Obviously, we have $\|A\|\ge \sup\{ \|Af\| : f\ge 0, \|f\| = 1\}$. Now let $f\in E$ with $\|f\| = 1$. Then $|f|\in E$, $|f|\ge 0$ and $\| |f| \| = 1$. Since $A$ is positive, we have $Af\le A|f|$ (because $f\le |f|$) and $-Af\le A|f|$ (because $-f\le |f|$). This implies that $|Af|\le A|f|$. Thus, $\|Af\|^p = \int |Af|^p\, d\mu \le \int (A|f|)^p\, d\mu = \| A|f| \|^p$, and so $\|Af\|\le \|A|f|\|\le \sup\{ \|Ag\| : g\ge 0, \|g\| = 1 \}$.