Renormalization of Composite Chiral Superfields On page 24 of these lecture notes  http://arxiv.org/abs/hep-th/0309149 it is stated that products of chiral superfields do not suffer from short distance singularities. In other words, if I want to calculate the dimension of a composite operator built out of chiral superfields, I may just add the dimensions of the individual field operators that form the product (in other words, it is sufficient to just renormalize the individual operators). This is definitely not the case in ordinary QFT. Can someone please explain why this is the case? Is the same true for products of vector superfields (the wording in the notes seems ambiguous)?
 A: The Green's function of a product of chiral superfields is independent of space-time as a consequence of the SUSY algebra. This can be shown by noting that (roughly)
$$
\bar{\sigma}_\mu^{\alpha\dot{\alpha}} \partial^\mu \langle0|\Phi_1(x_1),\cdots,\Phi_n(x_n)|0\rangle= \langle0|\Phi_1(x_1),\cdots,\{Q,\Phi_i(x_i)\},\Phi_n(x_n)|0\rangle=0
$$
where it is assumed that $Q|0\rangle=0$. 
This independence of the space-time means that short distance singularities will not affect the product of chiral superfields (perturbatively).
I hope this helps
A: I've found the answer to this question, and it does basically follow from holomorphicity. The holomorphicity property is essentially the statement that the operator is BPS: $\Phi$ is annihilated by a supercharge $Q$. This is also the reason the property does not hold for vector superfields, which are not BPS.
The reason that the dimension of the operator is protected from quantum corrections is due to the structure of the superconformal algebra. The superconformal algebra contains, in addition to the supercharges $Q_\alpha$, the conformal supercharges $S_\alpha$ that come from the commutator $[Q,K]\approx S$. The commutation relations of the algebra show that S lowers dimensions by $1/2$, while $K_\mu$ still lowers dimensions by 1. The multiplets of the superconformal algebra are larger than those of the conformal algebra. The highest weight states are superconformal primaries that are annihilated by both $S$ and $K$. Starting with such an operator $\Phi$ satisfying
$[S,\Phi(0)]=[K,\Phi(0)]=0$
you may generate the rest of the representation by acting on $\Phi$ with both $Q$ and $P^\mu$. The point is that for special operators, called chiral primaries (also called BPS operators), $[Q,\Phi(0)]=0$ and you get a short multiplet. You can use this fact along with the commutation relations of the superconformal algebra (specifically $\{ Q,S   \} \approx R + D$ where R is the R charge and I am assuming a scalar operator for simplicity) to show that the dimension of a chiral primary is determined solely by it's R-charge. Note that in a superconformal theory R-charge is well defined since it cannot be anomalous without breaking superconformal invariance. 
The point is that these chiral primaries cannot pick up anomalous dimensions. Basically, we have the dimension of the operator in the free theory where the coupling is 0. Regardless of the coupling, the chiral primary is still annihilated by the supercharge, it remains a chiral primary in the interacting theory. If this were not the case, continuously varying the coupling would lead to a discontinuity in the number of operators of a given dimension (this must be an integer, it cannot vary continuously with the coupling). So the operator in the interacting theory is still a chiral primary, its dimension is related to its R-charge. 
In the case I asked about, the chiral superfield is a chiral primary. Products of chiral superfields are still chiral, so the composite operator is still BPS, and its dimension is related to its R-charge. But the R-charge of the operator is the sum of the R-charges of the individual operators, so we can simply add dimensions and not worry about any anomalous dimension.
The answer is a little more complicated than I had expected, but I am almost certain that this is the correct explanation. 
A: You might want to look here: arxiv.org/abs/hep-ph/9309335
Your idea about holomorphicity of the super-potential is essentially correct.
