0
$\begingroup$

Does continuity of a function $f:[a,b]\to\mathbb{R}$ at a point $c$, $a<c<b$ imply that it achieves it supremum/infimum around that point?

Let's say we for an $\varepsilon$ chose a $\delta$ such that $d(x,c)<\delta$, does this imply that the supremum on that interval is attained?

I know that this is the case when the interval is compact, but in this case it's open $(c-\delta,c+\delta)$ and, I surmise, not compact.

$\endgroup$
2
  • $\begingroup$ Why the downvote? $\endgroup$ Dec 30 '16 at 16:37
  • $\begingroup$ Don't know, but I'm upvoting $\endgroup$
    – MPW
    Dec 30 '16 at 23:04
2
$\begingroup$

No this can't be.

Take $f(x) = x$ on a unit interval $[0,1]$. Then if we look at the open set around $c = 0.1$, so for example $(0.09, 0.11)$ one can see that the supremum of $f$ on that interval is the same as the supremum of that interval which is $0.11$. Yet $0.11 \not\in (0.09, 0.11)$.

$\endgroup$
1
$\begingroup$

Yes, but not necessarily on that interval. It may be achieved on the boundary, hence compactness is needed as you mention.

Consider $f(x)=x$ on $(a,b)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.