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Consider the eigenfunction $\varphi_R>0$ $$L\varphi_R=\lambda_R\varphi_R, \ \ \ in \ \ B_R,$$ and $$\varphi_R=0, \ \ \ in \ \ \partial B_R,$$ where $L$ is a elliptic operator and $\lambda_R$ is the corresponding principal eigenvalue. We can normalize $\varphi_R$ such that $\varphi_R(0)=1$. Then, for the "Harnack inequality" in the ball $B_{2R}$, exists a constant $\delta_R>0$ such that $$\delta_R\leq\varphi_R\leq\delta_R^{-1}, \ \ \ in \ \ \overline{B_{3R/2}}.$$

Where can I find this Harnack inequality?

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    I think some of your $R$s and $2R$s are mixed up...2012-11-22
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    What this means?2012-11-22
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    How can you expect $\varphi_R$ to satisfy a Harnack inequality on $B_{3R/2}$ if it's only a solution in $B_R$?2012-11-22
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    After you fix the ratio of the ball you can try: Corollary 9.25, page 250 from the book of Gilbard Trudinger2012-11-22
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    Nate, i think that $\varphi_R$ is defined in a domain bigger than $B_R$. But the first equality is only satisfies in the ball.2012-11-22
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    Well in that case your inequality can't hold. If, say, $B_R = B(x_0,R)$ and $\varphi_R$ satisfies your assumptions then so does $$\varphi(x) = \begin{cases} \varphi_R(x), & |x-x_0| \le R \\ 1/(|x-x_0|-R), & |x-x_0| > R \end{cases} $$ which is unbounded on $B(x,3R/2)$ and so cannot satisfy any such inequality there.2012-11-22

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