I'm trying to prove the following:
$$ F(\alpha x) = \sum_{m=1}^M \bigl( \sum_{i=1}^N (\alpha x_i)^{\omega_m}\bigl)^{1/{\omega_m}} = \alpha F(x) $$
But I can't quite understand how to pull $\alpha$ out. I got to the following step. But if ${\omega_m}$ depends on $m$ from the outermost summation, is it possible to pull $\alpha$ out again?
$$ \sum_{m=1}^M (\alpha)^{\omega_m} \bigl(\sum_{i=1}^N x_i^{\omega_m}\bigl)^{1/{\omega_m}} $$
When you pull the $\alpha$ from the inner summation, I think you've misplaced your parenthesis:
$$ \sum_{m=1}^M \bigl(\alpha^{\omega_m} \sum_{i=1}^N x_i^{\omega_m}\bigl)^{1/\omega_m} $$