Is a finite function bounded on a closed bounded set?

Question

Given a finite function $f$ in $\boldsymbol{R}^n$, if set $A$ is a closed bounded set, could I say $\{f(\boldsymbol{x}), \boldsymbol{x} \in A\}$ is bounded?

The definition of finite function is $-\infty < f(\boldsymbol{x}) < +\infty, \forall \boldsymbol{x} \in \boldsymbol{R}^n$.

What is a finite function? – José Carlos Santos Sep 20 '19 at 08:14 — José Carlos Santos, Sep 20 '19 at 08:14
The finite function is described in this problem – Zenan Li Sep 20 '19 at 08:26 — Zenan Li, Sep 20 '19 at 08:26
It should be part of your question. – José Carlos Santos Sep 20 '19 at 08:28 — José Carlos Santos, Sep 20 '19 at 08:28

score 2 · Answer 1 · answered Sep 20 '19 at 08:15

2

$f(x)=\frac 1 x$ for $0<x\leq 1$ and $f(0)=0$ is a counterexample. I am taking $n=1$.

answered Sep 20 '19 at 08:15

Kavi Rama Murthy

359,332

thank you, but could I add some conditions to make the function bounded? – Zenan Li Sep 20 '19 at 08:17
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Continuity should be enough if by finite you mean bounded. – Charles Hudgins Sep 20 '19 at 08:19
@Ze-NanLi Continuity and monotonicity are both sufficient in the case $n=1$. Monotonicity in each variable would be fine for $\mathbb R^{n}$. – Kavi Rama Murthy Sep 20 '19 at 08:20
I know continuity is enough, but maybe this condition is too strict? – Zenan Li Sep 20 '19 at 08:20
@Kavi Rama Murthy, thanks, but i also want to know, if the function is lower semi-continuous and convex? could it be bounded? – Zenan Li Sep 20 '19 at 08:24
@Ravi You seem to know what a finite function is! Would be good if you can shed light on the topic for us! – mathcounterexamples.net Sep 20 '19 at 08:24
I think OP is saying finite function for a real valued function as opposed to an extended real valued function. It is not uncommon for people to use the term 'finite valued'. – Kavi Rama Murthy Sep 20 '19 at 08:31

Is a finite function bounded on a closed bounded set?

1 Answers1