Prove $x^2+x$ is uniformly continuous in $(0,1)$ using the $\epsilon , \delta$ method. My try:

Let $\epsilon>0$ and $\delta=\frac{\epsilon}3$. Then pick $x,y\in(0,1)$ s.t. $|x-y|<\delta$, so we have $|x^2+x-y^2-y|=|(x^2-y^2)+(x-y)|\leq|x^2-y^2|+|x-y|.$

Note that $|x^2-y^2|=|(x-y)(x+y)|$ and $x,y\in(0,1)$ so $(x+y)>0$$\space$ and $<2$. So it's equal to $|x-y|(x+y)<2\delta$.

Coming back we have: $|x^2-y^2|+|x-y|=|x-y|(x+y)+|x-y|<3\delta=\epsilon.$

It's this okay? How could I write it better? Am I wrong somewhere? Thanks.

  • 2
    $\begingroup$ Yes. Your argument is OK and I think you've written it as well as possible already. $\endgroup$ – stressed out Jan 29 at 18:01
  • $\begingroup$ That is, x(x+1) is uniformly continuous. It is suffficient to show that the product of uniformly continuous functions is uniformly continuous. $\endgroup$ – Jacob Wakem Jan 29 at 18:14
  • $\begingroup$ @Alephnull But it's not possible to show that (unless you also assume the two functions are bounded, or something along those lines). For example, $f(x) = g(x) = x$ is uniformly continuous on $\mathbb{R}$ but their product isn't. $\endgroup$ – Daniel Schepler Jan 29 at 20:43
  • $\begingroup$ It is well-known that x^2 is uniformly continuous. You can use the same delta (or is it epsilon?) . $\endgroup$ – Jacob Wakem Jan 30 at 17:00
  • $\begingroup$ x^2+x is between x^2 and (x+1)^2 . It is well-known x^2 is uniformlycontinuous and thus by graph-similarity (x+1)^2 is uniformly continuous. Thus x^2+x is uniformly continuous. $\endgroup$ – Jacob Wakem Feb 8 at 20:16

You are correct. The same result can be obtained in a easier way by using the Mean Value Theorem: if $f(x)=x^2+x$ then for $x,y\in (0,1)$ there is $t\in (0,1)$ such that $$|f(x)-f(y)|=|f'(t)||x-y|=|2t+1||x-y|\leq 3|x-y|.$$ More generally a differentiable function whose derivative is bounded in an interval $I$ is also uniformly continuous in $I$.

  • 1
    $\begingroup$ @MayureshL Here we are talking about uniform continuity. $\endgroup$ – Robert Z Jan 29 at 18:08
  • $\begingroup$ Robert.Gives the result from a different angle.Nice! $\endgroup$ – Peter Szilas Jan 29 at 18:13

Your proof is fine. A shortcut would be to note that $f$ is continuous on the compact set $[0,1]$, and so uniformly continuous there; hence, on $(0,1),$ too.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.