Find strictly subharmonic function that vanishes at infinity

Question

I am not sure about the term "strictly" subharmonic.

What I want is a function $\psi\in C^{\infty}(\mathbb{R}^3)$ with $\Delta\psi>0$ and $\lim\limits_{|x|\rightarrow\infty}\psi(x)=0$.

I tried several times but still failed at the origin. I took $\psi=\exp\left(-\frac{1}{|x|}\right)-1$ with $$\Delta\psi=\frac{1}{|x|^4}\exp\left(-\frac{1}{|x|}\right)>0$$

at every point $x\not=0$, and the boundary value does vanish, but $\Delta\psi(0)=0$.

Also, is this possible in $\mathbb{R}^2$? I want to use the comparison principle for the elliptic operator Laplacian $\Delta$ to disprove it, but that can only tell me $\psi<0$. How can I make use of the dimension? Or is this also possible?

Any help is desired.

The question is also posted on MSE

Answer 1

For an explicit example on $\mathbb{R}^n$ with $n > 2$:

Let $\phi(x) = -(\sqrt{1 + |x|^2})^{2-n}$, then

$$ \nabla \phi(x) = (n-2) x(\sqrt{1 + |x|^2})^{-n} $$

and

$$ \triangle \phi = (n-2)n (\sqrt{1+|x|^2})^{-n} - (n-2)n |x|^2 (\sqrt{1+|x|^2})^{-n-2} = (2-n)n (\sqrt{1 + |x|^2})^{-n -2} > 0$$

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Here's a simple proof that when $n = 2$ the sort of function you are looking for is not possible.

Suppose for contradiction $\psi$ is such a function. Let $\phi$ be the spherical average of $\psi$:

$$ \phi(x) = \int_{\mathbb{S}^1} \psi(R_\theta x) ~d\theta $$

here $R_\theta: \mathbb{R}^2 \to \mathbb{R}^2$ is the linear transformation rotating the plane by angle $\theta$. So $\phi$ is radially symmetric. Using that the Laplacian is rotationally invariant, you get

$$ \triangle \phi(x) = \int_{\mathbb{S}^1} \triangle \psi(R_\theta x) ~d\theta > 0 $$

and also by inspection you have $\lim_{|x|\to\infty} \phi(x) = 0$ since this holds for $\psi$.

In polar coordinates this means

$$ \partial_r (r \partial_r\phi) > 0 $$

which means that $r\partial_r \phi$ is strictly increasing. This function is $0$ at the origin, and so you have that $r\partial_r\phi > 0$ for all $|x| > 0$. But this means $\partial_r\phi|_{|x| = R} > \frac{1}{R} \partial_r \phi|_{|x| = 1}$. Integrating radially you find that $\phi$ must grow at least logarithmically, and this contradicts $\phi \to 0$.