Given a sequence of real numbers $c\_k\to-\infty$, is there always a $C^\infty$ subharmonic function $f$ on $\mathbb R^2$ and a sequence $z\_k\to\infty$ with $|z\_k|<k$ such that 
$$\displaystyle\limsup\_{z\to\infty} \frac{f(z)}{\log |z|}<\infty\ \ \text{and}\ \ f(z\_k)<c\_k\ ?$$ 


I do not even know the answer for non-smooth $f$. In Hayman's "Subharmonic functions", volume 2, in example 7.26 on page 449 there is a weaker result that there is subharmonic $f$ of sub-logarithmic growth (in the sense above) and $z\_k\to\infty$ such that $f(z\_k)<-\log|z\_k|$. 


The question came up in research but it seems like an interesting question by itself.



Oscillation of subharmonic functions of slow growth
Yes, of course. Take any convergent series with positive terms $a\_k$.
Consider the function $u(z)=\sum a\_k\log|1-z/z\_k|$. This is a subharmonic function,
$u(z\_k)=-\infty$, and satisfies $u(z)=O(\log|z|)$ because the Riesz mass is finite.
It is continuous (in the extended sense, $-\infty$ is allowed).


To make it smooth, take a symmetric positive cap with bounded support, and take a convolution with this cap. When the support does not contain $z\_k$ this convolution does not change the function (because it is harmonic there and has the average property).
This allows to to take the cap support smaller and smaller with $k$.


More formally, let $\phi\_k$ be a positive symmetric $C^\infty$ cap supported in the disk of radius less than $1/4$ of the distance from $z\_k$ to the rest of $z\_j$. In a neighborhood of $z\_j$, replace $u$ by $u\*\phi\_k$. The result is the $C^\infty$ function you need.


