Does local convergence of analytic functions extend to a cut?
Let $Omegasubsetmathbb{C}$ be a region (=open and connected set) which is divided by a simple curve $Gamma$ into two disjoint regions. Suppose further that ${f_{n}}$ and $f$ are functions analytic in $Omega$ and
$$lim_{ntoinfty}f_{n}=f$$
locally uniformly in $OmegasetminusGamma$. Is it then true that
$$lim_{ntoinfty}f_{n}=f$$
locally uniformly in $Omega$? That is, the limit relation can be extended to the cut $Gamma$.
complex-analysis
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Let $Omegasubsetmathbb{C}$ be a region (=open and connected set) which is divided by a simple curve $Gamma$ into two disjoint regions. Suppose further that ${f_{n}}$ and $f$ are functions analytic in $Omega$ and
$$lim_{ntoinfty}f_{n}=f$$
locally uniformly in $OmegasetminusGamma$. Is it then true that
$$lim_{ntoinfty}f_{n}=f$$
locally uniformly in $Omega$? That is, the limit relation can be extended to the cut $Gamma$.
complex-analysis
add a comment |
Let $Omegasubsetmathbb{C}$ be a region (=open and connected set) which is divided by a simple curve $Gamma$ into two disjoint regions. Suppose further that ${f_{n}}$ and $f$ are functions analytic in $Omega$ and
$$lim_{ntoinfty}f_{n}=f$$
locally uniformly in $OmegasetminusGamma$. Is it then true that
$$lim_{ntoinfty}f_{n}=f$$
locally uniformly in $Omega$? That is, the limit relation can be extended to the cut $Gamma$.
complex-analysis
Let $Omegasubsetmathbb{C}$ be a region (=open and connected set) which is divided by a simple curve $Gamma$ into two disjoint regions. Suppose further that ${f_{n}}$ and $f$ are functions analytic in $Omega$ and
$$lim_{ntoinfty}f_{n}=f$$
locally uniformly in $OmegasetminusGamma$. Is it then true that
$$lim_{ntoinfty}f_{n}=f$$
locally uniformly in $Omega$? That is, the limit relation can be extended to the cut $Gamma$.
complex-analysis
complex-analysis
edited Nov 28 '18 at 16:52
asked Nov 28 '18 at 16:20
Twi
20719
20719
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add a comment |
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