Let Dolbeault cohomology and the corresponding $L^2$-cohomology be denoted by $H^{p,q}(X) $ 
 and $H^{p,q}\_{(2)}(X)$ respectively.  

As is well known, on a compact complex manifold $X$, $H^{p,q}(X) \cong H^{p,q}\_{(2)}(X)$ by the Hodge isomorphism.


My question is : **On a noncompact complex manifold, can we compare this 2 type of cohomolog**y. 
Does $H^{p,q}(X) \subset H^{p,q}\_{(2)}(X)$ holds (in the sense of isomorphism of groups ) in general ?



Relations between Dolbeault cohomology and the corresponding $L^2$-cohomology
I don't know if that was a typo (did you want isomorphism, or did you really mean inclusion/injection?). In any case, the question in either interpretation has a negative answer. Let $X$ be a compact Riemann surface $\overline{X}$ minus a postive but finite number of points $p\_1, p\_2,\ldots$. 
An element of $H^{1,0}\_{(2)}(X)$ is represented by a holomorphic $1$-form $\omega$, such that $\int\_X\omega\wedge\overline{\omega}<\infty$. One can check by calculating in polar coordinates near each $p\_i$ that this forces $\omega$ to extend to $\overline{X}$. Therefore the $L^2$ space is finite dimensional. On the other hand, $H^{1,0}(X)$ is infinite dimensional.


To put this example in a broader context, various authors have shown/conjectured that in good cases $L^2$-cohomology coincdes with intersection cohomology, so it would be finite dimensional. For this particular case, one can see this directly as above.


