Bounded Harmonic Functions on the Disk












6












$begingroup$


Denote by $mathbb{D}$ the open unit disk in $mathbb{R}^2$.
Is it possible to find a bounded harmonic function $u : mathbb{D} to mathbb{R}$ that is not uniformly continuous?



I tried using functions that oscillate near $partial mathbb{D}$ but was unable to get anything substantial.






real-analysis complex-analysis pde harmonic-functions uniform-continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    But aren't bounded harmonic functions constant ?
    $endgroup$
    – Yadati Kiran
    Dec 5 '18 at 6:07








  • 2




    $begingroup$
    Wouldn't you need $u$ to be harmonic in all $mathbb{R}^2$ for this?
    $endgroup$
    – ImNotThereRightNow_
    Dec 5 '18 at 6:13










  • $begingroup$
    For one we can use the inclusion map to make it harmonic on all of $mathbb{R}^2$. I am not quite sure whether all the bounded harmonic functions on open disk are constant.
    $endgroup$
    – Yadati Kiran
    Dec 5 '18 at 6:26








  • 4




    $begingroup$
    @YadatiKiran: Bounded harmonic functions in $Bbb C$ are constant. $u(x, y) = x$ is harmonic and bounded in $Bbb D$, but not constant.
    $endgroup$
    – Martin R
    Dec 5 '18 at 6:28


















6












$begingroup$


Denote by $mathbb{D}$ the open unit disk in $mathbb{R}^2$.
Is it possible to find a bounded harmonic function $u : mathbb{D} to mathbb{R}$ that is not uniformly continuous?



I tried using functions that oscillate near $partial mathbb{D}$ but was unable to get anything substantial.






real-analysis complex-analysis pde harmonic-functions uniform-continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    But aren't bounded harmonic functions constant ?
    $endgroup$
    – Yadati Kiran
    Dec 5 '18 at 6:07








  • 2




    $begingroup$
    Wouldn't you need $u$ to be harmonic in all $mathbb{R}^2$ for this?
    $endgroup$
    – ImNotThereRightNow_
    Dec 5 '18 at 6:13










  • $begingroup$
    For one we can use the inclusion map to make it harmonic on all of $mathbb{R}^2$. I am not quite sure whether all the bounded harmonic functions on open disk are constant.
    $endgroup$
    – Yadati Kiran
    Dec 5 '18 at 6:26








  • 4




    $begingroup$
    @YadatiKiran: Bounded harmonic functions in $Bbb C$ are constant. $u(x, y) = x$ is harmonic and bounded in $Bbb D$, but not constant.
    $endgroup$
    – Martin R
    Dec 5 '18 at 6:28
















6












6








6


3



$begingroup$


Denote by $mathbb{D}$ the open unit disk in $mathbb{R}^2$.
Is it possible to find a bounded harmonic function $u : mathbb{D} to mathbb{R}$ that is not uniformly continuous?



I tried using functions that oscillate near $partial mathbb{D}$ but was unable to get anything substantial.






real-analysis complex-analysis pde harmonic-functions uniform-continuity







share|cite|improve this question











$endgroup$




Denote by $mathbb{D}$ the open unit disk in $mathbb{R}^2$.
Is it possible to find a bounded harmonic function $u : mathbb{D} to mathbb{R}$ that is not uniformly continuous?



I tried using functions that oscillate near $partial mathbb{D}$ but was unable to get anything substantial.






real-analysis complex-analysis pde harmonic-functions uniform-continuity




real-analysis complex-analysis pde harmonic-functions uniform-continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 17:54









zhw.

71.9k43075




71.9k43075










asked Dec 5 '18 at 5:57









ImNotThereRightNow_ImNotThereRightNow_

998




998












  • $begingroup$
    But aren't bounded harmonic functions constant ?
    $endgroup$
    – Yadati Kiran
    Dec 5 '18 at 6:07








  • 2




    $begingroup$
    Wouldn't you need $u$ to be harmonic in all $mathbb{R}^2$ for this?
    $endgroup$
    – ImNotThereRightNow_
    Dec 5 '18 at 6:13










  • $begingroup$
    For one we can use the inclusion map to make it harmonic on all of $mathbb{R}^2$. I am not quite sure whether all the bounded harmonic functions on open disk are constant.
    $endgroup$
    – Yadati Kiran
    Dec 5 '18 at 6:26








  • 4




    $begingroup$
    @YadatiKiran: Bounded harmonic functions in $Bbb C$ are constant. $u(x, y) = x$ is harmonic and bounded in $Bbb D$, but not constant.
    $endgroup$
    – Martin R
    Dec 5 '18 at 6:28




















  • $begingroup$
    But aren't bounded harmonic functions constant ?
    $endgroup$
    – Yadati Kiran
    Dec 5 '18 at 6:07








  • 2




    $begingroup$
    Wouldn't you need $u$ to be harmonic in all $mathbb{R}^2$ for this?
    $endgroup$
    – ImNotThereRightNow_
    Dec 5 '18 at 6:13










  • $begingroup$
    For one we can use the inclusion map to make it harmonic on all of $mathbb{R}^2$. I am not quite sure whether all the bounded harmonic functions on open disk are constant.
    $endgroup$
    – Yadati Kiran
    Dec 5 '18 at 6:26








  • 4




    $begingroup$
    @YadatiKiran: Bounded harmonic functions in $Bbb C$ are constant. $u(x, y) = x$ is harmonic and bounded in $Bbb D$, but not constant.
    $endgroup$
    – Martin R
    Dec 5 '18 at 6:28


















$begingroup$
But aren't bounded harmonic functions constant ?
$endgroup$
– Yadati Kiran
Dec 5 '18 at 6:07






$begingroup$
But aren't bounded harmonic functions constant ?
$endgroup$
– Yadati Kiran
Dec 5 '18 at 6:07






2




2




$begingroup$
Wouldn't you need $u$ to be harmonic in all $mathbb{R}^2$ for this?
$endgroup$
– ImNotThereRightNow_
Dec 5 '18 at 6:13




$begingroup$
Wouldn't you need $u$ to be harmonic in all $mathbb{R}^2$ for this?
$endgroup$
– ImNotThereRightNow_
Dec 5 '18 at 6:13












$begingroup$
For one we can use the inclusion map to make it harmonic on all of $mathbb{R}^2$. I am not quite sure whether all the bounded harmonic functions on open disk are constant.
$endgroup$
– Yadati Kiran
Dec 5 '18 at 6:26






$begingroup$
For one we can use the inclusion map to make it harmonic on all of $mathbb{R}^2$. I am not quite sure whether all the bounded harmonic functions on open disk are constant.
$endgroup$
– Yadati Kiran
Dec 5 '18 at 6:26






4




4




$begingroup$
@YadatiKiran: Bounded harmonic functions in $Bbb C$ are constant. $u(x, y) = x$ is harmonic and bounded in $Bbb D$, but not constant.
$endgroup$
– Martin R
Dec 5 '18 at 6:28






$begingroup$
@YadatiKiran: Bounded harmonic functions in $Bbb C$ are constant. $u(x, y) = x$ is harmonic and bounded in $Bbb D$, but not constant.
$endgroup$
– Martin R
Dec 5 '18 at 6:28












3 Answers
3






active

oldest

votes


















6





+50







$begingroup$

For any bounded measurable function $fin L^infty(S^1)$, the Poisson integral $P[f]$ gives a bounded harmonic functions in $B_1$. This function is continuous (equivalently, uniformly continuous) if and only if $f$ itself is continuous.



Here you can find what the Poisson kernel is and how it's used to build harmonic functions in the ball.



Said otherwise, $fmapsto P[f]$ is a linear isometry of $L^p(S^1)$ onto $h^p(B_1)$. We used the case $p=infty$, where $h^infty(B_1)$ stands for the the bounded harmonic functions in the ball. You can read more about it here.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If $f$ is discontinuous on $S^1$, then $P[f]$ is discontinuous when viewed as a function on $bar{mathbb{D}}$. But $P[f]$ is still harmonic and continuous in $mathbb{D}$? I'm confused as to how I can conclude from this that $P[f]$ is not uniformly continuous on $mathbb{D}$.
    $endgroup$
    – ImNotThereRightNow_
    Dec 7 '18 at 18:00






  • 1




    $begingroup$
    If $P[f]$ were uniformly continuous in $mathbb D$, it would extend continuously to the closure, but it can't, because its extension is discontinuous on $S^1$
    $endgroup$
    – Federico
    Dec 7 '18 at 18:03






  • 1




    $begingroup$
    Yes you are right in saying that $P[f]$ is discontinuous in $bar{mathbb D}$ and continuous in $mathbb D$. The point is that it cannot be uniformly continuous in $mathbb D$, otherwise it would be in $bar{mathbb D}$ too.
    $endgroup$
    – Federico
    Dec 7 '18 at 18:05












  • $begingroup$
    Thanks, I had forgotten about this property.
    $endgroup$
    – ImNotThereRightNow_
    Dec 7 '18 at 18:07






  • 1




    $begingroup$
    I should probably have written it better :)
    $endgroup$
    – Federico
    Dec 7 '18 at 18:08



















1












$begingroup$

Let $log$ denote the principal value $log.$ Then $log(1+z)$ is holomorphic in $mathbb D.$ Hence its imaginary part, $u(z)=arg (1+z),$ is harmonic in $mathbb D.$ We have $|u|<pi/2$ in the disc, so $u$ is bounded there. For small $r>0,$ $-1+re^{ipi/4}in mathbb D.$ For such $r,$



$$u(-1+re^{ipi/4})- u(-1+r) = pi/4-0 = pi/4.$$



But $(-1+re^{ipi/4})-(-1+r) to 0$ as $rto 0.$ This shows $u$ cannot be uniformly continuous in $mathbb D.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Are you considering the principal value $log$ where $thetainleft(dfrac{-pi}{2},dfrac{pi}{2}right]$ ?
    $endgroup$
    – Yadati Kiran
    Dec 10 '18 at 17:19










  • $begingroup$
    @YadatiKiran Yes, but that should be $(-pi,pi).$
    $endgroup$
    – zhw.
    Dec 10 '18 at 17:25










  • $begingroup$
    If its $(-pi,pi)$ then how do you say $|u|<pi/2$ in the disc? Am I missing something here?
    $endgroup$
    – Yadati Kiran
    Dec 10 '18 at 17:29










  • $begingroup$
    @YadatiKiran Because $1+z$ is in the right half plane
    $endgroup$
    – zhw.
    Dec 10 '18 at 17:41










  • $begingroup$
    Oh Yes ! I get it.
    $endgroup$
    – Yadati Kiran
    Dec 10 '18 at 17:43



















0












$begingroup$

Start with
$$
phi(z)=frac{1+z}{1-z},;;; zinmathbb{C},; |z| < 1.
$$
This function is holomorphic with
begin{align}
Re phi(z) &= Refrac{1+z}{1-z}frac{1-overline{z}}{1-overline{z}} \
&=Refrac{1+z-overline{z}-|z|^2}{|1-z|^2} \
&= frac{1-|z|^2}{|1-z|^2} > 0.
end{align}
$psi(z)=e^{-phi(z)}$ is bounded and holomorphic in $|z| < 1$, but $psi$ is not uniformly continuous in the open disk.






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6





    +50







    $begingroup$

    For any bounded measurable function $fin L^infty(S^1)$, the Poisson integral $P[f]$ gives a bounded harmonic functions in $B_1$. This function is continuous (equivalently, uniformly continuous) if and only if $f$ itself is continuous.



    Here you can find what the Poisson kernel is and how it's used to build harmonic functions in the ball.



    Said otherwise, $fmapsto P[f]$ is a linear isometry of $L^p(S^1)$ onto $h^p(B_1)$. We used the case $p=infty$, where $h^infty(B_1)$ stands for the the bounded harmonic functions in the ball. You can read more about it here.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      If $f$ is discontinuous on $S^1$, then $P[f]$ is discontinuous when viewed as a function on $bar{mathbb{D}}$. But $P[f]$ is still harmonic and continuous in $mathbb{D}$? I'm confused as to how I can conclude from this that $P[f]$ is not uniformly continuous on $mathbb{D}$.
      $endgroup$
      – ImNotThereRightNow_
      Dec 7 '18 at 18:00






    • 1




      $begingroup$
      If $P[f]$ were uniformly continuous in $mathbb D$, it would extend continuously to the closure, but it can't, because its extension is discontinuous on $S^1$
      $endgroup$
      – Federico
      Dec 7 '18 at 18:03






    • 1




      $begingroup$
      Yes you are right in saying that $P[f]$ is discontinuous in $bar{mathbb D}$ and continuous in $mathbb D$. The point is that it cannot be uniformly continuous in $mathbb D$, otherwise it would be in $bar{mathbb D}$ too.
      $endgroup$
      – Federico
      Dec 7 '18 at 18:05












    • $begingroup$
      Thanks, I had forgotten about this property.
      $endgroup$
      – ImNotThereRightNow_
      Dec 7 '18 at 18:07






    • 1




      $begingroup$
      I should probably have written it better :)
      $endgroup$
      – Federico
      Dec 7 '18 at 18:08
















    6





    +50







    $begingroup$

    For any bounded measurable function $fin L^infty(S^1)$, the Poisson integral $P[f]$ gives a bounded harmonic functions in $B_1$. This function is continuous (equivalently, uniformly continuous) if and only if $f$ itself is continuous.



    Here you can find what the Poisson kernel is and how it's used to build harmonic functions in the ball.



    Said otherwise, $fmapsto P[f]$ is a linear isometry of $L^p(S^1)$ onto $h^p(B_1)$. We used the case $p=infty$, where $h^infty(B_1)$ stands for the the bounded harmonic functions in the ball. You can read more about it here.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      If $f$ is discontinuous on $S^1$, then $P[f]$ is discontinuous when viewed as a function on $bar{mathbb{D}}$. But $P[f]$ is still harmonic and continuous in $mathbb{D}$? I'm confused as to how I can conclude from this that $P[f]$ is not uniformly continuous on $mathbb{D}$.
      $endgroup$
      – ImNotThereRightNow_
      Dec 7 '18 at 18:00






    • 1




      $begingroup$
      If $P[f]$ were uniformly continuous in $mathbb D$, it would extend continuously to the closure, but it can't, because its extension is discontinuous on $S^1$
      $endgroup$
      – Federico
      Dec 7 '18 at 18:03






    • 1




      $begingroup$
      Yes you are right in saying that $P[f]$ is discontinuous in $bar{mathbb D}$ and continuous in $mathbb D$. The point is that it cannot be uniformly continuous in $mathbb D$, otherwise it would be in $bar{mathbb D}$ too.
      $endgroup$
      – Federico
      Dec 7 '18 at 18:05












    • $begingroup$
      Thanks, I had forgotten about this property.
      $endgroup$
      – ImNotThereRightNow_
      Dec 7 '18 at 18:07






    • 1




      $begingroup$
      I should probably have written it better :)
      $endgroup$
      – Federico
      Dec 7 '18 at 18:08














    6





    +50







    6





    +50



    6




    +50



    $begingroup$

    For any bounded measurable function $fin L^infty(S^1)$, the Poisson integral $P[f]$ gives a bounded harmonic functions in $B_1$. This function is continuous (equivalently, uniformly continuous) if and only if $f$ itself is continuous.



    Here you can find what the Poisson kernel is and how it's used to build harmonic functions in the ball.



    Said otherwise, $fmapsto P[f]$ is a linear isometry of $L^p(S^1)$ onto $h^p(B_1)$. We used the case $p=infty$, where $h^infty(B_1)$ stands for the the bounded harmonic functions in the ball. You can read more about it here.






    share|cite|improve this answer











    $endgroup$



    For any bounded measurable function $fin L^infty(S^1)$, the Poisson integral $P[f]$ gives a bounded harmonic functions in $B_1$. This function is continuous (equivalently, uniformly continuous) if and only if $f$ itself is continuous.



    Here you can find what the Poisson kernel is and how it's used to build harmonic functions in the ball.



    Said otherwise, $fmapsto P[f]$ is a linear isometry of $L^p(S^1)$ onto $h^p(B_1)$. We used the case $p=infty$, where $h^infty(B_1)$ stands for the the bounded harmonic functions in the ball. You can read more about it here.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 7 '18 at 17:54

























    answered Dec 7 '18 at 17:46









    FedericoFederico

    4,889514




    4,889514












    • $begingroup$
      If $f$ is discontinuous on $S^1$, then $P[f]$ is discontinuous when viewed as a function on $bar{mathbb{D}}$. But $P[f]$ is still harmonic and continuous in $mathbb{D}$? I'm confused as to how I can conclude from this that $P[f]$ is not uniformly continuous on $mathbb{D}$.
      $endgroup$
      – ImNotThereRightNow_
      Dec 7 '18 at 18:00






    • 1




      $begingroup$
      If $P[f]$ were uniformly continuous in $mathbb D$, it would extend continuously to the closure, but it can't, because its extension is discontinuous on $S^1$
      $endgroup$
      – Federico
      Dec 7 '18 at 18:03






    • 1




      $begingroup$
      Yes you are right in saying that $P[f]$ is discontinuous in $bar{mathbb D}$ and continuous in $mathbb D$. The point is that it cannot be uniformly continuous in $mathbb D$, otherwise it would be in $bar{mathbb D}$ too.
      $endgroup$
      – Federico
      Dec 7 '18 at 18:05












    • $begingroup$
      Thanks, I had forgotten about this property.
      $endgroup$
      – ImNotThereRightNow_
      Dec 7 '18 at 18:07






    • 1




      $begingroup$
      I should probably have written it better :)
      $endgroup$
      – Federico
      Dec 7 '18 at 18:08


















    • $begingroup$
      If $f$ is discontinuous on $S^1$, then $P[f]$ is discontinuous when viewed as a function on $bar{mathbb{D}}$. But $P[f]$ is still harmonic and continuous in $mathbb{D}$? I'm confused as to how I can conclude from this that $P[f]$ is not uniformly continuous on $mathbb{D}$.
      $endgroup$
      – ImNotThereRightNow_
      Dec 7 '18 at 18:00






    • 1




      $begingroup$
      If $P[f]$ were uniformly continuous in $mathbb D$, it would extend continuously to the closure, but it can't, because its extension is discontinuous on $S^1$
      $endgroup$
      – Federico
      Dec 7 '18 at 18:03






    • 1




      $begingroup$
      Yes you are right in saying that $P[f]$ is discontinuous in $bar{mathbb D}$ and continuous in $mathbb D$. The point is that it cannot be uniformly continuous in $mathbb D$, otherwise it would be in $bar{mathbb D}$ too.
      $endgroup$
      – Federico
      Dec 7 '18 at 18:05












    • $begingroup$
      Thanks, I had forgotten about this property.
      $endgroup$
      – ImNotThereRightNow_
      Dec 7 '18 at 18:07






    • 1




      $begingroup$
      I should probably have written it better :)
      $endgroup$
      – Federico
      Dec 7 '18 at 18:08
















    $begingroup$
    If $f$ is discontinuous on $S^1$, then $P[f]$ is discontinuous when viewed as a function on $bar{mathbb{D}}$. But $P[f]$ is still harmonic and continuous in $mathbb{D}$? I'm confused as to how I can conclude from this that $P[f]$ is not uniformly continuous on $mathbb{D}$.
    $endgroup$
    – ImNotThereRightNow_
    Dec 7 '18 at 18:00




    $begingroup$
    If $f$ is discontinuous on $S^1$, then $P[f]$ is discontinuous when viewed as a function on $bar{mathbb{D}}$. But $P[f]$ is still harmonic and continuous in $mathbb{D}$? I'm confused as to how I can conclude from this that $P[f]$ is not uniformly continuous on $mathbb{D}$.
    $endgroup$
    – ImNotThereRightNow_
    Dec 7 '18 at 18:00




    1




    1




    $begingroup$
    If $P[f]$ were uniformly continuous in $mathbb D$, it would extend continuously to the closure, but it can't, because its extension is discontinuous on $S^1$
    $endgroup$
    – Federico
    Dec 7 '18 at 18:03




    $begingroup$
    If $P[f]$ were uniformly continuous in $mathbb D$, it would extend continuously to the closure, but it can't, because its extension is discontinuous on $S^1$
    $endgroup$
    – Federico
    Dec 7 '18 at 18:03




    1




    1




    $begingroup$
    Yes you are right in saying that $P[f]$ is discontinuous in $bar{mathbb D}$ and continuous in $mathbb D$. The point is that it cannot be uniformly continuous in $mathbb D$, otherwise it would be in $bar{mathbb D}$ too.
    $endgroup$
    – Federico
    Dec 7 '18 at 18:05






    $begingroup$
    Yes you are right in saying that $P[f]$ is discontinuous in $bar{mathbb D}$ and continuous in $mathbb D$. The point is that it cannot be uniformly continuous in $mathbb D$, otherwise it would be in $bar{mathbb D}$ too.
    $endgroup$
    – Federico
    Dec 7 '18 at 18:05














    $begingroup$
    Thanks, I had forgotten about this property.
    $endgroup$
    – ImNotThereRightNow_
    Dec 7 '18 at 18:07




    $begingroup$
    Thanks, I had forgotten about this property.
    $endgroup$
    – ImNotThereRightNow_
    Dec 7 '18 at 18:07




    1




    1




    $begingroup$
    I should probably have written it better :)
    $endgroup$
    – Federico
    Dec 7 '18 at 18:08




    $begingroup$
    I should probably have written it better :)
    $endgroup$
    – Federico
    Dec 7 '18 at 18:08











    1












    $begingroup$

    Let $log$ denote the principal value $log.$ Then $log(1+z)$ is holomorphic in $mathbb D.$ Hence its imaginary part, $u(z)=arg (1+z),$ is harmonic in $mathbb D.$ We have $|u|<pi/2$ in the disc, so $u$ is bounded there. For small $r>0,$ $-1+re^{ipi/4}in mathbb D.$ For such $r,$



    $$u(-1+re^{ipi/4})- u(-1+r) = pi/4-0 = pi/4.$$



    But $(-1+re^{ipi/4})-(-1+r) to 0$ as $rto 0.$ This shows $u$ cannot be uniformly continuous in $mathbb D.$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Are you considering the principal value $log$ where $thetainleft(dfrac{-pi}{2},dfrac{pi}{2}right]$ ?
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:19










    • $begingroup$
      @YadatiKiran Yes, but that should be $(-pi,pi).$
      $endgroup$
      – zhw.
      Dec 10 '18 at 17:25










    • $begingroup$
      If its $(-pi,pi)$ then how do you say $|u|<pi/2$ in the disc? Am I missing something here?
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:29










    • $begingroup$
      @YadatiKiran Because $1+z$ is in the right half plane
      $endgroup$
      – zhw.
      Dec 10 '18 at 17:41










    • $begingroup$
      Oh Yes ! I get it.
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:43
















    1












    $begingroup$

    Let $log$ denote the principal value $log.$ Then $log(1+z)$ is holomorphic in $mathbb D.$ Hence its imaginary part, $u(z)=arg (1+z),$ is harmonic in $mathbb D.$ We have $|u|<pi/2$ in the disc, so $u$ is bounded there. For small $r>0,$ $-1+re^{ipi/4}in mathbb D.$ For such $r,$



    $$u(-1+re^{ipi/4})- u(-1+r) = pi/4-0 = pi/4.$$



    But $(-1+re^{ipi/4})-(-1+r) to 0$ as $rto 0.$ This shows $u$ cannot be uniformly continuous in $mathbb D.$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Are you considering the principal value $log$ where $thetainleft(dfrac{-pi}{2},dfrac{pi}{2}right]$ ?
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:19










    • $begingroup$
      @YadatiKiran Yes, but that should be $(-pi,pi).$
      $endgroup$
      – zhw.
      Dec 10 '18 at 17:25










    • $begingroup$
      If its $(-pi,pi)$ then how do you say $|u|<pi/2$ in the disc? Am I missing something here?
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:29










    • $begingroup$
      @YadatiKiran Because $1+z$ is in the right half plane
      $endgroup$
      – zhw.
      Dec 10 '18 at 17:41










    • $begingroup$
      Oh Yes ! I get it.
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:43














    1












    1








    1





    $begingroup$

    Let $log$ denote the principal value $log.$ Then $log(1+z)$ is holomorphic in $mathbb D.$ Hence its imaginary part, $u(z)=arg (1+z),$ is harmonic in $mathbb D.$ We have $|u|<pi/2$ in the disc, so $u$ is bounded there. For small $r>0,$ $-1+re^{ipi/4}in mathbb D.$ For such $r,$



    $$u(-1+re^{ipi/4})- u(-1+r) = pi/4-0 = pi/4.$$



    But $(-1+re^{ipi/4})-(-1+r) to 0$ as $rto 0.$ This shows $u$ cannot be uniformly continuous in $mathbb D.$






    share|cite|improve this answer











    $endgroup$



    Let $log$ denote the principal value $log.$ Then $log(1+z)$ is holomorphic in $mathbb D.$ Hence its imaginary part, $u(z)=arg (1+z),$ is harmonic in $mathbb D.$ We have $|u|<pi/2$ in the disc, so $u$ is bounded there. For small $r>0,$ $-1+re^{ipi/4}in mathbb D.$ For such $r,$



    $$u(-1+re^{ipi/4})- u(-1+r) = pi/4-0 = pi/4.$$



    But $(-1+re^{ipi/4})-(-1+r) to 0$ as $rto 0.$ This shows $u$ cannot be uniformly continuous in $mathbb D.$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 8 '18 at 18:55

























    answered Dec 8 '18 at 17:29









    zhw.zhw.

    71.9k43075




    71.9k43075












    • $begingroup$
      Are you considering the principal value $log$ where $thetainleft(dfrac{-pi}{2},dfrac{pi}{2}right]$ ?
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:19










    • $begingroup$
      @YadatiKiran Yes, but that should be $(-pi,pi).$
      $endgroup$
      – zhw.
      Dec 10 '18 at 17:25










    • $begingroup$
      If its $(-pi,pi)$ then how do you say $|u|<pi/2$ in the disc? Am I missing something here?
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:29










    • $begingroup$
      @YadatiKiran Because $1+z$ is in the right half plane
      $endgroup$
      – zhw.
      Dec 10 '18 at 17:41










    • $begingroup$
      Oh Yes ! I get it.
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:43


















    • $begingroup$
      Are you considering the principal value $log$ where $thetainleft(dfrac{-pi}{2},dfrac{pi}{2}right]$ ?
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:19










    • $begingroup$
      @YadatiKiran Yes, but that should be $(-pi,pi).$
      $endgroup$
      – zhw.
      Dec 10 '18 at 17:25










    • $begingroup$
      If its $(-pi,pi)$ then how do you say $|u|<pi/2$ in the disc? Am I missing something here?
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:29










    • $begingroup$
      @YadatiKiran Because $1+z$ is in the right half plane
      $endgroup$
      – zhw.
      Dec 10 '18 at 17:41










    • $begingroup$
      Oh Yes ! I get it.
      $endgroup$
      – Yadati Kiran
      Dec 10 '18 at 17:43
















    $begingroup$
    Are you considering the principal value $log$ where $thetainleft(dfrac{-pi}{2},dfrac{pi}{2}right]$ ?
    $endgroup$
    – Yadati Kiran
    Dec 10 '18 at 17:19




    $begingroup$
    Are you considering the principal value $log$ where $thetainleft(dfrac{-pi}{2},dfrac{pi}{2}right]$ ?
    $endgroup$
    – Yadati Kiran
    Dec 10 '18 at 17:19












    $begingroup$
    @YadatiKiran Yes, but that should be $(-pi,pi).$
    $endgroup$
    – zhw.
    Dec 10 '18 at 17:25




    $begingroup$
    @YadatiKiran Yes, but that should be $(-pi,pi).$
    $endgroup$
    – zhw.
    Dec 10 '18 at 17:25












    $begingroup$
    If its $(-pi,pi)$ then how do you say $|u|<pi/2$ in the disc? Am I missing something here?
    $endgroup$
    – Yadati Kiran
    Dec 10 '18 at 17:29




    $begingroup$
    If its $(-pi,pi)$ then how do you say $|u|<pi/2$ in the disc? Am I missing something here?
    $endgroup$
    – Yadati Kiran
    Dec 10 '18 at 17:29












    $begingroup$
    @YadatiKiran Because $1+z$ is in the right half plane
    $endgroup$
    – zhw.
    Dec 10 '18 at 17:41




    $begingroup$
    @YadatiKiran Because $1+z$ is in the right half plane
    $endgroup$
    – zhw.
    Dec 10 '18 at 17:41












    $begingroup$
    Oh Yes ! I get it.
    $endgroup$
    – Yadati Kiran
    Dec 10 '18 at 17:43




    $begingroup$
    Oh Yes ! I get it.
    $endgroup$
    – Yadati Kiran
    Dec 10 '18 at 17:43











    0












    $begingroup$

    Start with
    $$
    phi(z)=frac{1+z}{1-z},;;; zinmathbb{C},; |z| < 1.
    $$
    This function is holomorphic with
    begin{align}
    Re phi(z) &= Refrac{1+z}{1-z}frac{1-overline{z}}{1-overline{z}} \
    &=Refrac{1+z-overline{z}-|z|^2}{|1-z|^2} \
    &= frac{1-|z|^2}{|1-z|^2} > 0.
    end{align}
    $psi(z)=e^{-phi(z)}$ is bounded and holomorphic in $|z| < 1$, but $psi$ is not uniformly continuous in the open disk.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Start with
      $$
      phi(z)=frac{1+z}{1-z},;;; zinmathbb{C},; |z| < 1.
      $$
      This function is holomorphic with
      begin{align}
      Re phi(z) &= Refrac{1+z}{1-z}frac{1-overline{z}}{1-overline{z}} \
      &=Refrac{1+z-overline{z}-|z|^2}{|1-z|^2} \
      &= frac{1-|z|^2}{|1-z|^2} > 0.
      end{align}
      $psi(z)=e^{-phi(z)}$ is bounded and holomorphic in $|z| < 1$, but $psi$ is not uniformly continuous in the open disk.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Start with
        $$
        phi(z)=frac{1+z}{1-z},;;; zinmathbb{C},; |z| < 1.
        $$
        This function is holomorphic with
        begin{align}
        Re phi(z) &= Refrac{1+z}{1-z}frac{1-overline{z}}{1-overline{z}} \
        &=Refrac{1+z-overline{z}-|z|^2}{|1-z|^2} \
        &= frac{1-|z|^2}{|1-z|^2} > 0.
        end{align}
        $psi(z)=e^{-phi(z)}$ is bounded and holomorphic in $|z| < 1$, but $psi$ is not uniformly continuous in the open disk.






        share|cite|improve this answer









        $endgroup$



        Start with
        $$
        phi(z)=frac{1+z}{1-z},;;; zinmathbb{C},; |z| < 1.
        $$
        This function is holomorphic with
        begin{align}
        Re phi(z) &= Refrac{1+z}{1-z}frac{1-overline{z}}{1-overline{z}} \
        &=Refrac{1+z-overline{z}-|z|^2}{|1-z|^2} \
        &= frac{1-|z|^2}{|1-z|^2} > 0.
        end{align}
        $psi(z)=e^{-phi(z)}$ is bounded and holomorphic in $|z| < 1$, but $psi$ is not uniformly continuous in the open disk.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 4:40









        DisintegratingByPartsDisintegratingByParts

        58.8k42579




        58.8k42579






























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