
> 
> Let $\phi$ be a real smooth superharmonic function on unit disc $D$ in $\mathbb C$; i.e. $\triangle \phi\le 0$.
>  Then there is a curve $\gamma$ from the center of $D$ to its boundary such that 
>  $$\int\limits\_\gamma e^\phi<\infty.$$
> 
> 
> 


The question came from my failed answer [to this question](https://mathoverflow.net/questions/43614).
I know that the answer is YES, but I do not see a direct proof.



Curve integral of exponent of superharmonic function.
I'll assume that $\varphi=-\psi$ where $\psi$ is subharmonic and not too weird (say, with isolated non-degenerate critical points; it seems like you can always add something bounded to achieve it but I haven't checked the details). Take any piece of some level curve of $\psi$ inside the disk parameterized by length $\ell$ and start the gradient accent from each point parameterized by the level $t$ of $\psi$. All but countably many of those escape to the boundary. Let $v(t)$ be the absolute value of the gradient and $S(t)$ be the "cross-section factor". Then $S(t)v(t)$ is non-decreasing (divergence of the gradient field is positive), the length element is $1/v(t)dt$ and the area element is $S(t)/v(t)d\ell dt$. Note that the total area of the disk is finite and the gradient curves cannot meet (we use the non-negativity of the divergence again here). Thus $\int S(t)/v(t)dt<+\infty$ most of the time. But then, since $Sv$ is non-decreasing, we also have $\int 1/v(t)^2dt<+\infty$ while we need just $\int e^{-t}/v(t)<+\infty$ and Cauchy-Schwartz ends the story.


