uniformly continuous function on unbounded set

Question

Let $f:[0,+\infty]\longrightarrow \mathbb{R}$ continuous. I want to compare these properties:

1. $f$ is uniformly continuous (i.e $\forall\varepsilon>0 \,\ \,\ \exists\delta>0\,\ :\,\ \lvert f(x)-f(y)\rvert<\varepsilon \,\ \,\ \forall x,y\ge 0\,\ ,\,\ \lvert x-y\rvert<\delta$)

2 $f$ has a asymptote $y=mx+q$ (i.e. $\lim\limits_{x\to+\infty} (f(x)-mx-q)=0$)

3 $\lim\limits_{x\to+\infty}\dfrac{f(x)}{x}$ exists finite.

4 $\exists a,b>0( \lvert f(x)\rvert\le ax+b \,\ \forall x)$

I know that:

2--> 1 but no conversely, for exemple $f(x)=\sqrt{x}$ is u.c. but no have asymptotes.

• If $f$ is u.c. and if $\lim\limits_{x\to+\infty}\dfrac{f(x)}{x}$ exist,then it is finite.

1-->4 and 3-->4 (because i supponed $f$ continuous and $f=O(x)$).

My question is if 3-->1? and if 4-->1? I think the answer is yes because i don't found a easy counterexemple.

Help me,please.

Answer 1

As dfnu has hinted in the comments, 3 $\implies$ 1 is certainly false, as you can let $f(x)=x+\sin(x^2)$ as a counterexample. 4 $\implies$ 1 also fails by the same counterexample. (and in any case, we have 3 $\implies$ 4, so we can't have 4 $\implies$ 1 as well, since we don't have 3 $\implies$ 1.)

Your counterexample for 1 $\implies$ 3 doesn't work, since $\sin(\frac{1}{x})\to 0$ as $x\to \infty$.

To get a correct counterexample for 1 $\implies$ 3, simply let $f(x)=\text{dist}(x,\{2^n\mid n\in \mathbb N\})$. Then $f$ is Lipschitz, hence certainly uniformly continuous, yet whenever $x=3(2^n)$, we have $\frac{f(x)}{x}=\frac{1}{3}$, and yet $\frac{f(x)}{x}=0$ whenever $x=2^n$.

Update

Regarding the remaining implications, since 2 implies both 1, 3, and 4, it cannot be implied by either 1 or 3, as then we would gain an implication 1 $\implies $ 3 or 3 $\implies$ 1, both of which have been disproven. Likewise it cannot be implied by 4, since 3 $\implies $4.

Finally, 4 certainly does not imply any of the other conditions, as you can take the $f(x)= \text{dist}(x,\{2^n\mid n\in \mathbb N\})+\sin(x^2)$, and it will fail 1, 2, and 3.

Answer 2

3 $\implies$ 1 is false, and we can take $f(x)=sin(x^{2})$, or $f(x)=x+sin(x^{2})$ (as pointed out in the comment), or whichever function will have a certain set of $\epsilon$ for which $\delta>0$ does not exist. $sin(x^{2})$ is not uniformly continuous on $[0,+\infty)$ despite being continuous there.

4 $\implies$ 1 is false: take any periodic function and change it so that, as x approaches infinity, its amplitude tends to infinity, but slower than $ax+b$. For instance, $f(x)=x*sin(x)$ and any linear function with large enough slope (i.e. derivative) and intercept like $f(x)=4x+3$.