Let $h:\mathbb{N}\rightarrow\mathbb{C}$ be a bounded multiplicative function with $h(p)=0$. The motivation for this question is just a general enquiry and, since I suppose it has already been considered, it seems worthwhile asking it here.



> 
> What is the best possible bound on the modulus of the sum 
>  $$S(x)=\sum\_{n\leq x}h(n),$$that is, for all such $h$?
> 
> 
> 


Using some basic facts about Euler products it follows that $S(x)\in O(x^{1/2+\epsilon})$, $\epsilon>0$, but can $\epsilon$ be replaced by something explicit?



Bound for sums of bounded multiplicative functions that are zero at primes
To expand on Lucia's comment, we note that we can assume without loss of generality that each coefficient of $h$ is bounded by $1$, in which case $S(x)$ is largest when $h(p^k) = 1$ for all $k \geq 2$ (and of course $h(p) = 0$). So
\[S(x) = \#\left\{n \leq x : p | n \implies p^2 | n\right\}.\]
This is the number of [powerful numbers](https://en.wikipedia.org/wiki/Powerful_number) less than $x$, which [Golomb](http://dx.doi.org/10.2307%2F2317020) has shown to be asymptotic to
\[\sum\_{n = 1}^{\infty}\frac{\mu^2(n)}{n^{3/2}} x^{1/2} = \frac{\zeta(3/2)}{\zeta(3)} x^{1/2},\]
with
\[\frac{\zeta(3/2)}{\zeta(3)} \approx 2.173.\]
One can see this directly by noting that the Dirichlet series with coefficients given by the multiplicative function $h(p^k) = 1$ for all $k \geq 2$, $h(p) = 0$, satisfies
\[\sum\_{n = 1}^{\infty} \frac{h(n)}{n^s} = \prod\_p \sum\_{\substack{k = 0 \\ k \neq 1}}^{\infty}\frac{1}{p^{ks}} = \prod\_p \left(\frac{1}{1 - p^{-s}} - \frac{1}{p^s}\right),\]
and some manipulation shows that this is equal to
\[\prod\_p \frac{1 - p^{-6s}}{(1 -p^{-2s})(1 - p^{-3s})} = \frac{\zeta(2s)\zeta(3s)}{\zeta(6s)},\]
which is holomorphic for $\Re(s) > 1/2$ and has simple poles at $s = 1/2$, $s = 1/3$, and then at the nontrivial zeroes of $\zeta(6s)$ in the strip $0 < \Re(s) < 1/6$, plus at the trivial zeroes of $\zeta(6s)$ to the left of $\Re(s) = 0$. So Cauchy's residue theorem implies that
\[\begin{split}
S(x) & = \frac{1}{2\pi i} \int^{\sigma + i\infty}\_{\sigma - i\infty} \frac{\zeta(2s)\zeta(3s)}{\zeta(6s)} \frac{x^s}{s} ds \\
& = \frac{\zeta(3/2)}{\zeta(3)} x^{1/2} + \frac{6 \zeta(2/3)}{\pi^2} x^{1/3} + o(x^{1/6}).
\end{split}\]


