leq k-1$.



Assume that there exist limits of derivatives:



$$
lim_{nrightarrow infty} P_n^{(j)}(x_0) (textrm{ for } j=0,1,...,m_0);
$$
$$
lim_{nrightarrow infty} P_n^{(j)}(x_1) (textrm{ for } j=0,1,...,m_1);
$$
...........................
$$
lim_{nrightarrow infty} P_n^{(j)}(x_s) (textrm{ for } j=0,1,...,m_s).
$$



I wish to know that $P_n(x)$ is uniformly convergent on compact intervals. It would be sufficient to show that there exist limits:
$$
lim_{nrightarrow infty} a_{nj} (textrm{ for } j=0,...,k-1).
$$



Maybe proof or references.



Thanks.






analysis polynomials convergence uniform-convergence







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Let $k,sin mathbb N$,



    let $x_0,x_1,...,x_s$ be given pairweise different real numbers,



    let $m_0,m_1,...,m_s$ be given nonnegative integers such that $sum_{i=0}^s m_i=k$,



    and let
    $$
    P_n(x)=a_{n0}+a_{n1}x+...+a_{n,k-1}x^{k-1} (textrm{ for } nin mathbb N, xin mathbb R)
    $$
    be a sequence of real polynomials of order $leq k-1$.



    Assume that there exist limits of derivatives:



    $$
    lim_{nrightarrow infty} P_n^{(j)}(x_0) (textrm{ for } j=0,1,...,m_0);
    $$
    $$
    lim_{nrightarrow infty} P_n^{(j)}(x_1) (textrm{ for } j=0,1,...,m_1);
    $$
    ...........................
    $$
    lim_{nrightarrow infty} P_n^{(j)}(x_s) (textrm{ for } j=0,1,...,m_s).
    $$



    I wish to know that $P_n(x)$ is uniformly convergent on compact intervals. It would be sufficient to show that there exist limits:
    $$
    lim_{nrightarrow infty} a_{nj} (textrm{ for } j=0,...,k-1).
    $$



    Maybe proof or references.



    Thanks.






    analysis polynomials convergence uniform-convergence







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      Let $k,sin mathbb N$,



      let $x_0,x_1,...,x_s$ be given pairweise different real numbers,



      let $m_0,m_1,...,m_s$ be given nonnegative integers such that $sum_{i=0}^s m_i=k$,



      and let
      $$
      P_n(x)=a_{n0}+a_{n1}x+...+a_{n,k-1}x^{k-1} (textrm{ for } nin mathbb N, xin mathbb R)
      $$
      be a sequence of real polynomials of order $leq k-1$.



      Assume that there exist limits of derivatives:



      $$
      lim_{nrightarrow infty} P_n^{(j)}(x_0) (textrm{ for } j=0,1,...,m_0);
      $$
      $$
      lim_{nrightarrow infty} P_n^{(j)}(x_1) (textrm{ for } j=0,1,...,m_1);
      $$
      ...........................
      $$
      lim_{nrightarrow infty} P_n^{(j)}(x_s) (textrm{ for } j=0,1,...,m_s).
      $$



      I wish to know that $P_n(x)$ is uniformly convergent on compact intervals. It would be sufficient to show that there exist limits:
      $$
      lim_{nrightarrow infty} a_{nj} (textrm{ for } j=0,...,k-1).
      $$



      Maybe proof or references.



      Thanks.






      analysis polynomials convergence uniform-convergence







      share|cite|improve this question











      $endgroup$




      Let $k,sin mathbb N$,



      let $x_0,x_1,...,x_s$ be given pairweise different real numbers,



      let $m_0,m_1,...,m_s$ be given nonnegative integers such that $sum_{i=0}^s m_i=k$,



      and let
      $$
      P_n(x)=a_{n0}+a_{n1}x+...+a_{n,k-1}x^{k-1} (textrm{ for } nin mathbb N, xin mathbb R)
      $$
      be a sequence of real polynomials of order $leq k-1$.



      Assume that there exist limits of derivatives:



      $$
      lim_{nrightarrow infty} P_n^{(j)}(x_0) (textrm{ for } j=0,1,...,m_0);
      $$
      $$
      lim_{nrightarrow infty} P_n^{(j)}(x_1) (textrm{ for } j=0,1,...,m_1);
      $$
      ...........................
      $$
      lim_{nrightarrow infty} P_n^{(j)}(x_s) (textrm{ for } j=0,1,...,m_s).
      $$



      I wish to know that $P_n(x)$ is uniformly convergent on compact intervals. It would be sufficient to show that there exist limits:
      $$
      lim_{nrightarrow infty} a_{nj} (textrm{ for } j=0,...,k-1).
      $$



      Maybe proof or references.



      Thanks.






      analysis polynomials convergence uniform-convergence




      analysis polynomials convergence uniform-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 2:29









      Alex Ravsky

      40.7k32282




      40.7k32282










      asked Dec 9 '18 at 15:15









      RichardRichard

      1,79511946




      1,79511946






















          1 Answer
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          2





          +50







          $begingroup$

          It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
          $$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$



          Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$



          It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by



          $$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$



          But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.



          Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
            $endgroup$
            – Lukas Geyer
            Dec 15 '18 at 13:51











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          1 Answer
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          1 Answer
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          active

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          active

          oldest

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          2





          +50







          $begingroup$

          It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
          $$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$



          Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$



          It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by



          $$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$



          But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.



          Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
            $endgroup$
            – Lukas Geyer
            Dec 15 '18 at 13:51
















          2





          +50







          $begingroup$

          It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
          $$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$



          Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$



          It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by



          $$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$



          But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.



          Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
            $endgroup$
            – Lukas Geyer
            Dec 15 '18 at 13:51














          2





          +50







          2





          +50



          2




          +50



          $begingroup$

          It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
          $$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$



          Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$



          It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by



          $$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$



          But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.



          Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.






          share|cite|improve this answer









          $endgroup$



          It suffices to have $sum_{i=0}^s m_i=k-s-1$. Indeed, consider a map $f$ from $Bbb R^k$ to $Bbb R^k$ defined as follows. Let $a=(a_0,a_1,dots,a_{k-1})inBbb R^k$. Consider a polynomial
          $$P_a(x)=a_{0}+a_{1}x+dots+a_{k-1}x^{k-1}.$$



          Put $$f(a)=(P^{(0)}(x_0), P^{(1)}(x_0),dots, P^{(m_0)}(x_0), P^{(0)}(x_1), P^{(1)}(x_1),dots P^{(m_1)}(x_1), P^{(0)}(x_2),dots, P^{(0)}(x_{k-1}), P^{(1)}(x_{k-1}),dots, P^{(m_{k-1})}(x_{k-1})).$$



          It is easy to check that $f$ is a continuous linear map. Let $ainoperatorname{ker} f$, that is $f(a)=0$. Then $P_a$ is divisible by



          $$Q(x)=(x-x_0)^{m_0+1}(x-x_1)^{m_1+1}cdots (x-x_{k-1})^{m_{k-1}+1}.$$



          But degree of the polynomial $Q$ is $sum_{i=0}^s (m_i+1)=k,$ a contradiction.



          Therefore, the map $f$ is injective. By Invariance of domain, $f$ is an open map. Since $f$ is linear, it is surjective. So $f$ is a homeomorphism. Therefore a sequence ${a^n}$ of points of $Bbb R^k$ converges to a point $ainBbb R^k$ iff a sequence ${f(a^n)}$ converges to a point $f(a)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 5:22









          Alex RavskyAlex Ravsky

          40.7k32282




          40.7k32282








          • 1




            $begingroup$
            You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
            $endgroup$
            – Lukas Geyer
            Dec 15 '18 at 13:51














          • 1




            $begingroup$
            You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
            $endgroup$
            – Lukas Geyer
            Dec 15 '18 at 13:51








          1




          1




          $begingroup$
          You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
          $endgroup$
          – Lukas Geyer
          Dec 15 '18 at 13:51




          $begingroup$
          You don't really need the invariance of domain argument at the end, elementary linear algebra gives the fact that $f$ is a linear isomorphism.
          $endgroup$
          – Lukas Geyer
          Dec 15 '18 at 13:51


















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