$f:\Bbb{R} \rightarrow \Bbb{R}$ uniformly continuous function such that the sequence $(f(n))$ is increasing .

Question

$f:\Bbb{R} \rightarrow \Bbb{R}$ uniformly continuous function such that the sequence $(f(n))$ is increasing but there is no interval $(a,b)$ with $a<b$ such that $\forall a<x<y<b$, $f(x)<f(y)$ holds.

For $a=1<x=2<y=3<b=4$, the above holds(that is $f(2)<f(3)$) but it is not for $a=1<x=2<y=2.5<b=3$ as ($f(2) ? f(2.5)$), so thinking that is false? seems like some monotone theorem/ intermediate value theorem will be used?

Answer 1

Such functions do exist. Let $f : [0, 1]$ be a continuous, nowhere differentiable function (such as the Weierstrass function). Naturally, $f$ is uniformly continuous, as it is defined on a compact interval.

Define $$g(n + x) = n + f(x) - f(0) + (f(0) + 1 - f(1))x$$ for $n \in \Bbb{Z}$ and $x \in [0, 1)$. Note that $f$ is still continuous, and non-differentiable at least on non-integer points. Moreover, one can show without too much effort that $g$ is still uniformly continuous, using the fact that $f$ was uniformly continuous.

If $g$ were increasing on any open interval, then by a theorem of Lebesgue, then $g$ should be differentiable almost everywhere on the interval. This contradicts $g$ being differentiable on some subset of the integers.

Answer 2

Hint: take a suitable nowhere-differentiable continuous function $f$ on $[0,1]$ with $f(0)=0$ and $f(1) > 0$, and define $f(n + x) = n f(1) + f(x)$ for $0 \le x < 1$.