# Proving $x^2+x$ uniformly continuous in (0,1) using $\epsilon, \delta$.

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.

• Yes. Your argument is OK and I think you've written it as well as possible already. – stressed out Jan 29 at 18:01
• That is, x(x+1) is uniformly continuous. It is suffficient to show that the product of uniformly continuous functions is uniformly continuous. – Jacob Wakem Jan 29 at 18:14
• @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. – Daniel Schepler Jan 29 at 20:43
• It is well-known that x^2 is uniformly continuous. You can use the same delta (or is it epsilon?) . – Jacob Wakem Jan 30 at 17:00
• 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. – 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 $$I$$.
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.