The theorem of the Chinese mathematician Chen Jingrun is currently one of the best results with respect to the binary Goldbach problem. It asserts that every even integer $n$ is in the sumset $P + P_2$, where $P$ denotes the set of primes and $P_2$ denotes the set of positive integers with at most two prime factors. However, I am not aware of this result asserting any control over the potential prime factors arising from the summand in $P_2$. In particular, I am interested in whether the following variant of Chen's theorem can be obtained:
Let $\mathcal{R}_f(N)$ denote the number of ways that $N$ can be written in the form $p + q_1 q_2$, where $p, q_1$ are primes and $q_2$ is either a prime or 1, and further $q_2 \ll f(N)$. Here $f : \mathbb{N} \rightarrow \mathbb{R}^+$ is a slowly increasing function that tends to infinity. Then for what choice of $f$ can we prove $\mathcal{R}_f(N) \rightarrow \infty$?
Chen's theorem would amount to the choice $f = N$ say and the result would be $r(N) = \mathcal{R}_f(N) \gg \mathfrak{S}(N) \frac{2N}{(\log N)^2}$, where $\mathfrak{S}(N) = \displaystyle \prod_{p > 2} \left(1 - \frac{1}{(p-1)^2}\right) \prod_{\substack{p | N \\ p > 2}} \frac{p-1}{p-2}.$
For instance, can we prove something analogous to Chen's theorem if we pick say $f = N^\theta$ for some $\theta \leq 1/8 - \epsilon$ for some $\epsilon > 0$? (See the edit for why the choice of $1/8 - \epsilon$ is chosen)
The motivation of course is if we can choose $f(N)$ so that $q_2 \leq 1$ for all sufficiently large $N$, then we would have proven the asymptotic binary Goldbach conjecture. If $q_2$ can be chosen to be bounded for all $N$ sufficiently large, that would still be a very tantalizing result.
Edit: I looked at the proof of Chen's theorem in Nathanson's book "Additive Number Theory: The Classical Bases" again and in that argument he (actually, the proof in Nathanson's book is due to Iwaniec) basically gives a lower bound for $r(N)$ by considering the case when $N = q_1 + q_2 q_3$, where $q_2, q_3$ are primes at least $N^{1/3}$. In light of this, it seems that my question is trivial when $f(N) \gg N^{1/3}$. Hence I modified the question by forcing $f$ to be smaller.
Edit 2: I misread the proof, the lower bound is actually $N^{1/8}$ instead of $N^{1/3}$.
