Supremum of minus a function
Actually you can't make any conclusion.
If one inequality always held for all f then it'd hold for $g(x) =  f(x)$ and that'd be a contradiction. (It'd also hold for $f(x) + c$ and $f(x)  c$ for all c; you see the problem there.)
(And as well, as $f$ could be any function how it have its upper bounds [if any] will have no requirements on how $f(x)$ will reach its upper bounds [if any]; in fact the highest values of $f$ will be the negative of the lowest values of $f$; and there need not be any relation between a functions upper bounds [if any] and its lower bounds [if any].)
What you can conclude, though, is if $\sup f = \alpha$ ($\alpha$ can be $\infty$) and $\inf f = \beta$ ($\beta$ can be $\infty$). Then $\sup f = \beta$ and $\inf f =  \alpha$.
Or in other words $\sup f =  \inf f$ and $\inf f = \sup f$.
(This should be intuitive but can be verified in that $\sup f$ (if it exists) $\ge f(x)$ for all x so $\sup f \le f(x)$ for all x, so $\sup f$ is lower bound of $f$. For any $y > \sup f$ then $y < \sup f$ so there is a $f(t) > y$ so $f(t) < y$ so $\sup f$ is greatest lower bound = $\inf f$.)
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TEX
Updated on August 01, 2022Comments

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Consider a realvalued function of $\theta$, $f(\cdot): \Theta \subset\mathbb{R}\rightarrow \mathbb{R}$. Could you help me to clarify the relation between $\sup_{\theta \in \Theta}f(\theta)$ and $\sup_{\theta \in \Theta}(f(\theta))$?
I think $\sup_{\theta \in \Theta}f(\theta) \geq \sup_{\theta \in \Theta}(f(\theta))$ but I would like to know whether there are other more significant relations.