Lemma related to proof of Montel's Theorem The Next CEO of Stack OverflowUsing Montel's Theorem to show locally uniform convergence of analytic functionsNormal sequences and Montel's TheoremAbout the proof of a corollary of Arzela-Ascoli Theorem.Show the converse of the Arzela-Ascoli theoremHow to understand Arzela-Ascoli theorem?Exercise: Applying Arzela-Ascoli to show uniform convergence on bounded subsets of $; mathbb R;$$f_n$ converges uniformly using Arzela AscoliA corollary of Arzela-AscoliSequence of functions $f_n(x)=frac2x^2x^2+(1-2nx)^2$, does it have a uniformly converging subsequence?Proof of Arzela's Theorem
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Lemma related to proof of Montel's Theorem
The Next CEO of Stack OverflowUsing Montel's Theorem to show locally uniform convergence of analytic functionsNormal sequences and Montel's TheoremAbout the proof of a corollary of Arzela-Ascoli Theorem.Show the converse of the Arzela-Ascoli theoremHow to understand Arzela-Ascoli theorem?Exercise: Applying Arzela-Ascoli to show uniform convergence on bounded subsets of $; mathbb R;$$f_n$ converges uniformly using Arzela AscoliA corollary of Arzela-AscoliSequence of functions $f_n(x)=frac2x^2x^2+(1-2nx)^2$, does it have a uniformly converging subsequence?Proof of Arzela's Theorem
$begingroup$
I have seen the following lemma in my Complex Analysis class:
Let $D subset mathbbC$ be open and connected, let $(f_n)_n in mathbbN$ be a sequence of functions holomorphic in $D$. Assume:
$(f_n)_n in mathbbN$ is local bounded
and there is a dense set $mathcalD subset D$ such that the sequence $(f_n(z))_n in mathbbN$ converges $forall z in mathcalD$. Then $(f_n)_n in mathbbN$ converges locally uniformly on $D$ to a holomorphic function $f$.
What I am not sure about is the first step of the proof. It says that it suffices to prove the following: let$z_0 in D$, pick $delta > 0$ such that $overlineB(z_0,2delta) subset D$. Then $forall varepsilon >0, exists N>0$ such that $forall n,m geq N$ and $forall Z in B(z_0,delta)$, we have $|f_n(z)-f_m(z)| leq varepsilon$ $(1)$
I understand the proof of $(1)$, but I am not sure how it implies the lemma. My attempt so far:
Set $B = overlineB(z_0,delta)$, $mathcalF = _B : n in mathbbN$. I have managed to show that $mathcalF$ is bounded and equicontinuous using $(1)$ and uniform boundedness, and so by Arzela-Ascoli it must be compact. This implies that there is a subsequence of $(f_n|_B)_n in mathbbN$ that converges uniformly on $B$ to some $f_z_0,delta : B to mathbbC$. But I am not where to go from there. I want to end up with a function $f$ defined on the whole of $D$, and such that $(f_n)_n in mathbbN$ converges to it locally uniformly (that is $forall K subset D$ compact, $f_n|_K to f|_K$ uniformly). Thank you for your help!
complex-analysis arzela-ascoli
asked Mar 19 at 21:26
ChloChlo
727
$endgroup$
add a comment |
$begingroup$
I have seen the following lemma in my Complex Analysis class:
Let $D subset mathbbC$ be open and connected, let $(f_n)_n in mathbbN$ be a sequence of functions holomorphic in $D$. Assume:
$(f_n)_n in mathbbN$ is local bounded
and there is a dense set $mathcalD subset D$ such that the sequence $(f_n(z))_n in mathbbN$ converges $forall z in mathcalD$. Then $(f_n)_n in mathbbN$ converges locally uniformly on $D$ to a holomorphic function $f$.
What I am not sure about is the first step of the proof. It says that it suffices to prove the following: let$z_0 in D$, pick $delta > 0$ such that $overlineB(z_0,2delta) subset D$. Then $forall varepsilon >0, exists N>0$ such that $forall n,m geq N$ and $forall Z in B(z_0,delta)$, we have $|f_n(z)-f_m(z)| leq varepsilon$ $(1)$
I understand the proof of $(1)$, but I am not sure how it implies the lemma. My attempt so far:
Set $B = overlineB(z_0,delta)$, $mathcalF = _B : n in mathbbN$. I have managed to show that $mathcalF$ is bounded and equicontinuous using $(1)$ and uniform boundedness, and so by Arzela-Ascoli it must be compact. This implies that there is a subsequence of $(f_n|_B)_n in mathbbN$ that converges uniformly on $B$ to some $f_z_0,delta : B to mathbbC$. But I am not where to go from there. I want to end up with a function $f$ defined on the whole of $D$, and such that $(f_n)_n in mathbbN$ converges to it locally uniformly (that is $forall K subset D$ compact, $f_n|_K to f|_K$ uniformly). Thank you for your help!
complex-analysis arzela-ascoli
asked Mar 19 at 21:26
ChloChlo
727
$endgroup$
add a comment |
$begingroup$
I have seen the following lemma in my Complex Analysis class:
Let $D subset mathbbC$ be open and connected, let $(f_n)_n in mathbbN$ be a sequence of functions holomorphic in $D$. Assume:
$(f_n)_n in mathbbN$ is local bounded
and there is a dense set $mathcalD subset D$ such that the sequence $(f_n(z))_n in mathbbN$ converges $forall z in mathcalD$. Then $(f_n)_n in mathbbN$ converges locally uniformly on $D$ to a holomorphic function $f$.
What I am not sure about is the first step of the proof. It says that it suffices to prove the following: let$z_0 in D$, pick $delta > 0$ such that $overlineB(z_0,2delta) subset D$. Then $forall varepsilon >0, exists N>0$ such that $forall n,m geq N$ and $forall Z in B(z_0,delta)$, we have $|f_n(z)-f_m(z)| leq varepsilon$ $(1)$
I understand the proof of $(1)$, but I am not sure how it implies the lemma. My attempt so far:
Set $B = overlineB(z_0,delta)$, $mathcalF = _B : n in mathbbN$. I have managed to show that $mathcalF$ is bounded and equicontinuous using $(1)$ and uniform boundedness, and so by Arzela-Ascoli it must be compact. This implies that there is a subsequence of $(f_n|_B)_n in mathbbN$ that converges uniformly on $B$ to some $f_z_0,delta : B to mathbbC$. But I am not where to go from there. I want to end up with a function $f$ defined on the whole of $D$, and such that $(f_n)_n in mathbbN$ converges to it locally uniformly (that is $forall K subset D$ compact, $f_n|_K to f|_K$ uniformly). Thank you for your help!
complex-analysis arzela-ascoli
asked Mar 19 at 21:26
ChloChlo
727
$endgroup$
I have seen the following lemma in my Complex Analysis class:
Let $D subset mathbbC$ be open and connected, let $(f_n)_n in mathbbN$ be a sequence of functions holomorphic in $D$. Assume:
$(f_n)_n in mathbbN$ is local bounded
and there is a dense set $mathcalD subset D$ such that the sequence $(f_n(z))_n in mathbbN$ converges $forall z in mathcalD$. Then $(f_n)_n in mathbbN$ converges locally uniformly on $D$ to a holomorphic function $f$.
What I am not sure about is the first step of the proof. It says that it suffices to prove the following: let$z_0 in D$, pick $delta > 0$ such that $overlineB(z_0,2delta) subset D$. Then $forall varepsilon >0, exists N>0$ such that $forall n,m geq N$ and $forall Z in B(z_0,delta)$, we have $|f_n(z)-f_m(z)| leq varepsilon$ $(1)$
I understand the proof of $(1)$, but I am not sure how it implies the lemma. My attempt so far:
Set $B = overlineB(z_0,delta)$, $mathcalF = _B : n in mathbbN$. I have managed to show that $mathcalF$ is bounded and equicontinuous using $(1)$ and uniform boundedness, and so by Arzela-Ascoli it must be compact. This implies that there is a subsequence of $(f_n|_B)_n in mathbbN$ that converges uniformly on $B$ to some $f_z_0,delta : B to mathbbC$. But I am not where to go from there. I want to end up with a function $f$ defined on the whole of $D$, and such that $(f_n)_n in mathbbN$ converges to it locally uniformly (that is $forall K subset D$ compact, $f_n|_K to f|_K$ uniformly). Thank you for your help!
complex-analysis arzela-ascoli
complex-analysis arzela-ascoli
asked Mar 19 at 21:26
ChloChlo
727
asked Mar 19 at 21:26
ChloChlo
727
asked Mar 19 at 21:26
ChloChlo
727
asked Mar 19 at 21:26
ChloChlo
727
asked Mar 19 at 21:26
ChloChlo
727
727
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(1) above implies $f_n$ is uniformly Cauchy sequence in $overlineB(z_0,fracdelta)2$, so in particular they converge uniformly to a continuous function $f$ there by basic convergence of functions theory, nothing to do with analyticity; Morera then shows $f$ analytic there since uniform convergence means you can integrate on small triangles inside; since $z_0$ is arbitrary you are done at least as small discs around any point go, so you get analytic $f$ and uniform convergence on such discs, and then any compact is a finite union of such so you are done; local boundness is crucial in proving (1) from the hypothesis
answered Mar 20 at 1:13
ConradConrad
1,32745
$endgroup$
$begingroup$
Thank you for your answer, indeed it was much simpler than I thought :)
$endgroup$
– Chlo
Mar 20 at 7:50
$begingroup$
You are welcome
$endgroup$
– Conrad
Mar 20 at 10:52
add a comment |
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1 Answer
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1 Answer
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$begingroup$
(1) above implies $f_n$ is uniformly Cauchy sequence in $overlineB(z_0,fracdelta)2$, so in particular they converge uniformly to a continuous function $f$ there by basic convergence of functions theory, nothing to do with analyticity; Morera then shows $f$ analytic there since uniform convergence means you can integrate on small triangles inside; since $z_0$ is arbitrary you are done at least as small discs around any point go, so you get analytic $f$ and uniform convergence on such discs, and then any compact is a finite union of such so you are done; local boundness is crucial in proving (1) from the hypothesis
answered Mar 20 at 1:13
ConradConrad
1,32745
$endgroup$
$begingroup$
Thank you for your answer, indeed it was much simpler than I thought :)
$endgroup$
– Chlo
Mar 20 at 7:50
$begingroup$
You are welcome
$endgroup$
– Conrad
Mar 20 at 10:52
add a comment |
$begingroup$
(1) above implies $f_n$ is uniformly Cauchy sequence in $overlineB(z_0,fracdelta)2$, so in particular they converge uniformly to a continuous function $f$ there by basic convergence of functions theory, nothing to do with analyticity; Morera then shows $f$ analytic there since uniform convergence means you can integrate on small triangles inside; since $z_0$ is arbitrary you are done at least as small discs around any point go, so you get analytic $f$ and uniform convergence on such discs, and then any compact is a finite union of such so you are done; local boundness is crucial in proving (1) from the hypothesis
answered Mar 20 at 1:13
ConradConrad
1,32745
$endgroup$
$begingroup$
Thank you for your answer, indeed it was much simpler than I thought :)
$endgroup$
– Chlo
Mar 20 at 7:50
$begingroup$
You are welcome
$endgroup$
– Conrad
Mar 20 at 10:52
add a comment |
$begingroup$
(1) above implies $f_n$ is uniformly Cauchy sequence in $overlineB(z_0,fracdelta)2$, so in particular they converge uniformly to a continuous function $f$ there by basic convergence of functions theory, nothing to do with analyticity; Morera then shows $f$ analytic there since uniform convergence means you can integrate on small triangles inside; since $z_0$ is arbitrary you are done at least as small discs around any point go, so you get analytic $f$ and uniform convergence on such discs, and then any compact is a finite union of such so you are done; local boundness is crucial in proving (1) from the hypothesis
answered Mar 20 at 1:13
ConradConrad
1,32745
$endgroup$
(1) above implies $f_n$ is uniformly Cauchy sequence in $overlineB(z_0,fracdelta)2$, so in particular they converge uniformly to a continuous function $f$ there by basic convergence of functions theory, nothing to do with analyticity; Morera then shows $f$ analytic there since uniform convergence means you can integrate on small triangles inside; since $z_0$ is arbitrary you are done at least as small discs around any point go, so you get analytic $f$ and uniform convergence on such discs, and then any compact is a finite union of such so you are done; local boundness is crucial in proving (1) from the hypothesis
answered Mar 20 at 1:13
ConradConrad
1,32745
answered Mar 20 at 1:13
ConradConrad
1,32745
answered Mar 20 at 1:13
ConradConrad
1,32745
answered Mar 20 at 1:13
ConradConrad
1,32745
1,32745
$begingroup$
Thank you for your answer, indeed it was much simpler than I thought :)
$endgroup$
– Chlo
Mar 20 at 7:50
$begingroup$
You are welcome
$endgroup$
– Conrad
Mar 20 at 10:52
add a comment |
$begingroup$
Thank you for your answer, indeed it was much simpler than I thought :)
$endgroup$
– Chlo
Mar 20 at 7:50
$begingroup$
You are welcome
$endgroup$
– Conrad
Mar 20 at 10:52
$begingroup$
Thank you for your answer, indeed it was much simpler than I thought :)
$endgroup$
– Chlo
Mar 20 at 7:50
$begingroup$
Thank you for your answer, indeed it was much simpler than I thought :)
$endgroup$
– Chlo
Mar 20 at 7:50
$begingroup$
You are welcome
$endgroup$
– Conrad
Mar 20 at 10:52
$begingroup$
You are welcome
$endgroup$
– Conrad
Mar 20 at 10:52
add a comment |
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