This question was asked in my assignment in number theory and I could not prove it.
Question : Define the multiplicative function w(n) such that $w(p^k) =0$ for $k\geq 2$ and w(p)= { $\frac{p} { f(p)} $ if $p\in P$, 0 if $p\notin P$}. Suppose that w(n) satisfies $ \sum_{p\leq z} \frac{w(p) log(p)} {p} = Klog z+O(1)$ where $K \geq 0$. Then show that $\prod_{p\leq z} (1- \frac{w(p)} {p}) \sim c_w \frac{e^{-\gamma K} } { (logz)^ K}$ where $c_w = \prod_{p} (1- \frac{w(p)} {p} ) ( 1- 1/p)^{-K}$ is a convergent product and $\gamma $ is a constant.
Attempt: By Merten's theorem , $ \prod_{p \leq z}( 1-1/p)^{K} \sim \frac {e^{-\gamma K}} { (log z)^K}$.
I can write $\prod_{p\leq z} (1- \frac{w(p) }{p}) $ as equal to $\prod_{ p\leq z} ( 1-1/p)^K (1-1/p)^{-K} ( 1- w(p)/p)$
But I don't have much ideas on how to proceed from here. Can you please help me with this?
By partial summation, one can deduce the existence of a constant $C_0$ such that
$$ \sum_{p\le z}{w(p)\over p}=K\log\log z+C_0+O\left(1\over\log z\right) $$
As $t\to0^+$, we know $-\log(1-t)=t+O(t^2)$, so we also know that there exists another constant $C_1$ such that
$$ -\sum_{p\le z}\log\left(1-{w(p)\over p}\right)=K\log\log z+C_1+O\left(1\over\log z\right) $$
Exponentiating both side gives
$$ \prod_{p\le z}\left(1-{w(p)\over p}\right)={e^{-C_1}\over\log^Kz}\left\{1+O\left(1\over\log z\right)\right\} $$
Juxtaposing this asymptotic formula with Mertens' theorem will yield what you expect.