Given is a positive definite matrix $\Sigma$ and vectors $\mu$. I'm considering the convex optimisation problem
$$ f_{\mu, \Sigma}(\sigma):= \sup_{w\in W, \sqrt{w^T\Sigma w}\le \sigma}\mu^Tw$$ where $W$ is a closed convex set in $\mathbb{R}^n$. it can be easily seen that $f_{\mu, \Sigma}$ is concave and increasing over $\sigma \ge \inf_w\sqrt{w^T\Sigma w}$. I would like to formally prove the following
$$ \sup_{w\in W}\frac{\mu^Tw- r}{\sqrt{w^T\Sigma w}} = \sup_{\sigma >0}\frac{f_{\mu, \Sigma}(\sigma)-r}{\sigma}$$ for $r<\sup_{w\in W}\mu^T w$.
$$ \begin{align} \sup_{\sigma >0}\frac{f_{\mu, \Sigma}(\sigma)-r}{\sigma} &= \sup_{\sigma >0} \frac{\sup_{w\in W : \sqrt{w^T\Sigma w}\le \sigma} \mu^T w-r}{\sigma} \\ &= \sup_{\sigma >0,w\in W : \sqrt{w^T\Sigma w}\le \sigma} \frac{\mu^T w-r}{\sigma} \\ &= \sup_{w\in W} \sup_{\sigma >0 : \sqrt{w^T\Sigma w}\le \sigma} \frac{\mu^T w-r}{\sigma} \\ &= \sup_{w\in W} \frac{\mu^T w-r}{\sqrt{w^T\Sigma w}} \end{align}$$ In the second step I used $\sigma>0$, while in the last step I used that the numerator is nonnegative ($r \leq \sup_{w\in W}\mu^T w$).