Fixed $a\in\mathbb{R}$ and let consider $$\nu(a)=\sup\{b\in\mathbb{R} : F(a, b)\ge 0\}$$ for a suitable function $F$.
My professor said that $\nu$ is a function. It is not clear for me how to see $\nu$ as a function. What associates and to whom?
I am sorry for my possibly dumb question, but I am a little bit confused.
I hope someone could help me.
Thank you in advance!
Without further information about $F$ one cannot really give an answer.
It seems that $F$ is a function $\mathbb R \times \mathbb R \to \mathbb R$, otherwise it does not make much sense to consider the expression $\sup\{b\in\mathbb{R} : F(a, b)\ge 0\}$ for each fixed $a \in \mathbb R$.
The function $\nu$ assigns to each $a \in \mathbb R$ the value $\nu(a) = \sup\{b\in\mathbb{R} : F(a, b)\ge 0\}$. But what is $\sup\{b\in\mathbb{R} : F(a, b)\ge 0\}$?
It could be that $\sup\{b\in\mathbb{R} : F(a, b)\ge 0\} = \infty$. This can only happen if $F$ is unbounded.
It could be that $\{b\in\mathbb{R} : F(a, b)\ge 0\} = \emptyset$ for some $a$ in which case $\nu(a) = - \infty$.
Thus in general $\nu$ is a function $$\nu : \mathbb R \to \mathbb R \cup \{-\infty,\infty\}$$ to the extended real line. With suitable assumptions on $F$ we can exclude that $\nu(a)$ takes values in $\{-\infty,\infty\}$. In that case we get a function $$\nu : \mathbb R \to \mathbb R .$$