a function $f:\mathbb{R}\rightarrow \mathbb{R}$ has at least one local min and at least one local max. Also all its local max values are less than its local min values. (i.e. if $f$ attains local max at $a$, and local min at $b$, then $f(a) < f(b)$). Show that $f$ must be discontinuous. Is it true?
A problem on discontinuous function
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1 Answers
The reasoning by @Parsa has a minor omission; however it is easily repaired. The issue concerns the definition of local extrema. The definition of local min at $a$ is that there is an open interval $I$ containing $a$ such that $f(x) \ge f(a)$, $\forall x \in I$ (ie, the inequality is not strict).
However, since all local max values are strictly less than the local min values, this means that if $f$ is continuous, then no other point in $I$ (other than $a$) can have the value $f(a)$ (for either a max or a min). To see this, suppose $a$ is a local min, and that a' \in I also satisfies f(a') = f(a). Then $f$ must have a local max between $a$ and a' whose value is not strictly less than $f(a)$, which is a contradiction. Consequently, $f(x) > f(a)$, $\forall x \in I \setminus \{a\}$. The same reasoning applies to a local max, mutatis mutandis.
After establishing this minor technicality, the remainder of the argument is as @Parsa described above.