Let $f$ be analytic on $mathbb{D}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$. Prove $|f'(0)| leq 4$.












1












$begingroup$


I am trying to solve the following problem:



Let $f$ be analytic on $mathbb{D} = {z in mathbb{C}: |z|<1}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$. Prove $|f'(0)| leq 4$.



I have an idea of how to do it, but I am stuck at one of the steps.



I'd like to use Schwarz' lemma. So I wish to find a function $h = g circ f, $ such that $h(0) = 0$, and where $g: f(mathbb{D}) rightarrow mathbb{D}$. Then $h$ would be a function from $mathbb{D}$ to $mathbb{D}$, and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$, I believe I would be done with the proof.



Right now, however, I am stuck on what $f(mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z)| < 1 + |f(z)|$. Simply by drawing a picture I've been able to conclude that if $Im(z) = 0$, then $Re(z)$ cannot be in $left[0, inftyright].$ But that's all I have at this point...






complex-analysis inequality







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am trying to solve the following problem:



    Let $f$ be analytic on $mathbb{D} = {z in mathbb{C}: |z|<1}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$. Prove $|f'(0)| leq 4$.



    I have an idea of how to do it, but I am stuck at one of the steps.



    I'd like to use Schwarz' lemma. So I wish to find a function $h = g circ f, $ such that $h(0) = 0$, and where $g: f(mathbb{D}) rightarrow mathbb{D}$. Then $h$ would be a function from $mathbb{D}$ to $mathbb{D}$, and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$, I believe I would be done with the proof.



    Right now, however, I am stuck on what $f(mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z)| < 1 + |f(z)|$. Simply by drawing a picture I've been able to conclude that if $Im(z) = 0$, then $Re(z)$ cannot be in $left[0, inftyright].$ But that's all I have at this point...






    complex-analysis inequality







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to solve the following problem:



      Let $f$ be analytic on $mathbb{D} = {z in mathbb{C}: |z|<1}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$. Prove $|f'(0)| leq 4$.



      I have an idea of how to do it, but I am stuck at one of the steps.



      I'd like to use Schwarz' lemma. So I wish to find a function $h = g circ f, $ such that $h(0) = 0$, and where $g: f(mathbb{D}) rightarrow mathbb{D}$. Then $h$ would be a function from $mathbb{D}$ to $mathbb{D}$, and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$, I believe I would be done with the proof.



      Right now, however, I am stuck on what $f(mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z)| < 1 + |f(z)|$. Simply by drawing a picture I've been able to conclude that if $Im(z) = 0$, then $Re(z)$ cannot be in $left[0, inftyright].$ But that's all I have at this point...






      complex-analysis inequality







      share|cite|improve this question











      $endgroup$




      I am trying to solve the following problem:



      Let $f$ be analytic on $mathbb{D} = {z in mathbb{C}: |z|<1}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$. Prove $|f'(0)| leq 4$.



      I have an idea of how to do it, but I am stuck at one of the steps.



      I'd like to use Schwarz' lemma. So I wish to find a function $h = g circ f, $ such that $h(0) = 0$, and where $g: f(mathbb{D}) rightarrow mathbb{D}$. Then $h$ would be a function from $mathbb{D}$ to $mathbb{D}$, and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$, I believe I would be done with the proof.



      Right now, however, I am stuck on what $f(mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z)| < 1 + |f(z)|$. Simply by drawing a picture I've been able to conclude that if $Im(z) = 0$, then $Re(z)$ cannot be in $left[0, inftyright].$ But that's all I have at this point...






      complex-analysis inequality




      complex-analysis inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 17:14







      j.eee

















      asked Dec 15 '18 at 17:04









      j.eeej.eee

      787




      787






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          $|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
          $$
          f(z) in Bbb C setminus [0, infty) , .
          $$
          It is a bit more convenient to consider
          $$
          -f(z) in Bbb C setminus (-infty, 0] = U , .
          $$
          The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.



          Then
          $$
          h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
          $$
          maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.



          Reversing the compositions we get
          $$
          f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
          f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
          f'(0) = -4 h'(0)
          $$
          and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
          $$
          f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
          $$
          for some $lambda in Bbb C$ with $|lambda| = 1$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            non-negative real multiple
            $endgroup$
            – mathworker21
            Dec 15 '18 at 17:28










          • $begingroup$
            @mathworker21: Yes, thanks!
            $endgroup$
            – Martin R
            Dec 15 '18 at 17:28











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041715%2flet-f-be-analytic-on-mathbbd-f0-1-and-1fz-1fz-prove%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          $|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
          $$
          f(z) in Bbb C setminus [0, infty) , .
          $$
          It is a bit more convenient to consider
          $$
          -f(z) in Bbb C setminus (-infty, 0] = U , .
          $$
          The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.



          Then
          $$
          h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
          $$
          maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.



          Reversing the compositions we get
          $$
          f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
          f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
          f'(0) = -4 h'(0)
          $$
          and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
          $$
          f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
          $$
          for some $lambda in Bbb C$ with $|lambda| = 1$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            non-negative real multiple
            $endgroup$
            – mathworker21
            Dec 15 '18 at 17:28










          • $begingroup$
            @mathworker21: Yes, thanks!
            $endgroup$
            – Martin R
            Dec 15 '18 at 17:28
















          2












          $begingroup$

          $|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
          $$
          f(z) in Bbb C setminus [0, infty) , .
          $$
          It is a bit more convenient to consider
          $$
          -f(z) in Bbb C setminus (-infty, 0] = U , .
          $$
          The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.



          Then
          $$
          h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
          $$
          maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.



          Reversing the compositions we get
          $$
          f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
          f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
          f'(0) = -4 h'(0)
          $$
          and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
          $$
          f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
          $$
          for some $lambda in Bbb C$ with $|lambda| = 1$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            non-negative real multiple
            $endgroup$
            – mathworker21
            Dec 15 '18 at 17:28










          • $begingroup$
            @mathworker21: Yes, thanks!
            $endgroup$
            – Martin R
            Dec 15 '18 at 17:28














          2












          2








          2





          $begingroup$

          $|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
          $$
          f(z) in Bbb C setminus [0, infty) , .
          $$
          It is a bit more convenient to consider
          $$
          -f(z) in Bbb C setminus (-infty, 0] = U , .
          $$
          The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.



          Then
          $$
          h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
          $$
          maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.



          Reversing the compositions we get
          $$
          f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
          f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
          f'(0) = -4 h'(0)
          $$
          and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
          $$
          f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
          $$
          for some $lambda in Bbb C$ with $|lambda| = 1$.






          share|cite|improve this answer











          $endgroup$



          $|1+f(z)| < 1+|f(z)|$ means that equality does not hold in the triangle inequality $$|1+f(z)| le 1+|f(z)|, ,$$ and that means that $f(z)$ is not a non-negative real multiple of $1$. In other words:
          $$
          f(z) in Bbb C setminus [0, infty) , .
          $$
          It is a bit more convenient to consider
          $$
          -f(z) in Bbb C setminus (-infty, 0] = U , .
          $$
          The “main branch” of the square root maps the slit domain $U$ conformally to the right-half plain, and that is mapped conformally to the unit disk with the Möbius transformation $T(w) = frac{w-1}{w+1}$.



          Then
          $$
          h(z) = frac{sqrt{-f(z)}-1}{sqrt{-f(z)}+1}
          $$
          maps $Bbb D$ into $Bbb D$ with $h(0) = 0$, and the Schwarz Lemma can be applied: $|h'(0)| le 1$.



          Reversing the compositions we get
          $$
          f(z) = - left( frac{1+h(z)}{1-h(z)} right)^2 \
          f'(z) = - 4 frac{1+h(z)}{(1-h(z))^3} h'(z) \
          f'(0) = -4 h'(0)
          $$
          and the desired conclusion $| f'(0) | le 4$ follows. Equality holds if and only if
          $$
          f(z) = - left( frac{1+lambda z}{1-lambda z} right)^2 \
          $$
          for some $lambda in Bbb C$ with $|lambda| = 1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 '18 at 18:21

























          answered Dec 15 '18 at 17:21









          Martin RMartin R

          29k33458




          29k33458








          • 1




            $begingroup$
            non-negative real multiple
            $endgroup$
            – mathworker21
            Dec 15 '18 at 17:28










          • $begingroup$
            @mathworker21: Yes, thanks!
            $endgroup$
            – Martin R
            Dec 15 '18 at 17:28














          • 1




            $begingroup$
            non-negative real multiple
            $endgroup$
            – mathworker21
            Dec 15 '18 at 17:28










          • $begingroup$
            @mathworker21: Yes, thanks!
            $endgroup$
            – Martin R
            Dec 15 '18 at 17:28








          1




          1




          $begingroup$
          non-negative real multiple
          $endgroup$
          – mathworker21
          Dec 15 '18 at 17:28




          $begingroup$
          non-negative real multiple
          $endgroup$
          – mathworker21
          Dec 15 '18 at 17:28












          $begingroup$
          @mathworker21: Yes, thanks!
          $endgroup$
          – Martin R
          Dec 15 '18 at 17:28




          $begingroup$
          @mathworker21: Yes, thanks!
          $endgroup$
          – Martin R
          Dec 15 '18 at 17:28


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041715%2flet-f-be-analytic-on-mathbbd-f0-1-and-1fz-1fz-prove%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more