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Determine if the following function is uniformly continuous in the given domain.

$f(x)=x^2 , \quad \text{in}\quad [0,\infty], [0,1]$

My try:

For the domain $[0,\infty]$. Let $(x_n)=n, (y_n)= n +\frac{1}{2n}$

Then $|n-n-\frac{1}{2n}|=\frac{1}{2n} < \frac{1}{n}$

But, $|(n)^2-(n+\frac{1}{2})^2| = 1 + \frac{1}{(2n)^2} \geq 1 = \epsilon _0$

Then $f(x)=x²$ is not uniformly continuous in the domain $[0,\infty]$

For the domain $[0,1]$. Let $(x_n)=\frac{1+n}{n}, (y_n)= \frac{1+2n}{2n}$

Then $|\frac{1+n}{n}-\frac{1+2n}{2n}| = \frac{1}{2n} < \frac{1}{n}$

But, $|(\frac{1+n}{n})^2-(\frac{1+2n}{2n})^2|=\frac{1}{n} + \frac{3}{4n^2} \geq 1 = \epsilon _0$

Then $f(x)=x²$ is not uniformly continuous in the domain $[0,1]$

I'm not sure if my method is correct. Any suggestions would be great!

Sofía Contreras
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    For $[0,1]$ your $x_n$ and $y_n$ are not in the domain; they are greater than $1$. And the conclusion is wrong: a continuous function is always uniformly continuous on a compact set such as $[0,1]$. –  Aug 17 '20 at 21:00
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    https://math.stackexchange.com/questions/1340552/show-that-fx-x2-is-not-uniformly-continuous-on-0-infty?noredirect=1&lq=1 – CountWolves Aug 17 '20 at 21:00
  • Btw, the other problem with your argument is the assertion that $$\frac{1}{n} + \frac{3}{4n^2} \geq 1.$$ This is obviously not true for large $n$ as the LHS converges to zero as $n \to \infty$. (In fact it's already false for $n=2$.) –  Aug 17 '20 at 21:48

3 Answers3

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Another way to see that the function is uniformly continuous on $[0,1]$ whithout using Heine's theorem is to prove that the definition of uniform continuity is satisfied.

Indeed, let $\varepsilon > 0$. Let $\eta = \varepsilon/2$. For all $x,y \in [0,1]$ such that $|x-y|<\eta$, you have $$|x^2-y^2| = |(x-y)(x+y)| \leq 2\eta = \varepsilon$$

So the definition is satisfied.

TheSilverDoe
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It's certainly uniformly continuous on $[0,1]$. In general, a continuous function will always be uniformly continuous on a compact set (as @Bungo pointed out in the comments).

To address the question in the comments:

For example, for any $\varepsilon$, if we just take $\delta=\frac{100}{4 \times 99} \varepsilon$, we have $$f\left(x+ \frac{99}{100} \delta\right) - f(x) \\ = f(x + \varepsilon/4) - f(x) \\ = x^2 + x\varepsilon/2 + \varepsilon^2/16 - x^2 \\ = x\varepsilon/2 + \varepsilon^2/16 \\ \leq \varepsilon/2 + \varepsilon^2/16 \\ \leq \varepsilon/2 + \varepsilon/16 \\ < \varepsilon$$

P.S. @TheSilverDoe 's answer is much cleaner, so I would check that one out :)

inavda
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    Thanks! What other argument I could use to show that is uniformly continuous in $[0,1]$? I feel that my profesor doesn´t want me to just use that theorem. – Sofía Contreras Aug 17 '20 at 21:14
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Your method for the domain $[0,\infty)$ is correct, and your result is correct as well. But for the domain $[0,1]$, it doesn't work, since your chosen $x_n,y_n$ aren't in the domain. Instead, you could use the fact that continuous functions on compact domains are uniformly continuous.

Vercassivelaunos
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