Global boundedness for weak solutions in Gilbarg/Trudinger [Chapter 8.5] I have a question concerning a step in the proof of Theorem 8.15 in Gilbarg/Trudinger "Elliptic PDEs of Second Order". 
I really hope someone might be familiar with this and would be so kind as to go through the trouble of reading the proof again. This may be a bit much to ask for, but it would certainly be greatly appreciated! 
Here is the statement of the theorem

Theorem 8.15: Let the operator $L$ satisfy conditions (8.5), (8.6) and suppose that $f^i \in L^q(\Omega)$, $i=1, \ldots, n$, $q\in L^{q/2}(\Omega)$ for some $q>n$. Then if $u$ is a $W^{1,2}(\Omega)$ subsolution of $Lu = g + D_if^i$ in $\Omega$ satisfying $u\le 0$ on $\partial \Omega$, we have 
  $$\sup_\Omega u \le C(\Vert u^+\Vert_2 + k)$$
  where $k=\lambda^{-1}(\Vert \mathbf{f}\Vert_q + \Vert g \Vert_{q/2})$ and $C = C(n,\nu, q, |\Omega|)$.

Conditions (8.5), (8.6) are strict ellipticity (with smallest eigenvalue $\lambda$) and uniform boundedness conditions on $L$.
Now to my question: I can follow the proof up to 
\begin{equation}\tag{8.36}
\Vert H(w) \Vert_{2\hat n/(\hat n - 2)} \le C \Vert H'(w)w \Vert_{2q/(q-2)}
\end{equation}
where $C = C(n, \nu, |\Omega|)$. With $H(w) = w^\beta - k^\beta$ for $k$ as in the statement of the theorem. But how can we deduce from this that 
\begin{equation}\tag{8.37}
\Vert w \Vert_{\beta\chi q^\ast} \le (C\beta)^{1/\beta}\Vert w \Vert_{\beta q^\ast}
\end{equation}
for $\beta\ge 1$, $q^\ast = 2q/(q-2)$ and $\chi = \hat n(q-2)/q(\hat n - 1)$?
It is clear to me that (8.36) is equivalent to
$$\Vert w^\beta - k^\beta \Vert_{\chi q^\ast}^{1/\beta}  \le (C\beta)^{1/\beta}\Vert w \Vert_{\beta q^\ast}$$
What happens to the summand $ -k^\beta$? 
At first I thought that some scaling argument could work, but as $w = u^+ + k$, I don't seem to be able to scale $w$ withouth also scaling $k$. And then I can't make $-k^\beta$ negligible...
Thanks for your help!
 A: Leonid Kovalev's approach works perfectly fine (with a minor adjustment). A big thank you from me again!
Everything written out explicitely (and throwing away the irrelevant information), one simply needs to prove that the assumption
$$\Vert w^\beta - k^\beta \Vert_{\chi q^\ast} \le C\beta \Vert w^\beta\Vert_{q^\ast}$$
for $w$ satisfying $w\ge k$ and $\beta \ge 1$ implies
$$\Vert w^\beta \Vert_{\chi q^\ast} \le C\beta \Vert w^\beta\Vert_{q^\ast}$$
On the one hand, if $w\le 2^{1/\beta} k$, then $$\Vert w^\beta \Vert_{\chi q^\ast}\le 2 \Vert k^\beta \Vert_{\chi q^\ast} = 2|\Omega|^{1/q^\ast\chi - 1/q^\ast}\Vert k^\beta\Vert_{q^\ast}\le 2|\Omega|^{1/q^\ast\chi - 1/q^\ast}  \Vert w^\beta \Vert_{q^\ast}$$ If on the other hand $w\ge 2^{1/\beta} k$, then 
$$w^\beta = \frac{1}{1-k^\beta/w^\beta}(w^\beta - k^\beta) \le 2(w^\beta - k^\beta)$$ 
So putting things together, we obtain $\Vert w^\beta \Vert_{\chi q^\ast} \le C\beta \Vert w^\beta \Vert_{q^\ast}$ for some new constant $C$, depending on the same parameters as the old constant.
A: I think it may also be viewed as a consequence of the triangular inequality together with the fact that $w\geq k$ by construction.
The triangular inequality gives
$$\Vert w^\beta\Vert_{\chi q*} \leq \Vert w^\beta - k^\beta\Vert_{\chi q*} + k^\beta |\Omega|^{1/\chi q*}$$
and the inequality $w\geq k$ (by construction if I don't mistake)
$$\kappa^\beta \leq \frac{\Vert w^\beta\Vert_{q*}}{|\Omega|^{1/q*}}$$
Therefore we get
$$\Vert w^\beta\Vert_{\chi q*} \leq \Vert w^\beta-\kappa^\beta\Vert_{\chi q*} + \Vert w^\beta\Vert_{q*} \big(|\Omega|^{(1-\frac{1}{\chi})}\big)^{1/q*}$$
When $|\Omega|<+\infty$, the proof is done, at least for large $\beta$.
You may notice that if $|\Omega|=+\infty$, we can apply the last computation on some subset $\Omega'\subseteq\Omega$ of finite Lebesgue measure, and then go to the limit using Fatou's lemma (if ever $w^\beta\in L^{q*}(\Omega)$).
