Function holomorphic in the units disk with different bound












2












$begingroup$


Suppose $f$ is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Moreover suppose that for $|z|=1$ we have:



$Re(z)leq0Rightarrow |f(z)|leq 1$



$Re(z)>0Rightarrow |f(z)|leq 2$



and I Need to prove $|f(0)|leq sqrt{2}$ I know from the maximum modulus principle we have that:



$$1leq max_{|z|=1}|f|=max_{bar{D}(0,1)}|f|leq 2$$



but I can't really see where the square root come from so I cannot go any further.






complex-analysis holomorphic-functions maximum-principle







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  • $begingroup$
    out of curiosity, is the bound tight?
    $endgroup$
    – AccidentalFourierTransform
    Dec 10 '18 at 18:11










  • $begingroup$
    @AccidentalFourierTransform As far as the text of my exercises says no
    $endgroup$
    – Renato Faraone
    Dec 11 '18 at 10:04
















2












$begingroup$


Suppose $f$ is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Moreover suppose that for $|z|=1$ we have:



$Re(z)leq0Rightarrow |f(z)|leq 1$



$Re(z)>0Rightarrow |f(z)|leq 2$



and I Need to prove $|f(0)|leq sqrt{2}$ I know from the maximum modulus principle we have that:



$$1leq max_{|z|=1}|f|=max_{bar{D}(0,1)}|f|leq 2$$



but I can't really see where the square root come from so I cannot go any further.






complex-analysis holomorphic-functions maximum-principle







share|cite|improve this question











$endgroup$












  • $begingroup$
    out of curiosity, is the bound tight?
    $endgroup$
    – AccidentalFourierTransform
    Dec 10 '18 at 18:11










  • $begingroup$
    @AccidentalFourierTransform As far as the text of my exercises says no
    $endgroup$
    – Renato Faraone
    Dec 11 '18 at 10:04














2












2








2





$begingroup$


Suppose $f$ is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Moreover suppose that for $|z|=1$ we have:



$Re(z)leq0Rightarrow |f(z)|leq 1$



$Re(z)>0Rightarrow |f(z)|leq 2$



and I Need to prove $|f(0)|leq sqrt{2}$ I know from the maximum modulus principle we have that:



$$1leq max_{|z|=1}|f|=max_{bar{D}(0,1)}|f|leq 2$$



but I can't really see where the square root come from so I cannot go any further.






complex-analysis holomorphic-functions maximum-principle







share|cite|improve this question











$endgroup$




Suppose $f$ is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Moreover suppose that for $|z|=1$ we have:



$Re(z)leq0Rightarrow |f(z)|leq 1$



$Re(z)>0Rightarrow |f(z)|leq 2$



and I Need to prove $|f(0)|leq sqrt{2}$ I know from the maximum modulus principle we have that:



$$1leq max_{|z|=1}|f|=max_{bar{D}(0,1)}|f|leq 2$$



but I can't really see where the square root come from so I cannot go any further.






complex-analysis holomorphic-functions maximum-principle




complex-analysis holomorphic-functions maximum-principle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 10:03







Renato Faraone

















asked Dec 10 '18 at 9:54









Renato FaraoneRenato Faraone

2,34111627




2,34111627












  • $begingroup$
    out of curiosity, is the bound tight?
    $endgroup$
    – AccidentalFourierTransform
    Dec 10 '18 at 18:11










  • $begingroup$
    @AccidentalFourierTransform As far as the text of my exercises says no
    $endgroup$
    – Renato Faraone
    Dec 11 '18 at 10:04


















  • $begingroup$
    out of curiosity, is the bound tight?
    $endgroup$
    – AccidentalFourierTransform
    Dec 10 '18 at 18:11










  • $begingroup$
    @AccidentalFourierTransform As far as the text of my exercises says no
    $endgroup$
    – Renato Faraone
    Dec 11 '18 at 10:04
















$begingroup$
out of curiosity, is the bound tight?
$endgroup$
– AccidentalFourierTransform
Dec 10 '18 at 18:11




$begingroup$
out of curiosity, is the bound tight?
$endgroup$
– AccidentalFourierTransform
Dec 10 '18 at 18:11












$begingroup$
@AccidentalFourierTransform As far as the text of my exercises says no
$endgroup$
– Renato Faraone
Dec 11 '18 at 10:04




$begingroup$
@AccidentalFourierTransform As far as the text of my exercises says no
$endgroup$
– Renato Faraone
Dec 11 '18 at 10:04










2 Answers
2






active

oldest

votes


















6












$begingroup$

First try. By Cauchy integral
$$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
=frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
Hence
$$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
But unfortunately $sqrt{2}<3/2$.



Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
$$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
Now apply the Cauchy integral to $F$:
$$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      First try. By Cauchy integral
      $$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
      =frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
      Hence
      $$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
      But unfortunately $sqrt{2}<3/2$.



      Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
      $$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
      Now apply the Cauchy integral to $F$:
      $$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$






      share|cite|improve this answer











      $endgroup$


















        6












        $begingroup$

        First try. By Cauchy integral
        $$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
        =frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
        Hence
        $$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
        But unfortunately $sqrt{2}<3/2$.



        Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
        $$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
        Now apply the Cauchy integral to $F$:
        $$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$






        share|cite|improve this answer











        $endgroup$
















          6












          6








          6





          $begingroup$

          First try. By Cauchy integral
          $$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
          =frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
          Hence
          $$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
          But unfortunately $sqrt{2}<3/2$.



          Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
          $$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
          Now apply the Cauchy integral to $F$:
          $$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$






          share|cite|improve this answer











          $endgroup$



          First try. By Cauchy integral
          $$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
          =frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
          Hence
          $$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
          But unfortunately $sqrt{2}<3/2$.



          Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
          $$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
          Now apply the Cauchy integral to $F$:
          $$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 10:20

























          answered Dec 10 '18 at 10:00









          Robert ZRobert Z

          97.7k1066137




          97.7k1066137























              4












              $begingroup$

              For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.






                  share|cite|improve this answer









                  $endgroup$



                  For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 10:44









                  user10354138user10354138

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