Take $z\in(M\setminus E)\cap B_{j}.$ Then

$$(u_{j,\delta}-u)(z)\leq(\frac{\chi(\delta)}{3}-2)\delta^{\alpha}$$

and therefore, by 2.5

$$(u_{j,\delta}-v)(z)\leq(\frac{\chi(\delta)}{3}-1)\delta^{\alpha}.$$

Since $u>1$ we get from this

$$(u_{\delta}-v)(z)\leq\max_{j:z\in B_{j}}(u_{j,\delta}-v)(z)\leq(\frac{\chi(\delta)}{3}-1)\delta^{\alpha}.$$

$2.6$

Again, by 2.5

$$(u_{j_{0},\delta}-v)(z_{0})\geq(\chi(\delta)-1)\delta^{\alpha}.$$

Therefore the definition of $u_{\delta}$ yields

$$(u_{\delta}-v)(z_{0})\geq(\chi(\delta)-1)\delta^{\alpha}-\eta(\delta)(2c_{1}||u||_{\infty}+1).$$

$2.7$

The Three Circles Theorem gives for $\delta$ small enough

$$(u_{j,N\delta}-u_{j,\delta})\geq\frac{\log N}{\log C}(u_{j,C\delta}-u_{j,\delta}).$$

It follows that, choosing $j$ so that

$$\eta(\delta)=\max_{z\in B_{j}}(u_{j,C\delta}-u_{j,\delta})(z)$$

we obtain

$$(N\delta)^{\alpha}\chi(N\delta)\geq\frac{\log N}{\log C}\eta(\delta).$$

Further, since $\chi(\delta)\geq\chi(N\delta)$, we get from 2.2 that

$$\delta^{\alpha}\chi(\delta)\geq\frac{\log N}{\log C}\eta(\delta)N^{-\alpha}>2\eta(\delta)(2c_{1}||u||_{\infty}+1).$$

Inserting this into 2.7 we finally arrive at

$$(u_{\delta}-v)(z_{0})>(\chi(\delta)/2-1)\delta^{\alpha}.$$

The proposition follows by comparing this inequality with 2.6.