0
$\begingroup$

Let $(X,d)$ be a complete metric space. I need some explanations about the class of all functions like $f$ which have $f:X \to \mathbb{R}\cup\{ +\infty\}$, be a lower bounded and, for all $y \in X$ we have the set $\{x \in X : f(x) \leq f(y)\}$ is closed.

I know this class is a Pure super-set of family of lower semi continuous lower bounded functions, and the subset of (sequentially) lower monotone lower bounded functions. I am not sure, if this class is a pure subset of (sequentially) lower monotone lower bounded or not.

I think it is better to mention my meaning by sequentially lower monotonicity : f is said to be (sequentially) lower monotone(lower semi continuous from above), if for all decreasing sequence $\{ f(x_n)\}_{n \in \mathbb{N} }$, the condition $\lim_{n\to \infty} x_n=x_0$ implies $f(x_0)\leq \liminf_{n \to \infty}f(x_n).$ Some references called this class to partially lower semi continuous, and some others called this lower semi continuous from above.

$\endgroup$
  • 1
    $\begingroup$ If I understand you correctly, the function $f:\mathbb{R}\to\mathbb{R}, x\mapsto x$ is not in your class (because it is not lower bounded). However, $f$ would be lower semi continuous. Wouldn't this be a contradiction to your statement, that your class is a superset of the family of lower semi continuous functions? $\endgroup$ – supinf Oct 25 '18 at 15:17
  • $\begingroup$ Thats right. I should correct my meaning. I meant, my class is superclass for the class of continuous lower bounded function. $\endgroup$ – M. Reza. K Oct 25 '18 at 16:49
0
$\begingroup$

I finally found a name for that mentioned class in some paper. J. Morgan & F. Scalzo have used this class in their paper( Pseudocontinuity in Optimization and Nonzero-Sum Games) in 2004, and called that class lower pseudocontinuous .

I should add this point that I am not sure if they have been first to use this expression or not. An example to show that this class is a proper subclass for lower monotone class is following:
$$f(x)= \left\{ \begin{array}{ll} e^{-x}+\frac{1}{2} & x\geq 0 \\ e^x & x<0 \\ \end{array} \right . $$ It is clear that $f$ is lower monotone, and $\{x \in \mathbb{R}: f(x)\leq f(\ln(2))\}$ is not closed. If we redefine $f(0):=\frac{5}{4}$, the results will hold, but this f is not upper semi continuous, even it is not upper monotone.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.