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.

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    $\begingroup$ 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
    Commented Oct 25, 2018 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
    Commented Oct 25, 2018 at 16:49

1 Answer 1


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.


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