For complex function : $f'(z)$ exists $implies$ $f$ continous in $z$?Singularity structure of function in the complex plane.Does every continuous function have a left and right derivative?Curves Bounding Complex FunctionA question regarding complex valued functionDifferentiation and continous functionsDifferences between the complex derivative and the multivariable derivative.Continuity of complex difference quotientFunction limiting its own derivative implies implies null functionIf $f$ is a complex function with constant derivativeDifferential Equations Proof for Equilibrium
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For complex function : $f'(z)$ exists $implies$ $f$ continous in $z$?
Singularity structure of function in the complex plane.Does every continuous function have a left and right derivative?Curves Bounding Complex FunctionA question regarding complex valued functionDifferentiation and continous functionsDifferences between the complex derivative and the multivariable derivative.Continuity of complex difference quotientFunction limiting its own derivative implies implies null functionIf $f$ is a complex function with constant derivativeDifferential Equations Proof for Equilibrium
$begingroup$
Consider a complex function $f(z): Asubsetmathbb C tomathbb C$.
If the derivative of $f$ exists then $f$ must necessarily be a continuous function?
Is the following true? $f'(z)$ exists $implies$ $f$ continuous in $z$
complex-analysis functions derivatives complex-numbers continuity
edited Mar 29 at 22:11
blub
3,241829
asked Mar 29 at 22:04
SørënSørën
1059
$endgroup$
add a comment |
$begingroup$
Consider a complex function $f(z): Asubsetmathbb C tomathbb C$.
If the derivative of $f$ exists then $f$ must necessarily be a continuous function?
Is the following true? $f'(z)$ exists $implies$ $f$ continuous in $z$
complex-analysis functions derivatives complex-numbers continuity
edited Mar 29 at 22:11
blub
3,241829
asked Mar 29 at 22:04
SørënSørën
1059
$endgroup$
5
$begingroup$
Yes, it is true.
$endgroup$
– Mark Viola
Mar 29 at 22:06
$begingroup$
To go a step further, not only is it continuous, it is analytic.
$endgroup$
– Clayton
Mar 29 at 22:13
1
$begingroup$
You can prove it directly from the definition of a derivative. Just consider $fracf(z+h)-f(z)hcdot h$ . You do it just like for the functions of real variable.
$endgroup$
– Jakobian
Mar 29 at 22:14
add a comment |
$begingroup$
Consider a complex function $f(z): Asubsetmathbb C tomathbb C$.
If the derivative of $f$ exists then $f$ must necessarily be a continuous function?
Is the following true? $f'(z)$ exists $implies$ $f$ continuous in $z$
complex-analysis functions derivatives complex-numbers continuity
edited Mar 29 at 22:11
blub
3,241829
asked Mar 29 at 22:04
SørënSørën
1059
$endgroup$
Consider a complex function $f(z): Asubsetmathbb C tomathbb C$.
If the derivative of $f$ exists then $f$ must necessarily be a continuous function?
Is the following true? $f'(z)$ exists $implies$ $f$ continuous in $z$
complex-analysis functions derivatives complex-numbers continuity
complex-analysis functions derivatives complex-numbers continuity
edited Mar 29 at 22:11
blub
3,241829
asked Mar 29 at 22:04
SørënSørën
1059
edited Mar 29 at 22:11
blub
3,241829
asked Mar 29 at 22:04
SørënSørën
1059
edited Mar 29 at 22:11
blub
3,241829
edited Mar 29 at 22:11
blub
3,241829
edited Mar 29 at 22:11
blub
3,241829
3,241829
asked Mar 29 at 22:04
SørënSørën
1059
asked Mar 29 at 22:04
SørënSørën
1059
asked Mar 29 at 22:04
SørënSørën
1059
1059
5
$begingroup$
Yes, it is true.
$endgroup$
– Mark Viola
Mar 29 at 22:06
$begingroup$
To go a step further, not only is it continuous, it is analytic.
$endgroup$
– Clayton
Mar 29 at 22:13
1
$begingroup$
You can prove it directly from the definition of a derivative. Just consider $fracf(z+h)-f(z)hcdot h$ . You do it just like for the functions of real variable.
$endgroup$
– Jakobian
Mar 29 at 22:14
add a comment |
5
$begingroup$
Yes, it is true.
$endgroup$
– Mark Viola
Mar 29 at 22:06
$begingroup$
To go a step further, not only is it continuous, it is analytic.
$endgroup$
– Clayton
Mar 29 at 22:13
1
$begingroup$
You can prove it directly from the definition of a derivative. Just consider $fracf(z+h)-f(z)hcdot h$ . You do it just like for the functions of real variable.
$endgroup$
– Jakobian
Mar 29 at 22:14
5
5
$begingroup$
Yes, it is true.
$endgroup$
– Mark Viola
Mar 29 at 22:06
$begingroup$
Yes, it is true.
$endgroup$
– Mark Viola
Mar 29 at 22:06
$begingroup$
To go a step further, not only is it continuous, it is analytic.
$endgroup$
– Clayton
Mar 29 at 22:13
$begingroup$
To go a step further, not only is it continuous, it is analytic.
$endgroup$
– Clayton
Mar 29 at 22:13
1
1
$begingroup$
You can prove it directly from the definition of a derivative. Just consider $fracf(z+h)-f(z)hcdot h$ . You do it just like for the functions of real variable.
$endgroup$
– Jakobian
Mar 29 at 22:14
$begingroup$
You can prove it directly from the definition of a derivative. Just consider $fracf(z+h)-f(z)hcdot h$ . You do it just like for the functions of real variable.
$endgroup$
– Jakobian
Mar 29 at 22:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A quick proof: Let $f:Atomathbb C$ and let $z_0in A$ be an accumulation point of $A$. Suppose $f'(z_0)$ exists, i.e. the limit
$$lim_zto z_0fracf(z)-f(z_0)z-z_0=L$$
exists and $Linmathbb C$. Then we have
$$lim_zto z_0f(z)-f(z_0)=lim_zto z_0(f(z)-f(z_0))=lim_zto z_0fracf(z)-f(z_0)z-z_0(z-z_0)=f'(z_0)cdot 0=0$$
Thus, $lim_zto z_0f(z)=f(z_0)$ and $f$ is continuous in $z_0$.
answered Mar 29 at 22:17
blubblub
3,241829
$endgroup$
$begingroup$
Albeit subtle, the assumption should be that the limit exists and is finite. Also, you repeated $lim_zto z_0(f(z)-f(z_0))$ twice although there were missing outer parentheses in the first occurrence (perhaps this is something you intended). Lastly, the proof relies on the theorem that the limit of a product of complex functions equals the product of the limits (when both limits exist and are finite) of the complex functions.
$endgroup$
– Mark Viola
Mar 29 at 22:27
1
$begingroup$
@MarkViola With "the limit exists" I mean that the limit is a member in the considered space, e.g. a real or complex number. Also, this is not a repetition, but the limit expression is meant to be read as any unary connective, i.e. for $lim_zto z_0f(z)=L$, the expression $lim_zto z_0f(z)-f(z_0)$ means $L-f(z_0)$. Lastly, I agree the proof relies on the properties of limits that the distribute over products and sums, if the corresponding inner limits exist.
$endgroup$
– blub
Mar 29 at 22:31
$begingroup$
Why would $A$ be open though. It would suffice if $z_0$ would be an accumulation point of $A$
$endgroup$
– Jakobian
Mar 29 at 22:34
$begingroup$
@Jakobian Indeed it would, out of habit I just projected my usual setting on the answer. But I'll edit that in thanks for the remark.
$endgroup$
– blub
Mar 29 at 22:35
$begingroup$
I agree with the first point, but you might just consider making it explicit just to tighten things a bit. As for the second point, I had commented parenthetically that you might have intended this step.
$endgroup$
– Mark Viola
Mar 29 at 22:47
|
show 2 more comments
Your Answer
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$begingroup$
A quick proof: Let $f:Atomathbb C$ and let $z_0in A$ be an accumulation point of $A$. Suppose $f'(z_0)$ exists, i.e. the limit
$$lim_zto z_0fracf(z)-f(z_0)z-z_0=L$$
exists and $Linmathbb C$. Then we have
$$lim_zto z_0f(z)-f(z_0)=lim_zto z_0(f(z)-f(z_0))=lim_zto z_0fracf(z)-f(z_0)z-z_0(z-z_0)=f'(z_0)cdot 0=0$$
Thus, $lim_zto z_0f(z)=f(z_0)$ and $f$ is continuous in $z_0$.
answered Mar 29 at 22:17
blubblub
3,241829
$endgroup$
$begingroup$
Albeit subtle, the assumption should be that the limit exists and is finite. Also, you repeated $lim_zto z_0(f(z)-f(z_0))$ twice although there were missing outer parentheses in the first occurrence (perhaps this is something you intended). Lastly, the proof relies on the theorem that the limit of a product of complex functions equals the product of the limits (when both limits exist and are finite) of the complex functions.
$endgroup$
– Mark Viola
Mar 29 at 22:27
1
$begingroup$
@MarkViola With "the limit exists" I mean that the limit is a member in the considered space, e.g. a real or complex number. Also, this is not a repetition, but the limit expression is meant to be read as any unary connective, i.e. for $lim_zto z_0f(z)=L$, the expression $lim_zto z_0f(z)-f(z_0)$ means $L-f(z_0)$. Lastly, I agree the proof relies on the properties of limits that the distribute over products and sums, if the corresponding inner limits exist.
$endgroup$
– blub
Mar 29 at 22:31
$begingroup$
Why would $A$ be open though. It would suffice if $z_0$ would be an accumulation point of $A$
$endgroup$
– Jakobian
Mar 29 at 22:34
$begingroup$
@Jakobian Indeed it would, out of habit I just projected my usual setting on the answer. But I'll edit that in thanks for the remark.
$endgroup$
– blub
Mar 29 at 22:35
$begingroup$
I agree with the first point, but you might just consider making it explicit just to tighten things a bit. As for the second point, I had commented parenthetically that you might have intended this step.
$endgroup$
– Mark Viola
Mar 29 at 22:47
|
show 2 more comments
$begingroup$
A quick proof: Let $f:Atomathbb C$ and let $z_0in A$ be an accumulation point of $A$. Suppose $f'(z_0)$ exists, i.e. the limit
$$lim_zto z_0fracf(z)-f(z_0)z-z_0=L$$
exists and $Linmathbb C$. Then we have
$$lim_zto z_0f(z)-f(z_0)=lim_zto z_0(f(z)-f(z_0))=lim_zto z_0fracf(z)-f(z_0)z-z_0(z-z_0)=f'(z_0)cdot 0=0$$
Thus, $lim_zto z_0f(z)=f(z_0)$ and $f$ is continuous in $z_0$.
answered Mar 29 at 22:17
blubblub
3,241829
$endgroup$
$begingroup$
Albeit subtle, the assumption should be that the limit exists and is finite. Also, you repeated $lim_zto z_0(f(z)-f(z_0))$ twice although there were missing outer parentheses in the first occurrence (perhaps this is something you intended). Lastly, the proof relies on the theorem that the limit of a product of complex functions equals the product of the limits (when both limits exist and are finite) of the complex functions.
$endgroup$
– Mark Viola
Mar 29 at 22:27
1
$begingroup$
@MarkViola With "the limit exists" I mean that the limit is a member in the considered space, e.g. a real or complex number. Also, this is not a repetition, but the limit expression is meant to be read as any unary connective, i.e. for $lim_zto z_0f(z)=L$, the expression $lim_zto z_0f(z)-f(z_0)$ means $L-f(z_0)$. Lastly, I agree the proof relies on the properties of limits that the distribute over products and sums, if the corresponding inner limits exist.
$endgroup$
– blub
Mar 29 at 22:31
$begingroup$
Why would $A$ be open though. It would suffice if $z_0$ would be an accumulation point of $A$
$endgroup$
– Jakobian
Mar 29 at 22:34
$begingroup$
@Jakobian Indeed it would, out of habit I just projected my usual setting on the answer. But I'll edit that in thanks for the remark.
$endgroup$
– blub
Mar 29 at 22:35
$begingroup$
I agree with the first point, but you might just consider making it explicit just to tighten things a bit. As for the second point, I had commented parenthetically that you might have intended this step.
$endgroup$
– Mark Viola
Mar 29 at 22:47
|
show 2 more comments
$begingroup$
A quick proof: Let $f:Atomathbb C$ and let $z_0in A$ be an accumulation point of $A$. Suppose $f'(z_0)$ exists, i.e. the limit
$$lim_zto z_0fracf(z)-f(z_0)z-z_0=L$$
exists and $Linmathbb C$. Then we have
$$lim_zto z_0f(z)-f(z_0)=lim_zto z_0(f(z)-f(z_0))=lim_zto z_0fracf(z)-f(z_0)z-z_0(z-z_0)=f'(z_0)cdot 0=0$$
Thus, $lim_zto z_0f(z)=f(z_0)$ and $f$ is continuous in $z_0$.
answered Mar 29 at 22:17
blubblub
3,241829
$endgroup$
A quick proof: Let $f:Atomathbb C$ and let $z_0in A$ be an accumulation point of $A$. Suppose $f'(z_0)$ exists, i.e. the limit
$$lim_zto z_0fracf(z)-f(z_0)z-z_0=L$$
exists and $Linmathbb C$. Then we have
$$lim_zto z_0f(z)-f(z_0)=lim_zto z_0(f(z)-f(z_0))=lim_zto z_0fracf(z)-f(z_0)z-z_0(z-z_0)=f'(z_0)cdot 0=0$$
Thus, $lim_zto z_0f(z)=f(z_0)$ and $f$ is continuous in $z_0$.
answered Mar 29 at 22:17
blubblub
3,241829
edited Mar 29 at 22:50
answered Mar 29 at 22:17
blubblub
3,241829
answered Mar 29 at 22:17
blubblub
3,241829
answered Mar 29 at 22:17
blubblub
3,241829
3,241829
$begingroup$
Albeit subtle, the assumption should be that the limit exists and is finite. Also, you repeated $lim_zto z_0(f(z)-f(z_0))$ twice although there were missing outer parentheses in the first occurrence (perhaps this is something you intended). Lastly, the proof relies on the theorem that the limit of a product of complex functions equals the product of the limits (when both limits exist and are finite) of the complex functions.
$endgroup$
– Mark Viola
Mar 29 at 22:27
1
$begingroup$
@MarkViola With "the limit exists" I mean that the limit is a member in the considered space, e.g. a real or complex number. Also, this is not a repetition, but the limit expression is meant to be read as any unary connective, i.e. for $lim_zto z_0f(z)=L$, the expression $lim_zto z_0f(z)-f(z_0)$ means $L-f(z_0)$. Lastly, I agree the proof relies on the properties of limits that the distribute over products and sums, if the corresponding inner limits exist.
$endgroup$
– blub
Mar 29 at 22:31
$begingroup$
Why would $A$ be open though. It would suffice if $z_0$ would be an accumulation point of $A$
$endgroup$
– Jakobian
Mar 29 at 22:34
$begingroup$
@Jakobian Indeed it would, out of habit I just projected my usual setting on the answer. But I'll edit that in thanks for the remark.
$endgroup$
– blub
Mar 29 at 22:35
$begingroup$
I agree with the first point, but you might just consider making it explicit just to tighten things a bit. As for the second point, I had commented parenthetically that you might have intended this step.
$endgroup$
– Mark Viola
Mar 29 at 22:47
|
show 2 more comments
$begingroup$
Albeit subtle, the assumption should be that the limit exists and is finite. Also, you repeated $lim_zto z_0(f(z)-f(z_0))$ twice although there were missing outer parentheses in the first occurrence (perhaps this is something you intended). Lastly, the proof relies on the theorem that the limit of a product of complex functions equals the product of the limits (when both limits exist and are finite) of the complex functions.
$endgroup$
– Mark Viola
Mar 29 at 22:27
1
$begingroup$
@MarkViola With "the limit exists" I mean that the limit is a member in the considered space, e.g. a real or complex number. Also, this is not a repetition, but the limit expression is meant to be read as any unary connective, i.e. for $lim_zto z_0f(z)=L$, the expression $lim_zto z_0f(z)-f(z_0)$ means $L-f(z_0)$. Lastly, I agree the proof relies on the properties of limits that the distribute over products and sums, if the corresponding inner limits exist.
$endgroup$
– blub
Mar 29 at 22:31
$begingroup$
Why would $A$ be open though. It would suffice if $z_0$ would be an accumulation point of $A$
$endgroup$
– Jakobian
Mar 29 at 22:34
$begingroup$
@Jakobian Indeed it would, out of habit I just projected my usual setting on the answer. But I'll edit that in thanks for the remark.
$endgroup$
– blub
Mar 29 at 22:35
$begingroup$
I agree with the first point, but you might just consider making it explicit just to tighten things a bit. As for the second point, I had commented parenthetically that you might have intended this step.
$endgroup$
– Mark Viola
Mar 29 at 22:47
$begingroup$
Albeit subtle, the assumption should be that the limit exists and is finite. Also, you repeated $lim_zto z_0(f(z)-f(z_0))$ twice although there were missing outer parentheses in the first occurrence (perhaps this is something you intended). Lastly, the proof relies on the theorem that the limit of a product of complex functions equals the product of the limits (when both limits exist and are finite) of the complex functions.
$endgroup$
– Mark Viola
Mar 29 at 22:27
$begingroup$
Albeit subtle, the assumption should be that the limit exists and is finite. Also, you repeated $lim_zto z_0(f(z)-f(z_0))$ twice although there were missing outer parentheses in the first occurrence (perhaps this is something you intended). Lastly, the proof relies on the theorem that the limit of a product of complex functions equals the product of the limits (when both limits exist and are finite) of the complex functions.
$endgroup$
– Mark Viola
Mar 29 at 22:27
1
1
$begingroup$
@MarkViola With "the limit exists" I mean that the limit is a member in the considered space, e.g. a real or complex number. Also, this is not a repetition, but the limit expression is meant to be read as any unary connective, i.e. for $lim_zto z_0f(z)=L$, the expression $lim_zto z_0f(z)-f(z_0)$ means $L-f(z_0)$. Lastly, I agree the proof relies on the properties of limits that the distribute over products and sums, if the corresponding inner limits exist.
$endgroup$
– blub
Mar 29 at 22:31
$begingroup$
@MarkViola With "the limit exists" I mean that the limit is a member in the considered space, e.g. a real or complex number. Also, this is not a repetition, but the limit expression is meant to be read as any unary connective, i.e. for $lim_zto z_0f(z)=L$, the expression $lim_zto z_0f(z)-f(z_0)$ means $L-f(z_0)$. Lastly, I agree the proof relies on the properties of limits that the distribute over products and sums, if the corresponding inner limits exist.
$endgroup$
– blub
Mar 29 at 22:31
$begingroup$
Why would $A$ be open though. It would suffice if $z_0$ would be an accumulation point of $A$
$endgroup$
– Jakobian
Mar 29 at 22:34
$begingroup$
Why would $A$ be open though. It would suffice if $z_0$ would be an accumulation point of $A$
$endgroup$
– Jakobian
Mar 29 at 22:34
$begingroup$
@Jakobian Indeed it would, out of habit I just projected my usual setting on the answer. But I'll edit that in thanks for the remark.
$endgroup$
– blub
Mar 29 at 22:35
$begingroup$
@Jakobian Indeed it would, out of habit I just projected my usual setting on the answer. But I'll edit that in thanks for the remark.
$endgroup$
– blub
Mar 29 at 22:35
$begingroup$
I agree with the first point, but you might just consider making it explicit just to tighten things a bit. As for the second point, I had commented parenthetically that you might have intended this step.
$endgroup$
– Mark Viola
Mar 29 at 22:47
$begingroup$
I agree with the first point, but you might just consider making it explicit just to tighten things a bit. As for the second point, I had commented parenthetically that you might have intended this step.
$endgroup$
– Mark Viola
Mar 29 at 22:47
|
show 2 more comments
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5
$begingroup$
Yes, it is true.
$endgroup$
– Mark Viola
Mar 29 at 22:06
$begingroup$
To go a step further, not only is it continuous, it is analytic.
$endgroup$
– Clayton
Mar 29 at 22:13
1
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You can prove it directly from the definition of a derivative. Just consider $fracf(z+h)-f(z)hcdot h$ . You do it just like for the functions of real variable.
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– Jakobian
Mar 29 at 22:14