Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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. I know that the answer is YES, but I do not see a direct proof.

share|improve this question
    
How about using "greedy algorithm" to choose γ? That is, let γ(0)=0,dγ=−∇ϕ(γ) so that $e^ϕ$ is minimized in each infinitesimal step. Can superharmonicity of ϕ guarantee that γ will touch the boundary eventually, to begin with? –  Syang Chen Mar 29 '11 at 21:14
    
@Xianghong, this way you at least decrease $\phi$ along $\gamma$, but you might get a curve of infinite length (I do not see why not). –  Anton Petrunin Mar 29 '11 at 23:35
    
@Anton, here is another "greedy curve" which will touch the boundary in finite time: let $|\gamma(r)|=r$ such that $\phi (\gamma(r))=$ min$_{|z|=r} \phi (z)$, where $ 0 \leq r < 1 $. But I am not sure if such $\gamma$ exists, and more importantly, if it has finite length. –  Syang Chen Mar 31 '11 at 7:30
1  
Alternatively, is it possible to show that(at least for those $\phi$ that you are interested in) $\int_{D} e^{\phi}dx<\infty$? If it is the case, then almost every radius could serve your purpose. As far as I know, $L^p$ integrability of $(\phi)_{+}$ always holds for some $p>0$. –  Syang Chen Mar 31 '11 at 7:31
add comment

2 Answers

up vote 5 down vote accepted

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.

share|improve this answer
add comment

Here is a simple answer in a special case when $\phi$ is harmonic.

Geometric proof: Let $g_0$ be the (incomplete) metric on the unit disc induced from $\mathbb R^2$. Then the metric $e^\phi g_0$ is complete if and only if the integral of $e^\phi$ over $\gamma$ is infinite for each $C^1$-curve $\gamma$ from the center to the boundary of the disk. Now the sectional curvature of $e^\phi g_0$ is $-\frac{1}{2}e^{-\phi}\Delta\phi$, which is zero since $\phi$ is assumed harmonic. Thus if $e^\phi g_0$ were complete, it would be isometric to the standard $\mathbb R^2$, but isometries are conformal, so we would get that the unit disk is conformal to the plane.

Sketch of complex analysis proof: If by a "direct proof" you mean then one that uses only complex analysis, then Huber's proof quoted in my question does just that. However, Huber's argument simplifies a lot when $\phi$ is harmonic. Indeed, harmonicity of $\phi$ allows you to find its harmonic conjugate, and since the disk is simply connected, $\phi$ becomes the real part of an analytic function $\tau$ on the disk. Thus $e^\phi=|e^\tau|$, and $e^\tau$ is analytic and nowhere vanishing. Now I think the simple argument in the note "Paths of rapid growth of entire functions" by Kaplan does the job (see middle of the first page).

Remark. Kaplan's note is in fact an elaboration of the Huber's proof, and it gives a simple answer to your question for all superharmonic functions of the form $\phi=-\log|f|$ such that $f$ does not vanish (and this is what you seem to care about by restricting to smooth superharmonic functions). For general superharmonic functions I do not see how to do better than Huber.

share|improve this answer
2  
Incidentally, applying the "geometric proof" to the plane (instead of the disk) we get the amusing conclusion that if $\phi$ is harmonic and non-constant, and $g_0$ is the standard metric on the plane, then $e^\phi g_0$ is not complete. This is because conformal automorphisms of the plane are affine maps, and pulling back the standard metric $g_0$ on the plane by an affine map yields the rescaling of $g_0$ by the conformal factor of the map. –  Igor Belegradek Apr 14 '11 at 16:07
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.