Abstract

We present an efficient explicit mechanism to convert bounds on $M(x)={\sum <!--FUNCTION\; APPLICATION-->}_{n\le x}\mu (n)$ to bounds on $m(x)={\sum <!--FUNCTION\; APPLICATION-->}_{n\le x}\mu (n)$$∕$$n$ and on $\check{m}(x)={\sum <!--FUNCTION\; APPLICATION-->}_{n\le x}(\mu (n)$$∕$$n)\log <!--FUNCTION\; APPLICATION-->(x$$∕$$n)$. We use this mechanism in three different ways. We first improve on existing bounds for $m(x)$ and $\check{m}(x)$; secondly, we compute the exact value of $\underset{x\ge 1}{\sup <!--FUNCTION\; APPLICATION-->}({\log <!--FUNCTION\; APPLICATION-->}^{2}x)$$|$$\check{m}(x)-1$$|$ and thirdly, we prove that $\overline{\lim <!--FUNCTION\; APPLICATION-->}$$|$$m(x)$$|$$\sqrt{x}>\sqrt{2}$. This establishes that the supremum of $|m(x)|\sqrt{x}$ is not reached for $x\to 2^{-}$, contrary to what the first numerical observations might suggest.

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