Skip to content

Commit

Permalink
Minor nits
Browse files Browse the repository at this point in the history
  • Loading branch information
jirilebl committed May 6, 2024
1 parent 3533b7f commit 5987286
Showing 1 changed file with 8 additions and 5 deletions.
13 changes: 8 additions & 5 deletions ca.tex
Original file line number Diff line number Diff line change
Expand Up @@ -14862,7 +14862,8 @@ \subsection{Isolated singularities}
For harmonic functions, we get the following classification of removable
singularities, which is sharp, that is, best possible.
The harmonic function $\log \sabs{z}$
has a nonremovable singularity at the origin, and any function
has a nonremovable singularity at the origin.
Any function
that blows up any slower than that, doesn't actually blow up and, in fact,
extends to be
harmonic at the origin.
Expand Down Expand Up @@ -14901,13 +14902,15 @@ \subsection{Isolated singularities}
\end{equation}
The estimate \eqref{eq:removableestimateharmonic} holds also when
$\sabs{z}=1$ as $g=0$ there.
The function $-\log\sabs{z}$ is harmonic outside of the origin, so
The functions $-\log\sabs{z}$ and $g$ are harmonic
outside of the origin, so
the maximum principle (the version in
\exerciseref{exercise:maxprincsecondharmonic}) implies that
\eqref{eq:removableestimateharmonic} holds also for $\delta < \sabs{z} < 1$,
and thus for all $z \in \D \setminus \{ 0 \}$.
As the estimate holds for all $\epsilon > 0$, we have $g(z) = 0$ for all
$z \in \D \setminus \{0\}$. So $u$ is the extension we are looking for.
$z \in \D \setminus \{0\}$. So $u$ is the extension near $0$ that
we are looking for.
\end{proof}

An isolated singularity of a harmonic function $g$ could be very wild, for
Expand Down Expand Up @@ -14972,8 +14975,8 @@ \subsection{Isolated singularities}
for some branch of the $\log$, where $\psi$ is a holomorphic function
on $\D \setminus \{ 0 \}$.
Taking the real part we get $f$, a well-defined function on
$\D \setminus \{ 0 \}$. That means that
$c_{-1} \log z$ has real part well-defined in $\D \setminus \{ 0 \}$,
$\D \setminus \{ 0 \}$. Therefore,
$c_{-1} \log z$ has real part that is well-defined in $\D \setminus \{ 0 \}$,
meaning $c_{-1} \in \R$.
So
\begin{equation*}
Expand Down

0 comments on commit 5987286

Please sign in to comment.