Within this entry, $\omega$ refers to the number of distinct prime factors function, $\lfloor \, \cdot \, \rfloor$ refers to the floor function, $\log$ refers to the natural logarithm, $p$ refers to a prime, and $k$ and $n$ refer to positive integers.

Proof. Since $y^{\omega(p^k)}=y$ for all $p$ and $k$ the real-valued nonnegative multiplicative function $y^{\omega(n)}$ satisfies the Wirsing condition with $c=y$ and $\lambda=1$ Thus:

$\displaystyle \sum_{n \le x} y^{\omega(n)}$	$\displaystyle =O_y \left( \frac{x}{\log x} \sum_{n \le x} \frac{y^{\omega(n)}}{n} \right)$
	$\displaystyle =O_y \left( \frac{x}{\log x} \prod_{p \le x} \left( 1+\sum_{k=1}^{\left\lfloor \frac{\log x}{\log p} \right\rfloor } \frac{y^{\omega(p^k)}}{p^k} \right) \right)$
	$\displaystyle =O_y \left( \frac{x}{\log x} \left( \exp \left( \sum_{p \le x} \sum_{k=1}^{\left\lfloor \frac{\log x}{\log p} \right\rfloor } \frac{y}{p^k} \right) \right) \right)$
	$\displaystyle =O_y \left( \frac{x}{\log x} \left( \exp \left( y \sum_{p \le x} \sum_{k=1}^{\left\lfloor \frac{\log x}{\log p} \right\rfloor } \frac{1}{p^k} \right) \right) \right)$
	$\displaystyle =O_y \left( \frac{x}{\log x} ( \exp (y(\log(\log x)+O(1)))) \right)$
	$\displaystyle =O_y \left( \frac{x}{\log x} ( \exp (\log (\log x)^y)) \right)$
	$\displaystyle =O_y \left( \frac{x}{\log x} (\log x)^y \right)$
	$\displaystyle =O_y(x(\log x)^{y-1})$