Is sup of max, same as max of sup?

Question

Let $\sigma_1, \sigma_2 \dots \sigma_n$ be functions of $\omega \in \mathbb{R}_+$.

Is $\sup_{\omega}(\max_{i=1:n} (\sigma_i))$ same as $\max_{i=1:n}( \sup_{\omega}(\sigma_i))$?

Could you also please give me more insight by giving some related references?

Answer 1

First, for each $i=1:n$ it is obvious that $\sup_{\omega}(\sigma_i)\le \sup_{\omega}(\max_{i=1:n} (\sigma_i))$ as the LHS sup is taken over a no greater function than in the RHS. And from this follows $\max_{i=1:n}( \sup_{\omega}(\sigma_i))\le \sup_{\omega}(\max_{i=1:n} (\sigma_i))$.

Second, it is also clear that for each $\omega$, we have $(\max_{i=1:n} (\sigma_i))\le \max_{i=1:n}( \sup_{\omega}(\sigma_i))$, because the max on LHS is taken over a set of numbers $\sigma_i(\omega)$, each of which is no greater than their counterpart in RHS, namely $\sup_\omega\sigma_i(\omega)$. Hence $\sup_\omega(\max_{i=1:n} (\sigma_i))\le \max_{i=1:n}( \sup_{\omega}(\sigma_i))$

Therefore the two sides equal.