# Limit of continuous function with lower bound but without a minimum when $x \to \infty$

Let $f: [0, \infty) \to R$ continuous function, that has lower limit:

$m = inf \{ f([0, \infty))\}$

but doesn't have global minimum.

Is it true that $lim_{x \to \infty}=m$? If yes, how to prove that?

• lower limit = liminf? – user370967 Feb 7 '18 at 22:24
• @Math_QED yes (15 chars) – Alon Gubkin Feb 7 '18 at 22:25
• @AlonGubkin Please remember that you can choose an aswer among the given is the OP is solved, more details here meta.stackexchange.com/questions/5234/… – user Mar 9 '18 at 22:42

You can prove by contradiction that if $$lim_{x \to \infty}\neq m$$
then by EVT $f$ should have a global minimum.
• But EVT is only true for closed bounds, isn't it? I can't use it for $[0,\infty)$ – Alon Gubkin Feb 7 '18 at 22:29
• We have that $f(0)=a>m$, if $lim_{x \to \infty}=L \neq m$ we have that L>m, thus exist $x_0 f(x_0)=b$ with $m<b<L$ then $f(x) in [0,x_0]$ has a minimum that is a global minimum for $f$. – user Feb 7 '18 at 22:44
• Why is the minimum in $[0, x_0]$ is the global minimum for $f$? – Alon Gubkin Feb 7 '18 at 23:25