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?

  • $\begingroup$ lower limit = liminf? $\endgroup$ – user370967 Feb 7 '18 at 22:24
  • $\begingroup$ @Math_QED yes (15 chars) $\endgroup$ – Alon Gubkin Feb 7 '18 at 22:25
  • $\begingroup$ @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/… $\endgroup$ – 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.

  • $\begingroup$ But EVT is only true for closed bounds, isn't it? I can't use it for $[0,\infty)$ $\endgroup$ – Alon Gubkin Feb 7 '18 at 22:29
  • $\begingroup$ You can also extend for continuos functions in [0,\infty) $\endgroup$ – user Feb 7 '18 at 22:33
  • $\begingroup$ Sorry, what do you mean? $\endgroup$ – Alon Gubkin Feb 7 '18 at 22:35
  • $\begingroup$ 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 $f$. $\endgroup$ – user Feb 7 '18 at 22:44
  • $\begingroup$ Why is the minimum in $[0, x_0]$ is the global minimum for $f$? $\endgroup$ – Alon Gubkin Feb 7 '18 at 23:25

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