Let $\varphi$ be a plurisubharmonic function in the unit ball $B_1\subset \mathbb{C}^n$ with $\varphi\le 0$. Suppose that the Lelong number $\nu(\varphi,0)<k$ for some $k>0$. Does it follow that there exists $\alpha>0$, possibly depending on $k$, such that $\int_{B_{\frac{1}{2}}}e^{-\alpha\varphi}dvol(z)\le C$?

Note that this does not immediately follow from the definition of Lelong number. Since $\nu(\varphi,0)=\lim\inf_{z\rightarrow0}\frac{\varphi(z)}{\log|z|}$, so $\nu(\varphi,0)<k$ apriori only implies $\varphi(z)-k\log|z|$ bounded from below along some sequence $z_i\rightarrow0$.

On the other hand, if we know $\nu(\varphi,0)=0$, then above is indeed true, since under this assumption, we would have $\varphi(0)$ bounded from below, and the result above follows from a Lemma of Hormander(Lemma 4.4 in his book on several complex variables).

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The answer is yes :

If the Lelong number of a plurisubharmonic function $\varphi$ at a point $a$ satisfies the condition $\nu(\varphi,a) < 2$, then the function $e^{-\varphi}$ is locally integrable with respect to the Lebesgue measure in a neighbourhood of $a$.

It follows from a result of Skoda, Proposition 7.1 in

H. Skoda, Sous-ensembles analytiques d’ordre fini ou infini dans $\mathbb{C}^n$, Bull. Soc. Math. de France 100 (1972), 353-408.

  • Thank you! I will take a look. – Jerry Oct 12 at 13:12

If you want $\alpha$ to not depend on more than $\nu(\varphi,0)$ and you want it to hold in $B_{1/2}(0)$, as I have interpreted your question now, then the answer is no:

In $\mathbb{C}$, you can take $\varphi_m = \log |z-1/4|^m-\log (5/4)^m$, which is $\leq 0$ on $B_1(0)$, and such that $\nu(\varphi_m,0) = 0$, while $e^{-\alpha\varphi_m}$ is integrable on $B_{1/2}(0)$ only for $\alpha < 2/m$.

If you on the other hand allow the smaller neighborhood to depend on the function, or depend on the Lelong number in the neighborhood, then the answer should be yes, basically by the answer of user111.

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