The proof follows the same argument as the uniformly elliptic case. From the inequality we know that for any $\zeta\in C_{c}^{\infty}(B_{1})$, with $\zeta\geq 0$, the following holds:

(6.15)

$$\int_{B_{1}}a^{ij}\partial_{j}u\partial_{i}\zeta dx\leq\int_{B_{1}}fu\zeta+g\zeta dx.$$

Now let $\eta\in C^{\infty}_{c}(B_{1})$, define $\bar{u}=u+1$. Take $\zeta=\eta^{2}\bar{u}^{\beta}$, for some $\beta>0$. We plug in this $\zeta$ and obtain

(6.16)

$$\begin{split}\int_{B_{1}}&\beta a_{ij}\partial_{i}\bar{u}\partial_{j}\bar{u}\bar{u}^{\beta-1}\eta^{2}\leq\int-a_{ij}\partial_{j}u\partial_{i}\eta\bar{u}^{\beta}2\eta+|f|\bar{u}^{\beta+1}\eta^{2}+|g|\bar{u}^{\beta}\eta^{2}\\ &\leq\int_{B_{1}}\frac{\beta}{2}a_{ij}\partial_{i}\bar{u}\partial_{j}\bar{u}\bar{u}^{\beta-1}\eta^{2}+\frac{4}{\beta}a_{ij}\partial_{i}\eta\partial_{j}\eta\bar{u}^{\beta+1}+(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}.\end{split}$$

Use the ellipticity condition to get:

(6.17)

$$\int_{B_{1}}\frac{\beta}{\lambda}|\nabla\bar{u}|^{2}\bar{u}^{\beta-1}\eta^{2}\leq\int_{B_{1}}\big(\frac{4\lambda}{\beta}|\nabla\eta|^{2}+|f|\eta^{2}+|g|\eta^{2}\big)\bar{u}^{\beta+1}.$$

This is equivalent to:

(6.18)

$$\int_{B_{1}}|\nabla(\bar{u}^{\frac{\beta+1}{2}})|^{2}\eta^{2}\frac{1}{\lambda}\leq\frac{(\beta+1)^{2}}{\beta^{2}}\int_{B_{1}}\lambda|\nabla\eta|^{2}\bar{u}^{\beta+1}+\frac{(\beta+1)^{2}}{4\beta}\int_{B_{1}}(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}.$$

Next observe

$$|\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)|^{2}\leq 2|\nabla(\bar{u}^{\frac{\beta+1}{2}})|^{2}\eta^{2}+2\bar{u}^{\beta+1}|\nabla\eta|^{2}.$$

Hence it follows from (6.18) that if $\beta\geq 1$,

(6.19)

$$\begin{split}\int_{B_{1}}|\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)|^{2}\frac{1}{\lambda}&\leq\int_{B_{1}}\big(\frac{2\lambda(\beta+1)^{2}}{\beta^{2}}+\frac{2}{\lambda}\big)\bar{u}^{\beta+1}|\nabla\eta|^{2}+\frac{(\beta+1)^{2}}{2\beta}\int_{B_{1}}(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}\\ &\leq\int_{B_{1}}10\lambda\bar{u}^{\beta+1}|\nabla\eta|^{2}+2\beta\int_{B_{1}}(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}.\end{split}$$

We would like to get rid of the $\lambda$ in the above estimate. Let $\varepsilon>0$ to be determined, then we have

(6.20)

$$||\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)||_{L^{2-\varepsilon}}^{2}\leq||\lambda||_{L^{\frac{2}{\varepsilon}-1}}^{\frac{2-\varepsilon}{2}}\int_{B_{1}}\frac{1}{\lambda}|\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)|^{2}.$$

On the other hand, we estimate the right hand side by Hölder’s inequality:

(6.21)			$\displaystyle\int_{B_{1}}\lambda\bar{u}^{\beta+1}|\nabla\eta|^{2}\leq||\lambda||_{L^{\frac{2}{\varepsilon}-1}}||\bar{u}^{\frac{\beta+1}{2}}|\nabla\eta|||_{L^{\frac{2-\varepsilon}{1-\varepsilon}}}^{2},$
(6.22)			$\displaystyle\int_{B_{1}}(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}\leq\big(||f||_{L^{p/2}}+||g||_{L^{p/2}}\big)||\bar{u}^{\frac{\beta+1}{2}}\eta||_{L^{\frac{2p}{p-2}}}^{2}.$

Therefore,

(6.23)

$$||\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)||_{L^{2-\varepsilon}}^{2}\leq||\lambda||_{L^{\frac{2}{\varepsilon}-1}}^{\frac{2-\varepsilon}{2}}\big(10||\lambda||_{L^{\frac{2}{\varepsilon}-1}}||\bar{u}^{\frac{\beta+1}{2}}|\nabla\eta|||^{2}_{L^{\frac{2-\varepsilon}{1-\varepsilon}}}+2\beta(||f||_{L^{p/2}}+||g||_{L^{p/2}})||\bar{u}^{\frac{\beta+1}{2}}\eta||_{L^{\frac{2p}{p-2}}}^{2}\big).$$

Now we choose $\varepsilon=\frac{2}{p+1}$, then $\frac{2}{\varepsilon}-1=p$. With this choice, we have $\frac{2-\varepsilon}{1-\varepsilon}<\frac{2p}{p-2}$ in the above, then we find for some constant $C_{6.1}$, depending on $||\lambda||_{L^{p}}$, $||f||_{L^{p/2}}$, $||g||_{L^{p/2}}$, such that

(6.24)

$$||\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)||_{L^{\frac{2p}{p+1}}}^{2}\leq C_{6.1}\big(||\bar{u}^{\frac{\beta+1}{2}}|\nabla\eta|||^{2}_{L^{\frac{2p}{p-2}}}+\beta||\bar{u}^{\frac{\beta+1}{2}}\eta||_{L^{\frac{2p}{p-2}}}^{2}\big).$$

Fix $\frac{1}{2}\leq r<R\leq 1$, Denote $r_{i}=r+2^{-i}(R-r)$, for $i\geq 0$. Note that $r_{0}=R$, and $r_{i}\rightarrow r$ as $i\rightarrow\infty$. We choose the cut-off function $\eta$ so that $0\leq\eta\leq 1$, $\eta\equiv 1$ on $B_{r_{i+1}}$, $supp\,\eta\subset B_{r_{i}}$, and $|\nabla\eta|\leq\frac{2}{r_{i}-r_{i+1}}=\frac{2^{i+2}}{R-r}$. Denote $\theta$ to be such that $\frac{1}{\theta}=\frac{p+1}{2p}-\frac{1}{d}$. Since $p>\frac{3d}{2}$, it follows that $\theta>\frac{2p}{p-2}$. Then apply the Sobolev inequality to get

(6.25)

$$||\bar{u}^{\frac{\beta+1}{2}}||_{L^{\theta}(B_{r_{i+1}})}\leq\frac{C_{6.2}\sqrt{\beta}2^{i+2}}{R-r}||\bar{u}^{\frac{\beta+1}{2}}||_{L^{\frac{2p}{p-2}}(B_{r_{i}})}.$$

This is equivalent to:

(6.26)

$$||\bar{u}||_{L^{\frac{\theta(\beta+1)}{2}}(B_{r_{i+1}})}\leq\frac{C_{6.2}^{\frac{2}{\beta+1}}\beta^{\frac{1}{\beta+1}}2^{\frac{2(i+2)}{\beta+1}}}{(R-r)^{\frac{2}{\beta+1}}}||\bar{u}||_{L^{\frac{2p}{p-2}\cdot\frac{\beta+1}{2}}(B_{r_{i}})}.$$

Now denote $\theta=\chi\cdot\frac{2p}{p-2}$ for some $\chi>1$, and choose $\beta$ to be $\frac{\beta+1}{2}=\chi^{i}$, then we obtain from (6.26):

(6.27)

$$||\bar{u}||_{L^{\frac{2p\chi^{i+1}}{p-2}}(B_{r_{i+1}})}\leq\big(\frac{2C_{6.2}}{R-r}\big)^{\frac{1}{\chi^{i}}}\chi^{\frac{i}{\chi^{i}}}2^{\frac{i+2}{\chi^{i}}}||\bar{u}||_{L^{\frac{2p\chi^{i}}{p-2}}(B_{r_{i}})},\textrm{ for $i\geq 0$.}$$

Iterating this inequality we obtain for any $\frac{1}{2}\leq r<R\leq 1$, and for some constant $C_{6.3}$ independent of $r$, $R$,

(6.28)

$$\begin{split}&||\bar{u}||_{L^{\infty}(B_{r})}\leq\frac{C_{6.3}}{(R-r)^{\sum_{i\geq 0}\chi^{-i}}}||\bar{u}||_{L^{\frac{2p}{p-2}}(B_{R})}=\frac{C_{6.3}}{(R-r)^{\frac{\chi}{\chi-1}}}||\bar{u}||_{L^{\frac{2p}{p-2}}(B_{R})}\\ &\leq\frac{C_{6.3}}{(R-r)^{\frac{\chi}{\chi-1}}}||\bar{u}||_{L^{1}(B_{R})}^{\frac{p-2}{2p}}||\bar{u}||_{L^{\infty}(B_{R})}^{\frac{p+2}{2p}}\leq\frac{1}{2}||\bar{u}||_{L^{\infty}(B_{R})}+\frac{2^{\frac{p+2}{p-2}}C_{6.3}^{\frac{2p}{p-2}}}{(R-r)^{\frac{2\chi p}{(\chi-1)(p-2)}}}||\bar{u}||_{L^{1}(B_{R})}.\end{split}$$

The desired conclusion now follows from the following lemma applied to $f(r)=||\bar{u}||_{L^{\infty}(B_{r})}$, which is a special case of Lemma 4.3 in [22]. ∎