I have a question: Is it true that $$ \underset{\omega}{\max}\; \omega^{\top}\cdot F(\omega') \leq \underset{\omega}{\max}\; \omega^{\top}\cdot F(\omega),\; \forall \omega' $$ where the elements of the vector $\omega$ satisfy $\sum\limits_{i} \omega_{i} = 1$.
Context: I was reading the mathematical proofs of the following paper https://arxiv.org/pdf/1908.08342.pdf on multi-objective reinforcement learning and, on page 16 in the Appendix, I stumbled across the proof of the theorem 2. For this theorem, the authors defined a pseudo-metric over the space of action-value functions $Q(s, a, \omega)$, where $s$ and $a$ are respectively the state of the environment and the action performed by an actor: $$ d(Q, Q') = \underset{s, a, \omega}{\max}\; \left|\omega^{\top}\cdot \left(Q(s, a, \omega) - Q'(s, a, \omega)\right)\right| $$
Then, the goal is to show that the Bellman optimality operator $(\mathcal{T}Q)$ defined below is a contraction, i.e. that $d(\mathcal{T}Q, \mathcal{T}Q') \leq \gamma d(Q, Q')$: $$ (\mathcal{T}Q)(s, a, \omega) = r(s, a) + \gamma \mathbb{E}_{s'\sim\mathcal{P}(s'|s, a)}\left[ \arg_{Q} \underset{a', \omega'}{\max}\; \omega^{\top}\cdot Q(s', a', \omega') \right] $$
The authors were able to show that: $$ (\mathcal{T}Q)(s, a, \omega) \leq \gamma\; \underset{s', a', \omega}{\max} \left|\omega^{\top}\cdot \left(Q(s', a', \omega') - Q'(s', a', \omega')\right)\right| = \gamma d(Q, Q') $$ completing their proof.
My Problem: My problem is in the equality saying $\underset{s', a', \omega}{\max} \left|\omega^{\top}\cdot \left(Q(s', a', \omega') - Q'(s', a', \omega')\right)\right| = d(Q, Q')$. Indeed, given the definition of the metric $d$ and since we have $\omega'$ as argument to the functions instead of $\omega$, I am thinking that the equality is false and that we have instead: $$ \underset{s', a', \omega}{\max} \left|\omega^{\top}\cdot \left(Q(s', a', \omega') - Q'(s', a', \omega')\right)\right| \neq \underset{s', a', \omega}{\max} \left|\omega^{\top}\cdot \left(Q(s', a', \omega) - Q'(s', a', \omega)\right)\right| = d(Q, Q') $$ which would not be good for the proof.
My solution: To make sense of the proof, I am proposing that maybe there is an omitted step where we would need the following relation: $$ \underset{s', a', \omega}{\max} \left|\omega^{\top}\cdot \left(Q(s', a', \omega') - Q'(s', a', \omega')\right)\right| \leq \underset{s', a', \omega}{\max} \left|\omega^{\top}\cdot \left(Q(s', a', \omega) - Q'(s', a', \omega)\right)\right| = d(Q, Q') $$ However, I don't know if such a relation is even valid, hence my question! My intuition tells me that the relation is true and that $\underset{\omega}{\max}\; \omega^{\top}\cdot F(\omega') \leq \underset{\omega}{\max}\; \omega^{\top}\cdot F(\omega),\; \forall \omega'$ is always valid, but I have no idea how to prove that, making me seriously doubt my intuition.
Summary of my questions:
(1) Is the relation $\underset{\omega}{\max}\; \omega^{\top}\cdot F(\omega') \leq \underset{\omega}{\max}\; \omega^{\top}\cdot F(\omega),\; \forall \omega'$ always true? And how can I show mathematically whether it is true or not?
(2) Am I right to assume that the proof is only valid if the above relation is true and used, i.e., that there is a problem in the original proof as given in the paper? Or did I miss something?
P.s.: I know that any person who wants to answer would still need to go to the paper, as not all details are necessarily included above. At first, I wanted to include every single details about the context, but then I found it would be rather lengthy. I hope that I can still get some help on the matter. Thank you!