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

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 :)

• Idk why ur looking at $\frac{\sin(x^3)}{x}$, cause that converges to $0$. Also, is it possible for the limit to just not exist (I'm excluding $\lim = +\infty$)? I feel like it's not, by a Baire Category theorem argument. Apr 10, 2020 at 18:54
• @mathworker21 What about $\sin x?$
– zhw.
Apr 10, 2020 at 19:04
• @zhw. does that satisfy $f(x+t)-f(x) \to 0$ as $x \to \infty$ for all $t$? Not to be rude, but it's a bit insulting if you were suggesting I was suggesting that it's not possible for an arbitrary function to not have a limit at infinity. Apr 10, 2020 at 19:05
• @mathworker21 I see, it has to work for all $t.$ I missed that.
– zhw.
Apr 10, 2020 at 19:08
• I have a counterexample if we have the condition for just $t=1$. Apr 10, 2020 at 19:09

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.

• sigh... i need to start using this form of the Baire Category Theorem. I always use a weaker one. anyways, nice solution +1 Apr 10, 2020 at 19:56
• @mathworker21 I always end up looking up the various forms of the Baire Category Theorem whenever I see a problem where it might apply - it's a strange theorem in that it's pretty clear when it might be helpful, and yet I can never remember exactly what it says. Apr 10, 2020 at 20:02
• Not to diminish your solution, but it really does trivialize this problem. It immediately gives all $t$ in an interval for a particular $X$. Apr 10, 2020 at 20:05

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).

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.