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!