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It is well known that for any non negative Harmonic function w ($\Delta w=0$, $w\geq 0$) in a ball, $B_1(0)$, $\exists$, C>0 such that $\forall y\in B_{1/2}(0)$ $$ Cw(0)\leq w (y) $$

It is a clear implication of Harnack's inequality.

I am trying to prove same inequality for the function w such that $\Delta w=1$ in $B_{1}(0)$. Assuming that the ball $B_1(0)$ is contained in the positivity set of $w$. i.e. $B_1(0)\subset \{w>0\}$.

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Write $w(x) = u(x) + c (1 - |x|^2)$ with $u \geqslant 0$ harmonic in $B_1(0)$, and apply standard Harnack's inequality: $$w(0) = u(0) + c \leqslant C^{-1} u(y) + \tfrac{4}{3} c (1 - |y|^2) \le \max(C^{-1}, \tfrac{4}{3}) w(y)$$ for $y \in B_{1/2}(0)$. (Note: with your notation, $C \in (0, 1)$).

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  • $\begingroup$ Thanks for the idea, but the function $w$ in the answer has laplacian negative , $\Delta w= -2nc$, I think it should have been $w(x)=u(x)+\frac{1}{2n}(|x|^2-1)$. In this case adjusting the constants is an issue. $\endgroup$
    – Harish
    Commented Sep 25, 2017 at 16:54
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    $\begingroup$ Whoops, of course you are right! But then there is no hope for Harnack's inequality, is there? The function $w(x) = \varepsilon + c |x - y|^2$ is a counterexample. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 25, 2017 at 17:43

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