I am trying to follow the proof of the main theorem of Part III of "Limits as $p \to \infty$ of $\Delta_p u_p = f$ and related extremal problems" by Bhattacharya, DiBenedetto, and Manfredi. In a simplified form, the theorem is:
Let $N \geq 2$. There exists $C > 0$ such that for every $2 \leq p < \infty$ and every $p$-harmonic function $u$ on $B_1 \subset \mathbb R^N$, $$\|\nabla u\|_{L^\infty(B_{1/2})} \leq C^{1/p} \|\nabla u\|_{L^p(B_1)}.$$
Unfortunately, I am unable to follow the key step of the proof. It seems likely that I am making some elementary error, and I hope that one of you can set me on the right course.
The proof is by de Giorgi iteration, and I can follow it up to the end of page 54, which I quote here for your convenience:
It follows from Lemma 5.6 of [LSU] p95 that there exists a constant $\gamma$ independent of $p$ such that if $k_o$ is chosen to satisfy $$k_o = \gamma \max\left(\left(\frac{p}{\tau R}\right)^N \int_{B_o} z^2 ~dx ; 1\right)$$ then $$\int_{B_{n + 1}} (z - k_n)_+^2 ~dx \to 0$$ as $n \to \infty$, ie $\|z\|_{L^\infty(B_\infty)} \leq k_o$. Recalling the definitions of $B_o$, $B_\infty$, and $z$ we have $$\|\nabla u\|_{L^\infty(B_{(1 - r)R})}^{\frac{p + 2}{2}} \leq \frac{\gamma p^N}{(\tau R)^{\frac{N}{2}}} \left(\int_{B_R} (1 + |\nabla u|)^{p + 2} ~dx\right)^{1/2}$$ and $\forall R > 0$ such that $B_R \subset \Omega$, $\forall \tau \in (0, 1)$ $$\|\nabla u\|_{L^\infty(B_{(1 - r)R})} \leq \frac{\gamma^{\frac{1}{p}}}{(\tau R)^{\frac{N}{p + 2}}} \left(\int_{B_R} (1 + |\nabla u|)^{p + 2} ~dx\right)^{\frac{1}{p + 2}}.$$
Here, we are assuming $N \geq 3$ for simplicity, $z = |\nabla u|^{\frac{p + 2}{2}}$, $\tau = (1 - r)$ (I think?) and $k_j, B_j$ are the usual cutoffs and balls used in de Giorgi iteration. [LSU] is Ladyzhenskaya--Solonnikov--Uraltseva's monograph on quasilinear parabolic PDE, and the cited lemma essentially assserts the desired $L^\infty$ bound from the first display. As stated, the bound does not pass the "dimensional analysis" smell check, since $k_o$ scales like $z^2$ but it is being used to bound $z$. However, when I run the argument myself, I get $$\|z\|_{L^\infty(B_{(1 - r) R})}^2 \lesssim \frac{p^{\frac{3N}{4}}}{R^{\frac{N}{2}}} \left(\int_{B_R} z^2 ~dx\right)^{\frac{1}{2}}$$ which does scale correctly, and does not affect what I am about to say. So this is not the essential point.
The essential point is that the $p^N$ (or $p^{3N/4}$, by my arithmetic) seems to completely disappear when we take $p + 2$th roots. Indeed, the penultimate display in my big block quote, and the final display, have very different behaviors as $p \to \infty$! Now I agree that the penultimate display is basically correct, but it is the final display which we are really interested in, as the point of the main theorem is that there is no dependence on $p$ in the constant.
Originally I thought that one could somehow use the failure of the bound which follows from the LSU citation to scale correctly to somehow eliminate the undesired $p^{O(1)}$. But since I get an estimate which scales correctly, this approach does not work.
Let me finally remark that in some applications, we really want the constant in the main theorem to be of the form $C^{1/p}$ and not $C^{1/p} p^{O(1)/p}$. In fact, there is a much shorter proof of the estimate with $p^{O(1)/p}$ loss by Moser iteration, in Daskalopoulos and Uhlenbeck's "Transverse measure and best Lipschitz and least gradient maps". One can read the importance of the lack of $p^{O(1)/p}$ off the proof of Theorem 4(v) of Katzourakis and Moser's "Minimisers of supremal functionals and mass-minimising 1-currents": it seems that if the sharp bound is of the form $C^{1/p}p^{\varepsilon/p}$ for some $\varepsilon > 0$, then their argument fails, essentially because they need fine control on the $p$th power of the Lipschitz constant.
