Entire function invariant by translation is constant [duplicate]
This question already has an answer here:
Proving that a doubly-periodic entire function $f$ is constant.
2 answers
I need to apply Liouville theorem ("entire bounded complex functions are constant") to prove that an entire function satisfying:
$$f(z)=f(z+1)=f(z+i)$$
for all complex numbers $z$ is constant. I'm really not sure on how to proceed, I've tried expanding in Taylor series but I've got confused with calculations and I can't proceed any further.
complex-analysis holomorphic-functions entire-functions
asked Dec 9 at 15:48
Renato Faraone
2,32911627
marked as duplicate by Martin R, user10354138, Paul Frost, Community♦ Dec 9 at 18:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Proving that a doubly-periodic entire function $f$ is constant.
2 answers
I need to apply Liouville theorem ("entire bounded complex functions are constant") to prove that an entire function satisfying:
$$f(z)=f(z+1)=f(z+i)$$
for all complex numbers $z$ is constant. I'm really not sure on how to proceed, I've tried expanding in Taylor series but I've got confused with calculations and I can't proceed any further.
complex-analysis holomorphic-functions entire-functions
asked Dec 9 at 15:48
Renato Faraone
2,32911627
marked as duplicate by Martin R, user10354138, Paul Frost, Community♦ Dec 9 at 18:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Proving that a doubly-periodic entire function $f$ is constant.
2 answers
I need to apply Liouville theorem ("entire bounded complex functions are constant") to prove that an entire function satisfying:
$$f(z)=f(z+1)=f(z+i)$$
for all complex numbers $z$ is constant. I'm really not sure on how to proceed, I've tried expanding in Taylor series but I've got confused with calculations and I can't proceed any further.
complex-analysis holomorphic-functions entire-functions
asked Dec 9 at 15:48
Renato Faraone
2,32911627
This question already has an answer here:
Proving that a doubly-periodic entire function $f$ is constant.
2 answers
I need to apply Liouville theorem ("entire bounded complex functions are constant") to prove that an entire function satisfying:
$$f(z)=f(z+1)=f(z+i)$$
for all complex numbers $z$ is constant. I'm really not sure on how to proceed, I've tried expanding in Taylor series but I've got confused with calculations and I can't proceed any further.
This question already has an answer here:
Proving that a doubly-periodic entire function $f$ is constant.
2 answers
complex-analysis holomorphic-functions entire-functions
complex-analysis holomorphic-functions entire-functions
asked Dec 9 at 15:48
Renato Faraone
2,32911627
asked Dec 9 at 15:48
Renato Faraone
2,32911627
asked Dec 9 at 15:48
Renato Faraone
2,32911627
asked Dec 9 at 15:48
Renato Faraone
2,32911627
asked Dec 9 at 15:48
Renato Faraone
2,32911627
2,32911627
marked as duplicate by Martin R, user10354138, Paul Frost, Community♦ Dec 9 at 18:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, user10354138, Paul Frost, Community♦ Dec 9 at 18:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
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By induction, $f(z)=f(z+m+in)$ for all integers $m,n$. This means that for each $zinmathbb{C}$ there is $win[0,1]times[0,1]$ such that $f(z)=f(w)$. Continuous functions are bounded on compact sets. It follows that $f$ is bounded on $mathbb{C}$. Now apply Liouville.
answered Dec 9 at 15:54
Ben W
1,413513
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
By induction, $f(z)=f(z+m+in)$ for all integers $m,n$. This means that for each $zinmathbb{C}$ there is $win[0,1]times[0,1]$ such that $f(z)=f(w)$. Continuous functions are bounded on compact sets. It follows that $f$ is bounded on $mathbb{C}$. Now apply Liouville.
answered Dec 9 at 15:54
Ben W
1,413513
add a comment |
By induction, $f(z)=f(z+m+in)$ for all integers $m,n$. This means that for each $zinmathbb{C}$ there is $win[0,1]times[0,1]$ such that $f(z)=f(w)$. Continuous functions are bounded on compact sets. It follows that $f$ is bounded on $mathbb{C}$. Now apply Liouville.
answered Dec 9 at 15:54
Ben W
1,413513
add a comment |
By induction, $f(z)=f(z+m+in)$ for all integers $m,n$. This means that for each $zinmathbb{C}$ there is $win[0,1]times[0,1]$ such that $f(z)=f(w)$. Continuous functions are bounded on compact sets. It follows that $f$ is bounded on $mathbb{C}$. Now apply Liouville.
answered Dec 9 at 15:54
Ben W
1,413513
By induction, $f(z)=f(z+m+in)$ for all integers $m,n$. This means that for each $zinmathbb{C}$ there is $win[0,1]times[0,1]$ such that $f(z)=f(w)$. Continuous functions are bounded on compact sets. It follows that $f$ is bounded on $mathbb{C}$. Now apply Liouville.
answered Dec 9 at 15:54
Ben W
1,413513
edited Dec 9 at 16:10
answered Dec 9 at 15:54
Ben W
1,413513
answered Dec 9 at 15:54
Ben W
1,413513
answered Dec 9 at 15:54
Ben W
1,413513
1,413513
add a comment |
add a comment |
