I am reading Han and Lin's Elliptic Partial Differential Equations.
PROPOSITION 2.27.Suppose $u \in C(\bar{\Omega}) \cap C^{2}(\Omega)$ satisfies $$ Q u \equiv a_{i j}(x, u, D u) D_{i j} u+b(x, u, D u)=0 \quad \text { in } \Omega $$ where $ a_{i j} \in C\left(\Omega \times \mathbb{R} \times \mathbb{R}^{n}\right) $ satisfies$$ a_{i j}(x, z, p) \xi_{i} \xi_{j}>0 \text { for any }(x, z, p) \in \Omega \times \mathbb{R} \times \mathbb{R}^{n} \text { and } \xi \in \mathbb{R}^{n} . $$ Suppose there exist nonnegative functions $g \in L_{\mathrm{loc}}^{n}\left(\mathbb{R}^{n}\right) \text { and } h \in L^{n}(\Omega) \text { such that }$ $$\frac{|b(x, z, p)|}{n D^{*}} \leq \frac{h(x)}{g(p)} \quad \text{for any} (x, z, p) \in \Omega \times \mathbb{R} \times \mathbb{R}^{n},$$ $$ \int_{\Omega} h^{n}(x) d x<\int_{\mathbb{R}^{n}} g^{n}(p) d p \equiv g_{\infty} $$ Then there holds $\sup _{\Omega} |u| \leq \sup _{\partial \Omega} |u|+C(h,g)\operatorname{diam}(\Omega)$.
But in his proof he only shows $\sup _{\Omega} u \leq \sup _{\partial \Omega} u^{+}+C(h,g)\operatorname{diam}(\Omega).$ I cannot figure out why $\sup _{\Omega} u \leq \sup _{\partial \Omega} u^{+}+M$ imply $\sup _{\Omega} |u| \leq \sup _{\partial \Omega} |u|+C(h,g)\operatorname{diam}(\Omega).$This seems not obvious to me.
Since if $sup _{\Omega} u<0$ then $\sup _{\Omega} u \leq \sup _{\partial \Omega} u^{+}+C(h,g)\operatorname{diam}(\Omega)$ certainly holds and why we can derive a stronger proposition that $\sup _{\Omega} |u| \leq \sup _{\partial \Omega} |u|+C(h,g)\operatorname{diam}(\Omega)$?
Any help will be thanked.
Note that $v = -u$ satisfies
$$ \bar Q u \equiv \bar a_{i j}(x, v, D v) D_{i j} v+\bar b(x, v, D v)=0 \quad \text { in } \Omega$$
where $\bar a(x,z,p) = a(x,-z,-p), \bar b(x, z, p) = b(x, -z, -p)$. Also, $\bar a_{ij}$ satisfies the same ellipticity condition and $\bar b$ satisfies $$\frac{|\bar b(x, z, p)|}{n D^{*}} \leq \frac{h(x)}{\bar g(p)} \quad \text{for any} (x, z, p) \in \Omega \times \mathbb{R} \times \mathbb{R}^{n},$$ and $h(x), \bar g(p) :=(g(-p)$ satisfies the same integral inequality. Thus
$$ \sup _\Omega v \le \sup _{\partial \Omega} v +C(h, \bar g)\operatorname{diam}(\Omega), $$ this would give (together with inequality on $u$) $$\sup_\Omega |u| \le \sup_{\partial \Omega} |u| + \tilde C \operatorname{diam}(\Omega), $$ with $\tilde C = \max\{ C(h, g), C(h, \bar g)\}$.