If $f$ is continuous at $z=a$ in a domain $D$ then $f$ is analytic on whole of $D$












1












$begingroup$



Let, $f$ be analytic in $Dsetminus {a}$. Also, $f$ is continuous at $z=a$. Then prove that $f$ is analytic on whole of $D$.




We know from Riemann's Theorem, if $f$ is not analytic at $a$ and in a nbd. of $a$, if $f$ is bounded then either $f$ is analytic at $a$ or $f$ has removable singularity at $a$. But here how can I use the continuity condition to show $f$ is analytic ?






complex-analysis analysis complex-numbers







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  • $begingroup$
    $f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
    $endgroup$
    – UserS
    Nov 24 '18 at 16:35


















1












$begingroup$



Let, $f$ be analytic in $Dsetminus {a}$. Also, $f$ is continuous at $z=a$. Then prove that $f$ is analytic on whole of $D$.




We know from Riemann's Theorem, if $f$ is not analytic at $a$ and in a nbd. of $a$, if $f$ is bounded then either $f$ is analytic at $a$ or $f$ has removable singularity at $a$. But here how can I use the continuity condition to show $f$ is analytic ?






complex-analysis analysis complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
    $endgroup$
    – UserS
    Nov 24 '18 at 16:35
















1












1








1





$begingroup$



Let, $f$ be analytic in $Dsetminus {a}$. Also, $f$ is continuous at $z=a$. Then prove that $f$ is analytic on whole of $D$.




We know from Riemann's Theorem, if $f$ is not analytic at $a$ and in a nbd. of $a$, if $f$ is bounded then either $f$ is analytic at $a$ or $f$ has removable singularity at $a$. But here how can I use the continuity condition to show $f$ is analytic ?






complex-analysis analysis complex-numbers







share|cite|improve this question











$endgroup$





Let, $f$ be analytic in $Dsetminus {a}$. Also, $f$ is continuous at $z=a$. Then prove that $f$ is analytic on whole of $D$.




We know from Riemann's Theorem, if $f$ is not analytic at $a$ and in a nbd. of $a$, if $f$ is bounded then either $f$ is analytic at $a$ or $f$ has removable singularity at $a$. But here how can I use the continuity condition to show $f$ is analytic ?






complex-analysis analysis complex-numbers




complex-analysis analysis complex-numbers






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Nov 25 '18 at 6:13







Topo

















asked Nov 24 '18 at 16:13









TopoTopo

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311214












  • $begingroup$
    $f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
    $endgroup$
    – UserS
    Nov 24 '18 at 16:35




















  • $begingroup$
    $f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
    $endgroup$
    – UserS
    Nov 24 '18 at 16:35


















$begingroup$
$f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
$endgroup$
– UserS
Nov 24 '18 at 16:35






$begingroup$
$f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
$endgroup$
– UserS
Nov 24 '18 at 16:35












1 Answer
1






active

oldest

votes


















1












$begingroup$

It is not restrictive to assume that $a=0$ (with a translation).



Consider $g(z)=z^2f(z)$, for $zin D$, $zne0$, and $g(0)=0$.



Prove that $g$ is differentiable at $0$. Thus it is differentiable over $D$, hence analytic over $D$. Hence it has a Taylor series expansion at $0$:
$$
g(z)=sum_{n=0}^{infty}b_nz^n
$$
but $b_0=b_1=0$ (prove it). Hence
$$
f(z)=sum_{n=0}^{infty}b_{n+2}z^n
$$
is analytic in a neighborhood of $0$.



(Adapted from Wikipedia)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Where you use the continuity?
    $endgroup$
    – Topo
    Nov 24 '18 at 17:04










  • $begingroup$
    @Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
    $endgroup$
    – egreg
    Nov 24 '18 at 17:52












  • $begingroup$
    Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
    $endgroup$
    – Topo
    Nov 24 '18 at 18:14










  • $begingroup$
    I've also doubt about my own question. Since $f$ is not analytic at $z=a$ and is continuous, so it has a singularity at $z=a$. As, $f$ is continuous at $z=a$ so the singularity must be removable. After removing that singularity we can redefine an analytic function in the whole domain $D$. But in that case the redefined function is the analytic extension of $f$. But how can I say about the analiticity of $f$ at $z=a$ ?
    $endgroup$
    – Topo
    Nov 24 '18 at 18:19












  • $begingroup$
    @Topo What's the doubt? Perhaps the phrasing can be ambiguous: we know that the function $f$ is analytic in $Dsetminus{a}$ and continuous at $a$; the claim is that $f$ is also analytic at $a$. The problem states this and not that $f$ is not analytic at $a$ (because it is analytic at $a$, as proved).
    $endgroup$
    – egreg
    Nov 24 '18 at 18:57













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1 Answer
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1 Answer
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active

oldest

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active

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active

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1












$begingroup$

It is not restrictive to assume that $a=0$ (with a translation).



Consider $g(z)=z^2f(z)$, for $zin D$, $zne0$, and $g(0)=0$.



Prove that $g$ is differentiable at $0$. Thus it is differentiable over $D$, hence analytic over $D$. Hence it has a Taylor series expansion at $0$:
$$
g(z)=sum_{n=0}^{infty}b_nz^n
$$
but $b_0=b_1=0$ (prove it). Hence
$$
f(z)=sum_{n=0}^{infty}b_{n+2}z^n
$$
is analytic in a neighborhood of $0$.



(Adapted from Wikipedia)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Where you use the continuity?
    $endgroup$
    – Topo
    Nov 24 '18 at 17:04










  • $begingroup$
    @Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
    $endgroup$
    – egreg
    Nov 24 '18 at 17:52












  • $begingroup$
    Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
    $endgroup$
    – Topo
    Nov 24 '18 at 18:14










  • $begingroup$
    I've also doubt about my own question. Since $f$ is not analytic at $z=a$ and is continuous, so it has a singularity at $z=a$. As, $f$ is continuous at $z=a$ so the singularity must be removable. After removing that singularity we can redefine an analytic function in the whole domain $D$. But in that case the redefined function is the analytic extension of $f$. But how can I say about the analiticity of $f$ at $z=a$ ?
    $endgroup$
    – Topo
    Nov 24 '18 at 18:19












  • $begingroup$
    @Topo What's the doubt? Perhaps the phrasing can be ambiguous: we know that the function $f$ is analytic in $Dsetminus{a}$ and continuous at $a$; the claim is that $f$ is also analytic at $a$. The problem states this and not that $f$ is not analytic at $a$ (because it is analytic at $a$, as proved).
    $endgroup$
    – egreg
    Nov 24 '18 at 18:57


















1












$begingroup$

It is not restrictive to assume that $a=0$ (with a translation).



Consider $g(z)=z^2f(z)$, for $zin D$, $zne0$, and $g(0)=0$.



Prove that $g$ is differentiable at $0$. Thus it is differentiable over $D$, hence analytic over $D$. Hence it has a Taylor series expansion at $0$:
$$
g(z)=sum_{n=0}^{infty}b_nz^n
$$
but $b_0=b_1=0$ (prove it). Hence
$$
f(z)=sum_{n=0}^{infty}b_{n+2}z^n
$$
is analytic in a neighborhood of $0$.



(Adapted from Wikipedia)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Where you use the continuity?
    $endgroup$
    – Topo
    Nov 24 '18 at 17:04










  • $begingroup$
    @Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
    $endgroup$
    – egreg
    Nov 24 '18 at 17:52












  • $begingroup$
    Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
    $endgroup$
    – Topo
    Nov 24 '18 at 18:14










  • $begingroup$
    I've also doubt about my own question. Since $f$ is not analytic at $z=a$ and is continuous, so it has a singularity at $z=a$. As, $f$ is continuous at $z=a$ so the singularity must be removable. After removing that singularity we can redefine an analytic function in the whole domain $D$. But in that case the redefined function is the analytic extension of $f$. But how can I say about the analiticity of $f$ at $z=a$ ?
    $endgroup$
    – Topo
    Nov 24 '18 at 18:19












  • $begingroup$
    @Topo What's the doubt? Perhaps the phrasing can be ambiguous: we know that the function $f$ is analytic in $Dsetminus{a}$ and continuous at $a$; the claim is that $f$ is also analytic at $a$. The problem states this and not that $f$ is not analytic at $a$ (because it is analytic at $a$, as proved).
    $endgroup$
    – egreg
    Nov 24 '18 at 18:57
















1












1








1





$begingroup$

It is not restrictive to assume that $a=0$ (with a translation).



Consider $g(z)=z^2f(z)$, for $zin D$, $zne0$, and $g(0)=0$.



Prove that $g$ is differentiable at $0$. Thus it is differentiable over $D$, hence analytic over $D$. Hence it has a Taylor series expansion at $0$:
$$
g(z)=sum_{n=0}^{infty}b_nz^n
$$
but $b_0=b_1=0$ (prove it). Hence
$$
f(z)=sum_{n=0}^{infty}b_{n+2}z^n
$$
is analytic in a neighborhood of $0$.



(Adapted from Wikipedia)






share|cite|improve this answer









$endgroup$



It is not restrictive to assume that $a=0$ (with a translation).



Consider $g(z)=z^2f(z)$, for $zin D$, $zne0$, and $g(0)=0$.



Prove that $g$ is differentiable at $0$. Thus it is differentiable over $D$, hence analytic over $D$. Hence it has a Taylor series expansion at $0$:
$$
g(z)=sum_{n=0}^{infty}b_nz^n
$$
but $b_0=b_1=0$ (prove it). Hence
$$
f(z)=sum_{n=0}^{infty}b_{n+2}z^n
$$
is analytic in a neighborhood of $0$.



(Adapted from Wikipedia)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 '18 at 16:36









egregegreg

180k1485202




180k1485202












  • $begingroup$
    Where you use the continuity?
    $endgroup$
    – Topo
    Nov 24 '18 at 17:04










  • $begingroup$
    @Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
    $endgroup$
    – egreg
    Nov 24 '18 at 17:52












  • $begingroup$
    Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
    $endgroup$
    – Topo
    Nov 24 '18 at 18:14










  • $begingroup$
    I've also doubt about my own question. Since $f$ is not analytic at $z=a$ and is continuous, so it has a singularity at $z=a$. As, $f$ is continuous at $z=a$ so the singularity must be removable. After removing that singularity we can redefine an analytic function in the whole domain $D$. But in that case the redefined function is the analytic extension of $f$. But how can I say about the analiticity of $f$ at $z=a$ ?
    $endgroup$
    – Topo
    Nov 24 '18 at 18:19












  • $begingroup$
    @Topo What's the doubt? Perhaps the phrasing can be ambiguous: we know that the function $f$ is analytic in $Dsetminus{a}$ and continuous at $a$; the claim is that $f$ is also analytic at $a$. The problem states this and not that $f$ is not analytic at $a$ (because it is analytic at $a$, as proved).
    $endgroup$
    – egreg
    Nov 24 '18 at 18:57




















  • $begingroup$
    Where you use the continuity?
    $endgroup$
    – Topo
    Nov 24 '18 at 17:04










  • $begingroup$
    @Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
    $endgroup$
    – egreg
    Nov 24 '18 at 17:52












  • $begingroup$
    Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
    $endgroup$
    – Topo
    Nov 24 '18 at 18:14










  • $begingroup$
    I've also doubt about my own question. Since $f$ is not analytic at $z=a$ and is continuous, so it has a singularity at $z=a$. As, $f$ is continuous at $z=a$ so the singularity must be removable. After removing that singularity we can redefine an analytic function in the whole domain $D$. But in that case the redefined function is the analytic extension of $f$. But how can I say about the analiticity of $f$ at $z=a$ ?
    $endgroup$
    – Topo
    Nov 24 '18 at 18:19












  • $begingroup$
    @Topo What's the doubt? Perhaps the phrasing can be ambiguous: we know that the function $f$ is analytic in $Dsetminus{a}$ and continuous at $a$; the claim is that $f$ is also analytic at $a$. The problem states this and not that $f$ is not analytic at $a$ (because it is analytic at $a$, as proved).
    $endgroup$
    – egreg
    Nov 24 '18 at 18:57


















$begingroup$
Where you use the continuity?
$endgroup$
– Topo
Nov 24 '18 at 17:04




$begingroup$
Where you use the continuity?
$endgroup$
– Topo
Nov 24 '18 at 17:04












$begingroup$
@Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
$endgroup$
– egreg
Nov 24 '18 at 17:52






$begingroup$
@Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
$endgroup$
– egreg
Nov 24 '18 at 17:52














$begingroup$
Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
$endgroup$
– Topo
Nov 24 '18 at 18:14




$begingroup$
Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
$endgroup$
– Topo
Nov 24 '18 at 18:14












$begingroup$
I've also doubt about my own question. Since $f$ is not analytic at $z=a$ and is continuous, so it has a singularity at $z=a$. As, $f$ is continuous at $z=a$ so the singularity must be removable. After removing that singularity we can redefine an analytic function in the whole domain $D$. But in that case the redefined function is the analytic extension of $f$. But how can I say about the analiticity of $f$ at $z=a$ ?
$endgroup$
– Topo
Nov 24 '18 at 18:19






$begingroup$
I've also doubt about my own question. Since $f$ is not analytic at $z=a$ and is continuous, so it has a singularity at $z=a$. As, $f$ is continuous at $z=a$ so the singularity must be removable. After removing that singularity we can redefine an analytic function in the whole domain $D$. But in that case the redefined function is the analytic extension of $f$. But how can I say about the analiticity of $f$ at $z=a$ ?
$endgroup$
– Topo
Nov 24 '18 at 18:19














$begingroup$
@Topo What's the doubt? Perhaps the phrasing can be ambiguous: we know that the function $f$ is analytic in $Dsetminus{a}$ and continuous at $a$; the claim is that $f$ is also analytic at $a$. The problem states this and not that $f$ is not analytic at $a$ (because it is analytic at $a$, as proved).
$endgroup$
– egreg
Nov 24 '18 at 18:57






$begingroup$
@Topo What's the doubt? Perhaps the phrasing can be ambiguous: we know that the function $f$ is analytic in $Dsetminus{a}$ and continuous at $a$; the claim is that $f$ is also analytic at $a$. The problem states this and not that $f$ is not analytic at $a$ (because it is analytic at $a$, as proved).
$endgroup$
– egreg
Nov 24 '18 at 18:57




















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