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I am trying to show that $f(x) = x\ln(x)$ is not uniformly continuous on the interval $(0,\infty)$. The solution given here Show $f(x)=x\ln x$ is not uniformly continuous does it by using $\epsilon-\delta$ but I want to do it by sequences if possible. The "x" term is messing things up because I cannot take $$x_n = \textrm e^{-n } $$ and $$y_n = \textrm e^{-n + 1}$$ because $|f(x) -f(y)|$ does not work out. Any ideas on sequences I can take?

Angina Seng
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Matt
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  • With those $x_n$ and $y_n$ you are going in the wrong direction: $f$ is uniformly continuous on $(0,1)$. You want sequences going off to $\infty$. – Angina Seng Nov 18 '18 at 04:23
  • Ohhhhh, so you look for sequences around where the derivative explodes right? – Matt Nov 18 '18 at 04:37

2 Answers2

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Take $x_n = e^{n^2}$ and $y_n = x_n + 1/n$. Note that $|x_n - y_n| \to 0$.

By the MVT there exists $\xi_n \in (x_n ,x_n + 1/n)$ such that

$$y_n \ln y_n - x_n \ln x_n = (1 + \ln \xi_n)/n > (1+\ln x_n)/n > n \to +\infty$$

More generally

A differentiable function $f:(0,\infty) \to \mathbb{R}$ is not uniformly continuous if $|f'(x)| \to +\infty$ as $x \to +\infty$. (It is not enough for $f'$ to be unbounded.)

For any $\delta > 0$ take $x \in (0,\infty)$ and $y = x + \delta/2$. We have $|x-y| < \delta$, but there exists $\xi \in (x,y)$ such that

$$|f(x) - f(y)| = |f'(\xi)||y-x| = |f'(\xi)|\delta/2$$ If $|f'(x)| \to +\infty$ then there exists $X$ such that for all $\xi > x > X$ we have $|f'(\xi)| > 2/\delta$ and, hence $|f(x) - f(y)| > 1$.

So for any $\delta > 0$ there exists $x,y \in (0,\infty)$ with $|x - y| < \delta$ and $|f(x) - f(y) > 1$, and $f$ is not uniformly continuous.

RRL
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  • The question here is about the uniform continuity of a function on an infinite interval , that happens to have a derivative where $f'(x) \to \infty$ as $x \to \infty$. Such a function can't be uniformly continuous. I'm simply (1) using this to construct the sequences OP requested, and (2) explaining the underlying principle. There are UC functions on $(0,\infty)$ with unbounded derivatives and UC functions on $[0,1]$ like $f(x) = \sqrt{x}$ where $f'(x) \to \infty$ as $x \to 0$ The principle I am using only pertains to the case where $|f'(x)| \to \infty$ as $x \to \infty$. – RRL Nov 18 '18 at 06:50
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    @bof: Yes if $f$ is UC on $[1,\infty)$, then $|f(x)| = \mathcal{O}(x)$ as $x \to \infty$. There are many ways to answer this question. I just took one approach. Although unless you already know the $\mathcal{O}(x)$ result it is not as easy to prove as the derivative result if you are starting from first principles. – RRL Nov 18 '18 at 07:02
  • thanks for the response!! @RRL , so in a general function that has a derivative that goes to infinity as x goes to infinity cant be uniformly continuous? – Matt Nov 18 '18 at 19:15
  • @Matt: Yes. If the derivative is bounded then it's uniformly continuous like $f(x) =x$. If the limit of the derivative as $x \to \infty$ is $\pm \infty$ then it's not uniformly continuous like $f(x) = x^2$. If it oscillates with no limit but is unbounded it may or may not be uniformly continuous. – RRL Nov 18 '18 at 19:37
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Let $x_n=n$ and $y_n=n+\frac 1 {\log n}$. Note that $\bigl(n+\frac 1 {\log n}\bigr) \log \bigl(n+\frac 1 {\log n}\bigr)-n \log n \geq \frac 1 {\log n} (\log n) \geq 1$.

The Amplitwist
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Kavi Rama Murthy
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