$x^{1+\epsilon}$ is not uniformly continuous on $[0,\infty)$

Question

There are two questions.

First: is the proof underneath correct?

Let $\epsilon>0$ and let $f(x)=x^{1+\epsilon}$. I aim to show that $f$ is not uniformly continuous on $[0,\infty)$.

We will show that for any $\delta>0$ there exists $x$ and $y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|\geq 1$.

Now let $\delta>0$ be given. Consider $f'(x)=(1+\epsilon)x^\epsilon$. We have that $f'\rightarrow\infty$ as $x\rightarrow \infty$ and that $f'$ is monotone increasing. Now choose an $x_0$ large enough so that $f'(x)$ is larger than $2/\delta$ for all $x\geq x_0-\delta$.

Now consider $|f(x_0)-f(y)|$ for some $y\in(x_0+\delta/2,x_0+\delta)$. The Mean Value Theorem guarantees us that there is a $\zeta\in(x_0, y)$ such that $|f(x_0)-f(y)|=|f'(\zeta)||x_0-y|$.

We thus have that $$ \begin{array}{rl} |f'(\zeta)||x_0-y|&=(1+\epsilon)\zeta^\epsilon|x_0-y|\\ &\geq(1+\epsilon)\zeta^\epsilon\delta/2\\ &\geq(1+\epsilon)x_0^\epsilon\delta/2\\ &\geq2/\delta\cdot\delta/2\\ &=1 \end{array} $$ Thus for this choice of $x_0$ and $y$, we have that $|f(x_0)-f(y)|\geq 1$ and $f$ is not uniformly continuous.

Second: If all is well and good, what other ways are there to prove this claim? I feel like the approach I took was a little bulky.

Answer 1

Your proof is fine, and the same proof works for any $f$ with $f' \to \infty$ as $x \to \infty$. Personally, I think your approach is excellent in that it is extremely clear. Perhaps you could have been more terse, but I like your style.

Another approach, however, if you insist on being as terse as possible, is to note that $f$ is uniformly continuous if and only if it maps Cauchy pairs of cofinal sequences to pairs of cofinal sequences. See Pedro's answer for an explanation.

(in fact, this function takes Cauchy sequences to Cauchy sequences but is not uniformly continuous. The condition of taking Cauchy sequences to Cauchy sequences is sufficient, however, when considering functions over bounded sets).

Answer 2

Yes the proof is correct, basically you use the fact that, for every nonnegative $u$, $$ (1+u)^{1+\varepsilon}-1\geqslant(1+\varepsilon)u, $$ hence, for every nonnegative $x$ and $\delta$, applying this to $u=\delta/x$ and multiplying by $x^{1+\varepsilon}$, one gets $$ (x+\delta)^{1+\varepsilon}-x^{1+\varepsilon}\geqslant(1+\varepsilon)x^\varepsilon\delta, $$ and that, for every positive $\delta$, the RHS is unbounded when $x\to\infty$.

Answer 3

An alternative solution is to find a pair of cofinal sequences $(x_n)$ and $(y_n)$ such that $(x_n^{1+\varepsilon})$ and $(y_n^{1+\varepsilon})$ are not cofinal. Can you do this?

Here cofinal means $d(x_n,y_n)\to 0$. For example, if $\varepsilon=1$; I can take $x_n=\sqrt n$ and $y_n=\sqrt{n+1}$. Then $|x_n-y_n|\to 0$ but $|x_n^2-y_n^2|=1\not\to 0$.