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I am looking for a super(sub) harmonic function for an elliptic operator.

Let $n$ be a positive integer. We denote by $(\cdot,\cdot)$ and $|\cdot|$ the standard inner product and norm on $\mathbb{R}^n$, respectively. We denote by $U \subset \mathbb{R}^n$ the open unit ball centered at the origin. The elliptic operator $\mathcal{L}$ is defined as follows: \begin{align*} \mathcal{L}f(x)=(1-|x|^2)\Delta f-c((x-\theta),\nabla f),\quad x \in U, \end{align*} where $\theta \in U$ and $c$ is a positive constant.

My question. Can we find a smooth and nonnegative function $f\colon U \to \mathbb{R}$ and $\varepsilon>0$ such that $\mathcal{L}f \ge \varepsilon$ on $U $ ? Needless to say, the function $f$ may depend on $\theta$. If necessarily, the ranges of $c$ and $|\theta|$ may be limited. If we find such a function, in a sense, we can understand the boundary behavior of the diffusion process associated with $\mathcal{L}$.

If $\theta=0$, we can find such a function. Indeed, if we set $f=\alpha^{-1}\{1-(1-|x|^2)^{\alpha}\}$, $\alpha \in (0,1)$ (there may be something simpler than this), we obtain that \begin{equation} \mathcal{L}f=4\{(1-c/2)-\alpha \}|x|^2(1-|x|^2)^{\alpha-1}+2n(1-|x|^2)^{\alpha}. \end{equation} Therefore, if $c<2$ and $\alpha \in (0,1-c/2)$, we find that $f$ possesses the desired property (in fact, $c=2$ is a border in a sense).

If $\theta \neq 0$, however, I could not find a function satisfying the above conditions.

If you find one, please let me know.

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  • $\begingroup$ Do you know what happens if $\theta=0$ and $c\geq 2$? $\endgroup$ Sep 8, 2020 at 12:35
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    $\begingroup$ @GiorgioMetafune Thank you for your comment. There will be no function satisfying the conditions if $c \ge 2$ and $\theta=0$. This is based on probabilistic considerations. $\endgroup$
    – sharpe
    Sep 8, 2020 at 12:56
  • $\begingroup$ It seems that your barrier works if $c(1+|\theta|) <2$; do you find the same? $\endgroup$ Sep 9, 2020 at 7:12
  • $\begingroup$ @GiorgioMetafune I don't find it. Could you tell me the reason? $\endgroup$
    – sharpe
    Sep 9, 2020 at 11:31
  • $\begingroup$ @GiorgioMetafune Your result is consistent with the case of $\theta=0$. $\endgroup$
    – sharpe
    Sep 9, 2020 at 12:51

1 Answer 1

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If you have a second order elliptic operator L on a smooth noncompact connected manifold then you can always find a smooth function f>0 such that Lf > 0 . See the paper by Napier and myself in L'Enseignment Mathematique vol 50 2004 pages 367-390 .

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  • $\begingroup$ Here the problem is the degeneracy of the elliptic part at the boundary. I guess that your result holds for uniformly elliptic (or strictly elliptic) operators, that is when the ellipticity constant is bounded above and below from 0. $\endgroup$ Sep 9, 2020 at 17:32
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    $\begingroup$ The result of Napier and Myself does not assume strict or uniform ellipticity. It uses a Runge style argument. $\endgroup$ Sep 9, 2020 at 18:40
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    $\begingroup$ I see, but then I do not understand the answer of @sharpe to a comment above, saying that such a function does not exist if $\theta=0, c \ge 2$. Unless he assumes boundedness and your construction yields (possibly) unbounded functions. $\endgroup$ Sep 9, 2020 at 20:39
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    $\begingroup$ Yes the function will be unbounded. In fact you can make it so that sublevel sets are relatively compact. $\endgroup$ Sep 9, 2020 at 21:36
  • $\begingroup$ @MohanRamachandran Thank you for your comment. If $\theta=0$ and $c\ge 2$, I thought that functions with the above conditions do not exit. But it was different. $\endgroup$
    – sharpe
    Sep 10, 2020 at 2:12

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