A bounded function on $\mathbb{R}$ with no local maximum and no local minimum at any point, upper semicontinuous at no point and lower semicontinuous at no point: compose the Hamel coefficient with a strictly increasing injection of $\mathbb{R}$ into $(0,1)$

Assume the Axiom of Choice (The Axiom of Choice, Zorn's lemma), which enters through Assuming the Axiom of Choice, $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$: there is $B\subseteq \mathbb{R}$ such that every real is a finite $\mathbb{Q}$-linear combination of elements of $B$ in exactly one way, and each basis vector carries a well-defined $\mathbb{Q}$-linear coefficient map. Let $f={\mathrm{\Lambda }}_{b_{\star }}:\mathbb{R}\to \mathbb{R}$ be the Hamel coefficient map of An additive $f:\mathbb{R}\to \mathbb{R}$ that is not $x\mapsto cx$: the coefficient of one fixed Hamel basis vector. It is unbounded above and below on every nondegenerate interval, its graph is dense in ${\mathbb{R}}^2$, and every nonempty level set is dense in $\mathbb{R}$, whose values are exactly the rationals and each of whose nonempty level sets is dense in $\mathbb{R}$. Define

$\phi :\mathbb{R}\to \mathbb{R},\phi (u):=\frac{1}{2}+\frac{u}{2(1+\mathrm{\mid }u\mathrm{\mid })},g:=\phi \circ f\mathrm{.}$

Say that $x$ is a local maximum point of $g$ when there is a real $\delta >0$ with $g(y)\le g(x)$ for every $y\in N_{\delta }(x)$ (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$), and a local minimum point when there is a real $\delta >0$ with $g(y)\ge g(x)$ for every $y\in N_{\delta }(x)$. Then:

1. $0<g(x)<1$ for every real $x$, so $g$ is bounded (Lower bound, bounded below, bounded set);
2. $g$ has no local maximum point and no local minimum point;
3. $g$ is upper semicontinuous at no point of $\mathbb{R}$ and lower semicontinuous at no point (Upper and lower semicontinuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$); in particular it is continuous at no point (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point).

Why $\phi$ and not a bijection onto $\mathbb{Q}\cap (0,1)$. All that is needed of $\phi$ is that it be strictly increasing, take values in $(0,1)$, and send rationals to rationals; the explicit formula above does all three and costs no countability argument.

• Boundedness is what makes the example surprising. A function with no local extremum anywhere is easy to arrange if it is allowed to be unbounded; here every value lies strictly inside $(0,1)$ and yet no point is even a local extremum, because arbitrarily close to any point the function takes both a strictly larger and a strictly smaller value.

• Everything comes from the level sets. The only property of $f$ used after step 1.1 is that its nonempty level sets are dense and indexed by the rationals (An additive $f:\mathbb{R}\to \mathbb{R}$ that is not $x\mapsto cx$: the coefficient of one fixed Hamel basis vector. It is unbounded above and below on every nondegenerate interval, its graph is dense in ${\mathbb{R}}^2$, and every nonempty level set is dense in $\mathbb{R}$); $\phi$ contributes only the bounding into $(0,1)$ and the preservation of strict order. Any function with countably many dense level sets, relabelled by a strictly increasing injection into a bounded interval, would do as well.

• The additivity of $f$ is not used here. It was used to prove that the level sets are dense, on the companion item; once that is known, $g$ has nothing to do with Cauchy's equation. In particular $g$ is not additive: it takes values in $(0,1)$ and $g(0)\mathrm{\ne }0$.