0.\tag 2.10$$ \endproclaim \demo{Proof of Assertion} Define $\Lambda_\infty=\{x:\ x \hbox{ is a limit point of a sequence }\{x_j\}$, where $ x_j\in\Lambda(\tilde u_j)\}$. Since $\tilde u_j\to \tilde u_0$ and $\nabla \tilde u_j\to \nabla \tilde u_0$ uniformly on $\overline{B_{(1/2)}}$, $$\Lambda_\infty\cap \overline{B_{(1/2)}}\subset \Lambda(\tilde u_0)\cap \overline{B_{(1/2)}}.\tag 2.11$$ Let $\delta >0$. Then by Lemma 2.3 there is an open set $O$, $O\supset(\Lambda_\infty\cap \overline{ B_{(1/2)}})$ and satisfies $$\hbox{CAP}(O)\leq \hbox{CAP}\left(\Lambda_\infty\cap \overline{ B_{(1/2)}}\right)+\delta.\tag 2.12$$ We claim there is $j_0$ such that $$ \left(\Lambda(\tilde u_j)\cap \overline{ B_{(1/2)}}\right)\subset O \qquad \hbox{for all }~j\geq j_0.\tag 2.13$$ Since, otherwise, there is a sequence $\{x_j\}\subset \left(\Lambda(\tilde u_j)\cap \overline{ B_{(1/2)}}\right) \setminus O$, with $x_j\to x\in \left(\Lambda_\infty \cap \overline{ B_{(1/2)}}\right)\subset O$. Since $O$ is open, $x_j\in O$ for all $j$ large enough. This is a contradiction. Now using (2.11)-(2.13), and letting $j\geq j_0$, we obtain $$\eqalign{\hbox{CAP}\left(\Lambda(\tilde u_0) \cap\overline{ B_{(1/2)}}\right) & \geq \hbox{CAP}\left(\Lambda_\infty\cap \overline{ B_{(1/2)}}\right)\geq\hbox{CAP}\left(\Lambda(\tilde u_j)\cap \overline{B_{(1/2)}}\right)-\delta \cr & =c(N){\hbox{CAP}\left(\Lambda(u_j)\cap\overline{B_{2^{k_j}}}\right)\over \hbox{CAP}(B_{2^{k_j}})}-\delta,\cr}$$ where $c(N)=\hbox{CAP}(B_1)/2^{(N-2)}$, for $N\geq 3$, and $c(2)=\hbox{CAP}(B_1)/2$ (for the last equality see e.g. [11; pp. 158-160]). Since $k_j\to \infty$ (by (b)), condition (c) implies (2.10), if we chose $\delta$ sufficiently small. This completes the proof of the Assertion and hence that of Lemma 2.2. \hfill\qed \enddemo \enddemo \demo{Proof of Theorem 1.2} Obviously we may assume $m>2$. For $u\in\Cal G(L,K,m,\varepsilon)$ let $\Bbb M' (u)$ be the maximal subset of $\Bbb N$ such that (2.4) holds. We first show that $$\Bbb M' (u)=\Bbb N.\tag 2.14$$ By Lemma 2.1', $\Bbb M(u)$ is infinite and by Lemma 2.2, $\Bbb M(u)\subset\Bbb M' (u)$ . Hence, $\Bbb M' (u)$ is an infinite subset of $\Bbb N$. Therefore in order to show (2.14), it suffices to show that if $j+1\in \Bbb M' (u)$, then $j\in \Bbb M' rime (u)$. Suppose not, i.e., there is $j\not\in \Bbb M' (u)$ such that $j+1\in \Bbb M' (u)$. By the maximality of both $\Bbb M (u)$ and $\Bbb M' (u)$, neither (2.2') nor (2.4) holds for elements outside $\Bbb M' (u)$. Hence $$\gather S(2^j,u) >K' \left(2^j\right)^2\cr S(2^{(j+1)},u) > {2^{(m+1)}S(2^j,u)}. \cr\endgather $$ Since $j+1\in \Bbb M' (u)$, we'll have by the latter inequalities, $$K' \left(2^{(j+1)}\right)^2\geq S(2^{(j+1)},u)>2^{(m+1)}S(2^j,u)>2^{(m+1)}K' \left(2^j\right)^2,$$ which implies $2^2>2^m$, contradicting $m>2$. Hence (2.4) holds for all $j\in \Bbb N$. For $2^j\leq r\leq 2^{(j+1)}$, we have $$S(r,u)\leq S(2^{(j+1)},u)\leq K' \left(2^{(j+1)}\right)^2\leq 4 K' r^2 $$ and the proof is complete. \hfill\qed \enddemo \head{\bf 3. Application to other elliptic operators} \endhead The method presented in the previous section can be applied to other elliptic operators. In this section we shall obtain the optimal growth of solutions to certain type of elliptic operators. In order to avoid complicated notations, we present two examples. The corresponding local estimate (1.6) will be treated elsewhere. Recall that Lemma 2.1' deals only with the doubling property of functions with polynomial growth. Therefore, to adopt the method of the previous section, we only need to deduce the corresponding "compactness" property in Lemma 2.2. \example{(i) A second order linear operator} Let $$\Cal L u=\Delta u+a_{ij}(x)\partial_i\partial_j u+b_i(x)\partial_i u+c(x)u,$$ where $\partial_i=\partial/\partial x_i$, $(\delta_{ij}+ a_{ij}(x))\xi_i\xi_j\geq\lambda |\xi |^2$ in ${\Bbb R}^N$ $(\lambda>0)$, $a_{ij},b_i,c\in C({\Bbb R}^N)$ and satisfy $$\lim_{x\to \infty}|a_{ij}(x)|=\lim_{x\to \infty}|x|~|b(x)|=\lim_{x\to\infty} |x|^2~|c(x)|=0.\tag 3.1$$ \proclaim{Theorem 3.1} If we replace the Laplace operator, in Definition 1.1, by the operator $\Cal L$ above. Then the conclusions of Theorem 1.2, as well as Corollary 1.3 remain true. \endproclaim \demo{Proof} Set $$\Cal L_r u=\Delta u+a_{ij}(rx)\partial_i\partial_j u+rb_i(rx)\partial_i u+r^2c(rx)u$$ and $$ C(r)=\sup_{B_1}\left(|a_{ij}(rx)|+|rb_i(rx)|+|r^2c(rx)|\right).$$ Then by (3.1) $$C(r)\to 0\qquad\hbox{as }~r\to\infty\,.$$ Hence the constant $C$, of the elliptic estimate $$\| v\|_{W^{2,p}(B_{(3/4)})}\leq C\left(\| v\|_{L^p(B_1)}+\| \Cal L_{r} v\|_{L^p(B_1)}\right) ,\qquad(1