Section D.1 Compactness in the proof of Lemma 5.8
Towards the middle of the proof of Lemma 5.8, there is a compactness argument which is gone over rather quickly. While I would not advocate revising this particular argument in detail as preparation for the exam, it is nevertheless a good exercise in Advanced Real Analysis, and there has been some interest in this during class in previous years. Below is one expanded version of the argument.
The compact set in question is \(\overline B \cap T_\mu\text{,}\) which is closed as it is the intersection of two closed sets, and bounded as \(B\) is bounded. We first claim that for each point \(x \in \overline B \cap T_\mu\text{,}\) there exists \(\delta_x \gt 0\) such that \(\partial_1 u \lt 0\) on \(B_{2\delta_x}(x) \cap B\text{.}\) Notice the factor of \(2\text{,}\) which will make our life easier later on! For \(x \in \partial B \cap T_\mu\) this is Lemma 5.6, while for \(x \in B
\cap T_\mu\) it follows from the continuity of \(\partial_1 u\) and the fact that \(\partial_1 u \lt 0\) on \(B \cap T_\mu\) (shown earlier in the proof).
The collection \(\{ B_{\delta_x}(x) : x \in \overline B \cap T_\mu \}\) is trivially an open cover of the compact subset \(\overline B \cap T_\mu\text{,}\) and so it has a finite subcover \(\{B_{\delta_{x_1}}(x_1),\ldots,B_{\delta_{x_n}}(x_n)\}\text{.}\) Let
\begin{equation*}
\varepsilon = \min\{\delta_{x_1},\ldots,\delta_{x_n}\} \gt 0\text{.}
\end{equation*}
We claim that
\begin{equation}
\Sigma(\mu-\varepsilon) \subset \Sigma(\mu) \cup B_{2\delta_{x_i}}(x_i) \cup \cdots \cup B_{2\delta_{x_n}}(x_n).\tag{D.1}
\end{equation}
Since \(\partial_1 u \lt 0\) on the sets \(B_{2\delta_{x_i}}(x_i) \cap B\text{,}\) this will imply that \(\partial_1 u \lt 0\) on \(\Sigma(\mu-\varepsilon)\text{,}\) which is what we are after.
To see that (D.1) holds, fix a point \(x \in \Sigma(\mu-\varepsilon)\text{,}\) i.e. a point \(x \in B = B_R(0)\) with \(x_1 \gt \mu-\varepsilon\text{.}\) If \(x_1 \gt
\mu\) then \(x \in \Sigma(\mu)\text{,}\) and so it suffices to consider the remaining case where \(\mu-\varepsilon \lt x_1 \le \mu\text{.}\) The nearest point on \(T_\mu\) to \(x\) is
\begin{equation*}
y = x + (\mu-x_1)e_1 \in T_\mu \cap B\text{.}
\end{equation*}
Since \(y \in T_\mu \cap \overline B\text{,}\) there exists \(i \in \{1,\ldots,n\}\) such that \(y \in B_{\delta_{x_i}}(x_i)\text{.}\) The triangle inequality then gives
\begin{align*}
\abs{x-x_i}
\amp \le \abs{x-y} + \abs{y-x_i}\\
\amp \lt (\mu-x_1) + \delta_{x_i}\\
\amp \lt \varepsilon + \delta_{x_i}\\
\amp \le 2\delta_{x_i}\text{,}
\end{align*}
where in the last step we have used the definition of \(\varepsilon\text{.}\) Thus \(x \in
B_{2\delta_{x_i}}(x_i)\text{,}\) and the argument is complete.
The use of this additional factor of \(2\) can be compared, for instance, to arguments from Advanced Real Analysis about the uniform continuity of continuous functions on compact sets.
