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Let $f : (0,1) → R$ be continuous on $(0,1)$ with

$\lim_{x\to 0} f(x) = \lim_{x\to 1} f(x) = 0$

and where $f(x) > 0$ for all $x ∈ (0,1)$.

Show that:

(a) there exists $z ∈ (0,1)$ such that $f(z)$ = sup {$f(x) : x ∈ (0,1)$},

(b) there does not exist $z ∈ (0,1)$ such that $f(z)$ = inf {$f(x) : x ∈ (0, 1)$}.

I can visualise this in my mind but can't figure out where to start.

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    $\begingroup$ The easier one is (b). It should be straightforward to show that $\inf_{x\in (0, 1)} f(x) = 0$. $\endgroup$ – Michael Lee Dec 12 '17 at 22:06
  • $\begingroup$ For (a), I'd probably let $0 < \epsilon < \sup_{x\in (0, 1)} f(x)$ and $0 < \delta < \frac{1}{2}$ such that for $0 < x < \delta$ or $1-\delta < x < 1$, we have $0 < f(x) < \epsilon$. Then, as $f$ is continuous, it maps $[\delta, 1-\delta]$ to a compact set which will therefore contain its supremum. There are a few steps missing here, but this is the outline of the first proof that occurs to me. $\endgroup$ – Michael Lee Dec 12 '17 at 22:09
  • $\begingroup$ Yes it is mainly (a) I am struggling with, trying to create a closed interval of [0,𝛿] but not sure that that is the correct direction. $\endgroup$ – Albert B Dec 12 '17 at 22:10
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a) Let be $g: [0,1] \to R$ the function given by $g(0) = g(1) = 0$ and $g(x) = f(x)$, $\forall x \in (0,1)$. Note that $g$ is continuos by hypothesis and, as range of compact set by continuos functions is compact, we have that there are $z \in [0,1]$ such that $g(z) = \sup \{ g(x) : x \in [0,1]\} = \sup \{ f(x) : x \in (0,1)\}$ . However, since $g(0) = g(1) = 0$ and $g(x) = f(x) > 0$ $\forall x \in (0,1)$, we have that $z \in (0,1)$.

b) From hypothesis, $0$ is a lower bounder of $\{ f(x) : x \in (0,1)\}$. By other side, as $\lim_{x \to 0} f(x) = 0$, given $\epsilon >0$, there are are $\delta > 0$ such that $|x| < \delta $ implies $ 0 < f(x) <\epsilon$. Follows characterization of the infimum, that $0 = \inf \{ f(x) : x \in (0,1)\}$.

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  • $\begingroup$ Thanks this has helped a lot. $\endgroup$ – Albert B Dec 12 '17 at 22:59
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Here is an answer to a) without extending the domain. By the definition of a limit we know that there exists a $\delta>0$ such that $x,y<\delta$ implies that $f(x),f(1-y)<1$. Hence $f$ is bounded on $(0,\delta)$ and $(1-\delta,1)$. As $f$ is continuous on the compact set $[\delta,1-\delta]$ we thus have that $f$ is bounded on its domain. Hence $\sup_{(0,1)}f$ exists.

Now by assumption $f(1/2)>0.$ So, again by the definition of a limit, there exists a $1/4>\delta_1>0$ such that $x,y<\delta_1$ implies that $f(x),f(1-y)<f(1/2)$. Furthermore, by compactness there exists a $z\in [\delta_1,1-\delta_1]$ such that $f(z)=\sup_{[\delta_1,1-\delta_1]}f$. We also know that $\sup_{(0,1)}f\geq f(1/2)$, and by assumption on the value of the end points and continuity on the interval that $\sup_{(0,\delta_1)\cup(1-\delta_1,1)}f\leq f(1/2)$, with equality only possible holding at $\delta_1$ and $1-\delta_1$.

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