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Recall that a complex manifold is Stein if it is holomorphically convex and separable. If we assume holomorphically convex alone, then there is Cartan-Remmert reduction to say how far it is from being Stein. If we assume holomorphic separation alone, I learnt from Th 9.9(b) in p41 of Demailly's lecture notes Here that:

"On a complex manifold X, if $O(X)$ locally separate points (i.e. for any $x \in X$, there is a nbhd of $U_x$ such that for any $y \in {U_x}-x$, there exists $f \in O(X)$ with $f(x) \neq f(y)$), then there exists a smooth nonnegative strictly plursisubharmonic function"

The proof looks like first construct global PSH functions which are locally strictly PSH on small nbhd, then adding them together after multiplying some constants. I don't understand why that sum converges. Since the proof looks very straightforward and I may miss sth simple here. I would appreciate if anyone can help. Thanks!

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In his proof, you have just to choose the multiplying constants converging very fast to 0. For instance, $\varepsilon_j=1/(2^j(1+\max_{\partial V_j}v_0))$ will work. –  diverietti Jun 19 '13 at 15:48
    
thanks for the suggestion. sorry I still can't figure it out, Say for a fixed point, those PSH functions $u_j$ may take very different values on this point, and they may grow large outside $V_j$, I did not get $\sum \epsilon_j u_j$ must converge everywhere. sorry if I made any simple misunderstanding. –  Bo_Y Jun 20 '13 at 6:07

1 Answer 1

up vote 1 down vote accepted

I realized that my comment was misleading, so I prefer to put a complete answer.

In Demailly's proof at some point you end up with a countable open cover $\{V_j\}$ of your complex manifold $X$ and a family of non-negative smooth psh functions $\{v_j\}$ such that $v_j$ is strictly psh on $V_j$. Then you want to sum them together in order to obtain a smooth non-negative strictly psh function $v$ on the whole of $X$.

At this point Demailly claims that you can choose a sequence of real numbers $\varepsilon_j>0$ converging to zero so fast that $$ \sum_{j=0}^{+\infty}\varepsilon_jv_j $$ converges in the $C^{\infty}$-topology. This of course would suffice to get the desired function $v$, right?

I shall give you all the details for the $C^0$-convergence and I'll leave you the task to fill it out for the derivatives.

Fix an exhaustion by compact sets $\{K_\nu\}$ of $X$ and set $$ \alpha_{j,\nu}:=\max_{K_\nu}|v_j|. $$ Now, for each $j\ge 1$ select a $\eta_j>\max_{1\le\nu\le j}\{\alpha_{j,\nu}\}$. Finally, set $\varepsilon_j:=1/(\eta_j2^j)$.

To check uniform convergence on compacta note that for any given compact set $K\subset X$ you have that $K\subset K_{\nu_0}$ for some $\nu_0$, so that, for $j\ge\nu_0$, you have

$$ \max_{K}|\varepsilon_j v_j|\le\max_{K_{\nu_0}}|\varepsilon_j v_j|=\frac{\alpha_{j,\nu_0}}{\eta_j 2^j}<\frac 1{2^j}. $$

To obtain the $C^\infty$-convergence, you have to deal with the topology generated by all seminorms $p_L^s$ when $s,L,\Omega$ vary, where $s\in\mathbb N$, $\Omega$ is a coordinate open set with coordinates $(x_1,\dots,x_m)$ and $L\subset \Omega$ is a compact subset. They are defined by $$ p^s_L(f)=\max_{x\in L}\max_{|\ell|\le s}|D^\ell f(x)|, $$ where $\ell=(\ell_1,\dots,\ell_m)\in\mathbb N^m$, $|\ell|=\ell_1+\cdots+\ell_m$ and $$ D^\ell f:=\frac{\partial^{|\ell|}f}{\partial x_1^{\ell_1}\cdots\partial x_m^{\ell_m}}. $$ The way to define a sequence $\{\varepsilon_j\}$ which works simultaneously for all derivatives is essentially the same, except for the flurry of subscripts...

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Oh, exactly! Thank you very much for your time! –  Bo_Y Jun 21 '13 at 23:01
    
You are welcome! And sorry again for the first misleading comment! –  diverietti Jun 22 '13 at 8:12

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