Suppose my progress is in Baby Rudin's chapter 4. Is it possible to discuss the uniform continuity of $x^t$ without using facts in later chapters?

Question

Let $f: [0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=x^t$. Prove that

1. If $t \in (0,1]$ then $f(x)=x^t$ is uniformly continuous on $[0, \infty)$.
2. If $t \in (1, \infty)$ then $f(x)=x^t$ is continuous but not uniformly continuous on $[0, \infty)$.

Question: Suppose my current progress is in Baby Rudin's chapter 4, where uniformly continuity is introduced. Is it possible to discuss the uniform continuity of the above function $f(x)=x^t$ without using facts in later chapters? (Remember the fallacy of using L'Hospital's rule without studying the proof in calculus courses.)

Edit1: Excuse me! In fact this exercise is given by a TA and is probably not from any part of analysis textbooks before derivative-related theorems are introduced. (In fact that TA used the MVT to solve before the professor rigorously introduce it.) So if there is no way to solve this exercise without using derivative-related theorems, then this exercise should better be put after the chapter on derivative or later.

Answer 1

No, the mean value theorem is not necessary to prove these. Neither of the following arguments uses the mean value theorem per se. The first relies on the concavity of $x^t$ for $0<t<1$, and the second uses that for positive $x,h$ and $t>1$, $(x+h)^t > x^t + tx^{t-1}h$. Both these facts can be demonstrated without using the MVT explicitly.

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First question:

Assume $|x-y| < \delta$. You can show, by using Jensen's inequality, that for $0<t<1$, $(x+h)^t \le \frac{x^t + h^t}{2^{1-t}} < x^t + h^t$. This leads to

$$|x^t - y^t| < |x-y|^t < \delta^t $$

So $\delta = \epsilon^{1/t}$ works for any $x,y$, demonstrating uniform continuity.

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For the second question:

I'll leave it to you to show continuity, the following only disproves uniform continuity.

Suppose $x^t$ is uniformly continuous.

Then, $\forall \epsilon >0, \exists \delta>0$ such that for every $x,y $ $|x-y| < \delta \implies |x^t - y^t| < \epsilon$. Choose $\epsilon = 1$, and let $y = x + \delta/2$ for the corresponding $\delta(1)$. This implies that for every $x$,

$$\left|x^t - \left(x + \frac{\delta}{2}\right)^t\right| < 1 \implies \left| tx^{t-1} \frac{\delta}{2}\right|<1$$

which is clearly not true, as I can pick a large enough $x$ to violate this.

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