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let $f\colon\mathbb{R}\to\mathbb{R}$ is continuous function.

For arbitrary two real number $a, b$ $(a<b)\\$

Set $S=\left\{f(x)\vert a<x<b\right\}$ has both maximum and minimum.

I find $f$ must have local minima, local maxima in interval$(a, b)$ and the local maximum must bigger than $f(a)$ (or $f(b)$) and local minimum must smaller than $f(a)$ (or $f(b)$).

But, to satisfy that conditions, $f$ must have infinitely many local minimum(maximun)s.

I found constant function $f$ satisfies conditions, but I wonder existence of non-constant function.

Is there non-constant $f$ which satisfy that conditions?

bFur4list
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    I don't think so. Let $M$ be the global maximum (on some non-degenerate interval) of $f$, attained at $x_1$ and $m$ be the global minimum, attained at $x_2$. Assume $x_1\ne x_2$. WLOG, assume $x_1<x_2$. There must be some $y$ in $(x_1,x_2)$ with $f(y)=M$. Let $z=\sup{ y\in(x_1,x_2) | f(y)=M}$. By continuity of $f$, it follows that $f(z)=M $. In order to avoid a contradiction, we must have $z=x_2$; and thus $m=M$. – David Mitra Oct 29 '19 at 16:58
  • @DavidMitra Isn't Weierstrass function (https://en.wikipedia.org/wiki/Weierstrass_function) an example of what OP is looking for? In case it isn't, why not? – dfnu Oct 29 '19 at 17:32
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    @dfnu It has local extrema on any non-degenerate interval; but, as far as I can gather, the OP's problem requires a global min and max on any $(a,b)$. – David Mitra Oct 29 '19 at 17:46
  • @DavidMitra You should make that an answer. – Milten Oct 30 '19 at 06:45

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