Every smooth manifold admits a complete Riemannian metric. In fact, every Riemannian metric is conformal to a complete Riemannian metric, see [this note](http://www.ams.org/journals/proc/1961-012-06/S0002-9939-1961-0133785-8/S0002-9939-1961-0133785-8.pdf). What about in the Kähler case?



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> Does a Kähler manifold always admit a complete Kähler metric?
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Of course, every metric on a compact manifold is complete, so the question is only of interest in the non-compact case.


One might hope that the proof in the aforementioned note will still be of use, but as is shown in [this question](https://mathoverflow.net/q/69792/21564), a metric conformal to a Kähler metric cannot be Kähler (with respect to the same complex structure) except in complex dimension one.


The only condition that I am aware of that ensures the existence of complete Kähler metrics is that the manifold is *weakly pseuodoconvex* (i.e. it admits a plurisubharmonic exhaustion function), see Demailly's [*Complex Algebraic and Analytic Geometry*](https://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf), Chapter VIII, Theorem 5.2.



Does a Kähler manifold always admit a complete Kähler metric?
Grauert proved that a relatively compact domain with real analytic boundary in C^n has a complete Kahler metric iff the boundary is pseudoconvex .For any
relatively compact domain in C^n which is the interior of its closure ,with a complete Kahler metric, Diederich and Pflug showed that it is locally Stein.
 See the paper of Demailly in Annales ENS vol 15 1982 page 487 .


