Holomorphic function on an annulus











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Let $f$ be a holomorphic function on the set $U={z in mathbb{C}: 1 leq |z| leq pi }$. Assume that $max_{|z|=1}|f(z)| leq 1$ and $max_{|z|=pi}|f(z)| leq pi^{pi}$.




How to prove that $max_{|z|=e}|f(z)| leq e^{pi}$?




Usually in such exercises one considers $g(z)$ of the form $alpha z^nf(z)$ where $alpha$ and $n$ are such that our assumptions on $f$ gives $|g(z)| leq 1$ on both circles being the boundary of $U$. However it is not guarantedd that $n$ would be integer thus in general we don't get a holomorphic function $g$ for which we could apply the maximum principle. If $n$ happens to be rational one can overcome this difficulty by considering $(alpha z^n f(z))^q$ where $n=frac{p}{q}$. However in our example $n$ turns out to be $pi$ and by considering $g(z)=f(z)/z^{pi}$ we don't get holomorphic function. How to handle this case?










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    Let $f$ be a holomorphic function on the set $U={z in mathbb{C}: 1 leq |z| leq pi }$. Assume that $max_{|z|=1}|f(z)| leq 1$ and $max_{|z|=pi}|f(z)| leq pi^{pi}$.




    How to prove that $max_{|z|=e}|f(z)| leq e^{pi}$?




    Usually in such exercises one considers $g(z)$ of the form $alpha z^nf(z)$ where $alpha$ and $n$ are such that our assumptions on $f$ gives $|g(z)| leq 1$ on both circles being the boundary of $U$. However it is not guarantedd that $n$ would be integer thus in general we don't get a holomorphic function $g$ for which we could apply the maximum principle. If $n$ happens to be rational one can overcome this difficulty by considering $(alpha z^n f(z))^q$ where $n=frac{p}{q}$. However in our example $n$ turns out to be $pi$ and by considering $g(z)=f(z)/z^{pi}$ we don't get holomorphic function. How to handle this case?










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      up vote
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      down vote

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      Let $f$ be a holomorphic function on the set $U={z in mathbb{C}: 1 leq |z| leq pi }$. Assume that $max_{|z|=1}|f(z)| leq 1$ and $max_{|z|=pi}|f(z)| leq pi^{pi}$.




      How to prove that $max_{|z|=e}|f(z)| leq e^{pi}$?




      Usually in such exercises one considers $g(z)$ of the form $alpha z^nf(z)$ where $alpha$ and $n$ are such that our assumptions on $f$ gives $|g(z)| leq 1$ on both circles being the boundary of $U$. However it is not guarantedd that $n$ would be integer thus in general we don't get a holomorphic function $g$ for which we could apply the maximum principle. If $n$ happens to be rational one can overcome this difficulty by considering $(alpha z^n f(z))^q$ where $n=frac{p}{q}$. However in our example $n$ turns out to be $pi$ and by considering $g(z)=f(z)/z^{pi}$ we don't get holomorphic function. How to handle this case?










      share|cite|improve this question













      Let $f$ be a holomorphic function on the set $U={z in mathbb{C}: 1 leq |z| leq pi }$. Assume that $max_{|z|=1}|f(z)| leq 1$ and $max_{|z|=pi}|f(z)| leq pi^{pi}$.




      How to prove that $max_{|z|=e}|f(z)| leq e^{pi}$?




      Usually in such exercises one considers $g(z)$ of the form $alpha z^nf(z)$ where $alpha$ and $n$ are such that our assumptions on $f$ gives $|g(z)| leq 1$ on both circles being the boundary of $U$. However it is not guarantedd that $n$ would be integer thus in general we don't get a holomorphic function $g$ for which we could apply the maximum principle. If $n$ happens to be rational one can overcome this difficulty by considering $(alpha z^n f(z))^q$ where $n=frac{p}{q}$. However in our example $n$ turns out to be $pi$ and by considering $g(z)=f(z)/z^{pi}$ we don't get holomorphic function. How to handle this case?







      complex-analysis optimization holomorphic-functions maximum-principle






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      asked 4 hours ago









      truebaran

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          As you pointed out, to define the function $zmapsto z^{-pi}$ holomorphically, a branch cut is necessary, and it cannot be defined properly on the whole of the domain. Hence, to avoid this technical issue, theory of subharmonic functions is used. To be specific, instead of considering $zmapsto z^{-pi}$, one considers a subharmonic function
          $$
          u(z) =-pilog|z| + log|f(z)| in[-infty, infty)
          $$
          defined on $1leq |z|leq pi$. Then, one can show that maximum principle applies to $u$, and one gets $$-pilog|z|+log|f(z)|leq 0, quadforall z,$$ as a consequence.



          Note: A subharmonic function $u:Omega to [-infty,infty)$ is an upper-semicontinuous function such that $u(x)$ is less than or equal to the average of $u$ on $partial B(x,r)$ for every $r>0$ such that $overline{B(x,r)}subsetOmega$. We can see that $log|f|$ is subharmonic by Jensen's formula (and if $f$ does not vanish, it is harmonic.) By the definition of subharmonic function, if maximum of $u$ occurs at the interior point $x$, then $u$ is a constant on a component containing $x$.






          share|cite|improve this answer























          • Being the real part of the locally holomorphic function $U(z)=-pi log(z)+log f(z)$ it is harmonic. Then the maximum modulus principle can be seen from that $U$ is locally analytic so $u(z) = u(z_0)+ Re(C(z-z_0)^n)+O(|z-z_0|^{n+1})$ which doesn't have a local maximum at $z_0$
            – reuns
            3 hours ago













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          up vote
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          As you pointed out, to define the function $zmapsto z^{-pi}$ holomorphically, a branch cut is necessary, and it cannot be defined properly on the whole of the domain. Hence, to avoid this technical issue, theory of subharmonic functions is used. To be specific, instead of considering $zmapsto z^{-pi}$, one considers a subharmonic function
          $$
          u(z) =-pilog|z| + log|f(z)| in[-infty, infty)
          $$
          defined on $1leq |z|leq pi$. Then, one can show that maximum principle applies to $u$, and one gets $$-pilog|z|+log|f(z)|leq 0, quadforall z,$$ as a consequence.



          Note: A subharmonic function $u:Omega to [-infty,infty)$ is an upper-semicontinuous function such that $u(x)$ is less than or equal to the average of $u$ on $partial B(x,r)$ for every $r>0$ such that $overline{B(x,r)}subsetOmega$. We can see that $log|f|$ is subharmonic by Jensen's formula (and if $f$ does not vanish, it is harmonic.) By the definition of subharmonic function, if maximum of $u$ occurs at the interior point $x$, then $u$ is a constant on a component containing $x$.






          share|cite|improve this answer























          • Being the real part of the locally holomorphic function $U(z)=-pi log(z)+log f(z)$ it is harmonic. Then the maximum modulus principle can be seen from that $U$ is locally analytic so $u(z) = u(z_0)+ Re(C(z-z_0)^n)+O(|z-z_0|^{n+1})$ which doesn't have a local maximum at $z_0$
            – reuns
            3 hours ago

















          up vote
          3
          down vote













          As you pointed out, to define the function $zmapsto z^{-pi}$ holomorphically, a branch cut is necessary, and it cannot be defined properly on the whole of the domain. Hence, to avoid this technical issue, theory of subharmonic functions is used. To be specific, instead of considering $zmapsto z^{-pi}$, one considers a subharmonic function
          $$
          u(z) =-pilog|z| + log|f(z)| in[-infty, infty)
          $$
          defined on $1leq |z|leq pi$. Then, one can show that maximum principle applies to $u$, and one gets $$-pilog|z|+log|f(z)|leq 0, quadforall z,$$ as a consequence.



          Note: A subharmonic function $u:Omega to [-infty,infty)$ is an upper-semicontinuous function such that $u(x)$ is less than or equal to the average of $u$ on $partial B(x,r)$ for every $r>0$ such that $overline{B(x,r)}subsetOmega$. We can see that $log|f|$ is subharmonic by Jensen's formula (and if $f$ does not vanish, it is harmonic.) By the definition of subharmonic function, if maximum of $u$ occurs at the interior point $x$, then $u$ is a constant on a component containing $x$.






          share|cite|improve this answer























          • Being the real part of the locally holomorphic function $U(z)=-pi log(z)+log f(z)$ it is harmonic. Then the maximum modulus principle can be seen from that $U$ is locally analytic so $u(z) = u(z_0)+ Re(C(z-z_0)^n)+O(|z-z_0|^{n+1})$ which doesn't have a local maximum at $z_0$
            – reuns
            3 hours ago















          up vote
          3
          down vote










          up vote
          3
          down vote









          As you pointed out, to define the function $zmapsto z^{-pi}$ holomorphically, a branch cut is necessary, and it cannot be defined properly on the whole of the domain. Hence, to avoid this technical issue, theory of subharmonic functions is used. To be specific, instead of considering $zmapsto z^{-pi}$, one considers a subharmonic function
          $$
          u(z) =-pilog|z| + log|f(z)| in[-infty, infty)
          $$
          defined on $1leq |z|leq pi$. Then, one can show that maximum principle applies to $u$, and one gets $$-pilog|z|+log|f(z)|leq 0, quadforall z,$$ as a consequence.



          Note: A subharmonic function $u:Omega to [-infty,infty)$ is an upper-semicontinuous function such that $u(x)$ is less than or equal to the average of $u$ on $partial B(x,r)$ for every $r>0$ such that $overline{B(x,r)}subsetOmega$. We can see that $log|f|$ is subharmonic by Jensen's formula (and if $f$ does not vanish, it is harmonic.) By the definition of subharmonic function, if maximum of $u$ occurs at the interior point $x$, then $u$ is a constant on a component containing $x$.






          share|cite|improve this answer














          As you pointed out, to define the function $zmapsto z^{-pi}$ holomorphically, a branch cut is necessary, and it cannot be defined properly on the whole of the domain. Hence, to avoid this technical issue, theory of subharmonic functions is used. To be specific, instead of considering $zmapsto z^{-pi}$, one considers a subharmonic function
          $$
          u(z) =-pilog|z| + log|f(z)| in[-infty, infty)
          $$
          defined on $1leq |z|leq pi$. Then, one can show that maximum principle applies to $u$, and one gets $$-pilog|z|+log|f(z)|leq 0, quadforall z,$$ as a consequence.



          Note: A subharmonic function $u:Omega to [-infty,infty)$ is an upper-semicontinuous function such that $u(x)$ is less than or equal to the average of $u$ on $partial B(x,r)$ for every $r>0$ such that $overline{B(x,r)}subsetOmega$. We can see that $log|f|$ is subharmonic by Jensen's formula (and if $f$ does not vanish, it is harmonic.) By the definition of subharmonic function, if maximum of $u$ occurs at the interior point $x$, then $u$ is a constant on a component containing $x$.







          share|cite|improve this answer














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          edited 2 hours ago

























          answered 3 hours ago









          Song

          1,57012




          1,57012












          • Being the real part of the locally holomorphic function $U(z)=-pi log(z)+log f(z)$ it is harmonic. Then the maximum modulus principle can be seen from that $U$ is locally analytic so $u(z) = u(z_0)+ Re(C(z-z_0)^n)+O(|z-z_0|^{n+1})$ which doesn't have a local maximum at $z_0$
            – reuns
            3 hours ago




















          • Being the real part of the locally holomorphic function $U(z)=-pi log(z)+log f(z)$ it is harmonic. Then the maximum modulus principle can be seen from that $U$ is locally analytic so $u(z) = u(z_0)+ Re(C(z-z_0)^n)+O(|z-z_0|^{n+1})$ which doesn't have a local maximum at $z_0$
            – reuns
            3 hours ago


















          Being the real part of the locally holomorphic function $U(z)=-pi log(z)+log f(z)$ it is harmonic. Then the maximum modulus principle can be seen from that $U$ is locally analytic so $u(z) = u(z_0)+ Re(C(z-z_0)^n)+O(|z-z_0|^{n+1})$ which doesn't have a local maximum at $z_0$
          – reuns
          3 hours ago






          Being the real part of the locally holomorphic function $U(z)=-pi log(z)+log f(z)$ it is harmonic. Then the maximum modulus principle can be seen from that $U$ is locally analytic so $u(z) = u(z_0)+ Re(C(z-z_0)^n)+O(|z-z_0|^{n+1})$ which doesn't have a local maximum at $z_0$
          – reuns
          3 hours ago




















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