Can the max of a polynomial be near a zero?












0












$begingroup$


The setup of this question is very similar to a recent question I asked.



Let $p(z)$ be a polynomial of degree $n$ with complex zeros ${z_i}_{i=1}^n subsetmathbb{C}$. Let $Zsubset mathbb{C}$ be the set near the zeros,
$$
Z = bigcup_{i=1}^n D_{(100n)^{-1}}(z_i),
$$

where $D_r(x)subset mathbb{C}$ is the disk of radius $r$ centered at $x$. Let $gamma$ be the line segment $[-1,1]$ on the real axis in the complex plane.



I am interested in whether the inequality
$$
sup_{gammacap Z} |p(x)| leq sup_{xin gammasetminus Z} |p(x)|
$$

holds. In other words, is it possible that the maximum value of $p$ on the line segment $gamma$ takes place in $Z$?



It seems that the maximum principle might be useful here, but I am only interested in what happens to the polynomial on a line segment (which does not contain any disks in the complex plane). I am not sure how to proceed otherwise.










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$endgroup$

















    0












    $begingroup$


    The setup of this question is very similar to a recent question I asked.



    Let $p(z)$ be a polynomial of degree $n$ with complex zeros ${z_i}_{i=1}^n subsetmathbb{C}$. Let $Zsubset mathbb{C}$ be the set near the zeros,
    $$
    Z = bigcup_{i=1}^n D_{(100n)^{-1}}(z_i),
    $$

    where $D_r(x)subset mathbb{C}$ is the disk of radius $r$ centered at $x$. Let $gamma$ be the line segment $[-1,1]$ on the real axis in the complex plane.



    I am interested in whether the inequality
    $$
    sup_{gammacap Z} |p(x)| leq sup_{xin gammasetminus Z} |p(x)|
    $$

    holds. In other words, is it possible that the maximum value of $p$ on the line segment $gamma$ takes place in $Z$?



    It seems that the maximum principle might be useful here, but I am only interested in what happens to the polynomial on a line segment (which does not contain any disks in the complex plane). I am not sure how to proceed otherwise.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      The setup of this question is very similar to a recent question I asked.



      Let $p(z)$ be a polynomial of degree $n$ with complex zeros ${z_i}_{i=1}^n subsetmathbb{C}$. Let $Zsubset mathbb{C}$ be the set near the zeros,
      $$
      Z = bigcup_{i=1}^n D_{(100n)^{-1}}(z_i),
      $$

      where $D_r(x)subset mathbb{C}$ is the disk of radius $r$ centered at $x$. Let $gamma$ be the line segment $[-1,1]$ on the real axis in the complex plane.



      I am interested in whether the inequality
      $$
      sup_{gammacap Z} |p(x)| leq sup_{xin gammasetminus Z} |p(x)|
      $$

      holds. In other words, is it possible that the maximum value of $p$ on the line segment $gamma$ takes place in $Z$?



      It seems that the maximum principle might be useful here, but I am only interested in what happens to the polynomial on a line segment (which does not contain any disks in the complex plane). I am not sure how to proceed otherwise.










      share|cite|improve this question









      $endgroup$




      The setup of this question is very similar to a recent question I asked.



      Let $p(z)$ be a polynomial of degree $n$ with complex zeros ${z_i}_{i=1}^n subsetmathbb{C}$. Let $Zsubset mathbb{C}$ be the set near the zeros,
      $$
      Z = bigcup_{i=1}^n D_{(100n)^{-1}}(z_i),
      $$

      where $D_r(x)subset mathbb{C}$ is the disk of radius $r$ centered at $x$. Let $gamma$ be the line segment $[-1,1]$ on the real axis in the complex plane.



      I am interested in whether the inequality
      $$
      sup_{gammacap Z} |p(x)| leq sup_{xin gammasetminus Z} |p(x)|
      $$

      holds. In other words, is it possible that the maximum value of $p$ on the line segment $gamma$ takes place in $Z$?



      It seems that the maximum principle might be useful here, but I am only interested in what happens to the polynomial on a line segment (which does not contain any disks in the complex plane). I am not sure how to proceed otherwise.







      real-analysis complex-analysis polynomials






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      asked Dec 24 '18 at 15:34









      felipehfelipeh

      1,581619




      1,581619






















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          $begingroup$

          The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
          $$
          x_k = cosleft(frac{2k-1}{2n}piright),
          $$

          and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.






          share|cite|improve this answer









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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
            $$
            x_k = cosleft(frac{2k-1}{2n}piright),
            $$

            and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
              $$
              x_k = cosleft(frac{2k-1}{2n}piright),
              $$

              and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
                $$
                x_k = cosleft(frac{2k-1}{2n}piright),
                $$

                and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.






                share|cite|improve this answer









                $endgroup$



                The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
                $$
                x_k = cosleft(frac{2k-1}{2n}piright),
                $$

                and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 2 at 18:11









                felipehfelipeh

                1,581619




                1,581619






























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