Give a proof or a counter-example. A function f is analytic in an open set implies f being bounded there.
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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
asked Jan 2 at 23:10
Lincon RibeiroLincon Ribeiro
556
$endgroup$
add a comment |
$begingroup$
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
asked Jan 2 at 23:10
Lincon RibeiroLincon Ribeiro
556
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 2 at 23:10
Lincon RibeiroLincon Ribeiro
556
$endgroup$
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
asked Jan 2 at 23:10
Lincon RibeiroLincon Ribeiro
556
asked Jan 2 at 23:10
Lincon RibeiroLincon Ribeiro
556
edited Jan 2 at 23:23
Lincon Ribeiro
asked Jan 2 at 23:10
Lincon RibeiroLincon Ribeiro
556
asked Jan 2 at 23:10
Lincon RibeiroLincon Ribeiro
556
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 2 at 23:33
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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oldest
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$begingroup$
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.
answered Jan 2 at 23:33
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 2 at 23:33
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 2 at 23:33
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
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.
answered Jan 2 at 23:33
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
answered Jan 2 at 23:33
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
answered Jan 2 at 23:33
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
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
add a comment |
$begingroup$
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
add a comment |
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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