Does the two-point function of free field reveal conformal anomaly? Consider free scalar field in two dimensions with the standard action written in complex coordinates $S=\int d^2z\, \partial \phi\bar{\partial}\phi$. The two-point correlation function is known to be
$$\left\langle\phi(z,\bar{z})\phi(w,\bar{w})\right\rangle = \alpha\ln|z-w|^2$$
here $\alpha$ is some unimportant constant. 
On the other hand, one expects from the conformal symmetry that the correlator must be invariant under any holomorphic variable charge. Apparentely, it is not. 
I am aware that the conformal symmetry is in some sense anomalous -- the symmetry algebra of the QFT is not the Witt algebra, but its central extension, the Virasoro algebra.
The question is whether this anomaly is seen right at the simplest example of the two-point correlation function for free scalar field? Or am I confused about something?
 A: The correlator in question is not invariant even under scaling transformations, and the $sl_2\mathbb{C}\times sl_2\mathbb{C}$ subalgebra of global conformal transformations (generated by $L_{\pm 1},L_0$ and their anti-holomorphic counterparts), which contains the scaling, of Virasoro algebra is always anomaly-free in the sense that you mention. (You can check that the central term in commutation relations vanishes for $n=\pm 1,0$.)
There is conformal anomaly on curved manifolds, but the correlator you write is in flat space.
So the answer is no, the issue you rise is not related to conformal anomaly in any obvious way. 
You can think of this as that the correlation functions of free scalar require a regularization, which cannot be removed completely and you get a scale dependence. It is somewhat similar to an anomaly, I guess, except that this correlator is most likely scheme dependent. In CFT language, the field $\phi$ itself is not a primary field, and its correlators are not required to have the standard transformation properties.
