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Based on the activity here and discussion in chat, I've enabled MathJax on this community.

I don't know Mathjax myself, so as verification, I'm copying some Math I found in [this Mathematics post](https://math.codidact.com/posts/292410/292418#answer-292418):

> In the proof of **Theorem 2**, we are aligning the $x_1$ axis with the vector $\gamma$. That is, $\frac{\partial}{\partial x_1} = \gamma \cdot \nabla$, the directional derivative in the $\gamma$ direction. Therefore, $\gamma \cdot \operatorname{grad} u = \gamma \cdot \nabla u = \frac{\partial u}{\partial x_1} = u_{x_1}$. One of the first consequences of **Theorem 2.1** is that $u_{x_1} < 0$ in $\Sigma=\Sigma_\gamma$. **Theorem 2.1** also has $\Omega = \Sigma \cup \Sigma' \cup (T_{\lambda_1}\cap \Omega)$ if $\gamma \cdot \nabla u = 0$ at some point of $\Omega \cap T_{\lambda_1}$ as a conclusion. For the annulus, the $T_{\lambda_1}$ for varying $\gamma$ will be the lines tangent to the circle halfway into the annulus. Clearly the (closure of the) maximal cap unioned with its reflection is not all of the annulus which is what **Theorem 2.1** implies in this case, thus it can't be the case that $\gamma\cdot\nabla u = 0$ even when $|x| = (R'+R)/2$ where before we only knew this for $|x| > (R'+R)/2$.