Is the Function Strictly Increasing on a Given Interval?

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Here's this week's problem!

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Problem. Suppose $f : (a,b) \to \Bbb R$ is a function that is strictly increasing at every point $c \in (a,b)$, i.e., for every $c\in (a,b)$, there exists a $\delta > 0$ such that $f(x) < f(c)$ whenever $c - \delta < x < c$, and $f(c) < f(x)$ whenever $c < x < c + \delta$. Prove that $f$ is strictly increasing on $(a,b)$.

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This week's problem was correctly solved by Opalg. Here is his solution.

Spoiler

Aiming for a contradiction, suppose that $f$ is not strictly increasing on $(a,b)$. Then there exist points $p$, $q$ with $a<p<q<b$ such that $f(q)\leqslant f( p).$

Let $S = \{x\in (p,b) : f(x) \leqslant f( p)\}.$ Then $S$ is nonempty (because $q\in S$) and bounded below by $p$. Thus $S$ has an infimum $r$, with $r\geqslant p$. Notice that $f( r) \leqslant f( p)$ because $f$ is continuous and weak inequalities are preserved under limits. There are now two possible cases.

Case 1: $r = p.$ Then for each $\delta>0$ there exists $s$ with $p<s<p+\delta$ such that $s\in S$ and so $f(s) \leqslant f( p)$. That contradicts the fact that $f$ is strictly increasing at $p$.

Case 2: $r > p.$ Then given $\delta>0$ we can choose $s \in (p,r)$ with $r-\delta < s < r$. Since $s<r$ it follows that $s \notin S$ and therefore $f(s) > f( p) \geqslant f( r)$. But that contradicts the fact that $f$ is strictly increasing at $r$.

The two contradictions together show that the initial assumption is untenable and therefore $f$ is increasing on $(a,b).$