Proving an integral formula containing the Poisson kernel












2














Specifically, the question is as follows:




Let $mathbb{D}$ be the open unit disk in $mathbb{C}$ and suppose that $f:bar{mathbb{D}}rightarrowmathbb{C}$ is a continuous function such that both $Re f$ and $Im f$ are harmonic.



(a) Show that
$$f(re^{itheta})=frac{1}{2pi}int_{-pi}^pi f(e^{it})P_r(theta-t)dt$$
for all $re^{itheta}$ in $mathbb{D}$.



(b) Show that $f$ is holomorphic in $mathbb{D}$ if and only if
$$int_{-pi}^pi f(e^{it})e^{int}dt=0$$
for all $ngeq1$.




I believe (a) is a direct consequence of Poisson's representation theorem for harmonic functions, stated as follows:



Let $mathbb{D}$ be the open unit disk and $f:partialmathbb{D}rightarrowmathbb{R}$ be continuous. Then there exists a continuous function $u:bar{mathbb{D}}rightarrowmathbb{R}$ such that $u(z)=f(z)$ if $zinpartialmathbb{D}$, $u$ is harmonic on $mathbb{D}$, and $u(re^{itheta})=frac{1}{2pi}int_{-pi}^pi P_r(theta-t)f(e^{it})dt$ and $u$ is unique.



It seems to me that this can just be applied to the real and imaginary parts of the function $f$ from the question to derive the desired result, but maybe I'm making a careless mistake?



For (b), it's not obvious to me how part (a) assists in the solution. This is a homework question, so I am looking for assistance with understanding it, but not a full solution. Thanks.






integration complex-analysis harmonic-functions







share|cite|improve this question



























    2














    Specifically, the question is as follows:




    Let $mathbb{D}$ be the open unit disk in $mathbb{C}$ and suppose that $f:bar{mathbb{D}}rightarrowmathbb{C}$ is a continuous function such that both $Re f$ and $Im f$ are harmonic.



    (a) Show that
    $$f(re^{itheta})=frac{1}{2pi}int_{-pi}^pi f(e^{it})P_r(theta-t)dt$$
    for all $re^{itheta}$ in $mathbb{D}$.



    (b) Show that $f$ is holomorphic in $mathbb{D}$ if and only if
    $$int_{-pi}^pi f(e^{it})e^{int}dt=0$$
    for all $ngeq1$.




    I believe (a) is a direct consequence of Poisson's representation theorem for harmonic functions, stated as follows:



    Let $mathbb{D}$ be the open unit disk and $f:partialmathbb{D}rightarrowmathbb{R}$ be continuous. Then there exists a continuous function $u:bar{mathbb{D}}rightarrowmathbb{R}$ such that $u(z)=f(z)$ if $zinpartialmathbb{D}$, $u$ is harmonic on $mathbb{D}$, and $u(re^{itheta})=frac{1}{2pi}int_{-pi}^pi P_r(theta-t)f(e^{it})dt$ and $u$ is unique.



    It seems to me that this can just be applied to the real and imaginary parts of the function $f$ from the question to derive the desired result, but maybe I'm making a careless mistake?



    For (b), it's not obvious to me how part (a) assists in the solution. This is a homework question, so I am looking for assistance with understanding it, but not a full solution. Thanks.






    integration complex-analysis harmonic-functions







    share|cite|improve this question

























      2












      2








      2


      1





      Specifically, the question is as follows:




      Let $mathbb{D}$ be the open unit disk in $mathbb{C}$ and suppose that $f:bar{mathbb{D}}rightarrowmathbb{C}$ is a continuous function such that both $Re f$ and $Im f$ are harmonic.



      (a) Show that
      $$f(re^{itheta})=frac{1}{2pi}int_{-pi}^pi f(e^{it})P_r(theta-t)dt$$
      for all $re^{itheta}$ in $mathbb{D}$.



      (b) Show that $f$ is holomorphic in $mathbb{D}$ if and only if
      $$int_{-pi}^pi f(e^{it})e^{int}dt=0$$
      for all $ngeq1$.




      I believe (a) is a direct consequence of Poisson's representation theorem for harmonic functions, stated as follows:



      Let $mathbb{D}$ be the open unit disk and $f:partialmathbb{D}rightarrowmathbb{R}$ be continuous. Then there exists a continuous function $u:bar{mathbb{D}}rightarrowmathbb{R}$ such that $u(z)=f(z)$ if $zinpartialmathbb{D}$, $u$ is harmonic on $mathbb{D}$, and $u(re^{itheta})=frac{1}{2pi}int_{-pi}^pi P_r(theta-t)f(e^{it})dt$ and $u$ is unique.



      It seems to me that this can just be applied to the real and imaginary parts of the function $f$ from the question to derive the desired result, but maybe I'm making a careless mistake?



      For (b), it's not obvious to me how part (a) assists in the solution. This is a homework question, so I am looking for assistance with understanding it, but not a full solution. Thanks.






      integration complex-analysis harmonic-functions







      share|cite|improve this question













      Specifically, the question is as follows:




      Let $mathbb{D}$ be the open unit disk in $mathbb{C}$ and suppose that $f:bar{mathbb{D}}rightarrowmathbb{C}$ is a continuous function such that both $Re f$ and $Im f$ are harmonic.



      (a) Show that
      $$f(re^{itheta})=frac{1}{2pi}int_{-pi}^pi f(e^{it})P_r(theta-t)dt$$
      for all $re^{itheta}$ in $mathbb{D}$.



      (b) Show that $f$ is holomorphic in $mathbb{D}$ if and only if
      $$int_{-pi}^pi f(e^{it})e^{int}dt=0$$
      for all $ngeq1$.




      I believe (a) is a direct consequence of Poisson's representation theorem for harmonic functions, stated as follows:



      Let $mathbb{D}$ be the open unit disk and $f:partialmathbb{D}rightarrowmathbb{R}$ be continuous. Then there exists a continuous function $u:bar{mathbb{D}}rightarrowmathbb{R}$ such that $u(z)=f(z)$ if $zinpartialmathbb{D}$, $u$ is harmonic on $mathbb{D}$, and $u(re^{itheta})=frac{1}{2pi}int_{-pi}^pi P_r(theta-t)f(e^{it})dt$ and $u$ is unique.



      It seems to me that this can just be applied to the real and imaginary parts of the function $f$ from the question to derive the desired result, but maybe I'm making a careless mistake?



      For (b), it's not obvious to me how part (a) assists in the solution. This is a homework question, so I am looking for assistance with understanding it, but not a full solution. Thanks.






      integration complex-analysis harmonic-functions




      integration complex-analysis harmonic-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 24 at 22:20









      Atsina

      791116




      791116






















          1 Answer
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          0














          Poisson's representation formula here assumes only a function on the boundary. Also, one does not know the relationship between $u:bar{mathbb{D}}rightarrowmathbb{R}$ from the theorem and $f:bar{mathbb{D}}rightarrowmathbb{R}$ from the question (unless the proof of the theorem is constructive, then maybe you can deduce it), so you cannot use it like this here.



          Hint for (a):




          Prove it first for holomorphic functions $g$, then use that $Re f$ is the real part of some holomorphic function $g$. Same for $Im f$.




          Hint for (b):




          Letting $rto 1$, we have for $tneq 0$ that $P_rto 0$. Now use the identity theorem.







          share|cite|improve this answer





















          • Thank you for the advice. So for (a), using the representation formula, since $mathbb{D}$ is simply connected, $u=Re(g)$ where $gin H(mathbb{D})$, so $g(z)=frac{1}{2pi}int_{-pi}^pifrac{e^{it}+z}{e^{it}-z}f(e^{it})dt$ (where $f$ is from the representation formula). But this isn't quite the desired result. Should I not be trying to use the representation formula at all?
            – Atsina
            Nov 25 at 0:57










          • For a), I think all you need to do is show that the result holds for trigonometric polynomials, then you can use a density argument plus the DCT.
            – Matematleta
            Nov 25 at 3:26












          • @Atsina To use my hint you don't need the representation formula, you only need to swap an integral and two series due to uniform convergence.
            – The Phenotype
            Nov 25 at 9:58











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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

          votes









          0














          Poisson's representation formula here assumes only a function on the boundary. Also, one does not know the relationship between $u:bar{mathbb{D}}rightarrowmathbb{R}$ from the theorem and $f:bar{mathbb{D}}rightarrowmathbb{R}$ from the question (unless the proof of the theorem is constructive, then maybe you can deduce it), so you cannot use it like this here.



          Hint for (a):




          Prove it first for holomorphic functions $g$, then use that $Re f$ is the real part of some holomorphic function $g$. Same for $Im f$.




          Hint for (b):




          Letting $rto 1$, we have for $tneq 0$ that $P_rto 0$. Now use the identity theorem.







          share|cite|improve this answer





















          • Thank you for the advice. So for (a), using the representation formula, since $mathbb{D}$ is simply connected, $u=Re(g)$ where $gin H(mathbb{D})$, so $g(z)=frac{1}{2pi}int_{-pi}^pifrac{e^{it}+z}{e^{it}-z}f(e^{it})dt$ (where $f$ is from the representation formula). But this isn't quite the desired result. Should I not be trying to use the representation formula at all?
            – Atsina
            Nov 25 at 0:57










          • For a), I think all you need to do is show that the result holds for trigonometric polynomials, then you can use a density argument plus the DCT.
            – Matematleta
            Nov 25 at 3:26












          • @Atsina To use my hint you don't need the representation formula, you only need to swap an integral and two series due to uniform convergence.
            – The Phenotype
            Nov 25 at 9:58
















          0














          Poisson's representation formula here assumes only a function on the boundary. Also, one does not know the relationship between $u:bar{mathbb{D}}rightarrowmathbb{R}$ from the theorem and $f:bar{mathbb{D}}rightarrowmathbb{R}$ from the question (unless the proof of the theorem is constructive, then maybe you can deduce it), so you cannot use it like this here.



          Hint for (a):




          Prove it first for holomorphic functions $g$, then use that $Re f$ is the real part of some holomorphic function $g$. Same for $Im f$.




          Hint for (b):




          Letting $rto 1$, we have for $tneq 0$ that $P_rto 0$. Now use the identity theorem.







          share|cite|improve this answer





















          • Thank you for the advice. So for (a), using the representation formula, since $mathbb{D}$ is simply connected, $u=Re(g)$ where $gin H(mathbb{D})$, so $g(z)=frac{1}{2pi}int_{-pi}^pifrac{e^{it}+z}{e^{it}-z}f(e^{it})dt$ (where $f$ is from the representation formula). But this isn't quite the desired result. Should I not be trying to use the representation formula at all?
            – Atsina
            Nov 25 at 0:57










          • For a), I think all you need to do is show that the result holds for trigonometric polynomials, then you can use a density argument plus the DCT.
            – Matematleta
            Nov 25 at 3:26












          • @Atsina To use my hint you don't need the representation formula, you only need to swap an integral and two series due to uniform convergence.
            – The Phenotype
            Nov 25 at 9:58














          0












          0








          0






          Poisson's representation formula here assumes only a function on the boundary. Also, one does not know the relationship between $u:bar{mathbb{D}}rightarrowmathbb{R}$ from the theorem and $f:bar{mathbb{D}}rightarrowmathbb{R}$ from the question (unless the proof of the theorem is constructive, then maybe you can deduce it), so you cannot use it like this here.



          Hint for (a):




          Prove it first for holomorphic functions $g$, then use that $Re f$ is the real part of some holomorphic function $g$. Same for $Im f$.




          Hint for (b):




          Letting $rto 1$, we have for $tneq 0$ that $P_rto 0$. Now use the identity theorem.







          share|cite|improve this answer












          Poisson's representation formula here assumes only a function on the boundary. Also, one does not know the relationship between $u:bar{mathbb{D}}rightarrowmathbb{R}$ from the theorem and $f:bar{mathbb{D}}rightarrowmathbb{R}$ from the question (unless the proof of the theorem is constructive, then maybe you can deduce it), so you cannot use it like this here.



          Hint for (a):




          Prove it first for holomorphic functions $g$, then use that $Re f$ is the real part of some holomorphic function $g$. Same for $Im f$.




          Hint for (b):




          Letting $rto 1$, we have for $tneq 0$ that $P_rto 0$. Now use the identity theorem.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 22:48









          The Phenotype

          4,77591732




          4,77591732












          • Thank you for the advice. So for (a), using the representation formula, since $mathbb{D}$ is simply connected, $u=Re(g)$ where $gin H(mathbb{D})$, so $g(z)=frac{1}{2pi}int_{-pi}^pifrac{e^{it}+z}{e^{it}-z}f(e^{it})dt$ (where $f$ is from the representation formula). But this isn't quite the desired result. Should I not be trying to use the representation formula at all?
            – Atsina
            Nov 25 at 0:57










          • For a), I think all you need to do is show that the result holds for trigonometric polynomials, then you can use a density argument plus the DCT.
            – Matematleta
            Nov 25 at 3:26












          • @Atsina To use my hint you don't need the representation formula, you only need to swap an integral and two series due to uniform convergence.
            – The Phenotype
            Nov 25 at 9:58


















          • Thank you for the advice. So for (a), using the representation formula, since $mathbb{D}$ is simply connected, $u=Re(g)$ where $gin H(mathbb{D})$, so $g(z)=frac{1}{2pi}int_{-pi}^pifrac{e^{it}+z}{e^{it}-z}f(e^{it})dt$ (where $f$ is from the representation formula). But this isn't quite the desired result. Should I not be trying to use the representation formula at all?
            – Atsina
            Nov 25 at 0:57










          • For a), I think all you need to do is show that the result holds for trigonometric polynomials, then you can use a density argument plus the DCT.
            – Matematleta
            Nov 25 at 3:26












          • @Atsina To use my hint you don't need the representation formula, you only need to swap an integral and two series due to uniform convergence.
            – The Phenotype
            Nov 25 at 9:58
















          Thank you for the advice. So for (a), using the representation formula, since $mathbb{D}$ is simply connected, $u=Re(g)$ where $gin H(mathbb{D})$, so $g(z)=frac{1}{2pi}int_{-pi}^pifrac{e^{it}+z}{e^{it}-z}f(e^{it})dt$ (where $f$ is from the representation formula). But this isn't quite the desired result. Should I not be trying to use the representation formula at all?
          – Atsina
          Nov 25 at 0:57




          Thank you for the advice. So for (a), using the representation formula, since $mathbb{D}$ is simply connected, $u=Re(g)$ where $gin H(mathbb{D})$, so $g(z)=frac{1}{2pi}int_{-pi}^pifrac{e^{it}+z}{e^{it}-z}f(e^{it})dt$ (where $f$ is from the representation formula). But this isn't quite the desired result. Should I not be trying to use the representation formula at all?
          – Atsina
          Nov 25 at 0:57












          For a), I think all you need to do is show that the result holds for trigonometric polynomials, then you can use a density argument plus the DCT.
          – Matematleta
          Nov 25 at 3:26






          For a), I think all you need to do is show that the result holds for trigonometric polynomials, then you can use a density argument plus the DCT.
          – Matematleta
          Nov 25 at 3:26














          @Atsina To use my hint you don't need the representation formula, you only need to swap an integral and two series due to uniform convergence.
          – The Phenotype
          Nov 25 at 9:58




          @Atsina To use my hint you don't need the representation formula, you only need to swap an integral and two series due to uniform convergence.
          – The Phenotype
          Nov 25 at 9:58


















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