For notational simplicity, we often use the notation ${}^AB$ to denote the set of all functions $f:A\to B.$
So, $R$ is a relation on ${}^{\Bbb R}\Bbb R$ given by $(f,g)\in R$ if and only if $f(x)-g(x)\ge0$ for all $x\in\Bbb R.$
In order for $R$ to be reflexive on ${}^{\Bbb R}\Bbb R,$ we must be able to say that $(f,f)\in R$ for all $f\in{}^{\Bbb R}\Bbb R,$ meaning that if $f:\Bbb R\to\Bbb R,$ then $f(x)-f(x)\ge 0$ for all $x\in\Bbb R.$ Is this true? If not, what would be a counterexample function?
In order for $R$ to be symmetric on ${}^{\Bbb R}\Bbb R,$ we must be able to say that if $(f,g)\in R$ for some $f,g\in{}^{\Bbb R}\Bbb R,$ then $(g,f)\in R.$ That is, if $f,g:\Bbb R\to\Bbb R$ are such that $f(x)-g(x)\ge0$ for all $x\in\Bbb R,$ then $g(x)-f(x)\ge 0$ for all $x\in\Bbb R.$ Is this true? If not, what would be a counterexample pair of functions?
In order for $R$ to be transitive on ${}^{\Bbb R}\Bbb R,$ we must be able to say that if $(f,g)\in R$ and $(g,h)\in R$ for some $f,g,h\in{}^{\Bbb R}\Bbb R,$ then $(f,h)\in R.$ That is, if $f,g,h:\Bbb R\to\Bbb R$ are such that $f(x)-g(x)\ge0$ and $g(x)-h(x)\ge0$ for all $x\in\Bbb R,$ then $f(x)-h(x)\ge 0$ for all $x\in\Bbb R.$ Is this true? If not, what would be a counterexample trio of functions?
For $R$ to be an equivalence relation on ${}^{\Bbb R}\Bbb R$ is the same as being reflexive, symmetric, and transitive. If it fails one or more of the above, then it is not an equivalence relation. If all of the above hold, then it is an equivalence relation.