Give a proof or a counter-example. A function f is analytic in an open set implies f being bounded there.












0












$begingroup$


Complete question: Give a proof or a counter-example. If $B^º$ is an open set of C and $f: B^ºrightarrow C$ is an analytic function, then f is bounded in $B^º$.



My attempt:



If $f$ is bounded, then $|f(z)|leq M, M in N$, $forall z in B^º$. Let $g(z)=e^z$, entire in C. Then, there exists an $M in N$, such that, $|e^z|leq M, forall z in B^º$.
$rightarrow ln(|e^z|)leq ln(M) Rightarrow zleq ln(M)$, but, as $B^ºsubset C$ is any open set, we could choose it as $B^ºcup { z_0 } $, where $z_0 > M$ which would led to a contradiction. Thus, $f$ being analytic does not imply being bounded in an open set.



Is this correct?






complex-analysis proof-verification proof-writing







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    what does "limited" mean?
    $endgroup$
    – zhw.
    Jan 2 at 23:22










  • $begingroup$
    ops, I've changed the word, it should be "bounded" instead of "limited".
    $endgroup$
    – Lincon Ribeiro
    Jan 2 at 23:24






  • 2




    $begingroup$
    Why not just look at $1/(1-z)$ in the open unit disc?
    $endgroup$
    – zhw.
    Jan 2 at 23:24










  • $begingroup$
    I tried with that function. Is the following argument correct? Let $f(z)=frac {1}{(1-z)}$ analytic in the open set $B^º(0,1)$. $lim_{xto 1^+} f(z)=-infty$ and $lim_{xto 1^-} f(z)=+infty$, thus, we could not find a M such that, $|f(z)|leq M, forall z in B^º$
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 12:34


















0












$begingroup$


Complete question: Give a proof or a counter-example. If $B^º$ is an open set of C and $f: B^ºrightarrow C$ is an analytic function, then f is bounded in $B^º$.



My attempt:



If $f$ is bounded, then $|f(z)|leq M, M in N$, $forall z in B^º$. Let $g(z)=e^z$, entire in C. Then, there exists an $M in N$, such that, $|e^z|leq M, forall z in B^º$.
$rightarrow ln(|e^z|)leq ln(M) Rightarrow zleq ln(M)$, but, as $B^ºsubset C$ is any open set, we could choose it as $B^ºcup { z_0 } $, where $z_0 > M$ which would led to a contradiction. Thus, $f$ being analytic does not imply being bounded in an open set.



Is this correct?






complex-analysis proof-verification proof-writing







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    what does "limited" mean?
    $endgroup$
    – zhw.
    Jan 2 at 23:22










  • $begingroup$
    ops, I've changed the word, it should be "bounded" instead of "limited".
    $endgroup$
    – Lincon Ribeiro
    Jan 2 at 23:24






  • 2




    $begingroup$
    Why not just look at $1/(1-z)$ in the open unit disc?
    $endgroup$
    – zhw.
    Jan 2 at 23:24










  • $begingroup$
    I tried with that function. Is the following argument correct? Let $f(z)=frac {1}{(1-z)}$ analytic in the open set $B^º(0,1)$. $lim_{xto 1^+} f(z)=-infty$ and $lim_{xto 1^-} f(z)=+infty$, thus, we could not find a M such that, $|f(z)|leq M, forall z in B^º$
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 12:34
















0












0








0





$begingroup$


Complete question: Give a proof or a counter-example. If $B^º$ is an open set of C and $f: B^ºrightarrow C$ is an analytic function, then f is bounded in $B^º$.



My attempt:



If $f$ is bounded, then $|f(z)|leq M, M in N$, $forall z in B^º$. Let $g(z)=e^z$, entire in C. Then, there exists an $M in N$, such that, $|e^z|leq M, forall z in B^º$.
$rightarrow ln(|e^z|)leq ln(M) Rightarrow zleq ln(M)$, but, as $B^ºsubset C$ is any open set, we could choose it as $B^ºcup { z_0 } $, where $z_0 > M$ which would led to a contradiction. Thus, $f$ being analytic does not imply being bounded in an open set.



Is this correct?






complex-analysis proof-verification proof-writing







share|cite|improve this question











$endgroup$




Complete question: Give a proof or a counter-example. If $B^º$ is an open set of C and $f: B^ºrightarrow C$ is an analytic function, then f is bounded in $B^º$.



My attempt:



If $f$ is bounded, then $|f(z)|leq M, M in N$, $forall z in B^º$. Let $g(z)=e^z$, entire in C. Then, there exists an $M in N$, such that, $|e^z|leq M, forall z in B^º$.
$rightarrow ln(|e^z|)leq ln(M) Rightarrow zleq ln(M)$, but, as $B^ºsubset C$ is any open set, we could choose it as $B^ºcup { z_0 } $, where $z_0 > M$ which would led to a contradiction. Thus, $f$ being analytic does not imply being bounded in an open set.



Is this correct?






complex-analysis proof-verification proof-writing




complex-analysis proof-verification proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 23:23







Lincon Ribeiro

















asked Jan 2 at 23:10









Lincon RibeiroLincon Ribeiro

556




556








  • 1




    $begingroup$
    what does "limited" mean?
    $endgroup$
    – zhw.
    Jan 2 at 23:22










  • $begingroup$
    ops, I've changed the word, it should be "bounded" instead of "limited".
    $endgroup$
    – Lincon Ribeiro
    Jan 2 at 23:24






  • 2




    $begingroup$
    Why not just look at $1/(1-z)$ in the open unit disc?
    $endgroup$
    – zhw.
    Jan 2 at 23:24










  • $begingroup$
    I tried with that function. Is the following argument correct? Let $f(z)=frac {1}{(1-z)}$ analytic in the open set $B^º(0,1)$. $lim_{xto 1^+} f(z)=-infty$ and $lim_{xto 1^-} f(z)=+infty$, thus, we could not find a M such that, $|f(z)|leq M, forall z in B^º$
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 12:34
















  • 1




    $begingroup$
    what does "limited" mean?
    $endgroup$
    – zhw.
    Jan 2 at 23:22










  • $begingroup$
    ops, I've changed the word, it should be "bounded" instead of "limited".
    $endgroup$
    – Lincon Ribeiro
    Jan 2 at 23:24






  • 2




    $begingroup$
    Why not just look at $1/(1-z)$ in the open unit disc?
    $endgroup$
    – zhw.
    Jan 2 at 23:24










  • $begingroup$
    I tried with that function. Is the following argument correct? Let $f(z)=frac {1}{(1-z)}$ analytic in the open set $B^º(0,1)$. $lim_{xto 1^+} f(z)=-infty$ and $lim_{xto 1^-} f(z)=+infty$, thus, we could not find a M such that, $|f(z)|leq M, forall z in B^º$
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 12:34










1




1




$begingroup$
what does "limited" mean?
$endgroup$
– zhw.
Jan 2 at 23:22




$begingroup$
what does "limited" mean?
$endgroup$
– zhw.
Jan 2 at 23:22












$begingroup$
ops, I've changed the word, it should be "bounded" instead of "limited".
$endgroup$
– Lincon Ribeiro
Jan 2 at 23:24




$begingroup$
ops, I've changed the word, it should be "bounded" instead of "limited".
$endgroup$
– Lincon Ribeiro
Jan 2 at 23:24




2




2




$begingroup$
Why not just look at $1/(1-z)$ in the open unit disc?
$endgroup$
– zhw.
Jan 2 at 23:24




$begingroup$
Why not just look at $1/(1-z)$ in the open unit disc?
$endgroup$
– zhw.
Jan 2 at 23:24












$begingroup$
I tried with that function. Is the following argument correct? Let $f(z)=frac {1}{(1-z)}$ analytic in the open set $B^º(0,1)$. $lim_{xto 1^+} f(z)=-infty$ and $lim_{xto 1^-} f(z)=+infty$, thus, we could not find a M such that, $|f(z)|leq M, forall z in B^º$
$endgroup$
– Lincon Ribeiro
Jan 3 at 12:34






$begingroup$
I tried with that function. Is the following argument correct? Let $f(z)=frac {1}{(1-z)}$ analytic in the open set $B^º(0,1)$. $lim_{xto 1^+} f(z)=-infty$ and $lim_{xto 1^-} f(z)=+infty$, thus, we could not find a M such that, $|f(z)|leq M, forall z in B^º$
$endgroup$
– Lincon Ribeiro
Jan 3 at 12:34












1 Answer
1






active

oldest

votes


















1












$begingroup$

Your argument is not correct but it becomes correct if you just choose $z$ to be real. For complex $z$ it is not true that $ln(|e^{z}|)=z$ and the inequality $z leq ln M$ does not make sense. Besides, you can take the open set $B^{0}$ to be $mathbb C$ in your argument.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I just thought something about it right now. When I add the $z_0$ to set $B^º$, wouldn't this change M again, so that the inequality $zleq ln(M)$ would still hold?
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 12:28








  • 1




    $begingroup$
    @LinconRibeiro To give a counter example you can take the open set $B^{0}$ to be $mathbb C$. Any real number $z_o >M$ can b e used in your argument.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 12:35











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Your argument is not correct but it becomes correct if you just choose $z$ to be real. For complex $z$ it is not true that $ln(|e^{z}|)=z$ and the inequality $z leq ln M$ does not make sense. Besides, you can take the open set $B^{0}$ to be $mathbb C$ in your argument.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I just thought something about it right now. When I add the $z_0$ to set $B^º$, wouldn't this change M again, so that the inequality $zleq ln(M)$ would still hold?
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 12:28








  • 1




    $begingroup$
    @LinconRibeiro To give a counter example you can take the open set $B^{0}$ to be $mathbb C$. Any real number $z_o >M$ can b e used in your argument.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 12:35
















1












$begingroup$

Your argument is not correct but it becomes correct if you just choose $z$ to be real. For complex $z$ it is not true that $ln(|e^{z}|)=z$ and the inequality $z leq ln M$ does not make sense. Besides, you can take the open set $B^{0}$ to be $mathbb C$ in your argument.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I just thought something about it right now. When I add the $z_0$ to set $B^º$, wouldn't this change M again, so that the inequality $zleq ln(M)$ would still hold?
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 12:28








  • 1




    $begingroup$
    @LinconRibeiro To give a counter example you can take the open set $B^{0}$ to be $mathbb C$. Any real number $z_o >M$ can b e used in your argument.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 12:35














1












1








1





$begingroup$

Your argument is not correct but it becomes correct if you just choose $z$ to be real. For complex $z$ it is not true that $ln(|e^{z}|)=z$ and the inequality $z leq ln M$ does not make sense. Besides, you can take the open set $B^{0}$ to be $mathbb C$ in your argument.






share|cite|improve this answer









$endgroup$



Your argument is not correct but it becomes correct if you just choose $z$ to be real. For complex $z$ it is not true that $ln(|e^{z}|)=z$ and the inequality $z leq ln M$ does not make sense. Besides, you can take the open set $B^{0}$ to be $mathbb C$ in your argument.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 23:33









Kavi Rama MurthyKavi Rama Murthy

63.9k42464




63.9k42464












  • $begingroup$
    I just thought something about it right now. When I add the $z_0$ to set $B^º$, wouldn't this change M again, so that the inequality $zleq ln(M)$ would still hold?
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 12:28








  • 1




    $begingroup$
    @LinconRibeiro To give a counter example you can take the open set $B^{0}$ to be $mathbb C$. Any real number $z_o >M$ can b e used in your argument.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 12:35


















  • $begingroup$
    I just thought something about it right now. When I add the $z_0$ to set $B^º$, wouldn't this change M again, so that the inequality $zleq ln(M)$ would still hold?
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 12:28








  • 1




    $begingroup$
    @LinconRibeiro To give a counter example you can take the open set $B^{0}$ to be $mathbb C$. Any real number $z_o >M$ can b e used in your argument.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 12:35
















$begingroup$
I just thought something about it right now. When I add the $z_0$ to set $B^º$, wouldn't this change M again, so that the inequality $zleq ln(M)$ would still hold?
$endgroup$
– Lincon Ribeiro
Jan 3 at 12:28






$begingroup$
I just thought something about it right now. When I add the $z_0$ to set $B^º$, wouldn't this change M again, so that the inequality $zleq ln(M)$ would still hold?
$endgroup$
– Lincon Ribeiro
Jan 3 at 12:28






1




1




$begingroup$
@LinconRibeiro To give a counter example you can take the open set $B^{0}$ to be $mathbb C$. Any real number $z_o >M$ can b e used in your argument.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 12:35




$begingroup$
@LinconRibeiro To give a counter example you can take the open set $B^{0}$ to be $mathbb C$. Any real number $z_o >M$ can b e used in your argument.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 12:35


















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