# Uniform continuity of some functions on $\mathbb{R}$

1) $$x(t)=\sqrt[3]{t^4+1}$$
2) $$x(t)=\sin e^t$$
3) $$x(t)=\sin (t^2-t+1)^{-1}$$
4) $$x(t)=\sin\ln(\lvert t \rvert +1)$$

My problem is to determine whether these functions are uniformly continuous on $$\mathbb{R}$$ or not. My work so far:

1) When I look at the graph of this function, it looks like $$t^2$$. Algebraically, I see that $$x(t)=\sqrt[3]{t^4+1}$$ approximately $$x(t)=\sqrt[3]{t^4}=t^{4/3}$$, which is not uniformly continuous. A similar argument to $$t^2$$ would work. But, How can I say that the above "approximately" part rigorously?

2) Looking at the graph, I see that it is periodic for the positive inputs but it grows very fast in these periods. I could not figure this out algebraically as in the first function. How can I see this? Moreover, I know that $$e^t$$ is not uniformly continuous. Is there a general fact about compositions of uniform continuous functions. For example, if you have one uniformly continuous and one which is not, then the composition is not uniformly continuous. Finally, can you give me a hint for this one also?

3) For this one, I checked the derivative of the function:
$$-\frac{2t-1}{(t^2-t+1)^2}\cdot \cos(t^2-t+1)^{-1}$$
The rational part of this function goes to zero as $$t \rightarrow \infty$$ and $$\cos$$ is always bounded, thus the derivative of this function is bounded. Is it reasoning correct, or am I missing something? Then, if this is correct, we have that the original function is uniformly continuous.

4) This function is even, so I will check only the positive reals. When I check the derivative as in the third question, I have:
$$\frac{1}{t+1}\cdot \cos\ln(t+1)$$
which is bounded by the same argument. Thus, the original function is uniformly continuous.

Actually, the problem I have for the third and the fourth ones is that I am not sure that checking only the limit when $$t \rightarrow \infty$$ is enough to conclude that the derivative is always bounded. Can you explain this part specifically in a more detailed way?

Lastly, if you know more intuitive/general or easier ways of solving these questions, please let me know. Thanks for any help.

## 1 Answer

1) It can't be uniformly continuous because a uniformly continuous on $$\mathbb{R}$$ must be bounded by a linear function.

2) ?

3) and 4) are uniformly continuous as you said, because both $$\frac{2t - 1}{t^2 - t + 1}$$ and $$\frac{1}{|t| + 1}$$ are bounded. But a function might vanish at infinity while being unbounded near some point. For example $$\frac{1}{t}$$, it's obviously unbounded near $$0$$.

• What did not you understand about 2? For 3 and 4, considering your counter-example to my argument, what else do you check in terms of boundedness in those functions except the limit as t tends to infinity? For the first one, is that a theorem? I do not know your argument. Thanks for the answer btw – Mahmut Esat Akın Jan 1 '20 at 20:19
• I just couldn't say anything meaningful about it. Regarding 3 and 4, you need to show that the denominators don't have real roots. And yes, it is a theorem. Let me state it more carefully so you can try and prove it: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be uniformly continuous. Then there are real numbers $a, b > 0$ such that $\forall x \in \mathbb{R}, |f(x)| \le a|x|+b$. – Lázaro Albuquerque Jan 1 '20 at 20:49