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Minor Math Jaxing (proper bracket scaling and $exp\to\exp$
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Daniele Tampieri
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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(-\frac{1}{|x|})-1$$\psi=\exp\left(-\frac{1}{|x|}\right)-1$ with $$\Delta\psi=\frac{1}{|x|^4}exp(-\frac{1}{|x|})>0$$$$\Delta\psi=\frac{1}{|x|^4}\exp\left(-\frac{1}{|x|}\right)>0$$

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

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 $\psi=exp(-\frac{1}{|x|})-1$ with $$\Delta\psi=\frac{1}{|x|^4}exp(-\frac{1}{|x|})>0$$

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

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

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MikeG
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Find strictly subharmonic function that vanishes at infinity

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 $\psi=exp(-\frac{1}{|x|})-1$ with $$\Delta\psi=\frac{1}{|x|^4}exp(-\frac{1}{|x|})>0$$

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