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I would like to know what is the precise definition of : $$\lim _{x\rightarrow a} \inf f(x) $$ when $f : R^n \rightarrow R$ in my course it is written that : $$\lim _{x\rightarrow a} \inf f(x) =\sup_{\epsilon>0}\inf_{x\neq a, \mid \mid x-a\mid \mid <\epsilon}f(x)$$ But I don't really understand what represents $\sup$ here ? We take the $\sup$ of what ?

Thanks,

Dicordi
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  • This definition is precise. Perhaps your question is for the definition of $\inf$ – pancini Nov 19 '19 at 19:53
  • yes, I don't understand this definition... the concept behind it... – Dicordi Nov 19 '19 at 19:55
  • You are taking the supremum of the rest of the expression over all $\epsilon >0$ – Nap D. Lover Nov 19 '19 at 19:56
  • Take the greatest lower bound of values $f(x)$ for $x$ such that $x\neq a$ and the norm-distance to $a$ is $<\epsilon$, then take the least upper bound of that GLB, over all $\epsilon$ i.e. as the distance shrinks – Nap D. Lover Nov 19 '19 at 19:58
  • yes but here we take the $\inf$ of all $f(x)$ such that $x\neq a$ and $x$ is in $B_{\epsilon}(a)$, so then, \inf of all this does not gives us a set, I am right ? – Dicordi Nov 19 '19 at 19:58
  • @Dicordi a value is given for each $\epsilon>0$, so you get a set indexed by $\epsilon >0$, no? – Nap D. Lover Nov 19 '19 at 19:59

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Consider the set $A(\epsilon) =\{x \in \mathbb R^n \mid \Vert x-a \Vert <\epsilon\}\setminus \{a\}$.

You can define $g(\epsilon)=\inf\limits_{x \in A(\epsilon)} f(x)$. $g$ depends on $\epsilon$.

Then $\liminf\limits_{x \to a} f(x) = \sup\limits_{\epsilon >0} g(\epsilon)$.

mathcounterexamples.net
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