Is a continuous function that satisfies a certain condition uniformly continuous?

Question

Let $f:[0,+\infty)$ be a continuous function that satisfies:

$f(x+q)$~$f(x)$ for $x\to\infty$ (for any $q$)

Does it follow that $f$ is uniformly continuous?

I have managed to show that if there exists $\space$ $\displaystyle\lim_{x\to\infty}\space f(x)=G\in\Bbb{R}$ $\space$ then the function must be uniformly continuos by for given $\epsilon$ picking an $N$ big enough that $\forall_{x>N} |f(x)-G|<\frac{\epsilon}{2}$ and then showing that the function is uniformly continuous on $[0,N]$ and satisfies the definition of unifom continuity for that $\epsilon$ on $[N,+\infty)$, thus proving it must be uniformly continuous,since we could have chosen any $\epsilon$.

However, that approach fails when we consider the cases where $\displaystyle\lim_{x\to\infty}\space f(x)$ is infinite or non-existent.

I've also tried to find a counterexample by experimenting with functions like $\frac{1}{x}\sin(x^{3})$ (which appeared promising since its derivative is unbounded) but so far I haven't found one and my intuition does not steer me to either of the answers.

I would appreciate any hints :)

Answer 1

Here's a counterexample if we only know that $\lim_{x \to +\infty} f(x+t)-f(x) = 0$ for all $t \in \mathbb{Q}$.

Let $T_n$ denote an isosceles triangle with width $\frac{1}{2n!}$ and height $n$. Let $T_n^{(j)}$ for $j=0,1,\dots,(n+1)!n!-n!n!$ be a slight morphing of $T_n$ into $T_{n+1}$ (i.e. very slightly shrinking the width and increasing the height from $j$ to $j+1$).

We let $f$ be $T_n^{(j)}$ with bottom left corner at $x=n!+\frac{j}{n!}$ ($f$ is a bunch of upward spikes).

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I suspect the result is true if we require $\lim_{x \to +\infty} f(x+t)-f(x) = 0$ for all $t \in \mathbb{R}$, and that a proof will use a Baire Category Theorem argument.

Answer 2

Yes, it does.

Firstly, let us define the following set:

$$S_{\varepsilon,X} = \{t \in \mathbb R_{\geq 0} : \text{if }x\geq X\text{ then }|f(x+t)-f(x)| \leq \varepsilon\}.$$ By hypothesis, for any fixed $\varepsilon$, we have $$\bigcup_{X\in\mathbb R_{\geq 0}}S_{\varepsilon,X} = \mathbb R_{\geq 0}.$$ Note that each set $S_{\varepsilon,X}$ is closed because for a fixed $x$, the set of values of $t$ such that $|f(x+t)-f(x)|\leq \varepsilon$ is closed and $S_{\varepsilon,X}$ is an intersection of these closed sets over all $x\geq X$.

Note that we could also say that $$\bigcup_{X\in\mathbb Z_{\geq 0}}S_{\varepsilon,X} = \mathbb R_{\geq 0}.$$ since the sets $S_{\varepsilon,X}$ increase with $X$ - giving a countable union of closed sets whose union is the whole space.

We can then apply the Baire Category Theorem to say that since a countable union of closed sets has non-empty interior, some element of the union must have an interior! In particular, for any $\varepsilon>0$, there must be some $X$ such that some interval $(a,b)$ is a subset of $S_{\varepsilon,X}$. However, then if we have that $x,y \geq X + a$ and $|x-y| < |b-a|$ we could choose some pair $a',b'\in (a,b)$ with $x-a' = y-b'$ and then observe that $$f(x)=f(x-a' + a')$$ $$f(y)=f(y-b' + b')$$ Then the distance from $f(x)$ to $f(x-a')$ is at most $\varepsilon$ as is the distance from $f(y)$ to $f(y-b')$ since $a',b'\in \subseteq S_{\varepsilon,X}$. Thus we find that if $x,y \geq X+a$ and $|x-y| < |b-a|$ we have $|f(x)-f(y)| \leq 2\varepsilon$ - and this works out for any choice of $\varepsilon>0$. This fact suffices to establish that $f$ is uniformly continuous with a small bit of further work.