We learned that proper functions are the functions $\ f(x):\!E \to \bar R$ where $\ domf \neq \emptyset$ and $ \ f(x) > - \infty \ \ \ \forall \ \ x\in \!E$. Do these conditions guarantee that there exists an affine minorant of $f$?
Do all proper functions have an affine minorant?
0
$\begingroup$
functional-analysis
optimization
affine-geometry
1 Answers
1
No, what about: $\mathbb R$, $f(x)=-x^2$ ?
-
0Just to be clear, the domain of the function is just the set of reals, and so it can never take the value $- \infty $ since x is never $- \infty$? (Also thanks for your answer!) – 2017-02-22