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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.
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 $0 = \inf \{ f(x) : x \in (0,1)\}$.
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 $1-\delta_1$.
