Can there be a non-isolated “pole” or “removable singularity”?
A pole or removable or even essential singularity must be isolated a priori. But still we can try to talk about the limit of the function at the point even on a disk removing some (countable amount of) points.
A well-know example of non-isolated singularity will be $z=0$ for $$f(z)=frac{1}{sin(frac{1}{z})}.$$
But $lim_{zto0}f(z)$ does not exist. My question is can there be a function with non-isolated singularity at $0$ with $lim_{zto0}|f(z)|=infty$ or even $lim_{zto0}f(z)=c$?
complex-analysis
asked Nov 25 '18 at 22:03
CO2
1529
add a comment |
A pole or removable or even essential singularity must be isolated a priori. But still we can try to talk about the limit of the function at the point even on a disk removing some (countable amount of) points.
A well-know example of non-isolated singularity will be $z=0$ for $$f(z)=frac{1}{sin(frac{1}{z})}.$$
But $lim_{zto0}f(z)$ does not exist. My question is can there be a function with non-isolated singularity at $0$ with $lim_{zto0}|f(z)|=infty$ or even $lim_{zto0}f(z)=c$?
complex-analysis
asked Nov 25 '18 at 22:03
CO2
1529
It seems you are calling "singularity" a point $a$ such that $f$ is meromorphic on $0 < |z-a| < epsilon$. In that case or $f$ is analytic on $0 < |z-a| < epsilon$ and one of the limit exists iff $f $ is meromorphic at $a$, or $f$ has infinitely many poles around $z=a$ and the limits don't exist (assume $lim_{zto a}|f(z)|=infty$ and look at $1/f$)
– reuns
Nov 25 '18 at 22:33
add a comment |
A pole or removable or even essential singularity must be isolated a priori. But still we can try to talk about the limit of the function at the point even on a disk removing some (countable amount of) points.
A well-know example of non-isolated singularity will be $z=0$ for $$f(z)=frac{1}{sin(frac{1}{z})}.$$
But $lim_{zto0}f(z)$ does not exist. My question is can there be a function with non-isolated singularity at $0$ with $lim_{zto0}|f(z)|=infty$ or even $lim_{zto0}f(z)=c$?
complex-analysis
asked Nov 25 '18 at 22:03
CO2
1529
A pole or removable or even essential singularity must be isolated a priori. But still we can try to talk about the limit of the function at the point even on a disk removing some (countable amount of) points.
A well-know example of non-isolated singularity will be $z=0$ for $$f(z)=frac{1}{sin(frac{1}{z})}.$$
But $lim_{zto0}f(z)$ does not exist. My question is can there be a function with non-isolated singularity at $0$ with $lim_{zto0}|f(z)|=infty$ or even $lim_{zto0}f(z)=c$?
complex-analysis
complex-analysis
asked Nov 25 '18 at 22:03
CO2
1529
asked Nov 25 '18 at 22:03
CO2
1529
edited Nov 25 '18 at 22:23
asked Nov 25 '18 at 22:03
CO2
1529
asked Nov 25 '18 at 22:03
CO2
1529
asked Nov 25 '18 at 22:03
CO2
1529
1529
It seems you are calling "singularity" a point $a$ such that $f$ is meromorphic on $0 < |z-a| < epsilon$. In that case or $f$ is analytic on $0 < |z-a| < epsilon$ and one of the limit exists iff $f $ is meromorphic at $a$, or $f$ has infinitely many poles around $z=a$ and the limits don't exist (assume $lim_{zto a}|f(z)|=infty$ and look at $1/f$)
– reuns
Nov 25 '18 at 22:33
add a comment |
It seems you are calling "singularity" a point $a$ such that $f$ is meromorphic on $0 < |z-a| < epsilon$. In that case or $f$ is analytic on $0 < |z-a| < epsilon$ and one of the limit exists iff $f $ is meromorphic at $a$, or $f$ has infinitely many poles around $z=a$ and the limits don't exist (assume $lim_{zto a}|f(z)|=infty$ and look at $1/f$)
– reuns
Nov 25 '18 at 22:33
It seems you are calling "singularity" a point $a$ such that $f$ is meromorphic on $0 < |z-a| < epsilon$. In that case or $f$ is analytic on $0 < |z-a| < epsilon$ and one of the limit exists iff $f $ is meromorphic at $a$, or $f$ has infinitely many poles around $z=a$ and the limits don't exist (assume $lim_{zto a}|f(z)|=infty$ and look at $1/f$)
– reuns
Nov 25 '18 at 22:33
It seems you are calling "singularity" a point $a$ such that $f$ is meromorphic on $0 < |z-a| < epsilon$. In that case or $f$ is analytic on $0 < |z-a| < epsilon$ and one of the limit exists iff $f $ is meromorphic at $a$, or $f$ has infinitely many poles around $z=a$ and the limits don't exist (assume $lim_{zto a}|f(z)|=infty$ and look at $1/f$)
– reuns
Nov 25 '18 at 22:33
add a comment |
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It's not entirely clear to me what you mean by "singularity". But the way in which you are using the term seems to indicate that you mean a point which is not in the domain of $f$ but which is in the closure of that domain. So, take $f$ to be the restriction of the identity function $z mapsto z$, where the restricted domain is $D - left({0} cup {2^{-n} mid n = 1,2,3,...}right)$. The singularity at $0$ is not isolated, because it is the limit of the singularities at $2^{-n}$. But $lim_{z to 0} f(z)=0$.
answered Nov 25 '18 at 22:33
Lee Mosher
48.1k33681
add a comment |
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It's not entirely clear to me what you mean by "singularity". But the way in which you are using the term seems to indicate that you mean a point which is not in the domain of $f$ but which is in the closure of that domain. So, take $f$ to be the restriction of the identity function $z mapsto z$, where the restricted domain is $D - left({0} cup {2^{-n} mid n = 1,2,3,...}right)$. The singularity at $0$ is not isolated, because it is the limit of the singularities at $2^{-n}$. But $lim_{z to 0} f(z)=0$.
answered Nov 25 '18 at 22:33
Lee Mosher
48.1k33681
add a comment |
It's not entirely clear to me what you mean by "singularity". But the way in which you are using the term seems to indicate that you mean a point which is not in the domain of $f$ but which is in the closure of that domain. So, take $f$ to be the restriction of the identity function $z mapsto z$, where the restricted domain is $D - left({0} cup {2^{-n} mid n = 1,2,3,...}right)$. The singularity at $0$ is not isolated, because it is the limit of the singularities at $2^{-n}$. But $lim_{z to 0} f(z)=0$.
answered Nov 25 '18 at 22:33
Lee Mosher
48.1k33681
add a comment |
It's not entirely clear to me what you mean by "singularity". But the way in which you are using the term seems to indicate that you mean a point which is not in the domain of $f$ but which is in the closure of that domain. So, take $f$ to be the restriction of the identity function $z mapsto z$, where the restricted domain is $D - left({0} cup {2^{-n} mid n = 1,2,3,...}right)$. The singularity at $0$ is not isolated, because it is the limit of the singularities at $2^{-n}$. But $lim_{z to 0} f(z)=0$.
answered Nov 25 '18 at 22:33
Lee Mosher
48.1k33681
It's not entirely clear to me what you mean by "singularity". But the way in which you are using the term seems to indicate that you mean a point which is not in the domain of $f$ but which is in the closure of that domain. So, take $f$ to be the restriction of the identity function $z mapsto z$, where the restricted domain is $D - left({0} cup {2^{-n} mid n = 1,2,3,...}right)$. The singularity at $0$ is not isolated, because it is the limit of the singularities at $2^{-n}$. But $lim_{z to 0} f(z)=0$.
answered Nov 25 '18 at 22:33
Lee Mosher
48.1k33681
answered Nov 25 '18 at 22:33
Lee Mosher
48.1k33681
answered Nov 25 '18 at 22:33
Lee Mosher
48.1k33681
answered Nov 25 '18 at 22:33
Lee Mosher
48.1k33681
48.1k33681
add a comment |
add a comment |
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It seems you are calling "singularity" a point $a$ such that $f$ is meromorphic on $0 < |z-a| < epsilon$. In that case or $f$ is analytic on $0 < |z-a| < epsilon$ and one of the limit exists iff $f $ is meromorphic at $a$, or $f$ has infinitely many poles around $z=a$ and the limits don't exist (assume $lim_{zto a}|f(z)|=infty$ and look at $1/f$)
– reuns
Nov 25 '18 at 22:33