Lower bound for sum of a multiplicative function based on lower bound on value of primes

Question

Let $f$ be a non-negative multiplicative function. Suppose we have some bound on $\sum_{p \le x} f(p)$, where the summation is over primes. Is it possible to give a lower bound on $\sum_{n \le x} f(n)$ that is better than the trivial bound $\sum_{n \le x} f(n) \ge \sum_{p \le x} f(p)$? Or perhaps if we assume some control of $f(p^k)$.

It seems like this type of problem ought to be have been investigated before, but I'm not sure where to look. I haven't been even able to save a factor of $\log x$.

Answer 1

Here's an example where I don't think you can save a factor of $\log x$. Let $f(p)=p!$, $f(p^k)=0$ for $k\ge2$. The sum over the primes is comparable to $x!$. The sum over the nonprimes is bounded by $(2x)(x/2)!$, which is tiny compared to the sum over the primes.

Answer 2

I'm no analytic number theorist, but here's a (very basic) attempt:

Let $F(x)=\sum_{n\leq x}f(x)$ and $P(x)=\sum'_{p<x} f(p)$ where the second summation is over primes. You want to say something non-trivial about the relationship between $F(x)$ and $P(x)$.

Two things: If your $f(p)$ are all non-negative, then you should be able to use that the prime powers in $[0,x]$ are quite scarce and probably just using products of two primes you should be able to save something.

That is, the contribution from numbers $n\in [0,x]$ that are products of two primes is lower bounded by $$\left(\sum_{0<p\leq \sqrt x}f(p)\right)^2-\sum_{0<p<\sqrt{x}}f(p)^2+\sum_{0<p\leq x}f(p^2) =\left[P(\sqrt x)\right]^2-\sum_{0<p<\sqrt{x}}f(p)^2+\sum_{0<p\leq x}f(p^2) $$

If the middle sum left at the very end is very small (for instance if $0<f(x)<1$), then this method might be useful. Or more generally, if you know some relation between $f(p)^2$ and $f(p^2)$, you can easily figure out if this heuristic will help. In particular, if the values are roughly the same, the $P(x)^2$ is your saving.