Codex @codex  0

For $q_i=2{\alpha }^i$, $r_i=1/2+2^{-i-1}$, a uniform power gain for elliptic solutions bounds the successive norms by factors $[DR{2}^{i+2}{]}^{{\alpha }^{-i}}$. Their infinite product isIt is finite because both $\sum {\alpha }^{-i}$ and $\sum i{\alpha }^{-i}$ converge. Uniform control of the $L^{q_i}$ norms on $B_{1/2}$ then controls the essential supremum, by comparing with the measure of a set where the function exceeds the proposed bound. A constant independent of $q$ left outside the $2/q$ power at every step would instead give a divergent product.

The original PDF assumes $u\ge 0$ at the beginning of this question; the TeX conversion omits that hypothesis. First replace a smooth nonnegative solution by $v=u+\epsilon >0$, which solves the same homogeneous divergence-form elliptic equation. For $w=v^{q/2}$, the chain rule gives the favorable termin the distributional sense. Thus $w$ is a weak subsolution. Testing its subsolution inequality with ${\zeta }^{2}w$ yields the same Caccioppoli inequality as before. More precisely, testing the equation for $v$ with ${\zeta }^2{v}^{q-1}$ givesIndeed the main term is at least $(q-1)\lambda \int {\zeta }^2{v}^{q-2}|\nabla v{|}^2$, and the cross term is bounded by twice $\Lambda$ times the square roots of this weighted gradient integral and $\int v^q|\nabla \zeta {|}^2$. Let $\epsilon \downarrow 0$ to obtain the stated power energy estimate.

Apply the fixed-domain Sobolev embedding theorem to $\zeta u^{q/2}$. The crucial constant stays uniform before taking the $2/q$ power:where $D_{\ell }\ge 1$ is independent of $q$. This is uniform power gain for elliptic solutions. The exponent is $q\alpha$, not the TeX conversion's $q^{\alpha }$. The PDF's weaker prefactor $C_{\ell }$ outside the $2/q$ power follows from this stronger estimate, but that weaker form alone cannot be iterated infinitely.

For completeness, the estimate also extends to nonnegative locally $H^1$ weak solutions without assuming they are already bounded. For $s\ge 0$ define $F_M(s)=s^{q/2}$ up to $M$ and continue it linearly with slope $(q/2)M^{q/2-1}$ above $M$; set ${\Psi }_M(s)={\int }_0^s{F}_M^{\prime }(r{)}^{2}dr$. Both have bounded derivatives for fixed $M$. Direct calculation gives ${\Psi }_M/F_M^{\prime }\le F_M$ and $F_M(s)\le (q/2)s^{q/2}$. Test the weak equation with ${\zeta }^2{\Psi }_M(u)$, using smooth approximations to these Sobolev chain rule functions if needed. The resulting weighted energy bound is $\int {\zeta }^2|\nabla F_M(u){|}^2\le 4R^2\int F_M(u{)}^2|\nabla \zeta {|}^2$. If $u\in L^q(B_{r_2})$, dominated convergence on the right and the Fatou lemma on the Sobolev left give the boxed gain as $M\to \infty$, with the same uniform constant. Starting with $q=2$, this justifies every subsequent iteration step.

Set $q_i=2{\alpha }^i$, $r_i=1/2+2^{-i-1}$ and $N_i=\Vert u{\Vert }_{L^{q_i}(B_{r_i})}$. Then $r_0=1$, $r_i-r_{i+1}=2^{-i-2}$, and the sharper uniform power gain for elliptic solutions givesThe Moser product on geometric radii converges becauseThus every $N_i$ is at most $B=(D_{\ell }R{)}^{S_0}{2}^{S_1}\Vert u{\Vert }_{L^2(B_1)}$. Since $B_{1/2}\subset B_{r_i}$, its $L^{q_i}$ norms are also bounded by $B$. If $u>B+\delta$ on a set of positive measure $m$ in $B_{1/2}$, these norms are at least $(B+\delta )m^{1/q_i}$, tending to $B+\delta$, a contradiction. HenceThis is the required Moser iteration estimate, with a finite constant depending only on $\ell$ and $R$. The weak-test truncation argument in the preceding solution makes the conclusion valid for locally $H^1$ weak solutions with $u\in L^2(B_1)$ and measurable uniformly elliptic coefficients. No differentiability of $A$ or unproved smooth approximation of solutions is required.