If $f$ is continuous at $z=a$ in a domain $D$ then $f$ is analytic on whole of $D$
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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
asked Nov 24 at 16:13
Topo
292214
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up vote
1
down vote
favorite
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
asked Nov 24 at 16:13
Topo
292214
$f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
– UserS
Nov 24 at 16:35
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 24 at 16:13
Topo
292214
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
asked Nov 24 at 16:13
Topo
292214
asked Nov 24 at 16:13
Topo
292214
edited Nov 25 at 6:13
asked Nov 24 at 16:13
Topo
292214
asked Nov 24 at 16:13
Topo
292214
asked Nov 24 at 16:13
Topo
292214
292214
$f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
– UserS
Nov 24 at 16:35
add a comment |
$f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
– UserS
Nov 24 at 16:35
$f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
– UserS
Nov 24 at 16:35
$f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
– UserS
Nov 24 at 16:35
add a comment |
1 Answer
1
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1
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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)
answered Nov 24 at 16:36
egreg
175k1383198
Where you use the continuity?
– Topo
Nov 24 at 17:04
@Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
– egreg
Nov 24 at 17:52
Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
– Topo
Nov 24 at 18:14
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 $z=a$ ?
– Topo
Nov 24 at 18:19
@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).
– egreg
Nov 24 at 18:57
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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)
answered Nov 24 at 16:36
egreg
175k1383198
Where you use the continuity?
– Topo
Nov 24 at 17:04
@Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
– egreg
Nov 24 at 17:52
Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
– Topo
Nov 24 at 18:14
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 $z=a$ ?
– Topo
Nov 24 at 18:19
@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).
– egreg
Nov 24 at 18:57
|
show 2 more comments
up vote
1
down vote
accepted
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)
answered Nov 24 at 16:36
egreg
175k1383198
Where you use the continuity?
– Topo
Nov 24 at 17:04
@Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
– egreg
Nov 24 at 17:52
Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
– Topo
Nov 24 at 18:14
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 $z=a$ ?
– Topo
Nov 24 at 18:19
@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).
– egreg
Nov 24 at 18:57
|
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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)
answered Nov 24 at 16:36
egreg
175k1383198
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)
answered Nov 24 at 16:36
egreg
175k1383198
answered Nov 24 at 16:36
egreg
175k1383198
answered Nov 24 at 16:36
egreg
175k1383198
answered Nov 24 at 16:36
egreg
175k1383198
175k1383198
Where you use the continuity?
– Topo
Nov 24 at 17:04
@Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
– egreg
Nov 24 at 17:52
Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
– Topo
Nov 24 at 18:14
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 $z=a$ ?
– Topo
Nov 24 at 18:19
@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).
– egreg
Nov 24 at 18:57
|
show 2 more comments
Where you use the continuity?
– Topo
Nov 24 at 17:04
@Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
– egreg
Nov 24 at 17:52
Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
– Topo
Nov 24 at 18:14
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 $z=a$ ?
– Topo
Nov 24 at 18:19
@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).
– egreg
Nov 24 at 18:57
Where you use the continuity?
– Topo
Nov 24 at 17:04
Where you use the continuity?
– Topo
Nov 24 at 17:04
@Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
– egreg
Nov 24 at 17:52
@Topo In showing $g$ is differentiable at $0$, boundedness of $f$ in a neighborhood of $0$ is needed.
– egreg
Nov 24 at 17:52
Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
– Topo
Nov 24 at 18:14
Thus we get an analytic extension of $f$ at $z=0$. But $f$ is not analytic at $z=0$.
– Topo
Nov 24 at 18:14
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 $z=a$ ?
– Topo
Nov 24 at 18:19
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 $z=a$ ?
– Topo
Nov 24 at 18:19
@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).
– egreg
Nov 24 at 18:57
@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).
– egreg
Nov 24 at 18:57
|
show 2 more comments
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$f$ is continuous at $aimplies f$ is bounded in a nbd of $a$.
– UserS
Nov 24 at 16:35