The proof of Hopf lemma for harmonic functions I would like to understand a passage from the proof of Hopf lemma. 
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In the second image above the author says:

Therefore by theorem 1.29 (Maximum Principle for Subharmonic
  Functions) $h_{\epsilon}$ assumes at the $x_{0}$ its maximum in A.
  This implies 
$\frac{\partial h_{\epsilon}}{\partial n}(x_{0}) \geq 0$ or
  $\frac{\partial u}{\partial n}(x_{0}) \geq -\epsilon \frac{\partial
> v}{\partial n}(x_{0})=2\alpha\epsilon e^{-\alpha}$

I do not understand this implication. Would this result remain valid if the partial derivative was evaluated in another vector?
The pictures are from the book Elliptic Partial Differential Equation - Qing Han and Fanghua Lin, AMS.
 A: The authors skipped the consideration of whether the normal derivative $\frac{\partial h_{\epsilon}}{\partial n}(x_{0})$ exists. For the moment, suppose that it exists. If this derivative were negative, then for sufficiently small $\delta>0$ we have 
$$ \frac{h_{\epsilon} ((1-\delta)x_{0}) - h_{\epsilon} (x_{0})}{-\delta} < 0 $$ by the definition of derivative. Then  $h_{\epsilon} ((1-\delta)x_{0})> h_{\epsilon} (x_{0})$, contradicting that $x_0$ is a point of maximum.  Hence, $\frac{\partial h_{\epsilon}}{\partial n}(x_{0})\ge 0$. Recalling how $h_\epsilon$ was defined, we get a lower bound on the derivative $\frac{\partial u}{\partial n}(x_{0})$.
In general, the normal derivative need not exist under the assumptions of  this theorem. The idea of the proof is still the same, but the conclusion is that 
$$\liminf_{\delta\to 0} \frac{h_{\epsilon} ((1-\delta)x_{0}) - h_{\epsilon} (x_{0})}{-\delta} \ge 0 $$
hence
$$\liminf_{\delta\to 0} \frac{u ((1-\delta)x_{0}) - u (x_{0})}{-\delta} \ge C(u(x_0) - u(0)) $$
One can see this as an inequality for "lower normal derivative".

And no, the result would not be valid at other points: we use the fact that $x_0$ is a point of maximum.
