$f$ analytic and $|f|leq1$ on a strip.












0












$begingroup$


Let $E$ be the strip ${zinmathbb{C}:0<Re z <1}$. Let $f$ be analytic on $E$ and continuous on $bar{E}$. Show that if $f$ is bounded on $E$ and $|f|leq1$ on the boundary of $E$, then $|f|leq1$ on $E$.



The hint that comes with the problem says to consider the analytic function $f_{epsilon}(z)=(1+epsilon z)^{-1}f(z)$ on open set ${zinmathbb{C}:0<Re z<1, -M<Im z<M}$ for $M$ large.



From some searching, it looks like I need to use Möbius transformation but I'm not sure how to integrate that in this case. Any input is appreciated.






complex-analysis maximum-principle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
    $endgroup$
    – Story123
    Dec 21 '18 at 18:58












  • $begingroup$
    Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
    $endgroup$
    – Song
    Dec 22 '18 at 12:36


















0












$begingroup$


Let $E$ be the strip ${zinmathbb{C}:0<Re z <1}$. Let $f$ be analytic on $E$ and continuous on $bar{E}$. Show that if $f$ is bounded on $E$ and $|f|leq1$ on the boundary of $E$, then $|f|leq1$ on $E$.



The hint that comes with the problem says to consider the analytic function $f_{epsilon}(z)=(1+epsilon z)^{-1}f(z)$ on open set ${zinmathbb{C}:0<Re z<1, -M<Im z<M}$ for $M$ large.



From some searching, it looks like I need to use Möbius transformation but I'm not sure how to integrate that in this case. Any input is appreciated.






complex-analysis maximum-principle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
    $endgroup$
    – Story123
    Dec 21 '18 at 18:58












  • $begingroup$
    Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
    $endgroup$
    – Song
    Dec 22 '18 at 12:36
















0












0








0





$begingroup$


Let $E$ be the strip ${zinmathbb{C}:0<Re z <1}$. Let $f$ be analytic on $E$ and continuous on $bar{E}$. Show that if $f$ is bounded on $E$ and $|f|leq1$ on the boundary of $E$, then $|f|leq1$ on $E$.



The hint that comes with the problem says to consider the analytic function $f_{epsilon}(z)=(1+epsilon z)^{-1}f(z)$ on open set ${zinmathbb{C}:0<Re z<1, -M<Im z<M}$ for $M$ large.



From some searching, it looks like I need to use Möbius transformation but I'm not sure how to integrate that in this case. Any input is appreciated.






complex-analysis maximum-principle







share|cite|improve this question











$endgroup$




Let $E$ be the strip ${zinmathbb{C}:0<Re z <1}$. Let $f$ be analytic on $E$ and continuous on $bar{E}$. Show that if $f$ is bounded on $E$ and $|f|leq1$ on the boundary of $E$, then $|f|leq1$ on $E$.



The hint that comes with the problem says to consider the analytic function $f_{epsilon}(z)=(1+epsilon z)^{-1}f(z)$ on open set ${zinmathbb{C}:0<Re z<1, -M<Im z<M}$ for $M$ large.



From some searching, it looks like I need to use Möbius transformation but I'm not sure how to integrate that in this case. Any input is appreciated.






complex-analysis maximum-principle




complex-analysis maximum-principle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 14:36









Song

18.6k21651




18.6k21651










asked Dec 20 '18 at 17:32









Ya GYa G

536211




536211












  • $begingroup$
    Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
    $endgroup$
    – Story123
    Dec 21 '18 at 18:58












  • $begingroup$
    Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
    $endgroup$
    – Song
    Dec 22 '18 at 12:36




















  • $begingroup$
    Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
    $endgroup$
    – Story123
    Dec 21 '18 at 18:58












  • $begingroup$
    Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
    $endgroup$
    – Song
    Dec 22 '18 at 12:36


















$begingroup$
Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
$endgroup$
– Story123
Dec 21 '18 at 18:58






$begingroup$
Another way one might approach this problem is by using a mobius transformation to map the right half plane to the unit disk then apply maximum modulus principle. The map should be begin{equation} frac{z+1}{z-1} end{equation} I would attempt this because we can apply the maximum modulus principle on the unit disk, so I want to transform his onto a problem on the unit disk.
$endgroup$
– Story123
Dec 21 '18 at 18:58














$begingroup$
Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
$endgroup$
– Song
Dec 22 '18 at 12:36






$begingroup$
Actually we cannot apply Maximum modulus principle in its raw form: the reason is that $f$ cannot be continuously extended to the closure of the domain. The problem arises from the fact that $xpm iinfty$ is mapped to $1$. Unless one deals with the behavior of $f$ near $1$, maximum modulus principle does not give the result. Most likely, it only makes the problem more complicated.
$endgroup$
– Song
Dec 22 '18 at 12:36












1 Answer
1






active

oldest

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1












$begingroup$

This is the Phragmen-Lindelof principle. Note that
$$
lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
$$ uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
$$
{x+iy;|;0leq xleq 1,;|y|leq M}
$$ for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
$$
|f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
$$ for all sufficiently large $M>0$ and thus
$$
|f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$ Finally take $epsilonto 0$ to get the desired bound
$$
|f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
$$






share|cite|improve this answer









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    1 Answer
    1






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    1












    $begingroup$

    This is the Phragmen-Lindelof principle. Note that
    $$
    lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
    $$     uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
    $$
    {x+iy;|;0leq xleq 1,;|y|leq M}
    $$     for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
    $$
    |f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
    $$     for all sufficiently large $M>0$ and thus
    $$
    |f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
    $$     Finally take $epsilonto 0$ to get the desired bound
    $$
    |f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      This is the Phragmen-Lindelof principle. Note that
      $$
      lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
      $$       uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
      $$
      {x+iy;|;0leq xleq 1,;|y|leq M}
      $$       for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
      $$
      |f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
      $$       for all sufficiently large $M>0$ and thus
      $$
      |f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
      $$       Finally take $epsilonto 0$ to get the desired bound
      $$
      |f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        This is the Phragmen-Lindelof principle. Note that
        $$
        lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
        $$         uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
        $$
        {x+iy;|;0leq xleq 1,;|y|leq M}
        $$         for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
        $$
        |f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
        $$         for all sufficiently large $M>0$ and thus
        $$
        |f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
        $$         Finally take $epsilonto 0$ to get the desired bound
        $$
        |f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
        $$






        share|cite|improve this answer









        $endgroup$



        This is the Phragmen-Lindelof principle. Note that
        $$
        lim_{|y|toinfty} |f_epsilon(x+iy)| = 0
        $$         uniformly on $xin [0,1]$. If we apply maximum modulus principle to the region
        $$
        {x+iy;|;0leq xleq 1,;|y|leq M}
        $$         for large $M>0$, we can see that the maximum modulus of $|f_epsilon|$ cannot occur on $y=pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives
        $$
        |f_epsilon(x+iy)|leq 1,quadforall xin[0,1],;|y|leq M,
        $$         for all sufficiently large $M>0$ and thus
        $$
        |f_epsilon(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
        $$         Finally take $epsilonto 0$ to get the desired bound
        $$
        |f(x+iy)|leq 1,quadforall xin [0,1], yin mathbb{R}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 17:47









        SongSong

        18.6k21651




        18.6k21651






























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