Let $A$ be a $\sigma$-unital C*-Algebra and $(x_{\lambda})_{\Lambda}$ be a norm bounded net in $\mathcal{M}(A)$ (the multiplier algebra of $A$). I proved that for a strictly positive element $h$ if $(x_{\lambda}h)_{\Lambda}$ and $(hx_{\lambda})_{\Lambda}$ are norm convergent in $A$ then $(x_{\lambda})_{\Lambda}$ is strictly convergent in $\mathcal{M}(A)$.
So here is my question: How can I show that the strict topology is metrizable on bounded subsets when $A$ is $\sigma$-unital.
Let $h$ be a strictly positive element. Define: $$d(x,y) = {\|h(x-y)\|+\|(x-y)h\|}.$$ You have checked that if $x_\alpha$ is a bounded net and $d(x_\alpha,x)\to0$ that then $x_\alpha\to x$ in the strict topology. Further if $x_\alpha\to x$ in the strict toplogy you have that $\|a(x_\alpha-x)\|\to0$ for all $a\in A$ and $\|(x_\alpha-x)a\|\to0$ for all $a$, in particular for $a=h$. Hence the notion of convergence given by this metric agrees on bounded sets with the convergence of the strict topology.