Function holomorphic in the units disk with different bound












2














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










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
















2














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
























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










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














2












2








2







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















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













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edited Dec 10 at 10:03

























asked Dec 10 at 9:54









Renato Faraone

2,32911627




2,32911627












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










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


















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










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
















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




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












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




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










2 Answers
2






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6














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































    4














    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














      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




























        6














        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


























          6












          6








          6






          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














          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 at 10:20

























          answered Dec 10 at 10:00









          Robert Z

          93.2k1061132




          93.2k1061132























              4














              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


























                4














                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
























                  4












                  4








                  4






                  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












                  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 at 10:44









                  user10354138

                  7,3942824




                  7,3942824






























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