# Supremum of minus a function

1,044

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 = $\inf -f$.)

Share:
1,044

Author by

### TEX

Updated on August 01, 2022

Consider a real-valued 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.